AP-Level Physics 1

A graph that catches almost everyone

A ball leaves your hand moving straight up at 4.1 m/s. A motion sensor above you records its height every twentieth of a second and prints a curve: rising from 0 at t = 0 s, flattening over a rounded peak of 0.86 m at t = 0.42 s, falling back to your hand at t = 0.84 s. Now the question that costs more marks than any other single question about graphs. What is the ball's acceleration at the instant it sits at the top?

The answer that feels right is zero. The ball is motionless up there, the curve is flat, nothing appears to be happening. Hold on to that answer for a few minutes. We are going to take it apart, but only after you own the two operations that settle it, because the point of this lesson is not the ball. It is that a motion graph contains more information than it appears to, and that getting it out is two moves, not twenty.

Three quantities, two operations

Kinematics describes motion without asking what causes it. It needs three quantities and they form a chain.

• Position x, measured in metres from an origin you choose. Position can be negative; that just means the other side of your origin.
• Velocity v, in m/s, the rate at which position changes, with a sign that says which way.
• Acceleration a, in m/s², the rate at which velocity changes, also signed.

Two operations connect them, and they run in opposite directions along the chain.

That is the whole toolkit. Before we use it, one distinction that the exam tests deliberately. Displacement is the change in position, final minus initial, and it carries a sign. Distance is the total path length travelled and never decreases. Walk 8.0 m east then 3.0 m west and your displacement is +5.0 m while the distance you covered is 11.0 m. Walk a full lap of a 400 m track and your displacement is 0 m, though your legs disagree.

Reading downward: slope, on real numbers

A low-friction cart is released on a ramp and a sensor records its position.

Average velocity over an interval is displacement divided by the time it took. Between 0.20 s and 0.40 s:

v_avg = (0.48 m - 0.12 m) / (0.40 s - 0.20 s) = 0.36 m / 0.20 s = 1.80 m/s

Between 0.40 s and 0.60 s:

v_avg = (1.08 m - 0.48 m) / (0.60 s - 0.40 s) = 0.60 m / 0.20 s = 3.00 m/s

The cart is getting faster. Doing the same for the other intervals gives 0.60, 1.80, 3.00, 4.20 and 5.40 m/s. Now a fact worth keeping: when acceleration is constant, the average velocity over an interval equals the instantaneous velocity at the midpoint time. So 1.80 m/s is the cart's actual speed at t = 0.30 s, not merely its average between 0.20 s and 0.40 s. Plot those five velocities against their midpoint times, 0.10, 0.30, 0.50, 0.70 and 0.90 s, and you get a straight line. Its slope is

a = (5.40 m/s - 0.60 m/s) / (0.90 s - 0.10 s) = 4.80 m/s / 0.80 s = 6.00 m/s2

Take the slope twice and a curved position graph becomes a single number. Notice what the shapes are doing. A position graph that curves upward means the slope is increasing, which means the velocity is increasing, which means positive acceleration. A straight position graph means constant velocity and zero acceleration. Curvature on an x-t graph is acceleration, and the direction it bends tells you the sign.

Key idea: on a position-time graph you read velocity from the slope and acceleration from the curvature. On a velocity-time graph you read acceleration from the slope and displacement from the area. Ask which graph you are looking at before you say anything about it.

Reading upward: area, including the negative kind

Here is a velocity-time graph for a delivery van, given in words because the numbers are what matter. From t = 0 to t = 4.0 s the velocity climbs in a straight line from 0 to 12 m/s. From 4.0 s to 10.0 s it is flat at 12 m/s. From 10.0 s to 13.0 s it falls in a straight line to 0.

Break the area into a triangle, a rectangle and a triangle:

• Triangle: (1/2)(4.0 s)(12 m/s) = 24 m
• Rectangle: (6.0 s)(12 m/s) = 72 m
• Triangle: (1/2)(3.0 s)(12 m/s) = 18 m

Total displacement is 24 m + 72 m + 18 m = 114 m. Average velocity for the whole trip is 114 m / 13.0 s = 8.8 m/s, which is not the average of 0 and 12 m/s, because the van spent longer at high speed than low. Watch the units do the work: (s)(m/s) = m. If your area comes out in the wrong unit you have read an axis wrong.

Now the case people get wrong. A cart is pushed up a ramp and rolls back down. Its velocity starts at +2.0 m/s at t = 0, falls in a straight line through zero at t = 1.0 s, and reaches -2.0 m/s at t = 2.0 s. The area from 0 to 1.0 s is (1/2)(1.0 s)(2.0 m/s) = +1.0 m. The area from 1.0 s to 2.0 s lies below the time axis, so it is -1.0 m. Net displacement: 0 m. The cart is back where it started. Total distance travelled: 2.0 m, because distance does not care about signs. The same graph answers both questions, as long as you know which one you were asked.

The area trick runs one step further up the chain. The area under an acceleration-time graph is the change in velocity. A rocket motor giving a constant 30 m/s² for 2.5 s changes the velocity by (30 m/s2)(2.5 s) = 75 m/s. Not the velocity: the change in it. You still need to know where you started.

Back to the ball, and why zero is wrong

Return to the toss. Sketch the velocity-time graph from the story rather than the position graph. The ball leaves your hand at +4.1 m/s. At the top, 0.42 s later, it is momentarily at rest, so v = 0. When it returns to your hand at 0.84 s it is moving downward at -4.1 m/s. Three points: (0, +4.1), (0.42, 0), (0.84, -4.1). They lie on a straight line, and its slope is

a = (-4.1 m/s - 4.1 m/s) / (0.84 s - 0 s) = -8.2 m/s / 0.84 s = -9.8 m/s2

A straight line has the same slope everywhere, including at t = 0.42 s. The acceleration at the top is -9.8 m/s², exactly what it is on the way up and on the way down. What is zero at the top is the velocity, not its rate of change. The x-t graph looked flat because the ball was not moving; the v-t graph is not flat at all, because gravity did not pause to let the ball turn around.

Check the area too. From 0 to 0.42 s the area under the velocity line is (1/2)(0.42 s)(4.1 m/s) = 0.86 m, which is the peak height the sensor recorded. From 0.42 s to 0.84 s the area is -0.86 m and the ball comes back. Two independent readings of the same graph agreeing is how you catch your own mistakes.

Signs: the four cases, and the one sentence that fixes them

Students lose points here in bulk, almost always by believing that negative acceleration means slowing down. It does not. Whether an object speeds up or slows down depends on whether a and v have the same sign.

One sentence covers all four: same signs means speeding up, opposite signs means slowing down. The sign of a on its own tells you nothing about speed, only the direction in which velocity is being pushed.

A graph-matching question, worked

The exam likes giving you one graph and asking for the other two. Try this one. An object's velocity-time graph: v = -3.0 m/s constant from t = 0 to t = 2.0 s, then rising in a straight line to +3.0 m/s at t = 4.0 s.

Acceleration first. From 0 to 2.0 s the line is flat, so a = 0. From 2.0 s to 4.0 s, a = (3.0 m/s - (-3.0 m/s)) / 2.0 s = 6.0 m/s / 2.0 s = +3.0 m/s2, constant. The a-t graph is therefore two flat segments, 0 then +3.0, with a jump at t = 2.0 s.

Position next, starting from x = 0. From 0 to 2.0 s the area is (-3.0 m/s)(2.0 s) = -6.0 m, so at t = 2.0 s the object is at x = -6.0 m and the x-t graph is a straight line of slope -3.0 sloping down. From 2.0 s to 3.0 s the velocity is still negative, passing through zero at t = 3.0 s; the area is (1/2)(1.0 s)(-3.0 m/s) = -1.5 m, so x = -7.5 m at t = 3.0 s, and the graph has reached its lowest point with a horizontal tangent. From 3.0 s to 4.0 s the area is +1.5 m, bringing the object back to x = -6.0 m. So the x-t graph falls in a straight line, curves through a minimum at (3.0 s, -7.5 m), and rises to (4.0 s, -6.0 m). The object moved backward, stopped, and came partway back, all while its acceleration was steadily positive.

Common misconceptions

• "At the top of its flight the ball's acceleration is zero." Its velocity is zero for one instant. Its acceleration is -9.8 m/s² throughout, which is precisely why the velocity does not stay at zero.
• "Negative acceleration means slowing down." Only when the velocity is positive. A ball falling with down chosen as negative has negative velocity and negative acceleration, and is speeding up.
• "A position-time graph shows the path the object took." It does not. A rising straight line on an x-t graph is an object moving at constant velocity along a straight road, not an object climbing a hill.
• "Steeper always means faster." On an x-t graph, yes. On a v-t graph steeper means a bigger acceleration, and an object can be slow and steep at the same time.
• "Where two motion graphs cross, the objects meet." Only on a position-time graph. Two velocity graphs crossing means the objects have the same speed at that moment, and they may be kilometres apart.

Where this leaves us

Everything in this lesson reduces to a chain and two moves along it. Slope takes you from position to velocity to acceleration; area takes you back, from acceleration to change in velocity to displacement, with area below the axis counted as negative. Curvature on a position graph is acceleration. Displacement is signed and can be zero while the distance walked is large. The signs of v and a together, never the sign of a alone, say whether something is speeding up. And the ball at the top of its flight has zero velocity and full acceleration, which is the one thing to remember if you remember nothing else: the flat part of a position graph tells you about v, and it is the v-t graph that tells you about a.

Sources

1. OpenStax. (2022). 2.8 Graphical analysis of one-dimensional motion. In College Physics 2e. openstax.org
2. OpenStax. (2022). 2.4 Acceleration. In College Physics 2e. openstax.org
3. The Physics Classroom. (n.d.). 1-D kinematics. physicsclassroom.com
4. Knight, R. D., Jones, B., and Field, S. (2019). College Physics: A Strategic Approach (4th ed.), ch. 2. Pearson.
5. College Board. (2024). AP Physics 1: Algebra-Based course and exam description, Unit 1. New York: College Board.

Thirty-eight metres of skid marks

A crash investigator arrives at a straight suburban road and measures 38 m of skid marks ending at a stopped car. A test with an identical car on the same surface that afternoon gives a braking deceleration of 7.2 m/s². Nobody timed the skid. Nobody saw the crash. The speed limit is 50 km/h and the driver says he was under it.

By the end of this section you will get his speed at the moment the brakes locked out of those two numbers alone, with no stopwatch anywhere in the calculation. That is what the fourth of the constant-acceleration equations is for, and rather than hand you a list of four formulas to memorise, this lesson builds all four out of a single trapezoid.

One graph, four equations

Constant acceleration means one thing on a velocity-time graph: the line is straight. Draw it. At t = 0 the velocity is v₀; at time t it is v; the line runs straight between them. Every one of the equations of motion comes from reading that line by slope or by area.

First, the slope. Acceleration is the slope of that line, so

a = (v - v0) / t, which rearranges to (1) v = v₀ + at

Second, the area. The region under the line between 0 and t is a trapezoid with parallel sides v₀ and v and width t. The area of a trapezoid is the average of the parallel sides times the width, so the displacement is

(2) x - x₀ = (1/2)(v₀ + v) t

Read that one out loud: displacement equals average velocity times time, where the average of the starting and finishing velocities is legitimate only because the graph is a straight line. If the acceleration is not constant, equation (2) is simply false, and so are the two that follow from it.

Third, substitute (1) into (2) to get rid of v:

x - x0 = (1/2)(v0 + v0 + at) t = (1/2)(2v0 + at) t

(3) x - x₀ = v₀t + (1/2)at²

Fourth, eliminate t instead. From (1), t = (v - v₀)/a. Put that into (2):

x - x0 = (1/2)(v0 + v) (v - v0) / a

The numerator is a difference of two squares, (v + v₀)(v - v₀) = v² - v₀², so

2a(x - x0) = v2 - v02, that is (4) v² = v₀² + 2a(x - x₀)

Four equations, one trapezoid, no memorisation required beyond the picture. Notice what each one leaves out, because that is how you pick.

The point: do not hunt for the right formula. List what you have and what you want, notice which quantity is missing from both lists, and take the equation that does not contain it.

Back to the skid marks

Set the origin where the skid begins, take the direction of travel as positive. Known: x - x₀ = 38 m, v = 0 (the car stopped), a = -7.2 m/s² (opposing the motion, so negative). Wanted: v₀. Missing from both lists: time. Equation (4).

v2 = v02 + 2a(x - x0)

0 = v02 + 2(-7.2 m/s2)(38 m)

v02 = 547.2 m2/s2

v0 = 23.4 m/s

Multiply by 3.6 to convert to km/h: 84 km/h in a 50 zone. The driver was wrong, and the arithmetic that showed it took four lines. One more thing worth noticing from equation (4): stopping distance goes as the square of the speed. Double the speed and you quadruple the skid. At 25 m/s the same car needs 43 m; at 50 m/s it needs 174 m.

Free fall, and why the mass cancels

Free fall is the special case where gravity is the only force worth counting. Near Earth's surface every freely falling object has the same acceleration, magnitude g = 9.80 m/s² for our purposes, directed downward. The internationally agreed conventional value, called standard gravity, is 9.80665 m/s²; the real local value varies from about 9.78 at the equator to 9.83 at the poles, which is why a careful experiment can tell you where on Earth it was done.

Why does mass not appear? Run it forward. The gravitational force on a mass m is F = mg. Newton's second law says a = F/m. So a = mg/m = g, and the m cancels. A bowling ball feels a much larger force than a marble and needs a proportionally larger force to accelerate it at the same rate, and the two effects cancel exactly. On 2 August 1971 David Scott stood on the Moon and released a 1.32 kg aluminium geological hammer and a falcon feather from the same height. With no atmosphere to interfere, they landed together, which was the point of doing it there.

So the four equations apply unchanged, with a = -9.80 m/s² if you call up positive. Nothing else changes.

A measurement you can check: how far is that window?

A student measures the sill of a first-floor window at 4.35 m above the pavement with a builder's tape, then drops a golf ball and times the fall five times with a phone stopwatch: 0.94, 0.97, 0.91, 0.95, 0.93 s. The mean is 0.940 s. Starting from rest, equation (3) gives

x - x0 = (1/2)at2, so g = 2(x - x0)/t2 = 2(4.35 m)/(0.940 s)2 = 8.70 m / 0.8836 s2 = 9.85 m/s2

That is within 0.5 percent of the accepted value, using a tape measure and a thumb. The scatter in the five times is about 0.02 s, roughly 2 percent, and because t is squared, a 2 percent error in time becomes a 4 percent error in g. Notice which measurement dominates the error, because in the next lesson that observation becomes a method.

One problem, worked end to end

A ball is thrown straight up at 8.0 m/s from a balcony rail 12.0 m above the ground. Find (a) the greatest height it reaches above the ground, (b) how long it takes to hit the ground, (c) how fast it is going when it lands. Ignore air resistance.

Set up. Up is positive. Origin at ground level, so y₀ = +12.0 m, v₀ = +8.0 m/s, a = -9.80 m/s² for the entire flight, including the moment at the top.

(a) At the highest point v = 0, and no time is involved, so equation (4):

0 = (8.0 m/s)2 + 2(-9.80 m/s2)(y - 12.0 m)

y - 12.0 m = 64.0 / 19.6 m = 3.27 m, so y = 15.3 m above the ground.

(b) Landing means y = 0, and we want t, so equation (3):

0 = 12.0 m + (8.0 m/s)t + (1/2)(-9.80 m/s2)t2

4.90t2 - 8.0t - 12.0 = 0

Quadratic formula, with the units carried along: t = [8.0 +/- sqrt(64.0 + 4(4.90)(12.0))] / (2 x 4.90) = [8.0 +/- sqrt(299.2)] / 9.80

t = (8.0 + 17.3)/9.80 = 2.58 s. The negative root, -0.95 s, is the time the ball would have left the ground had it been launched from below, which is real mathematics and a wrong answer to this question.

(c) Two independent routes. Equation (1): v = 8.0 m/s + (-9.80 m/s2)(2.58 s) = -17.3 m/s. Or equation (4), avoiding any dependence on part (b): v2 = 64.0 + 2(-9.80)(0 - 12.0) = 64.0 + 235.2 = 299.2, so v = 17.3 m/s downward. Agreement between two routes is the cheapest error check you will ever run.

Change one input and watch what flips

Same balcony, same 8.0 m/s, but the ball is thrown straight down. Now v₀ = -8.0 m/s. Equation (3) becomes

4.90t2 + 8.0t - 12.0 = 0, so t = [-8.0 + sqrt(299.2)] / 9.80 = 9.30/9.80 = 0.95 s

The fall takes 0.95 s instead of 2.58 s, which is no surprise. But now the impact speed, from equation (4): v2 = (-8.0)2 + 2(-9.80)(0 - 12.0) = 64.0 + 235.2 = 299.2, giving 17.3 m/s again. Identical.

That is not a coincidence and it is worth understanding rather than memorising. The ball thrown upward comes back down past the balcony rail moving at exactly 8.0 m/s downward, because the flight above the rail is symmetric: same height, same acceleration, same speed at the same height on the way down as on the way up. From that instant onward the two balls are in identical situations. Only 1.63 s of extra flight time separates them, spent entirely above the rail.

Common misconceptions

• "Heavier objects fall faster." Both the gravitational force and the inertia grow in proportion to mass, so the acceleration is the same. What actually separates a feather from a hammer on Earth is air resistance, which depends on shape and surface area, not weight.
• "g is a force." g is an acceleration of 9.80 m/s², or equivalently a field strength of 9.80 N/kg. The force is the weight, mg, which is different for every object.
• "You can use these equations any time an object accelerates." Only when the acceleration is constant. The trapezoid area that produced them assumes a straight velocity-time line, so a car with a varying throttle or a falling body near terminal speed is out of bounds.
• "A ball thrown up is weightless at the top." Its velocity is zero for an instant and its weight is unchanged. Weightlessness means no supporting force, which is true for the entire flight, not just the top.
• "The negative answer from a quadratic is a mistake." It is usually the other intersection of a parabola that has been extended backward in time. Reject it because it lies outside the situation described, not because it is negative.

What to carry forward

Four equations, all of them the same straight velocity-time line read two ways. Slope gives v = v₀ + at, the trapezoid gives x - x₀ = (1/2)(v₀ + v)t, and substituting one into the other gives the two you will use most: x - x₀ = v₀t + (1/2)at² when you know the time, and v² = v₀² + 2a(x - x₀) when nobody timed anything. Choose by the quantity that is missing. In free fall a = -9.80 m/s² throughout, mass cancels out of the problem entirely, and the whole apparatus is only valid while the acceleration stays constant. And a ball thrown up at 8.0 m/s and one thrown down at 8.0 m/s from the same rail hit the ground at exactly the same speed.

Sources

1. OpenStax. (2022). 2.5 Motion equations for constant acceleration in one dimension. In College Physics 2e. openstax.org
2. OpenStax. (2022). 2.7 Falling objects. In College Physics 2e. openstax.org
3. National Institute of Standards and Technology. (2019). Standard acceleration of gravity. CODATA fundamental physical constants. physics.nist.gov
4. Giancoli, D. C. (2016). Physics: Principles with Applications (7th ed.), ch. 2. Pearson.
5. College Board. (2024). AP Physics 1: Algebra-Based course and exam description, Unit 1. New York: College Board.

A lab report that says 11.3

Here is a student's whole procedure, copied out. I held a marble at the 1.00 m mark on a metre rule standing on the floor. I started the stopwatch as I let go and stopped it when I heard the marble hit. The time was 0.42 s. Using x = (1/2)gt², g = 2(1.00 m)/(0.42 s)² = 11.3 m/s². The accepted value is 9.8 m/s². The difference is due to human error.

The arithmetic is correct. 2 divided by 0.1764 really is 11.34. The answer is 16 percent high, the explanation explains nothing, and every single thing that is wrong here is a thing the AP exam asks about by name in its experimental design questions. So let us take this procedure apart piece by piece, because the repairs are the content of this lesson.

Flaw one: one trial is not a measurement

A single reading gives you no idea how much it would have moved if you did it again. Repeat this drop five times by hand and the times scatter over something like 0.38 s to 0.55 s, because a human thumb on a stopwatch is good to roughly 0.1 s to 0.2 s in each direction. The spread itself is data: it is your estimate of the random error, the unpredictable variation that pushes readings both ways and that averaging genuinely reduces.

Repeating is not optional bookkeeping. Without a spread you cannot state an uncertainty, and a result with no uncertainty cannot be compared with anything, including the accepted value.

Flaw two: the interval chosen is too short to measure that way

A fall of 1.00 m takes 0.45 s. If your timing is good to 0.1 s, that is a 22 percent uncertainty in t before you do anything at all. The fix is not to concentrate harder. The fix is to change the experiment so that the quantity you measure badly is large: drop from 2.50 m and the fall takes 0.71 s, or better, stop using a thumb and count frames in a 240 frames per second phone video, where one frame is 0.00417 s.

This matters more than it looks because of how the error propagates. Since g = 2x/t², the time enters squared, so a percentage error in t becomes twice that percentage in g. Here are the rules the exam expects you to use, and they are simpler than they sound.

Worked on the repaired drop: x = 1.200 +/- 0.005 m is 0.4 percent, and t = 0.496 +/- 0.008 s is 1.6 percent. For g = 2x/t² the total is 0.4 + 2(1.6) = 3.6 percent. On 9.8 m/s² that is 0.35, so the result is g = 9.8 +/- 0.4 m/s². That is a sentence you can defend. The student's 11.3 is not, and notice that 11.3 does not even lie within a generous uncertainty band of the accepted value, which is itself a signal that something systematic went wrong.

Flaw three: no variable was varied, so there is no graph

This is the big one, and it is the difference between a calculation and an experiment. With one height you get one number and no way to tell a good number from a lucky one. With five heights you get a line, and a line carries information a single point cannot.

But h = (1/2)gt² is not a straight line: plot h against t and you get a curve, and eyeballing a curve tells you nothing. So you linearise. Compare the relationship with the equation of a line, y = mx + b, and choose what to plot so the two match up:

h = (g/2) t2 against y = m x + b gives y = h, x = t2, m = g/2, b = 0

So plot h on the vertical axis against t² on the horizontal, and the slope will be g/2. Here is the repaired experiment, five heights, each time the mean of five frame-counted videos:

Those five points lie on a straight line. Taking the best-fit line through them gives

slope = 4.91 m/s2, so g = 2 x slope = 9.82 m/s2

You can get the same thing without a regression calculation: take the first and last points, which is what the exam means by drawing a best-fit line and reading two points off the line, not off the data. (2.000 m - 0.400 m) / (0.4083 s2 - 0.0824 s2) = 1.600 / 0.3259 = 4.91 m/s2. Same answer, 0.2 percent from the accepted value, from a phone and a tape measure.

What matters here: the slope has units and a meaning, and you should always state both. Here the slope is in m/s² and it is half of g. A slope whose units do not work out is a sign you linearised wrongly.

What the intercept tells you that the slope cannot

Theory says the line should pass through the origin: zero time, zero fall. Suppose it does not. Suppose your five points give a good straight line with slope 4.55 m/s² and a vertical intercept of -0.10 m. The slope would give g = 9.10 m/s², 7 percent low, and the intercept is physically impossible: it claims the marble had already fallen 10 cm backward before the clock started.

That impossible intercept is the most useful number in the experiment, because it tells you the error is systematic: something shifts every reading in the same direction by roughly the same amount, in this case a timing offset of about 0.04 s from stopping the clock late on every trial. Random error scatters points about a line. Systematic error moves the whole line. Averaging more trials kills the first and does nothing whatever to the second, which is exactly why a single-point experiment can be extremely precise and still be wrong by 16 percent.

The linearising table you should be able to reconstruct

Almost every experiment in this course is one of these. The skill is always the same: rearrange until the relationship looks like y = mx + b, then read off what to plot.

Designing one from scratch: a spring constant

The exam asks you to design a procedure, and a good answer names the apparatus, says what is measured and with what, says what is varied and what is held fixed, says what is plotted, and says how the target quantity comes out of the graph. Five sentences, in that order, and no more.

Apparatus: a spring hung from a clamped rod, a set of slotted masses from 100 g to 500 g, a metre rule clamped vertically beside the spring, and a balance to check the masses rather than trusting their labels. Measure: the position of the bottom of the spring with no load, then with each mass hung on it, reading the rule at eye level to avoid parallax. Vary: the hanging mass in five steps; hold fixed the spring, the clamp position and the rule. Plot: the weight F = mg in newtons on the vertical axis against the extension x in metres on the horizontal. Extract: by Hooke's law F = kx the graph is a straight line through the origin of slope k.

Real data from a classroom spring: masses of 0.100, 0.200, 0.300, 0.400 and 0.500 kg give extensions of 0.049, 0.098, 0.144, 0.196 and 0.243 m. The weights are 0.981, 1.962, 2.943, 3.924 and 4.905 N. Two points on the best-fit line give

k = (4.905 N - 0.981 N) / (0.243 m - 0.049 m) = 3.924 N / 0.194 m = 20.2 N/m

Two design details earn points and are easy to forget. Measure the extension, the change from the unloaded position, not the total length of the spring, or your line will have a large false intercept. And use five masses rather than one, because a single mass gives a ratio that cannot reveal whether the spring obeys Hooke's law at all; the straightness of the line is the evidence that it does.

Flaw four: "human error" is not a source of error

The phrase names no mechanism, gives no direction, and suggests no repair, which is why it scores nothing. Every acceptable answer does three things: names the specific source, says which way it pushes the result, and proposes a change to the procedure. Compare.

The third row carries the habit worth stealing: when you name a source of error, estimate how big it is. An error you can show is 0.1 percent cannot explain a 16 percent gap, and saying so is worth more than listing five vague suspects.

Common misconceptions

• "Taking more trials fixes the error." More trials reduce random scatter and do nothing at all to a systematic offset. A clock that always starts 0.04 s late is just as wrong after fifty trials, and only a graph with an unexpected intercept will tell you.
• "Precise and accurate mean the same thing." Five readings of 11.30, 11.32, 11.31, 11.33 and 11.31 m/s² are precise and wrong. Precision is agreement among your own readings; accuracy is agreement with the true value.
• "More decimal places means a better measurement." Writing g = 11.34567 m/s² from a stopwatch reading of 0.42 s claims seven-figure precision from a two-figure measurement. Your result cannot be more precise than your worst input.
• "An odd-looking point should be deleted." Record it, plot it, and investigate it. If you can identify what went wrong on that trial you may exclude it and say why; deleting points because they spoil the line is the one thing that turns a mistake into misconduct.
• "The best-fit line should pass through the data points." It should pass through the trend, with points scattered either side, and you read your two points off the line you drew, never off two raw data points.

The short version

A measurement is a number, an uncertainty and a procedure, and a result missing any of the three is not yet a result. One trial gives no spread and so no uncertainty. Choose the experiment so that the quantity you measure worst comes out large, and remember that powers multiply percentage uncertainties: a 1.6 percent error in t is 3.2 percent in t². Vary something, plot a line, and linearise first by rearranging the relationship until it matches y = mx + b, because a slope averages all your points and an unexpected intercept exposes a systematic error that no amount of repetition would have caught. State what the slope means and in what units. And when you account for a discrepancy, name the mechanism, give its direction, and estimate its size.

Sources

1. OpenStax. (2022). 1.3 Accuracy, precision, and significant figures. In College Physics 2e. openstax.org
2. National Institute of Standards and Technology. (n.d.). Uncertainty of measurement results. NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
3. OpenStax. (2016). 1.6 Significant figures. In University Physics Volume 1. openstax.org
4. Knight, R. D., Jones, B., and Field, S. (2019). College Physics: A Strategic Approach (4th ed.), ch. 1 and appendix on data analysis. Pearson.
5. College Board. (2024). AP Physics 1: Algebra-Based course and exam description, science practices 3 and 4. New York: College Board.

Eighty metres across, and where you actually land

A river is 80.0 m wide and the current runs at 1.20 m/s. You have a boat that does 2.50 m/s through still water, and there is a jetty directly opposite your launch point. Point the bow straight at the jetty, open the throttle, and you will not reach it. You will land 38.4 m downstream of it, every time, and no amount of steering effort during the crossing changes that as long as the bow stays pointed across.

Working out where you land, and what it costs to land at the jetty instead, is a vector problem. This lesson builds the machinery on that river and then compares three ways of making the crossing, because the comparison is where the ideas stop being definitions.

Scalars, vectors, and why 5 plus 8 can be 11.4

A scalar has size only: 3.0 kg, 12 s, 250 J. A vector has size and direction, and the ones in this course are displacement, velocity, acceleration, force, momentum and impulse. Speed is the scalar partner of velocity; distance is the scalar partner of displacement.

Vectors do not add like numbers. Walk 5.0 m east, then 8.0 m in a direction 60 degrees north of east. You have walked 13.0 m of pavement, but you are not 13.0 m from where you started. To find out where you are, break each vector into an east part and a north part, add the parts separately, and rebuild.

For a vector of magnitude A at angle theta measured from the positive x axis:

Ax = A cos(theta) and Ay = A sin(theta)

So the first leg is (5.0, 0) m. The second is (8.0 cos 60 degrees, 8.0 sin 60 degrees) = (4.0, 6.93) m. Add componentwise:

Rx = 5.0 + 4.0 = 9.0 m, Ry = 0 + 6.93 = 6.93 m

Rebuild the resultant with Pythagoras and the inverse tangent:

R = sqrt((9.0 m)2 + (6.93 m)2) = sqrt(81.0 + 48.0) = sqrt(129.0) = 11.4 m

theta = arctan(6.93 / 9.0) = arctan(0.770) = 37.6 degrees north of east

Eleven point four, not thirteen. The difference is not an approximation or an error; it is what "direction matters" means arithmetically. Two vectors add to the sum of their magnitudes only when they point the same way, and to the difference when they point in exactly opposite directions. Everything between those extremes gives something in between.

Remember: add components, never magnitudes. The only time you may add magnitudes directly is along a single line, and even then you must keep the signs.

Getting the direction right, which is where the marks go

Your calculator's inverse tangent returns an angle between -90 and +90 degrees, so it cannot tell the difference between a vector pointing up and right and one pointing down and left. Both give the same ratio. The repair is to look at the signs of the components before you trust the number.

Safer still, and what most examiners prefer: sketch the two components as the legs of a right triangle, work with the acute angle, and describe the direction in words. "37.6 degrees north of east" cannot be misread. "37.6 degrees" on its own can.

Relative velocity: the one equation, and how to read the subscripts

Relative velocity problems all reduce to adding two velocity vectors, and the bookkeeping trick is in the subscripts. Write v_BW for the velocity of the boat with respect to the water and v_WG for the velocity of the water with respect to the ground. Then

vBG = vBW + vWG

The inner subscripts match and cancel, and the outer ones survive: boat-with-respect-to-water plus water-with-respect-to-ground gives boat-with-respect-to-ground. If you ever need to reverse one, v_WB = -v_BW: the water moves backward past the boat exactly as fast as the boat moves forward through the water.

Three crossings, three costs

Now the river, with x across the river and y downstream. The boat does 2.50 m/s through the water, the current does 1.20 m/s downstream, and the river is 80.0 m wide.

Strategy A: point the bow straight across. Then v_BW = (2.50, 0) and v_WG = (0, 1.20), so v_BG = (2.50, 1.20) m/s. Here is the part that surprises people: the downstream current has no x component, so it cannot slow the crossing at all. The time to cross is set by the across component alone.

t = 80.0 m / 2.50 m/s = 32.0 s

drift = 1.20 m/s x 32.0 s = 38.4 m downstream

ground speed = sqrt(2.502 + 1.202) = sqrt(7.69) = 2.77 m/s, at arctan(1.20/2.50) = 25.6 degrees downstream of straight across

Strategy B: land at the jetty. Now you need the resultant to point straight across, which means the boat's upstream component must exactly cancel the current. If the bow is angled at theta upstream of the across direction, then

2.50 sin(theta) = 1.20, so sin(theta) = 0.480 and theta = 28.7 degrees upstream

What is left for crossing is the other component:

vacross = 2.50 cos(28.7 degrees) = 2.50 x 0.877 = 2.19 m/s

t = 80.0 m / 2.19 m/s = 36.5 s

Strategy C: split the difference, say 15.0 degrees upstream. Across component 2.50 cos 15 = 2.41 m/s, upstream component 2.50 sin 15 = 0.647 m/s, so the net downstream drift rate is 1.20 - 0.647 = 0.553 m/s. Time 80.0/2.41 = 33.2 s, drift 0.553 x 33.2 = 18.4 m.

Read the table and one general principle falls out. The minimum crossing time is always achieved by pointing straight across, whatever the current does, because only the across component shortens the trip. The minimum drift needs the bow angled upstream, and it always costs time. There is no strategy that wins both, and the reason is geometry rather than seamanship: a fixed-length vector split between two perpendicular jobs cannot be maximal at both.

One more thing the table settles. If the current were 2.50 m/s or faster, strategy B would be impossible, because sin(theta) = 2.50/2.50 = 1 needs the bow pointed straight upstream, leaving nothing for crossing. In a current faster than your boat you cannot reach the jetty at all, only choose how far downstream you land.

The same arithmetic, six kilometres up

A light aircraft flies at 220 km/h through the air, pointed due north. A wind blows from the west at 60.0 km/h. Take east as +x and north as +y.

vPA = (0, 220) km/h, vAG = (60.0, 0) km/h, so vPG = (60.0, 220) km/h

ground speed = sqrt(60.02 + 2202) = sqrt(3600 + 48400) = sqrt(52000) = 228 km/h

direction = arctan(60.0/220) = 15.3 degrees east of north

The aircraft is faster over the ground than through the air and is going the wrong way. To actually track due north, the pilot must turn into the wind by arcsin(60.0/220) = 15.8 degrees west of north, and the ground speed then drops to 220 cos(15.8 degrees) = 212 km/h. That angle has a name in aviation, the wind correction angle, and it is strategy B with wings.

Common misconceptions

• "The current makes the crossing take longer." Not when the bow points straight across. A downstream current has no component across the river, and perpendicular components do not talk to each other. It changes where you land, not when you arrive.
• "A boat pointed at the jetty will reach the jetty." The bow direction is the boat's velocity relative to the water. What matters is the resultant relative to the ground, and it points 25.6 degrees downstream of the bow in our example.
• "Vector magnitudes add: 5.0 m and 8.0 m give 13.0 m." Only if they point the same way. At 60 degrees apart the resultant is 11.4 m, and at right angles it would be 9.4 m.
• "The components are separate motions happening one after the other." They happen at once. The split into x and y is a bookkeeping device that works because perpendicular directions are independent, not a claim about what the object does first.
• "An inverse tangent always gives the right direction." It gives an angle between -90 and +90 degrees and cannot distinguish opposite directions. Check the signs of both components, or describe the angle in words relative to a named direction.

Putting it together

A vector is size plus direction, and the only reliable way to combine vectors is through components: A_x = A cos(theta), A_y = A sin(theta), add the x parts and the y parts separately, then rebuild with R = sqrt(R_x² + R_y²) and an angle you state in words. Relative velocities add as vectors with matching inner subscripts, v_BG = v_BW + v_WG. Perpendicular components are independent, which is why a 1.20 m/s current cannot lengthen a crossing made with the bow pointed across, and why the fastest crossing and the shortest crossing are different crossings. That independence is the whole idea behind the next lesson, where the two components are horizontal and vertical and the thing being thrown is a ball.

Sources

1. OpenStax. (2022). 3.2 Vector addition and subtraction: graphical methods. In College Physics 2e. openstax.org
2. OpenStax. (2022). 3.5 Addition of velocities. In College Physics 2e. openstax.org
3. The Physics Classroom. (n.d.). Vectors and projectiles. physicsclassroom.com
4. Giancoli, D. C. (2016). Physics: Principles with Applications (7th ed.), ch. 3. Pearson.

Two balls, one bench, one click

Put a steel ball on a bench 0.780 m above the floor and flick it off the edge at 1.60 m/s. Hold a second, identical ball at the bench edge and release it at the same instant. Two things happen that people find hard to believe until they hear them. The balls hit the floor at the same moment, one click and not two. And the flicked ball lands 0.64 m out from the foot of the bench.

Both facts come from one statement, and it is the only new idea in this lesson: the horizontal and vertical motions of a projectile are completely independent, and share nothing but the clock. Gravity pulls down. It has no sideways component, so it cannot change the horizontal velocity, and the horizontal velocity cannot change how fast the ball falls. This lesson works three projectile problems end to end with that one idea, and each one is set up the same way.

The setup that never changes

Take up as positive and the launch direction as positive x. Ignore air resistance, which the exam does throughout. Then:

The launch components come from trigonometry: v0x = v0 cos(theta) and v0y = v0 sin(theta), with theta measured from the horizontal.

The procedure, every time: write the two columns, find the time from the column you have most information about (almost always the vertical one), then carry that time across to the other column. Time is the bridge. Nothing else crosses.

Problem one: off the bench

A ball leaves a bench 0.780 m high, horizontally, at 1.60 m/s. Find the time of flight, the horizontal distance, and the speed and direction at impact.

Vertical first, because a horizontal launch means v_0y = 0 and the vertical problem is a pure drop.

0 = 0.780 m + 0 - 4.90t2, so t2 = 0.780/4.90 = 0.1592 s2, giving t = 0.399 s

Notice that v_0x never entered that calculation, which is exactly why the dropped ball and the flicked ball land together. Flick it at 10 m/s and it still takes 0.399 s to fall.

Horizontal next, using that time.

x = v0xt = (1.60 m/s)(0.399 s) = 0.638 m

Impact velocity. Horizontally, still 1.60 m/s. Vertically, vy = 0 - (9.80 m/s2)(0.399 s) = -3.91 m/s. Combine them as components of one vector:

v = sqrt((1.60)2 + (3.91)2) = sqrt(2.56 + 15.29) = sqrt(17.85) = 4.22 m/s

angle = arctan(3.91/1.60) = 67.7 degrees below the horizontal

This problem is also a measurement in disguise. Measure the bench height and the landing distance and you can recover the launch speed without ever timing anything: v_0x = x / sqrt(2h/g).

Problem two: a kick from level ground

A ball is kicked at 18.0 m/s at 35.0 degrees above the horizontal from level ground. Find the time of flight, the maximum height and the range.

Components. v0x = 18.0 cos(35.0) = 14.7 m/s, v0y = 18.0 sin(35.0) = 10.3 m/s

Time to the top, where v_y = 0: tup = 10.3/9.80 = 1.05 s. Since the launch and landing heights are equal, the flight is symmetric and the total time is twice that: t = 2.11 s.

Maximum height, from the vertical column with no time: 0 = (10.3)2 + 2(-9.80)(y - 0), so y = 106 / 19.6 = 5.43 m.

Range, from the horizontal column: x = (14.7 m/s)(2.11 s) = 31.1 m.

Assemble those steps in symbols and you get a formula worth knowing. The flight time is 2v₀sin(theta)/g and the range is v₀cos(theta) times that, so

R = 2v02 sin(theta) cos(theta) / g = v02 sin(2 theta) / g

using the double-angle identity. Check it: R = (18.0)2 sin(70.0 degrees)/9.80 = 324 x 0.940/9.80 = 31.1 m. It agrees, and it applies only when the landing height equals the launch height, which is why it is a shortcut and not a substitute for the columns.

Two consequences drop straight out of it. Since sin(2 theta) is largest when 2 theta = 90 degrees, the maximum range on level ground is at 45 degrees, giving R = v₀²/g = 33.1 m here. And since sin(70 degrees) = sin(110 degrees), a kick at 35 degrees and a kick at 55 degrees travel exactly the same distance. Complementary angles tie: the steep one goes higher and hangs longer, the shallow one moves faster horizontally, and the two effects cancel to the metre.

The upshot: the time of flight is set entirely by the vertical problem, and the range is that time multiplied by an unchanging horizontal speed. If a question changes the horizontal speed, the time does not move.

Problem three: thrown off a cliff, which the shortcut cannot touch

A ball is thrown from the top of a 25.0 m cliff at 12.0 m/s at 40.0 degrees above the horizontal, and lands on the beach below. Find the flight time, how far from the cliff base it lands, and its impact speed.

The launch and landing heights differ, so the flight is not symmetric and the range formula is simply wrong here. Back to the columns.

v0x = 12.0 cos(40.0) = 9.19 m/s, v0y = 12.0 sin(40.0) = 7.71 m/s

Vertical, with the beach at y = 0 and the launch at y₀ = 25.0 m:

0 = 25.0 + 7.71t - 4.90t2

4.90t2 - 7.71t - 25.0 = 0

t = [7.71 +/- sqrt(59.4 + 4(4.90)(25.0))] / 9.80 = [7.71 +/- sqrt(549)] / 9.80 = (7.71 + 23.4)/9.80 = 3.18 s

The negative root, -1.60 s, describes the parabola extended backward below the cliff and is discarded.

Horizontal: x = (9.19 m/s)(3.18 s) = 29.2 m from the base of the cliff.

Impact speed. The vertical component is vy = 7.71 - 9.80(3.18) = -23.5 m/s, and the horizontal is still 9.19 m/s, so

v = sqrt(9.192 + 23.52) = sqrt(84.5 + 552) = sqrt(636) = 25.2 m/s, at arctan(23.5/9.19) = 68.6 degrees below the horizontal

Check it another way, using the vertical equation that contains no time: vy2 = (7.71)2 + 2(-9.80)(0 - 25.0) = 59.4 + 490 = 549, so v_y = 23.4 m/s downward and the total speed is 25.2 m/s. Two routes, one answer.

Here is a result worth carrying: because the vertical equation with no time depends only on the drop and the initial vertical speed, the impact speed of that ball would be identical if it had been thrown at 12.0 m/s at 40 degrees below the horizontal, or straight down, or straight up. Only the flight time and the landing distance would change. Launch speed and height fix the landing speed; the angle redistributes it between time and distance.

Common misconceptions

• "A projectile's horizontal velocity gradually decreases." Not in this model. Gravity acts straight down and has no horizontal component, so v_x is the same at launch, at the top and at impact. Real air resistance does slow it, which is why a struck golf ball falls short of the ideal range.
• "A ball fired horizontally stays in the air longer than one dropped." They land together. The vertical problem does not know or care what is happening horizontally, which is what the bench experiment demonstrates in one audible click.
• "At the top of the arc the projectile is momentarily at rest." Only the vertical component is zero. A ball kicked at 18.0 m/s at 35 degrees is still moving at 14.7 m/s horizontally at the peak, and its acceleration is 9.80 m/s² downward there as everywhere.
• "The range formula always works." R = v₀²sin(2 theta)/g assumes the projectile lands at its launch height. Use it for a kick on a flat field, never for a throw off a cliff or a shot from a raised platform.
• "Steeper is always further." Beyond 45 degrees on level ground, extra hang time no longer compensates for the lost horizontal speed, so the range falls. Every angle above 45 degrees is matched by a shallower one that goes the same distance.

What you now know

Projectile motion is two one-dimensional problems that share a stopwatch. Horizontally the acceleration is zero and the velocity never changes, so x = v_0xt. Vertically the acceleration is a steady 9.80 m/s² downward and all four kinematic equations apply. Solve the vertical column for the time, then carry that time into the horizontal column. For a launch and landing at the same height the flight is symmetric and R = v₀²sin(2 theta)/g, which peaks at 45 degrees and gives equal ranges for complementary angles; for anything else, including every cliff and every table, use the quadratic. At the top of the arc the vertical velocity is zero and the acceleration is not, and at impact the speed is set by the launch speed and the height dropped, not by the launch angle.

Sources

1. OpenStax. (2022). 3.4 Projectile motion. In College Physics 2e. openstax.org
2. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Projectile motion. phet.colorado.edu
3. Wikipedia contributors. (n.d.). Projectile motion. Wikipedia.
4. Knight, R. D., Jones, B., and Field, S. (2019). College Physics: A Strategic Approach (4th ed.), ch. 3. Pearson.
5. College Board. (2024). AP Physics 1: Algebra-Based course and exam description, Unit 1. New York: College Board.

A bathroom scale in a lift reads 715 N

A student with a mass of 65.0 kg stands on a bathroom scale in a hotel lift. Standing still in the lobby the scale reads 637 N. The doors close, and for about two seconds the scale reads 715 N. What is the lift doing?

The tempting answer is that the lift is going up. That is not what the scale measures. Work it properly and you will find the lift is accelerating upward at 1.20 m/s², which it could be doing while moving up and gaining speed, or while moving down and slowing to a halt. The scale reads the force it pushes up on your feet with, and that force depends on the acceleration, not on the velocity. Getting from 715 N to 1.20 m/s² takes one diagram and one equation, and this lesson is about drawing that diagram correctly, because almost every lost mark in mechanics traces back to a bad one.

The three laws, stated in the form you will use them

Newton's laws of motion are three sentences that you will use in very different proportions.

1. First law. An object with zero net force keeps a constant velocity, which includes staying at rest. Motion does not need a cause; changes in motion do.
2. Second law. The net force equals mass times acceleration, Fnet = ma, as a vector equation, which means it holds separately in each direction.
3. Third law. If A pushes B, then B pushes A with an equal and opposite force of the same type. The two forces act on different objects, which is the whole content of the law and the part most often dropped.

The second law does almost all of the work, and the phrase that matters in it is "net". One force on a list tells you nothing; the vector sum is what accelerates things.

Drawing the diagram, with the rules that keep it honest

A free-body diagram is a single dot or box representing one object, with one labelled arrow for each force acting on it, drawn from the dot outward. Six rules, and breaking any of them produces wrong answers that look tidy.

• Choose one object and state which. A diagram for "the blocks" is a diagram for nothing.
• Only forces on that object. The force the block exerts on the table belongs on the table's diagram, never on the block's.
• Name the agent of every force. Not "F" but "the rope's pull" or "the table's push". If you cannot name what is doing the pushing, the force does not exist.
• No ma arrow. Mass times acceleration is the result of the forces, not one of them. Drawing it means counting the same thing twice.
• No force of motion. A sliding puck has no forward arrow. Nothing pushes it forward; it is simply not being stopped quickly.
• Arrow lengths should roughly match magnitudes, so that a glance at the diagram already suggests the answer.

In this course the cast of forces is short: weight (the Earth's pull, magnitude mg, always straight down), the normal force (a surface's push, always perpendicular to the surface), tension (a rope's pull, along the rope, away from the object), friction (along the surface, opposing relative sliding), a spring force, and whatever someone applies by hand.

In short: the normal force is whatever it needs to be to stop the object sinking into the surface, and it equals mg only in the special case of a horizontal surface with nothing else pushing vertically.

The lift, solved

The object is the student. Two forces act on her: her weight, 637 N down, and the scale's normal push, N up. Take up as positive.

Fnet = N - mg = ma

715 N - 637 N = (65.0 kg)a

a = 78 N / 65.0 kg = 1.20 m/s2 upward

Run the same equation for every case and you get the whole family of lift problems in one table. The student's mass never changes; what the scale reports is the normal force, which is why people describe it loosely as apparent weight.

The last row is worth a pause. In free fall the scale reads zero, and the student floats: that is what weightlessness actually is. Not the absence of gravity, which is still pulling her at 637 N, but the absence of a supporting force. Astronauts on the space station are in exactly that condition, falling freely around the Earth with nothing pressing on their feet.

Tilting the axes, which is the one trick for inclines

A 4.00 kg block sits on a frictionless inclined plane at 25.0 degrees and is released. Find its acceleration and the normal force.

Drawn vertically and horizontally this problem is ugly, because two of the three forces have components in both directions. Tilt the axes instead: let x run down the slope, in the direction the block will actually accelerate, and y run perpendicular to the surface. Now the normal force lies entirely along y and the acceleration lies entirely along x. Only the weight needs resolving.

Drop a perpendicular from the weight vector to each axis and the geometry gives, for an incline of angle theta,

weight component along the slope = mg sin(theta)

weight component into the surface = mg cos(theta)

Which one gets the sine catches people out, so sanity-check it at the extremes. At theta = 0 the surface is flat and nothing should pull the block along it: sin(0) = 0, correct. At theta = 90 degrees the surface is vertical and the block should simply fall with nothing pressing on the surface: cos(90) = 0 gives zero normal force, correct.

Perpendicular to the surface, the block does not accelerate, so the components balance:

N - mg cos(theta) = 0, so N = (4.00 kg)(9.80 m/s2) cos(25.0 degrees) = 39.2 x 0.906 = 35.5 N

Note that N is less than the 39.2 N weight, and grows smaller as the slope steepens. That result matters later: friction depends on N, so a steeper slope offers less grip as well as more pull.

Along the surface, with x down the slope:

mg sin(theta) = ma, so a = g sin(theta) = (9.80)(0.423) = 4.14 m/s2 down the slope

The mass cancelled, exactly as in free fall, and for the same reason. On a frictionless incline every object accelerates at g sin(theta) whatever its mass, which is the observation Galileo used to slow gravity down to a speed his water clocks could measure.

Add friction and nothing about the method changes

Same block, same slope, now with a coefficient of kinetic friction of 0.200 between block and ramp. The block slides down, so friction acts up the slope, opposing the sliding.

The perpendicular equation is untouched: N = 35.5 N. The friction force is

f = (0.200)(35.5 N) = 7.10 N

Along the slope, taking down-slope as positive:

mg sin(theta) - f = ma

(39.2 N)(0.423) - 7.10 N = (4.00 kg)a

16.6 N - 7.10 N = 9.50 N = (4.00 kg)a, so a = 2.37 m/s2 down the slope

Friction has cut the acceleration roughly in half. And here is a question worth carrying into the next lesson: how shallow would the slope have to be for the block to slide at constant speed, or not to start at all?

Common misconceptions

• "Motion requires a continuous force." This is the single most persistent wrong idea in mechanics and it is older than Newton: Aristotle taught it. A hockey puck sliding across ice has no forward force on it at all. It keeps going because nothing substantial is stopping it, and it eventually stops because friction is.
• "The normal force always equals the weight." It equals mg cos(theta) on an incline, more than mg in an upward-accelerating lift, less in a downward-accelerating one, and zero in free fall. It is a response, not a twin.
• "An object moving at constant velocity has no forces on it." It has zero net force. A car at a steady 30 m/s has engine thrust, drag, friction, weight and normal force all acting, and they cancel.
• "Mass and weight are the same thing." Mass is in kilograms and is the same everywhere; weight is a force in newtons and depends on where you are. A 65.0 kg student weighs 637 N on Earth and 105 N on the Moon.
• "Weightlessness means gravity has switched off." It means no surface is supporting you. The astronaut and the falling lift passenger are both fully subject to gravity, which is precisely why they are falling.

Pulling it together

Newton's second law is a procedure, not a formula: pick one object, draw every force acting on it with the agent named, choose axes that make the acceleration lie along one of them, then write F_net = ma once per direction and solve. On a slope, tilt the axes so x runs down the slope, and resolve only the weight, into mg sin(theta) along the surface and mg cos(theta) into it. The normal force is whatever the surface must supply, which is mg cos(theta) on an incline and 715 N rather than 637 N in a lift accelerating up at 1.20 m/s². A frictionless incline gives a = g sin(theta) for every mass. And nothing needs a force to keep moving, which is the idea to overwrite if you overwrite only one.

Sources

1. OpenStax. (2022). 4.3 Newton's second law of motion: concept of a system. In College Physics 2e. openstax.org
2. OpenStax. (2022). 4.2 Newton's first law of motion: inertia. In College Physics 2e. openstax.org
3. The Physics Classroom. (n.d.). Newton's laws. physicsclassroom.com
4. Giancoli, D. C. (2016). Physics: Principles with Applications (7th ed.), ch. 4. Pearson.

An answer that says the crate rolls over the person pushing it

Here is a worked solution from a student's homework. The problem: a 40.0 kg crate sits on a concrete floor. The coefficient of static friction is 0.500 and the coefficient of kinetic friction is 0.350. You push horizontally with 150 N. Find the friction force on the crate and its acceleration.

The student writes: The normal force is N = mg = (40.0)(9.80) = 392 N. Friction is f = (0.500)(392 N) = 196 N, opposing my push. So the net force is 150 N - 196 N = -46 N, and a = -46/40.0 = -1.15 m/s². The crate accelerates backward at 1.15 m/s².

Every number in that solution is arithmetically correct, and the conclusion is that a stationary crate on a level floor starts moving toward you because you pushed it away from you. Nothing in the universe does that. One line is wrong, and finding it is the content of this lesson, because the same line is wrong in a third of all friction answers.

The wrong line, and the inequality that fixes it

The error is f = (0.500)(392 N) = 196 N. That is not the static friction force. It is the largest static friction force the surfaces can produce. Static friction obeys an inequality, not an equation:

fs is less than or equal to musN

Within that limit, static friction takes exactly the value needed to keep the object from sliding, and not a newton more. Push with 40 N and static friction is 40 N. Push with 150 N and static friction is 150 N. The crate does not move, the net force is zero, and the acceleration is zero. Only when your push exceeds 196 N does the surface run out of grip and the object break loose.

Kinetic friction is different in kind. Once sliding is happening, the force settles to a fixed value that does not adjust itself:

fk = mukN, an equality

So the correct answer to the homework problem is: friction is 150 N backward, the net force is zero, and the crate does not accelerate at all. The 196 N never appears in the answer; it appears in the test you perform to find out which regime you are in.

Bottom line: with static friction, first compute mu_sN and compare it with the applied force. If the push is smaller, friction equals the push and nothing moves. If the push is larger, the object slides and you switch to mu_kN.

The lurch, which is what mu_k being smaller feels like

For almost every pair of materials mu_k is less than mu_s, and the consequence is physical rather than algebraic. Keep increasing your push on the crate. At 196 N it breaks free, and at that instant the resisting force drops from 196 N to

fk = (0.350)(392 N) = 137 N

If you are still pushing with 196 N, the net force suddenly becomes 196 N - 137 N = 59 N, and

a = 59 N / 40.0 kg = 1.48 m/s2

The crate lurches away from you and you stumble forward. That is the experience of moving furniture, and it is the graph below in words: plot the friction force against the applied force and you get a straight line of slope 1 rising to (196, 196), then a vertical drop to 137, then a horizontal line at 137 however hard you push. The peak is the breakaway point, and the step down is why a stuck drawer flies open.

A few properties of friction that the model builds in, all of them approximations good enough for this course:

• It does not depend on contact area. A brick on its side and the same brick on its end need the same force to drag. Doubling the area halves the pressure, and the two effects cancel.
• Kinetic friction does not depend much on speed. Sliding at 0.5 m/s or 5 m/s costs about the same force.
• It is proportional to the normal force. This is why loading a van makes it harder to skid and easier to stop, and why a cyclist leaning into a turn has no more grip than one sitting upright.
• The coefficients are dimensionless, being a ratio of two forces, and typically run from 0.04 for waxed skis on snow to about 1.0 for rubber on dry concrete.

The ramp that measures mu_s for you

Put the crate on a plank and slowly raise one end. At some angle it begins to slide. That angle, the angle of repose, gives you mu_s directly, and the derivation takes three lines.

At the instant sliding begins, the block is still stationary, so along the slope the friction is at its maximum and exactly balances the weight component:

mg sin(theta) = mus N

Perpendicular to the slope, N = mg cos(theta). Substitute:

mg sin(theta) = mus mg cos(theta)

The mg cancels on both sides, leaving

tan(theta) = mus

For mu_s = 0.500, the block starts to slide at theta = arctan(0.500) = 26.6 degrees. Two features of that result are worth more than the number. First, the mass cancelled, so a heavy crate and a light one slip at the same angle, which is easy to test and surprises people who expect the heavy one to grip better. Second, you now have an instrument: tilt, read the angle with a protractor or a phone's inclinometer, take the tangent, and you have measured a coefficient of friction without measuring a single force.

A braking distance, and why anti-lock brakes exist

A car travelling at 25.0 m/s brakes hard on dry asphalt where mu_s between tyre and road is 0.800 and mu_k is 0.600. How far does it take to stop, and does locking the wheels help?

First, a point that catches people: for a rolling tyre that is not skidding, the relevant coefficient is the static one, because the patch of rubber in contact with the road is not sliding along it. Braking at the limit of grip therefore uses mu_s.

Wheels rolling, at the limit of static friction. The only horizontal force is friction, so

ma = musmg, giving a = musg = (0.800)(9.80) = 7.84 m/s2

The mass cancels again, which is why a loaded lorry and a small car stop in similar distances on the same surface. Then from the kinematic equation with no time in it:

0 = (25.0)2 + 2(-7.84)d, so d = 625 / 15.68 = 39.9 m

Wheels locked and skidding. Now the rubber really is sliding, so mu_k applies: a = (0.600)(9.80) = 5.88 m/s2 and d = 625 / 11.76 = 53.1 m.

Thirteen metres, which is three car lengths, purely from letting the wheels lock. That difference is what anti-lock braking systems are built to recover, by releasing and reapplying the brakes many times a second to keep each tyre just below the point of skidding. A locked wheel is also a steered wheel that no longer steers, which is the other half of the argument.

Friction as the force that moves you

One more correction to the intuition. Friction is usually introduced as the thing that slows objects down, and then it turns out to be the force that propels almost everything that moves on land.

When you walk, your shoe pushes backward on the ground and static friction pushes forward on your shoe. That forward force is what accelerates you. On wet ice it vanishes and you cannot start walking at all, which is the experiment that proves the point. A car's driven wheels do the same: the tyre pushes backward on the road, the road pushes forward on the tyre. Friction opposes relative sliding at the contact, which in these cases means it points forward in the direction you are going.

Common misconceptions

• "Static friction equals mu_sN." It equals whatever is needed, up to that maximum. The product mu_sN is a ceiling you test against, and using it as the actual force produces answers like a crate accelerating into the person pushing it.
• "Friction always opposes motion." It opposes relative sliding between surfaces. The friction on your shoe, on a driven tyre, and on the feet of a sprinter all point forward, and they are the forces that produce the motion.
• "A heavier object slides at a smaller ramp angle." The angle of repose satisfies tan(theta) = mu_s, with no mass in it. A 1 kg block and a 20 kg block on the same plank slip at the same angle.
• "Wider tyres give more friction because of the larger area." In this model friction is independent of contact area. Wide tyres help for reasons outside it: heat dissipation, wear, and keeping rubber temperature in its grippy range.
• "Locking the wheels stops a car fastest." Skidding switches you from mu_s to the smaller mu_k, lengthening the stop from 39.9 m to 53.1 m from 25 m/s, and it removes your steering at the same time.

The takeaway

Two regimes and one test. Static friction adjusts itself to prevent sliding and can reach at most mu_sN, so compare the applied force with that ceiling first: below it nothing moves and friction equals the push, above it the object slides and friction becomes the fixed mu_kN. Because mu_k is smaller, an object that has just broken loose accelerates suddenly, which is why furniture lurches. The coefficients are dimensionless, do not depend on contact area, and enter the answer only through N, so masses often cancel: tan(theta) = mu_s at the angle of repose, and a = mu g for a car braking at the limit. And friction is what propels walkers and cars forward, not only what slows them down.

Sources

1. OpenStax. (2022). 5.1 Friction. In College Physics 2e. openstax.org
2. OpenStax. (2022). 5.2 Drag forces. In College Physics 2e. openstax.org
3. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Forces and motion: basics. phet.colorado.edu
4. Knight, R. D., Jones, B., and Field, S. (2019). College Physics: A Strategic Approach (4th ed.), ch. 5. Pearson.
5. College Board. (2024). AP Physics 1: Algebra-Based course and exam description, Unit 2. New York: College Board.

Two students, one cart, and an argument that has run for three centuries

Maya says: a horse pulls a cart with 800 N. By Newton's third law the cart pulls back on the horse with 800 N. Those forces are equal and opposite, so they add to zero, so the net force is zero, so the cart cannot possibly accelerate. Since carts obviously do accelerate, the third law must be an approximation that fails when things move.

Ravi says: the third law is fine, but it must only apply when nothing is accelerating. A horse standing still pushes on the cart and the cart pushes back equally. Once the horse starts pulling harder than the cart pulls back, the pair is broken and the cart moves.

Both students are wrong, and they are wrong in ways that are worth more than a correct answer, because each mistake is a specific misreading of the law. This lesson settles the argument, and then uses the settled version to solve the connected-object problems the exam is built from.

What the law says, word by word

Newton's third law: if object A exerts a force on object B, then B exerts a force on A that is equal in magnitude, opposite in direction, of the same type, and acting at the same instant.

Four clauses, and the fatal one for both students is hidden in plain sight: the two forces act on different objects. Forces only add up when they act on the same object. The horse's pull acts on the cart; the cart's pull acts on the horse. They are on different free-body diagrams and can never appear on the same one, so they cannot cancel any more than your rent and your neighbour's rent cancel because they are equal.

Maya's error is adding two forces that live on different diagrams. Ravi's error is subtler and more common: he imagines the pair can become unequal. It cannot. The pair is exactly equal at every instant, whether the objects are at rest, accelerating, or being demolished.

The core of it: to find out whether something accelerates, draw the diagram for that object alone and add only the forces on it. Third-law partners never appear together on one diagram.

Settling the cart

Draw the cart's diagram: the horse's forward pull of 800 N, and friction from the ground backward, say 600 N. Net force forward, 200 N. The cart accelerates. The cart's pull on the horse is nowhere on this diagram, because it is not a force on the cart.

Now draw the horse's diagram: the cart's backward pull of 800 N, and the ground's forward friction on the horse's hooves, which is what the horse generates by pushing backward on the ground. If that friction exceeds 800 N, the horse accelerates. If the ground is icy and cannot supply it, the horse's hooves slip and nothing moves, which is not a failure of the third law but a shortage of friction.

That last row is the one people never think about. The Earth pulls a 20 kg cart down with 196 N, and the cart pulls the Earth up with 196 N. The Earth does accelerate toward the cart, at 196 N divided by about 6 x 10²⁴ kg, which is why nobody notices. Equal forces, wildly unequal accelerations, because F = ma and the masses differ by twenty-three orders of magnitude.

The same arithmetic settles the classic question about a lorry hitting a mosquito. The forces during the impact are exactly equal. The mosquito's acceleration is enormous and the lorry's is negligible, entirely because a = F/m. Nothing about the collision is unfair; only the masses are.

Two blocks and a string, done twice

A 3.00 kg block and a 2.00 kg block sit on a frictionless table, joined by a light string. You pull the 3.00 kg block horizontally with 20.0 N. Find the acceleration and the string's tension.

Route one: treat the pair as one system. Legal here because the string is inextensible, so both blocks have the same acceleration, and because tension is internal to the system and therefore cancels in the sum.

a = F / (m1 + m2) = 20.0 N / 5.00 kg = 4.00 m/s2

Route two: one diagram per block, which is the only route that gives the tension. The 2.00 kg block has exactly one horizontal force on it, the string's pull:

T = m2a = (2.00 kg)(4.00 m/s2) = 8.00 N

Check against the other block: F - T = 20.0 - 8.00 = 12.0 N, and m1a = (3.00)(4.00) = 12.0 N. Consistent.

Now change one thing. Move the 20.0 N to the other end so that you pull the 2.00 kg block and it drags the 3.00 kg block. The acceleration is unchanged at 4.00 m/s², because the total mass and the total external force are the same. But the tension is now

T = m1a = (3.00 kg)(4.00 m/s2) = 12.0 N

Half as much again. The tension is always whatever it takes to accelerate everything behind the string, so pulling the lighter block first means the string has more mass to drag. Which end you pull matters to the rope even when it does not matter to the motion, which is why couplings on trains are rated for the load behind them.

The Atwood machine

Two masses hang from a light string over a frictionless pulley of negligible mass. This is an Atwood machine, built by George Atwood in 1784 for exactly the reason we still use it: it dilutes gravity to a manageable acceleration you can time by hand.

Take m₁ = 3.00 kg and m₂ = 5.00 kg. The string is inextensible, so as m₂ descends by some distance, m₁ rises by the same distance in the same time: the magnitudes of their accelerations are equal even though their directions differ. That single geometric fact is what makes the problem solvable.

Choose the positive direction to follow the motion: down for m₂, up for m₁. Then each block gets one equation.

m2: m2g - T = m2a

m1: T - m1g = m1a

Add the two equations. The tension cancels, which is the whole reason for adding them:

(m2 - m1)g = (m1 + m2)a

a = (m2 - m1)g / (m1 + m2) = (2.00 kg)(9.80 m/s2) / 8.00 kg = 2.45 m/s2

Read that formula for a second. The driving force is only the difference in the weights, but the inertia being accelerated is the sum of the masses. Make the two masses nearly equal and the acceleration becomes tiny, which is how Atwood turned a fall of a few metres into an event lasting several seconds.

Now the tension, from either equation:

T = m1(g + a) = (3.00 kg)(9.80 + 2.45) = (3.00)(12.25) = 36.8 N

Check with the other: T = m2(g - a) = (5.00)(9.80 - 2.45) = (5.00)(7.35) = 36.8 N. Agreement.

Sanity-check the size of that number, because it is where wrong answers announce themselves. The two weights are 29.4 N and 49.0 N, and the tension is 36.8 N, comfortably between them. It must be: the tension is bigger than the light block's weight, or that block would not rise, and smaller than the heavy block's weight, or that block would not fall. Any tension outside that window is an arithmetic error, and a tension equal to either weight would mean no acceleration at all.

The modified Atwood, which is the one you will meet in a lab

A 2.00 kg cart on a frictionless horizontal track is joined by a string over a pulley at the table edge to a 0.500 kg hanging mass. The same method applies, with the cart's equation written horizontally and the hanger's vertically.

cart: T = mca

hanger: mhg - T = mha

Adding: mhg = (mc + mh)a, so a = (0.500)(9.80)/2.50 = 4.90/2.50 = 1.96 m/s2

T = (2.00 kg)(1.96 m/s2) = 3.92 N

Notice that T = 3.92 N is less than the hanger's weight of 4.90 N. It has to be, because the hanger is accelerating downward, and if the string pulled up as hard as gravity pulled down the hanger would hang still. Students who answer T = 4.90 N have assumed equilibrium in a problem that explicitly is not in equilibrium, and it is the single commonest error in this topic.

Common misconceptions

• "Action and reaction cancel, so nothing can accelerate." They act on different objects and never appear on the same free-body diagram. What accelerates the cart is the sum of the forces on the cart alone.
• "The bigger object exerts the bigger force." A lorry and a mosquito exert equal and opposite forces on each other. The accelerations differ enormously because a = F/m and the masses differ enormously.
• "The tension in an Atwood machine equals the heavier weight." It lies strictly between the two weights, at 36.8 N for a 3.00 kg and 5.00 kg pair. Equal to either weight would mean zero acceleration.
• "The heavier mass falls at g." It falls at (m₂ - m₁)g/(m₁ + m₂), which is 2.45 m/s² here, because the string is holding it back while it drags the other mass up.
• "The system shortcut works on any group of objects." It works only when every part shares the same magnitude of acceleration, and it can never give you an internal force such as tension. For that you must go back to individual diagrams.

Summing up

Third-law pairs are equal at every instant, act on two different objects, and therefore never cancel; the cart accelerates because of what is on the cart's diagram, and the horse because of what is on the horse's. Equal forces produce unequal accelerations whenever the masses differ, which covers lorries, mosquitoes and the Earth. For connected objects, use the system as one body to get the shared acceleration, then go back to a single object to get the tension, which depends on how much mass the string has to drag and so changes when you pull from the other end. An Atwood machine accelerates at (m₂ - m₁)g/(m₁ + m₂), driven by a difference and resisted by a sum, and its tension always lies strictly between the two weights.

Sources

1. OpenStax. (2022). 4.4 Newton's third law of motion: symmetry in forces. In College Physics 2e. openstax.org
2. OpenStax. (2022). 4.5 Normal, tension, and other examples of forces. In College Physics 2e. openstax.org
3. Wikipedia contributors. (n.d.). Atwood machine. Wikipedia.
4. Knight, R. D., Jones, B., and Field, S. (2019). College Physics: A Strategic Approach (4th ed.), ch. 5 and 7. Pearson.
5. College Board. (2024). AP Physics 1: Algebra-Based course and exam description, Unit 2. New York: College Board.

A curve, a speed, and whether the car makes it

A car enters an unbanked bend of radius 45.0 m at 22.0 m/s on dry asphalt where the coefficient of static friction is 0.850. Does it get round?

The instinct is to compare 22.0 m/s with some speed limit, but the road does not know the speed limit. What matters is whether friction can supply the acceleration the turn demands. The turn demands

ac = v2/r = (22.0 m/s)2 / 45.0 m = 484 / 45.0 = 10.8 m/s2

and the road can supply at most

amax = musg = (0.850)(9.80 m/s2) = 8.33 m/s2

Demand exceeds supply, so the car slides off the bend, and the mass never entered the comparison. The fastest it could have taken that curve is v = sqrt(musgr) = sqrt(8.33 x 45.0) = sqrt(375) = 19.4 m/s, about 70 km/h. Everything in this lesson is a variation on that comparison, so the first job is to earn the v²/r.

Where v squared over r comes from, with no calculus

Move at a constant speed v around a circle of radius r. Speed is constant; velocity is not, because its direction changes continuously, and a change in velocity is an acceleration. That is the whole reason circular motion is accelerated motion even when the speedometer never moves.

To find how big the acceleration is, draw two pictures. In the first, the position vectors to the object at two nearby instants, separated by a small angle theta, with the arc between them of length s. In the second, the two velocity vectors from those instants, drawn tail to tail. Because velocity is always perpendicular to the radius, the angle between the two velocity vectors is the same theta, and the two triangles are similar: one has two sides of length r with a chord between them, the other two sides of length v with the change in velocity between them.

Similar triangles give equal ratios of corresponding sides:

|change in v| / v = s / r

Rearrange and divide by the time interval:

|change in v| / (time) = (v / r)(s / time) = (v / r)(v) = v2 / r

because s divided by time is just the speed. So

ac = v2 / r, directed toward the centre of the circle

The direction falls out of the same picture: as the angle shrinks, the change-in-velocity vector turns to point at right angles to the velocity, which is straight at the centre. That is what centripetal means, from the Latin for centre-seeking.

Two other forms are worth having. If the object takes a time T for one revolution, then v = 2 pi r / T, and substituting gives

ac = 4 pi2 r / T2

Use the first form when you know the speed, the second when you know the period, and never both at once.

Centripetal force is a job, not a new force

Here is the idea that stops most of the errors in this topic. There is no such thing as "the centripetal force" in the way there is a gravitational force or a normal force. Centripetal force is the name for the net inward force, whatever real forces happen to be producing it. Newton's second law, applied along the inward radial direction, reads

Fnet, inward = m v2 / r

So the procedure is unchanged from Module 3: draw the free-body diagram, identify which real forces have inward components, and set their sum equal to mv²/r. Never add an extra arrow labelled "centripetal force"; that would count the same physics twice, exactly as drawing ma on a diagram does.

Worth holding on to: write the radial equation with inward as positive, list only forces that are actually there, and let mv²/r sit alone on the right-hand side as the answer, never as an entry in the list.

Banking a curve so friction is not needed

Highway engineers tilt bends inward so that the road itself does the turning. Consider a car on a frictionless bank at angle theta. Two forces act: the weight mg straight down, and the normal force N perpendicular to the road surface, therefore tilted by theta from the vertical.

Resolve N into vertical and horizontal parts. Vertically the car does not accelerate, so

N cos(theta) = mg

Horizontally, the inward component of N supplies the whole centripetal requirement:

N sin(theta) = m v2 / r

Divide the second equation by the first. Both N and m cancel, which is the elegant part:

tan(theta) = v2 / (r g)

For a bend of radius 120 m designed for 25.0 m/s:

tan(theta) = (25.0)2 / [(120)(9.80)] = 625 / 1176 = 0.531, so theta = 28.0 degrees

Read what cancelled. The mass is gone, so the same bank angle works for a motorcycle and a loaded lorry. And the equation contains one speed, the design speed: at exactly 25.0 m/s the car needs no friction at all and would hold the bend on sheet ice. Below that speed it tends to slide down the bank and friction must act up the slope; above it, friction must act down the slope. That is why banked bends carry advisory speeds rather than limits, and why they are treacherous in ice at any speed but the design one.

The vertical circle, where the answer is a minimum speed

Swing a bucket of water in a vertical circle on a 0.800 m rope. At the top, both the rope's tension and the weight point downward, that is, toward the centre. So

T + mg = m v2 / r

The rope can pull but not push, so the smallest possible tension is zero. Set T = 0 to find the slowest speed at which the water still makes it:

mg = m v2 / r, and the mass cancels, leaving vmin = sqrt(gr) = sqrt((9.80)(0.800)) = sqrt(7.84) = 2.80 m/s

Below 2.80 m/s at the top, gravity alone provides more inward force than the circular path requires, and the water leaves the circle, which is to say it lands on you. The corresponding period is Tperiod = 2 pi r / v = 5.03 / 2.80 = 1.80 s, so a swing of about a revolution every 1.8 seconds is the threshold.

Note what is not happening at the top. Nothing is holding the water up. The water is falling, at 9.80 m/s² like everything else, and the bucket is falling out of its way just as fast. Exactly the same account explains a rider at the top of a loop and an astronaut in orbit.

Common misconceptions

• "Something throws you outward in a turn." The centrifugal force is not a force in the ground frame. Your body continues in a straight line, as the first law requires, while the car curves away beneath you; the door then pushes you inward. The sensation of being flung out is your inertia meeting the door, and there is no agent exerting an outward force on you at all.
• "Centripetal force is an extra force to draw on the diagram." It is the name of the net inward force, made up of real forces you have already drawn. Adding it as a separate arrow double-counts.
• "Constant speed means no acceleration." Velocity includes direction. A car going round a bend at a steady 22.0 m/s is accelerating at 10.8 m/s², which is more than g.
• "A heavier car takes a bend worse." The maximum speed on a flat bend is sqrt(mu_sgr), with no mass in it, and the design speed for a bank is sqrt(rg tan(theta)), also with no mass in it. Both the demand and the supply scale with mass.
• "Water stays in the bucket because the centrifugal force holds it in." It stays because the bucket bottom keeps getting in its way. The water is in free fall at the top; the bucket is simply falling on the same curve.

What to remember

Turning is accelerating, because velocity has a direction. Similar triangles give the size of that acceleration as v²/r directed at the centre, or 4 pi²r/T² if you know the period instead of the speed. Centripetal force is a role filled by real forces, so the method never changes: draw the diagram, set the net inward force equal to mv²/r, and solve. On a flat bend friction plays the role and the maximum speed is sqrt(mu_sgr), mass-independent, which is why a 22.0 m/s car on a 45.0 m bend with mu_s = 0.850 leaves the road. Banking lets the normal force do the job, with tan(theta) = v²/(rg) at the design speed. At the top of a vertical circle the minimum speed is sqrt(gr), the point at which the support goes to zero and the object is simply falling along the path it was going to take anyway.

Sources

1. OpenStax. (2022). 6.2 Centripetal acceleration. In College Physics 2e. openstax.org
2. OpenStax. (2022). 6.3 Centripetal force. In College Physics 2e. openstax.org
3. The Physics Classroom. (n.d.). Circular and satellite motion. physicsclassroom.com
4. Knight, R. D., Jones, B., and Field, S. (2019). College Physics: A Strategic Approach (4th ed.), ch. 6. Pearson.

The calculation Newton did not publish for twenty years

Some time around 1666 Isaac Newton asked whether the force that pulls an apple down is the same force that keeps the Moon in its orbit. The test he devised is one any student in this course can now repeat, and it rests on a guess: that the force weakens as the square of the distance.

The Moon sits about sixty Earth radii from the centre of the Earth. If gravity falls as one over r squared, then at the Moon's distance it should be weaker by a factor of 60² = 3600, so the Moon's acceleration should be

a = g / 3600 = 9.80 / 3600 = 2.72 x 10-3 m/s2

Now compute the Moon's actual acceleration from its orbit, using the centripetal relation from the last lesson. The Moon's orbital radius is 3.84 x 10⁸ m and its period is 27.3 days, which is 2.36 x 10⁶ s.

ac = 4 pi2r / T2 = (39.48)(3.84 x 108) / (2.36 x 106)2 = 1.52 x 1010 / 5.57 x 1012 = 2.73 x 10-3 m/s2

Two numbers obtained from completely different data, a falling apple and a lunar calendar, agreeing to three figures. That agreement is the evidence that one law governs both, and the rest of this lesson follows the consequence all the way out to a satellite parked over the equator.

The law itself

Newton's law of universal gravitation: every pair of masses attracts, along the line joining them, with a force

F = G m1 m2 / r2

where G = 6.674 x 10^-11 N m²/kg² and r is the distance between the centres, not between the surfaces. Three features of that equation are worth noticing before using it.

• The force is a third-law pair. The Earth pulls you with 637 N and you pull the Earth with 637 N. Same force, hugely different accelerations.
• G is tiny, which is why you feel no attraction to the person sitting next to you. Two 70 kg people 1.0 m apart attract with about 3 x 10^-7 N, roughly the weight of a grain of fine sand.
• Distance is measured to the centre. A uniform sphere pulls external objects exactly as if all its mass sat at its centre, a result Newton had to invent new mathematics to prove and which is why r for a satellite is the Earth's radius plus the altitude.

That last point gives g immediately. For an object of mass m on the surface, its weight mg must equal the gravitational force, so

mg = GMEm / RE2, and cancelling m, g = GME / RE2

Check it with the Earth's mass 5.97 x 10²⁴ kg and radius 6.371 x 10⁶ m:

g = (6.674 x 10-11)(5.97 x 1024) / (6.371 x 106)2 = 3.984 x 1014 / 4.059 x 1013 = 9.81 m/s2

The product GM_E = 3.986 x 10¹⁴ m³/s² is known far more precisely than either factor separately, and it is the number to keep on hand for every orbit calculation.

A circular orbit, in four lines

A satellite of mass m orbits at radius r. Gravity is the only force acting, and it points at the centre, so gravity is playing the centripetal role:

G ME m / r2 = m v2 / r

Cancel one r and the satellite's mass m, which appears on both sides:

v = sqrt(G ME / r)

The satellite's own mass has vanished. A 400 kg communications relay and the 420,000 kg space station in the same orbit move at exactly the same speed, for the same reason a feather and a hammer fall together. Orbital speed is set by what you are orbiting and how far out you are, and by nothing about you.

Now substitute v = 2 pi r / T and rearrange:

G ME / r = 4 pi2 r2 / T2

T2 = (4 pi2 / G ME) r3

That is Kepler's third law: the square of the period is proportional to the cube of the orbital radius, with a constant of proportionality that depends only on the mass of the central body. Johannes Kepler extracted that pattern from Tycho Brahe's observations in 1619 without knowing why it held; Newton derived it in four lines sixty-eight years later.

Why this matters: T² proportional to r³ means you can weigh a planet you have never visited. Measure the period and radius of any one of its moons, and M = 4 pi²r³/(GT²) follows.

The pattern in the solar system

Measure a in astronomical units and T in years and the constant becomes 1, so T² should equal a³ for every planet.

The last two columns agree to a fraction of a percent across a range of 80 in distance and 680 in period, which is what a real law looks like when you test it against data it was not fitted to. Kepler's other two laws complete the picture: orbits are ellipses with the Sun at one focus, and a planet sweeps equal areas in equal times, which is why Mercury races through perihelion and crawls through aphelion.

Putting a satellite over Jakarta

A geostationary satellite must hang over one point on the equator, so its period has to match the Earth's rotation relative to the stars, 23 h 56 min 4 s, which is 86,164 s. Find its orbital radius, its altitude and its speed.

Rearrange Kepler's third law for r:

r3 = G ME T2 / (4 pi2)

r3 = (3.986 x 1014)(86,164)2 / 39.48 = (3.986 x 1014)(7.424 x 109) / 39.48

r3 = 2.959 x 1024 / 39.48 = 7.496 x 1022 m3

r = 4.216 x 107 m = 42,160 km

Subtract the Earth's radius to get the altitude:

altitude = 42,160 km - 6,371 km = 35,790 km

That is the number every television and weather satellite sits at, and it came out of one equation and a clock. The orbital speed follows:

v = 2 pi r / T = (6.283)(4.216 x 107) / 86,164 = 2.649 x 108 / 86,164 = 3,074 m/s

Contrast the geostationary orbit with the space station's, 420 km up, so r = 6.791 x 10⁶ m:

v = sqrt(G ME / r) = sqrt(3.986 x 1014 / 6.791 x 106) = sqrt(5.870 x 107) = 7,662 m/s

T = 2 pi r / v = (6.283)(6.791 x 106) / 7,662 = 5,569 s = 92.8 minutes

Closer in means faster and quicker round, which is what T² proportional to r³ says. The station laps the Earth roughly sixteen times a day while the satellite over Jakarta does not appear to move at all.

Why the astronauts float, which is not what most people think

Work out g at the station's altitude:

g = G ME / r2 = 3.986 x 1014 / (6.791 x 106)2 = 3.986 x 1014 / 4.612 x 1013 = 8.64 m/s2

That is 88 percent of the surface value. Gravity at the space station is almost as strong as it is in your kitchen, and a 70 kg astronaut is pulled toward the Earth with 605 N. She floats for the reason the passenger in the falling lift floated: nothing is supporting her. Both she and the station are in free fall, and they fall together along the same curved path. Orbit is not an escape from gravity; it is a fall arranged so that the ground keeps curving away at exactly the rate you drop.

Common misconceptions

• "There is no gravity in space." At the space station g is 8.64 m/s², down by only 12 percent. At the Moon's distance it is still 2.7 x 10^-3 m/s², which is enough to hold the Moon in orbit indefinitely.
• "A heavier satellite must orbit faster, or lower." The satellite's mass cancels out of both v = sqrt(GM/r) and Kepler's third law. Mass matters for the rocket that lifts it, not for the orbit it then keeps.
• "Satellites stay up because they are beyond the pull of the Earth." They stay up because they are moving sideways fast enough that their fall curves around the planet. Stop a geostationary satellite dead and it would drop straight down.
• "Gravity acts from the surface, so r is the altitude." A uniform sphere attracts as though its mass were concentrated at its centre, so r is the Earth's radius plus the altitude. Using 420 km instead of 6,791 km would give an orbital speed wrong by a factor of four.
• "Kepler explained why the orbits follow his laws." He found the patterns in Brahe's data and searched for a cause without finding one. The explanation is Newton's, and it is the inverse-square law plus the second law of motion.

Looking back

One law, F = Gm₁m₂/r², with distances measured between centres, accounts for the apple and the Moon: 9.80 divided by 3600 and 4 pi²r/T² agree to three figures. At a planet's surface it gives g = GM/R², which is how a planet gets weighed. Setting gravity equal to mv²/r loses the orbiting mass entirely and gives v = sqrt(GM/r), and substituting v = 2 pi r/T gives Kepler's third law, T² = (4 pi²/GM)r³, which holds across the solar system to a fraction of a percent. A 24-hour period puts a satellite at a radius of 42,160 km, an altitude of 35,790 km, and a speed of 3.07 km/s; the space station at 420 km does 7.66 km/s and circles in 92.8 minutes. And the astronauts inside it are not beyond gravity, which is still 8.64 m/s² there; they are falling, with nothing in the way.

Sources

1. OpenStax. (2022). 6.5 Newton's universal law of gravitation. In College Physics 2e. openstax.org
2. OpenStax. (2022). 6.6 Satellites and Kepler's laws. In College Physics 2e. openstax.org
3. NASA Science. (n.d.). Orbits and Kepler's laws. science.nasa.gov
4. National Institute of Standards and Technology. (2019). Newtonian constant of gravitation. CODATA fundamental physical constants. physics.nist.gov
5. Giancoli, D. C. (2016). Physics: Principles with Applications (7th ed.), ch. 5. Pearson.

Why the skid marks quadruple

A 1200 kg car travelling at 24.0 m/s locks its brakes on a road where the coefficient of kinetic friction is 0.800. It leaves a skid 36.7 m long. Now double the speed. At 48.0 m/s the skid is not 73 m. It is 147 m.

Module 1 can already get that number. Friction gives a = mukg = (0.800)(9.80 m/s2) = 7.84 m/s2, and v2 = v02 + 2ad with a final speed of zero rearranges to d = v02/(2a) = (24.0)2/(2 x 7.84) = 576/15.68 = 36.7 m. The squared initial speed on top is the factor of four.

The route this lesson builds gets the same answer and names what changed hands. The car arrives carrying something the road has to take away, all of it, before the car can stop. That something is kinetic energy, and the taking away is called work.

Work is a force times a displacement, with a cosine in it

When a constant force F acts on an object that moves a displacement d, and the angle between the force and the displacement is theta, the work done by that force is

W = F d cos(theta)

The unit is the joule: 1 J = 1 N m = 1 kg m²/s². Work is a scalar. It has a sign but no direction, and that is exactly why energy methods are often easier than force methods: you add numbers, not arrows.

The cosine is where the meaning sits. It picks out the part of the force that lies along the motion, and throws the rest away. Three cases are worth memorising by feel rather than by formula.

• theta = 0. Push a box along the floor in the direction it is going. cos(0) = 1 and the work is the full Fd. Positive work adds energy.
• theta = 90 degrees. cos(90) = 0 and the work is exactly zero. Carry a 15 kg suitcase horizontally down a corridor and you do no work on it at all, however tired you are, because your upward force is perpendicular to its horizontal motion. The same reasoning makes the normal force do zero work on every sliding block in Module 3, and gravity do zero work on a satellite in a circular orbit.
• theta = 180 degrees. cos(180) = -1 and the work is negative. Friction on a sliding crate, and the brake pads on our car, take energy out.

Work the car skid this way. The friction force is fk = mukmg = (0.800)(1200 kg)(9.80 m/s2) = 9408 N, pointing backwards, so theta = 180 degrees and the work friction does over a distance d is -9408d joules.

Kinetic energy, and the theorem that links it to work

Start from the kinematic relation v2 = v02 + 2ad and do one piece of algebra. Multiply every term by one half of the mass:

(1/2)mv2 = (1/2)mv02 + mad

But ma is the net force, by the second law, so mad is the net work. Rearranged:

Wnet = (1/2)mv2 - (1/2)mv02 = KEfinal - KEinitial

That is the work-energy theorem, and the quantity KE = (1/2)mv2 is kinetic energy. Notice that nothing was assumed except constant acceleration along a line, and notice what the theorem does not contain: no time, and no direction. It relates a force acting over a distance to a change in speed, which is precisely the question "how far to stop" and "how fast at the bottom".

Our car carries KE = (1/2)(1200 kg)(24.0 m/s)2 = (600)(576) = 345,600 J, about 346 kJ. The road must do -345,600 J of work to bring it to rest, and it does -9408 J for every metre skidded, so

d = 345,600 J / 9408 N = 36.7 m

At 48.0 m/s the kinetic energy is (600)(2304) = 1,382,400 J, four times as much, so the distance is four times as long: 147 m. Doubling your speed does not double your stopping distance. It quadruples it, because the speed is squared and the friction force never changed.

So what?: a force decides how quickly the energy comes out, per metre. The energy decides how many metres there have to be.

A force that changes: work as the area under a graph

The cosine formula assumes F is constant. A spring is the standard case where it is not. Hooke's law says the force you must apply to stretch a spring by x is F = kx, growing steadily from zero, where k is the spring constant in newtons per metre.

Plot that applied force against x and you get a straight line through the origin with slope k. Work is the area under a force-displacement graph, exactly as displacement was the area under a velocity-time graph in Module 1, and the area here is a triangle:

W = (1/2)(base)(height) = (1/2)(x)(kx) = (1/2)kx2

Take a spring with k = 240 N/m stretched 0.150 m. The force at the end is (240)(0.150) = 36.0 N, but the average force over the stretch is only half of that, 18.0 N, so the work is (18.0 N)(0.150 m) = 2.70 J, and (1/2)(240)(0.150)2 = (120)(0.0225) = 2.70 J agrees.

The area rule works for any shape of graph, which is what makes it worth more than the formula. Here is a force that rises, holds and falls as a machine pushes a 4.0 kg cart along 6.0 m of track.

Total work 120 J. If the cart started at rest and nothing else acted along the track, then (1/2)(4.0)v2 = 120, so v2 = 60 and v = 7.75 m/s. No kinematics were used, and none could have been: the acceleration was changing the whole way.

A problem where the theorem beats the free-body diagram

You drag a 25.0 kg crate 4.00 m across a floor by a rope held at 30.0 degrees above the horizontal, pulling with 120 N. The coefficient of kinetic friction is 0.200. How fast is the crate moving at the end, starting from rest?

Take the four forces one at a time and add up their work.

1. The rope. W = Fd cos(theta) = (120 N)(4.00 m)cos(30.0) = (480)(0.866) = 416 J.
2. Gravity. The crate does not rise, so theta = 90 degrees and W = 0.
3. The normal force. Perpendicular to the motion, so W = 0. But its size matters for friction, and it is not mg here, because the rope lifts: N = mg - F sin(theta) = (25.0)(9.80) - (120)(0.500) = 245 - 60.0 = 185 N.
4. Friction. fk = mukN = (0.200)(185) = 37.0 N, opposite the motion, so W = -(37.0)(4.00) = -148 J.

Net work: 416 - 148 = 268 J. Then (1/2)(25.0)v2 = 268, so v2 = 21.4 and v = 4.63 m/s.

The free-body route would have needed components, a net force, an acceleration and then a kinematic equation. Four lines of arithmetic replaced it, and the only thing that had to be got right was which forces had a component along the motion.

Power, and why your kettle is not your car

Power is the rate of doing work:

Paverage = W / t, measured in watts, where 1 W = 1 J/s

If the force is along the motion and the speed is steady, substituting W = Fd gives the form worth remembering:

P = Fd/t = Fv

A 60.0 kg student runs up a staircase of vertical height 3.60 m in 4.20 s. The work done against gravity is mgh = (60.0)(9.80)(3.60) = 2117 J, so the power is 2117/4.20 = 504 W. That is two thirds of a horsepower (1 hp = 745.7 W) and it is a genuine effort; a fit adult can sustain roughly 100 to 150 W for long periods and several hundred watts for a few seconds.

The P = Fv form explains a fact about cars that the work formula hides. A car cruising at a steady 30.0 m/s against 600 N of drag and rolling resistance needs P = (600)(30.0) = 18,000 W = 18 kW, about 24 hp. Drag grows roughly with the square of speed, so the power to overcome it grows roughly with the cube: at 40.0 m/s the drag is near 600 x (40/30)2 = 1067 N and the power near (1067)(40.0) = 42.7 kW. A third more speed costs well over twice the fuel per second.

One last unit, because energy bills use it. A kilowatt hour is a power of 1000 W maintained for 3600 s, so 1 kWh = (1000 J/s)(3600 s) = 3.6 x 106 J. It is a unit of energy, not of power, despite containing the word watt.

Common misconceptions

• "If I am exerting a force, I am doing work." Hold a 20 kg box still at chest height for a minute. Your muscles burn fuel because they cannot hold tension without twitching, but the physics work on the box is zero: no displacement, no work. A table holds it for a century and burns nothing.
• "Work is always positive." A force opposing the motion does negative work and removes energy. Brakes, friction, air resistance and a catcher's mitt all do negative work, which is how anything ever slows down.
• "Twice the speed, twice the stopping distance." Kinetic energy goes as v squared, so it is four times the distance. This one has a body count, which is why 30 km/h zones exist outside schools.
• "Power and energy are the same thing." Power is energy per second. A 2 kW kettle running for 3 minutes uses 360 kJ; a 100 W bulb needs an hour to use the same. Confusing the two makes every electricity bill unreadable.
• "Work is force times distance, full stop." Only when the force lies along the displacement. The cosine is not decoration, and forgetting it is the single commonest error in this topic on exam scripts.

Recap

Work is W = Fd cos(theta), a scalar measured in joules, positive when the force has a component along the motion and negative when it opposes it. When the force varies, work is the area under the force-displacement graph, which for a spring stretched x gives (1/2)kx2. The work-energy theorem, Wnet = (1/2)mv2 - (1/2)mv02, follows from one line of algebra applied to v2 = v02 + 2ad, and it turns any question about force over a distance into arithmetic with no vectors in it. A 1200 kg car at 24.0 m/s carries 346 kJ and needs 36.7 m of skid to shed it; at twice the speed it carries four times as much and needs 147 m. Power is work per second, P = W/t = Fv, in watts, and the kilowatt hour is 3.6 MJ of energy rather than any amount of power at all.

Sources

1. OpenStax. (2022). 7.1 Work: the scientific definition. In College Physics 2e. openstax.org
2. OpenStax. (2022). 7.2 Kinetic energy and the work-energy theorem. In College Physics 2e. openstax.org
3. OpenStax. (2022). 7.7 Power. In College Physics 2e. openstax.org
4. The Physics Classroom. (n.d.). Work, energy and power. physicsclassroom.com
5. Giancoli, D. C. (2016). Physics: Principles with Applications (7th ed.), ch. 6. Pearson.

A 3.20 m drop, and a question forces cannot answer

A 58.0 kg skateboarder rolls over the lip of a half-pipe 3.20 m deep and starts from rest. How fast is she moving at the bottom?

Try it with Newton's second law and you stop almost at once. The wall of a half-pipe is curved, so the angle of the slope changes continuously, so the component of gravity along the surface changes continuously, so the acceleration is never constant and not one of the equations from Module 1 applies. You would need the shape of the curve, and then calculus, which this course does not use.

Now try it the other way. Whatever the shape, the only force doing work on her is gravity, since the normal force from the ramp is perpendicular to her motion everywhere on it. Gravity does work mgh as she descends a vertical height h, and the work-energy theorem turns that into speed:

mgh = (1/2)mv2, so v = sqrt(2gh) = sqrt(2 x 9.80 x 3.20) = sqrt(62.7) = 7.92 m/s

Her mass cancelled, the shape of the ramp never entered, and the whole calculation was one line. That is the trade this lesson is about: energy methods throw away information about the path and about time, and in exchange they answer questions the path would have made impossible.

Potential energy, and the free choice of zero

Lift a book of mass m through a height h and you do mgh of work against gravity. Let it go and gravity gives all of it back as kinetic energy. Because the giving-back is exact and does not depend on the route the book took, we can bank the work as gravitational potential energy:

PEgrav = mgh

The h is measured from a level you choose. Any level. Put it at the bottom of the half-pipe, at the skateboarder's waist, or at sea level, and the answers for speed come out identical, because only changes in potential energy ever appear in an equation. Pick the level that makes one of your terms zero and write it down before you start; an unstated reference level is the commonest source of sign errors in this topic.

A stretched or compressed spring banks energy the same way. The work done against Hooke's law in Lesson 11 was the triangle (1/2)kx2, and a spring gives it all back, so

PEelastic = (1/2)kx2

with x the distance from the spring's natural length, positive whether you stretched or squashed it, since it is squared.

Two families of force, and why only one gets a potential energy

The property that made both of those work is path independence. Carry a 2.0 kg book from the floor to a shelf 1.5 m up: straight, by a ladder, or all round the room first, and gravity does the same -29.4 J every time, returning it if you come back down. Drag the same book those three routes across a rough floor and friction takes 10 J, 30 J or 300 J depending on how far you went. That difference sorts forces into two families.

Key idea: a potential energy exists only for a force whose work is a function of position. Friction fails that test, so there is no such thing as frictional potential energy, and friction must always be handled as work.

The conservation equation, and a spring that launches a ball

Define the total mechanical energy as E = KE + PE. When only conservative forces do work, E is the same at every instant, which is the practical form of the conservation of energy:

(1/2)mvi2 + mghi + (1/2)kxi2 = (1/2)mvf2 + mghf + (1/2)kxf2

Choose two instants, write every term that is not zero, cross out the ones that are, and solve. Here is a toy launcher: a spring of constant 320 N/m compressed 0.120 m fires a 0.045 kg ball straight up. How high does it go?

1. Instant one, fully compressed and at rest. KE = 0, take h = 0 there, and PEelastic = (1/2)(320)(0.120)2 = (160)(0.0144) = 2.30 J.
2. Instant two, at the top of the flight. The ball is momentarily at rest, so KE = 0, the spring is relaxed so its term is 0, and the store is all mgh = (0.045)(9.80)h = 0.441h.
3. Equate. 2.30 = 0.441h, so h = 5.22 m.

Notice how much never had to be calculated: the speed leaving the spring, the time in the air, the acceleration during the launch. Energy skipped straight from one end of the problem to the other. For the record, the launch speed follows if you want it: (1/2)(0.045)v2 = 2.30 gives v = 10.1 m/s.

Where the missing joules went: putting friction in the equation

A 2.00 kg block is released from rest 1.50 m up a smooth curved ramp, reaches the floor, and then slides along a rough horizontal floor where the coefficient of kinetic friction is 0.250. How far does it go before stopping?

At the top it has mgh = (2.00)(9.80)(1.50) = 29.4 J, all of which becomes kinetic energy at the bottom. Friction on the floor is fk = mukmg = (0.250)(2.00)(9.80) = 4.90 N, and it removes fkd joules over a distance d. Everything is gone when

29.4 J = (4.90 N)d, so d = 6.00 m

Do it symbolically and the result is prettier: mgh = mukmgd gives d = h/muk = 1.50/0.250 = 6.00 m. Mass, and g, both cancel. A lead block and a wooden block of the same shape released from the same height slide exactly the same distance, which surprises most people the first time.

The general bookkeeping, and the version to write on an exam, is

KEi + PEi = KEf + PEf + fkd

The last term is not energy destroyed. It is energy that has left the mechanical account and gone into the thermal energy of the block and the floor, warming both by a fraction of a degree. Rub your hands together for five seconds and you can feel that term.

Reading an energy bar chart

The AP exam asks for bar charts as often as for numbers, and drawing one prevents most algebra errors, because it forces you to decide what is zero before you write anything. Imagine three bars at each instant: kinetic, gravitational, elastic, plus a fourth column for energy dispersed by friction. The rule is that the total height of the bars is the same at every instant unless friction has a bar of its own.

For the block above, in joules:

The right-hand column is the point of the whole chapter.

The loop the loop, where energy and circular motion meet

A marble runs down a track and round a vertical loop of radius r. From what minimum height must it be released to stay on the track at the top? This needs both Module 4 and this lesson, which is why it is a favourite exam question.

Module 4 gave the condition at the top: the slowest speed for which the track can still be in contact is vtop2 = gr. Now conserve energy between release at height h and the top of the loop, which sits at height 2r:

mgh = (1/2)mvtop2 + mg(2r) = (1/2)m(gr) + 2mgr

Divide through by mg:

h = r/2 + 2r = 2.5r

For a loop of radius 0.200 m you must release from at least 0.500 m. Build it from track and a marble and you will find you need rather more than that, because a real marble rolls rather than slides and some of its energy goes into spin, which is exactly the correction Module 6 will make.

Common misconceptions

• "Energy is lost to friction." Nothing is lost. The joules are in the thermal energy of two warm surfaces, and if you could measure the temperature rise precisely enough you would find every one of them. What is lost is their usefulness for doing mechanical work.
• "A heavier object slides further, or reaches the bottom faster." Both mgh and the friction force are proportional to mass, so it cancels: d = h/muk contains no m at all, and the speed at the bottom of a smooth ramp is sqrt(2gh) for a pebble and for a boulder.
• "Potential energy needs a correct zero level." It needs a stated one. Physics only ever uses differences in PE, so any level works as long as you use the same one on both sides of the equation.
• "Conservation of energy means the speed is the same everywhere." It means the total is the same. The split between kinetic and potential shifts constantly, and that shifting is the motion.
• "If the ramp is steeper, the object arrives faster." For a smooth ramp, arrival speed depends only on the vertical drop, so a steep ramp and a gentle one of the same height give the same speed. The steep one gets there sooner, which is a statement about time, not about speed at the bottom.

Where this leaves us

Gravity and spring forces are conservative: their work depends only on where you start and finish, so it can be banked as potential energy, mgh and one half k x squared. Friction and drag are not, so they never get a potential energy and must be entered as work. When only conservative forces act, KE + PE is the same at every instant, which lets you jump between two moments of a problem and ignore everything in between, including curved ramps that no constant-acceleration equation can handle. A 58.0 kg skater in a 3.20 m half-pipe arrives at 7.92 m/s, and so does a 30 kg child. When friction acts, add f_kd to the final side and the books balance again: a block from 1.50 m slides 6.00 m on a floor with mu_k = 0.250, whatever its mass. And when circular motion is involved, the two chapters combine: a loop of radius r needs a release height of 2.5r.

Sources

1. OpenStax. (2022). 7.3 Gravitational potential energy. In College Physics 2e. openstax.org
2. OpenStax. (2022). 7.6 Conservation of energy. In College Physics 2e. openstax.org
3. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Energy skate park: basics. phet.colorado.edu
4. Knight, R. D., Jones, B., and Field, S. (2019). College Physics: A Strategic Approach (4th ed.), ch. 10. Pearson.

An egg, a floor, and a factor of a hundred

Drop a 58 g egg from 1.20 m. It arrives at v = sqrt(2gh) = sqrt(2 x 9.80 x 1.20) = 4.85 m/s, and on tile it stops in something like 1.0 ms. On a pillow it stops in about 0.10 s. The landing speed is identical. The stopping time is not, and the egg only survives one of them.

Put numbers on that. The quantity the floor has to remove is momentum, p = mv = (0.058 kg)(4.85 m/s) = 0.281 kg m/s, and it removes it by exerting a force for a time. On tile:

Favg = (change in p)/(time) = 0.281 / 0.0010 = 281 N

On the pillow, the same 0.281 kg m/s spread over 0.10 s gives Favg = 2.81 N. The egg's own weight is only 0.57 N, so the tile hits it with about 500 times its weight and the pillow with about five. Airbags, crumple zones, running shoes, a boxer rolling with a punch and a cricketer drawing the hands back to catch all do the same single thing: they stretch the time, because the momentum change is fixed by the collision and only the time is negotiable.

Momentum, impulse, and the second law written properly

Momentum is mass times velocity, p = mv, in kg m/s. It is a vector, so in one dimension it carries a sign, and getting that sign right is most of the work in a collision problem.

Impulse is the effect of a force acting for a time:

J = Favg (change in t) = change in p = mvf - mvi

Its unit, N s, is the same as kg m/s, as it must be. This is the impulse-momentum theorem, and it is the second law in the form Newton actually wrote: force is the rate of change of momentum. Compare the pair and the parallel is exact. Work is force over a distance and changes kinetic energy, a scalar. Impulse is force over a time and changes momentum, a vector. Choose energy when the question mentions distances and speeds; choose momentum when it mentions times, forces or a collision.

The point: in any collision the change in momentum is set by the before and after velocities. The peak force is set by how long you take to make that change.

Impulse as an area, read off a graph

Real collision forces are not constant. A force sensor on a cart gives a curve rising to a peak and falling, and the impulse is the area under it, exactly as work was the area under a force-displacement graph. Here is a set of readings from a 0.500 kg cart struck by a spring plunger.

Total impulse J = 0.45 + 1.35 + 0.675 = 2.475 N s. If the cart began at rest, its final speed is v = J/m = 2.475/0.500 = 4.95 m/s. The average force over the whole 0.080 s is 2.475/0.080 = 30.9 N, well below the 45 N peak, and a question that asks for the average force when you have found the peak is testing exactly that difference.

The procedure: one collision, worked end to end

Two cars meet at a junction, and we will do this one in full, in the order you should always use.

The setup. A 1500 kg car moving east at 20.0 m/s runs into a stationary 1000 kg car. The bumpers lock and the wreck slides away together. What is the wreck's speed, and how much kinetic energy is gone?

1. Draw the system and choose a positive direction. Take east as positive. Both cars are inside the system; the road and the air are outside it.
2. Check the system is isolated. External forces exist, chiefly friction from the road, but the collision lasts around 0.1 s and friction removes very little momentum in that time compared with the huge internal forces between bumpers. Momentum is conserved to a good approximation during the impact, and that argument is worth writing out on an exam.
3. Write total momentum before. pbefore = (1500)(20.0) + (1000)(0) = 30,000 kg m/s, east.
4. Write total momentum after. They move as one object of mass 2500 kg at an unknown v: pafter = 2500v.
5. Set them equal and solve. 2500v = 30,000, so v = 12.0 m/s east.
6. Now check the energy, separately. Before: (1/2)(1500)(20.0)2 = 300,000 J. After: (1/2)(2500)(12.0)2 = 180,000 J. Missing: 120,000 J, that is 40 percent, gone into bending steel, sound and heat.

A collision in which the objects stick together is perfectly inelastic, and it loses the most kinetic energy any collision can while still conserving momentum. Momentum was conserved anyway. That is the crucial asymmetry of this topic: momentum is conserved in every collision in an isolated system, whatever happens to the shape of the objects, while kinetic energy usually is not.

Change one input and the answer flips

Now fit the same two cars with perfect spring bumpers so that nothing is dented and no energy is lost. The collision is elastic, and now two equations hold at once: momentum, and kinetic energy.

Solving the pair in general is tedious, so use the standard result for a one-dimensional elastic collision with a stationary target:

v1' = [(m1 - m2)/(m1 + m2)] v1 and v2' = [2m1/(m1 + m2)] v1

With m₁ = 1500 kg, m₂ = 1000 kg, v₁ = 20.0 m/s:

v1' = (500/2500)(20.0) = 4.00 m/s and v2' = (3000/2500)(20.0) = 24.0 m/s

Check both books. Momentum: (1500)(4.00) + (1000)(24.0) = 6000 + 24,000 = 30,000 kg m/s, unchanged. Kinetic energy: (1/2)(1500)(16.0) + (1/2)(1000)(576) = 12,000 + 288,000 = 300,000 J, also unchanged. The struck car leaves faster than the striking car arrived, which sounds wrong until you notice that the heavy car kept going forward too.

Three special cases fall out of those formulas and are worth knowing cold. Equal masses, m1 = m2: the first term is zero, so the moving object stops dead and the target leaves at the original speed, which is what a Newton's cradle and a clean pool shot do. A light object hitting a very heavy one: v1' approaches -v1, a bounce straight back, which is a ball off a wall. A heavy object hitting a very light one: v1' is nearly unchanged and v2' approaches 2v1, a lorry and a fly.

The centre of mass keeps its head

The centre of mass of a set of objects along a line is the mass-weighted average of their positions:

xcm = (m1x1 + m2x2 + ...)/(m1 + m2 + ...)

Put a 3.00 kg mass at x = 0 and a 1.00 kg mass at x = 2.00 m. Then xcm = [(3.00)(0) + (1.00)(2.00)]/4.00 = 0.500 m, one quarter of the way along, close to the heavy end. It is the balance point: the place where a ruler carrying the two would sit level on a finger.

The velocity of the centre of mass is the total momentum divided by the total mass, vcm = ptotal/M, and because momentum is conserved, that velocity does not change during a collision. For our two cars vcm = 30,000/2500 = 12.0 m/s before, during and after, in both the elastic and the inelastic version. That is why the wreck's speed came out at exactly 12.0 m/s in the sticky case: sticking together means both objects end up moving with the centre of mass. Throw a spanner spinning across a room and its centre of mass traces a clean parabola while the ends of the tool tumble around it, which is the same statement about an isolated system with only gravity acting.

Common misconceptions

• "Momentum is only conserved in elastic collisions." It is the other way round. Momentum survives every collision in an isolated system, including the ones that shatter things. Kinetic energy is what elastic collisions additionally preserve.
• "A crumple zone reduces the impulse." It cannot. The car must lose the same momentum either way. Crumpling stretches the stopping time from milliseconds to tenths of a second, which cuts the peak force by the same factor.
• "The heavier vehicle experiences the bigger force in a crash." The third law forbids it: the forces on the two vehicles are equal and opposite at every instant. What differs is the acceleration, because the same force divided by a small mass gives a larger number, which is why the occupants of the small car are hurt more.
• "Momentum is just another word for kinetic energy." One is mv and a vector; the other is one half mv squared and a scalar. Two identical balls thrown in opposite directions at equal speed have zero total momentum and plenty of kinetic energy.
• "The centre of mass is always inside the object." The centre of mass of a doughnut is in the hole, and the centre of mass of a high jumper arched over the bar can pass underneath it while every part of her passes over.

Putting it together

Momentum is mv, a vector, and the impulse-momentum theorem says a force acting over a time changes it: Favg(change in t) = change in p, with impulse readable as the area under a force-time graph. That single relation is why an egg survives a pillow and not a tile at the same landing speed of 4.85 m/s: 0.281 kg m/s removed in 1.0 ms needs 281 N, and removed in 0.10 s needs 2.81 N. In an isolated system total momentum is the same before and after any collision, so a 1500 kg car at 20.0 m/s striking a stationary 1000 kg car gives a 2500 kg wreck at 12.0 m/s, having shed 40 percent of its kinetic energy. Fit ideal bumpers to the same pair and energy is conserved too, giving 4.00 m/s and 24.0 m/s. The centre of mass is the mass-weighted average position, its velocity is the total momentum over the total mass, and it sails through the collision at 12.0 m/s without noticing that anything happened.

Sources

1. OpenStax. (2022). 8.1 Linear momentum and force. In College Physics 2e. openstax.org
2. OpenStax. (2022). 8.3 Conservation of momentum. In College Physics 2e. openstax.org
3. OpenStax. (2022). 8.4 Elastic collisions in one dimension. In College Physics 2e. openstax.org
4. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Collision lab. phet.colorado.edu
5. Halliday, D., Resnick, R., and Walker, J. (2018). Fundamentals of Physics (11th ed.), ch. 9. Wiley.

One wheel on a repair stand

A bicycle wheel is clamped in a workshop stand, free to spin. Rim and tyre together weigh 0.95 kg and sit at a radius of 0.340 m; the spokes and hub are light enough to ignore. You spin it up from rest with your hand on the tyre, and 6.00 s later it is turning at 4.00 revolutions per second. Everything in this lesson comes out of that wheel: how fast it is turning, what you had to do to get it there, and how much energy is now stored in it.

Start with the angle. Rotational motion is measured in radians, defined so that an arc of length s at radius r subtends theta = s/r. Because the full circumference is 2 pi r, one complete turn is 2 pi radians, about 6.28. The radian is the reason every formula in this lesson is clean; work in degrees and factors of pi over 180 infest everything.

So 4.00 revolutions per second is omega = (4.00)(2 pi) = 25.1 rad/s, where angular velocity omega is the rate of change of angle. Reaching it from rest in 6.00 s means an angular acceleration

alpha = (change in omega)/(change in t) = (25.1 - 0)/6.00 = 4.19 rad/s2

The same five equations, with new letters

Constant angular acceleration obeys equations identical in form to Module 1, because the algebra that produced them never cared what the symbols meant.

How many turns did the wheel make while speeding up? Use the average angular velocity: theta = (1/2)(omega0 + omega)t = (1/2)(0 + 25.1)(6.00) = 75.4 rad, and dividing by 2 pi gives 12.0 revolutions.

Three bridges connect the two columns, and every one of them needs radians:

s = r theta, v = r omega, atangential = r alpha

A point on the tyre is moving at v = (0.340)(25.1) = 8.53 m/s, about 31 km/h. A point halfway to the hub shares the same omega but moves at half that speed, which is why omega and not v is the honest description of a rotating rigid body.

Torque: where you push matters as much as how hard

Push a door at the handle and it swings. Push just as hard two centimetres from the hinge and nothing happens. The quantity that measures turning effectiveness is torque:

torque = r F sin(theta)

where r is the distance from the pivot to where the force acts and theta is the angle between the force and that radial line. The combination r sin(theta) is the lever arm: the perpendicular distance from the pivot to the line along which the force acts. Extending the force's line of action on a sketch and dropping a perpendicular onto it from the pivot is the quickest way to get this right.

The unit is the newton metre. It is deliberately not called a joule even though the units match, because a torque is not an energy.

Undo a tight nut with a 0.250 m spanner and 180 N of force. Pull at right angles and torque = (0.250)(180)(1) = 45.0 N m. Pull at 60.0 degrees to the spanner and torque = (0.250)(180)(0.866) = 39.0 N m, a loss of 13 percent for an angle that looks nearly right. Pull straight along the spanner, at 0 degrees, and the torque is zero however hard you heave.

What matters here: a force applied through the pivot, or along a line that passes through it, produces no torque at all. That is why the hinge force never appears in a door problem and why choosing the pivot cleverly deletes unknown forces from an exam question.

Balancing: torques that cancel

An object not rotating, or rotating at a steady rate, has zero net torque. Take a uniform seesaw pivoted at its centre. A 25.0 kg child sits 1.20 m from the pivot on the left. Where must a 35.0 kg child sit on the right?

Anticlockwise torque from the left child: (25.0)(9.80)(1.20) = 294 N m. Clockwise torque from the right child at distance d: (35.0)(9.80)d = 343d. Setting them equal:

343d = 294, so d = 0.857 m

Notice that g appears on both sides and cancels, so the balance condition is just m1d1 = m2d2, and the seesaw would balance identically on the Moon. The plank's own weight produced no torque because its centre of mass sits directly above the pivot, with zero lever arm. Move the pivot off centre and that term returns, acting at the plank's centre of mass, which is exactly how a question is made harder.

Rotational inertia: mass, and where you put it

Mass resists a change in velocity. For rotation, what resists a change in angular velocity is rotational inertia, also called the moment of inertia:

I = sum of m r2 over all the particles of the body

The r is squared, so a gram at twice the radius counts four times as much. Rotational inertia is not a property of the object alone; it depends on the axis you choose. The same rod is three times harder to spin about one end than about its centre.

Read the coefficients as a statement about where the mass sits. The hoop has all of it at the rim and scores 1. The solid disc spreads mass from the centre outward and scores 0.5. The sphere packs even more of itself near the axis and scores 0.4. Nothing here needs memorising beyond that ordering, and the exam supplies the formulas, but knowing which is largest tells you the answer to most conceptual questions before any arithmetic.

Back to the wheel: torque, force and stored energy

The rim and tyre are a hoop, so I = MR2 = (0.95 kg)(0.340 m)2 = (0.95)(0.1156) = 0.110 kg m2.

Newton's second law for rotation gives the torque you had to supply:

torque = I alpha = (0.110)(4.19) = 0.461 N m

Your hand pressed on the tyre at radius 0.340 m, tangentially, so the force was

F = torque/r = 0.461/0.340 = 1.36 N

about the weight of a small apple, held for six seconds. And the rotational kinetic energy now stored is

KE = (1/2)I omega2 = (1/2)(0.110)(25.1)2 = (0.055)(630) = 34.7 J

Check that against the work you did: torque times angle, (0.461 N m)(75.4 rad) = 34.8 J. The two agree to rounding, which is the rotational work-energy theorem doing exactly what its straight-line ancestor did in Lesson 11.

One design consequence follows immediately. Move that same 0.95 kg inward to an average radius of 0.170 m and I drops to (0.95)(0.0289) = 0.0275 kg m2, a quarter of what it was, and the wheel accelerates four times as readily for the same torque. Cyclists pay real money for light rims and not for light hubs, and now you know the exponent that justifies the price.

Common misconceptions

• "Torque is just another word for force." A force has units of newtons and changes velocity; a torque has units of newton metres and changes angular velocity. The same force gives a large torque at the door handle and none at all at the hinge.
• "Rotational inertia is a property of the object." It is a property of the object and an axis. A metre rule spun about its middle has I = ML squared over 12; the identical rule spun about its end has four times that.
• "Every point on a spinning wheel moves at the same speed." Every point shares the same angular velocity, but v = r omega, so the tyre moves at 8.53 m/s while a point halfway in moves at 4.27 m/s.
• "Newton metres and joules are the same, so torque is energy." The units coincide because a torque acting through an angle gives energy, but a torque by itself is not energy, and the two are never added.
• "Degrees and radians are interchangeable if you are careful." The bridges v = r omega and a = r alpha are only true in radians, because the radian is defined by arc over radius. This is the most common source of a wrong answer by a factor of 57.3.

The takeaway

Rotation reuses every idea in the course with new symbols: angle in radians for displacement, omega for velocity, alpha for acceleration, and four kinematic equations of identical form. The bridges are s = r theta, v = r omega and a = r alpha, valid in radians only. Torque, r F sin(theta), is force times lever arm, and it vanishes for any force whose line of action passes through the pivot, which is what makes a well-chosen pivot delete unknowns. An object in rotational equilibrium has clockwise and anticlockwise torques equal, so a 25.0 kg child at 1.20 m balances a 35.0 kg child at 0.857 m with g cancelling from both sides. Rotational inertia, the sum of m r squared, replaces mass, and the r squared is why a hoop scores MR squared while a solid sphere scores only 0.4 MR squared. Then torque = I alpha and KE = one half I omega squared complete the translation: a 0.95 kg rim at 0.340 m needs 0.461 N m to reach 25.1 rad/s in six seconds, and ends up holding 34.7 J.

Sources

1. OpenStax. (2022). 10.2 Kinematics of rotational motion. In College Physics 2e. openstax.org
2. OpenStax. (2022). 10.3 Dynamics of rotational motion: rotational inertia. In College Physics 2e. openstax.org
3. OpenStax. (2022). 10.4 Rotational kinetic energy: work and energy revisited. In College Physics 2e. openstax.org
4. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Balancing act. phet.colorado.edu
5. Knight, R. D., Jones, B., and Field, S. (2019). College Physics: A Strategic Approach (4th ed.), ch. 7. Pearson.

Three objects at the top of a plank

Tilt a 1.50 m plank to about 15 degrees and hold three things at the top: a roll of masking tape, a jar lid lying like a wheel, and a glass marble. Release all three at the same instant. The marble reaches the bottom first, the lid second, the tape roll last, and it happens that way every time whatever the three things weigh.

Ask five people why and you get three answers, each of which sounds reasonable.

• Prediction one: the heaviest wins. Gravity pulls harder on more mass, so more mass means more force means a faster arrival. This is a good instinct: it is how most people explain falling, and it is wrong for the same reason it was wrong in Module 1, because heavier things also take more force to accelerate.
• Prediction two: the biggest wins. A larger radius means the object covers more ground per turn, so a big wheel should outrun a small one. This also sounds right, and it is the reasoning behind larger wheels on a bicycle.
• Prediction three: only the shape matters. Neither mass nor radius has any effect; what decides the race is how the mass is distributed about the axis. This sounds like the least likely of the three, and it is the one the plank keeps voting for.

Deciding between them takes one new quantity and one old one, so start with the new one.

Angular momentum: the rotational version of mv

For a rigid body turning about a fixed axis, angular momentum is

L = I omega, in kg m²/s

which is the pattern of the whole module: swap m for I and v for omega. The rotational version of the impulse-momentum theorem follows the same rule: a net external torque acting for a time changes the angular momentum, (net torque)(change in t) = change in L. So when the net external torque on a system is zero, L cannot change. That is the conservation of angular momentum, and it is a separate conservation law from momentum and from energy, not a consequence of either.

The phrase that does the work is external. A skater's arms pull inward with large forces, but those forces are internal to the skater and their torques about her own spin axis cancel in pairs. The ice can exert almost no torque about a vertical axis, because it is slippery. So L is fixed, and anything the skater does to I must be paid for by omega.

The skater, with the numbers filled in

A skater spins with her arms and one leg extended, and in that shape her rotational inertia about the vertical axis is 3.60 kg m². She is turning at 1.20 revolutions per second, which is omega1 = (1.20)(2 pi) = 7.54 rad/s. Her angular momentum is

L = I1 omega1 = (3.60)(7.54) = 27.1 kg m2/s

She pulls everything in tight, and her rotational inertia falls to 0.900 kg m², a quarter of what it was, because the r in the sum of m r squared has roughly halved for her arms. With no external torque, L is unchanged:

omega2 = L/I2 = 27.1/0.900 = 30.2 rad/s, which is 30.2/(2 pi) = 4.80 revolutions per second

Four times the spin rate, for the same angular momentum. That is exactly what a televised spin looks like, and you can reproduce it on an office chair in the activity below.

Where the extra energy came from

Now check the energy, because this is the part that catches people out. Before:

KE1 = (1/2)I1omega12 = (1/2)(3.60)(7.54)2 = (1.80)(56.9) = 102 J

After:

KE2 = (1/2)(0.900)(30.2)2 = (0.450)(912) = 410 J

The kinetic energy has quadrupled. Nothing is broken: energy was not supposed to be conserved here, because the skater's muscles did work. Pulling her arms inward means pulling them against the outward tendency of objects moving in a circle, over a real distance, and every joule of that 308 J difference came out of her. Let the arms back out slowly and she gets the work back, slowing to 1.20 rev/s again. Let them fly out freely and the energy goes into the arms' motion and then into her shoulders.

Writing kinetic energy as KE = L2/(2I) makes it obvious: with L fixed, halving I doubles the energy.

Why this matters: in one process, momentum may be conserved, energy may not, and angular momentum may be. They are independent books. Deciding which one is balanced, by asking whether there is an external force, an external torque, or a dissipative force, is most of the skill in this half of the course.

Rolling: two motions happening at once

Back to the plank. A rolling object is doing two things: its centre of mass is sliding down the slope, and the body is spinning about that centre. Rolling without slipping locks the two together, because the contact point does not skid:

v = R omega

so the kinetic energy has two terms:

KEtotal = (1/2)mv2 + (1/2)I omega2

Write every rotational inertia as I = cMR2, where c is the coefficient from the table in Lesson 14: 1 for a hoop, 0.5 for a solid disc, 0.4 for a solid sphere. Substituting omega = v/R:

KEtotal = (1/2)mv2 + (1/2)(cmR2)(v2/R2) = (1/2)mv2(1 + c)

The R cancelled. Keep that in view; it decides the race.

Settling the plank

Conserve energy from a drop of height h. Rolling without slipping means the friction acts at a point that is instantaneously at rest, so it does no work and nothing is dissipated:

mgh = (1/2)mv2(1 + c), so v = sqrt(2gh/(1 + c))

Both m and R are gone. The only thing left that varies between the three objects is c, which is a statement about shape alone. Prediction three wins, and the other two were testing a variable that cancels.

For a drop of h = 0.500 m:

The reason is a kind of tax. Every rolling object has to spend some of its gravitational energy on spin rather than on getting down the slope, and the fraction it must spend is c/(1 + c): a fifth for a marble, a third for a lid, a half for a roll of tape. The hoop keeps all its mass at the rim, so it needs the most spin energy for a given speed and arrives last.

What would settle it in your kitchen

The three predictions make different testable claims, and you can separate them in ten minutes.

• Race a heavy hoop against a light hoop. Prediction one says the heavy one wins. It does not: they finish together, dead level, because m cancelled.
• Race a large marble against a small marble. Prediction two says the large one wins. It does not: R cancelled too.
• Race a marble against a roll of tape. Prediction three says the marble wins by a margin set by c, about 2.65 against 2.21 m/s, and it does.

One warning about the second test: a very small ball can slip rather than roll on a smooth surface, and a slipping ball beats a rolling one because it does not have to pay the spin tax. If your marbles do not finish level, roughen the track.

Common misconceptions

• "The skater speeds up because she pushes off something." There is nothing to push. Her angular momentum is unchanged from start to finish; only its division between I and omega moved, and the arithmetic is a single division.
• "Angular momentum conservation means energy is conserved too." The skater ends with four times the kinetic energy she started with, supplied by her arms. Conserved quantities have to be checked one at a time.
• "Heavier things roll down faster." A cannonball and a ball bearing of the same shape arrive together, since v = sqrt(2gh/(1 + c)) has no mass in it. Only air resistance, which does depend on size and mass, breaks the tie in real races.
• "Friction must be doing work on a rolling wheel, or it would not roll." In rolling without slipping, the contact point is instantaneously at rest, so static friction does no work. That is why energy conservation could be used at all, and it is also why a rolling wheel does not warm up the way a skidding one does.
• "A sliding block and a rolling ball arrive together." The block has no spin to pay for and arrives faster, at 3.13 m/s against the marble's 2.65 m/s from a half-metre drop. Ice a track and race a cube against a marble to see it.

What to carry forward

Angular momentum is L = I omega, and when no external torque acts on a system it is fixed. A skater whose rotational inertia falls from 3.60 to 0.900 kg m² must therefore spin four times as fast, from 1.20 to 4.80 revolutions per second, and her kinetic energy rises from 102 J to 410 J because her muscles did 308 J of work pulling inward. Momentum, energy and angular momentum are separate accounts, and each has its own condition for staying balanced. For an object rolling without slipping, v = R omega ties the two motions together and the kinetic energy becomes one half m v squared times (1 + c), with c the shape coefficient. Conserving energy down a drop h then gives v = sqrt(2gh/(1 + c)), in which mass and radius have both vanished. So a rolling race is decided entirely by shape: marble, then lid, then tape roll, always, and a frictionless block that does not have to spend anything on spin beats all three.

Sources

1. OpenStax. (2022). 10.5 Angular momentum and its conservation. In College Physics 2e. openstax.org
2. OpenStax. (2022). 10.4 Rotational kinetic energy: work and energy revisited. In College Physics 2e. openstax.org
3. Wikipedia contributors. (n.d.). List of moments of inertia. en.wikipedia.org
4. Halliday, D., Resnick, R., and Walker, J. (2018). Fundamentals of Physics (11th ed.), ch. 11. Wiley.

A hex nut, half a metre of thread, and 1.42 seconds

Tie a hex nut to 0.500 m of sewing thread, hold the top still, pull the nut 5 cm aside and let go. It will come back to the same place every 1.42 s, and it will keep doing that for a few hundred swings while the arc shrinks toward nothing.

You can write down 1.42 s before you tie the knot. You do not need the mass of the nut, and you do not need to know how far aside you pulled it. The only measurement that matters is the length, and that is the strange and useful fact this lesson is built around. By the end you will have predicted that number, understood why the nut's mass is irrelevant, and used the same relation backwards to measure g on your own kitchen floor.

What makes an oscillation simple

Many things repeat. Only some of them are simple harmonic motion, and the entry condition is one line: the restoring force must be proportional to the displacement and directed back toward equilibrium.

F = -kx

The minus sign is the whole physical content. Pull the object right and the force pushes left; push it left and the force pushes right. The size grows in exact proportion to how far you displaced it, which is Hooke's law from Lesson 11. When that condition holds, three things follow that hold for no other kind of repetition, and they are what make this topic worth a lesson.

• The motion is a sine or cosine in time, so x = A cos(2 pi t/T) starting from full displacement.
• The period does not depend on the amplitude. A wide swing and a narrow one take the same time, which is why this is the basis of every mechanical clock.
• The acceleration is largest at the ends, where the displacement is greatest, and zero at the centre, where the speed is greatest. This is the reverse of most people's intuition.

Three words need fixing before the algebra. Amplitude A is the maximum displacement from equilibrium, in metres, not the full width of the swing. Period T is the time for one complete round trip, in seconds. Frequency f is the number of round trips per second, in hertz, and f = 1/T.

The procedure: a mass on a spring, worked end to end

Hang a 0.250 kg mass on a spring of stiffness 40.0 N/m, pull it down 0.080 m and release. Work every quantity in order.

1. Period. T = 2 pi sqrt(m/k) = 2 pi sqrt(0.250/40.0) = 2 pi sqrt(0.00625) = 2 pi (0.0791) = 0.497 s.
2. Frequency. f = 1/T = 1/0.497 = 2.01 Hz, so it bobs about twice a second.
3. Total energy. At the moment of release everything is elastic: E = (1/2)kA2 = (1/2)(40.0)(0.080)2 = (20.0)(0.0064) = 0.128 J.
4. Maximum speed. At the centre all of that energy is kinetic: (1/2)mvmax2 = 0.128, so vmax2 = 1.024 and vmax = 1.01 m/s. Equivalently vmax = A sqrt(k/m) = (0.080)(12.6) = 1.01 m/s.
5. Maximum acceleration. At the ends the force is largest: F = kA = (40.0)(0.080) = 3.20 N, so amax = F/m = 3.20/0.250 = 12.8 m/s2, more than g.
6. Speed partway. At x = 0.040 m, half the amplitude: elastic energy is (1/2)(40.0)(0.0016) = 0.032 J, so kinetic energy is 0.128 - 0.032 = 0.096 J, giving v = sqrt(2 x 0.096/0.250) = sqrt(0.768) = 0.876 m/s. At half the amplitude the object is still doing 87 percent of its top speed.

Now change one input. Quadruple the mass to 1.00 kg. The period becomes 2 pi sqrt(1.00/40.0) = 2 pi (0.158) = 0.993 s, exactly double, because the mass sits under a square root. Quadrupling the stiffness instead would halve the period. And doubling the amplitude changes the period not at all, although it doubles the maximum speed and quadruples the stored energy.

Remember: for a spring the period is set by mass and stiffness only. A vertical spring has the same period as a horizontal one, because gravity shifts the equilibrium position downward by mg/k and then plays no further part.

The pendulum, and where the mass goes

A pendulum is not obviously a spring, so it takes an argument. Displace a bob of mass m by a small angle theta from the vertical on a string of length L. The restoring force is the component of gravity along the arc:

F = -mg sin(theta)

For small angles, and in radians, sin(theta) is very nearly theta, and the displacement along the arc is x = L theta, so theta = x/L and

F = -mg(x/L) = -(mg/L)x

That has exactly the form F = -kx with an effective stiffness k = mg/L. Substituting into the spring period:

T = 2 pi sqrt(m/k) = 2 pi sqrt(mL/(mg)) = 2 pi sqrt(L/g)

The mass cancels, and that is the answer to the puzzle in the opening paragraph. A heavier bob is pulled harder, but it is also harder to accelerate, in exactly the same proportion, which is the same cancellation that makes all objects fall together.

For the hex nut on 0.500 m of thread: T = 2 pi sqrt(0.500/9.80) = 2 pi sqrt(0.0510) = 2 pi (0.226) = 1.42 s.

Two lengths are worth carrying around. A pendulum of L = 0.248 m has a period of 1.00 s. A pendulum of L = 0.993 m has a period of 2.00 s, so it ticks once per second in each direction, which is why longcase clocks are about a metre tall inside.

How small is a small angle?

The approximation sin(theta) equal to theta is the only shaky step in the derivation, so it deserves a number rather than a reassurance. To a good first correction, the true period exceeds the simple formula by a factor of about 1 + theta02/16, with the amplitude theta₀ in radians.

So a pendulum is only approximately isochronous, and the swing has to be kept narrow for the claim to hold. Keep amplitudes under about 10 degrees in the laboratory and the error is smaller than your timing uncertainty, which is the practical rule.

Energy going round in a circle

Through one oscillation the total stays fixed at E = (1/2)kA2 and the split shifts continuously.

The energy is quadratic in both x and v, which is why the speed falls off so slowly at first: an object at half amplitude has given up only a quarter of its energy. There is a second, prettier way to see all of this. Watch a point moving steadily round a circle of radius A from directly beside the circle, edge on. Its shadow on the wall performs exact simple harmonic motion of amplitude A, and the period of the oscillation is the period of the revolution. That projection is where the 2 pi in every formula comes from.

Measuring g on your kitchen floor

Turn the pendulum formula round:

T2 = (4 pi2/g)L, so g = 4 pi2L/T2

Written as T2 against L, this is a straight line through the origin with slope 4 pi2/g, which is the linearising trick from Module 1 applied to a new relationship: plot the square of the period, not the period, and read g off the slope as g = 4 pi2/slope.

A worked run: L = 1.000 m, and 20 complete swings timed at 40.14 s, so T = 40.14/20 = 2.007 s. Then

g = (4)(9.8696)(1.000)/(2.007)2 = 39.478/4.028 = 9.80 m/s2

Timing twenty swings rather than one divides your reaction-time uncertainty by twenty, which is the single most valuable habit in this experiment. The internationally agreed standard value is 9.80665 m/s² exactly, fixed by definition rather than measured, while the true local value varies by a few parts in a thousand with latitude and altitude.

Common misconceptions

• "A heavier pendulum bob swings more slowly." The mass cancels: T = 2 pi sqrt(L/g) contains no m. Tie a steel nut and a paper clip to threads of the same length and start them together; they stay in step for dozens of swings.
• "A bigger swing takes longer." Not for small angles, and that independence is the whole reason pendulums became clocks. At 30 degrees it does take about 1.7 percent longer, so the claim has a limit, and a real clock keeps its amplitude constant to avoid it.
• "The acceleration is greatest where the object moves fastest." The opposite. At the centre the restoring force is zero, so the acceleration is zero while the speed is maximum; at the turning points the object is momentarily still while the acceleration is at its largest.
• "Adding mass to a spring makes it oscillate faster, because it stretches further." It stretches further and it oscillates more slowly: four times the mass doubles the period. The extra stretch only moves the equilibrium point.
• "A pendulum's period depends on the strength of the push you give it." The push sets the amplitude, and for small angles the amplitude does not appear in the period at all.

Summing up

Simple harmonic motion is what happens whenever the restoring force obeys F = -kx, and it brings with it a sinusoidal position in time, an acceleration largest at the ends and zero at the centre, and a period independent of amplitude. For a mass on a spring, T = 2 pi sqrt(m/k): 0.497 s for 0.250 kg on a 40.0 N/m spring, doubling if the mass quadruples, unchanged if the amplitude doubles. The energy total is one half k A squared, 0.128 J in that example, shuttling between elastic and kinetic so that the object is still at 87 percent of top speed when it is halfway out. A pendulum qualifies because sin(theta) is nearly theta for small angles, giving an effective stiffness mg/L and a period T = 2 pi sqrt(L/g) with the mass cancelling: 1.42 s for 0.500 m of thread, 2.00 s for the 0.993 m of a clock. Run that relation backwards, plotting T squared against L and taking g as 4 pi squared over the slope, and a nut on a string measures the acceleration due to gravity to better than one percent.

Sources

1. OpenStax. (2022). 16.3 Simple harmonic motion: a special periodic motion. In College Physics 2e. openstax.org
2. OpenStax. (2022). 16.4 The simple pendulum. In College Physics 2e. openstax.org
3. OpenStax. (2022). 16.5 Energy and the simple harmonic oscillator. In College Physics 2e. openstax.org
4. National Institute of Standards and Technology. (n.d.). Standard acceleration of gravity, 9.80665 m s^-2 (exact). NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
5. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Pendulum lab. phet.colorado.edu

Four questions, ninety-five minutes, and one that is not a calculation

The AP Physics 1 exam, as the College Board describes it at the time of writing, is in two halves worth 50 percent each: Section I is 42 multiple-choice questions in 85 minutes, and Section II is four free-response questions in 95 minutes. The four free-response questions are of fixed types, and they are announced in advance: mathematical routines, translation between representations, experimental design and analysis, and qualitative/quantitative translation. A calculator is permitted throughout, and a reference sheet of equations is supplied, so nothing in this course needs to be memorised as a formula.

Check the current Course and Exam Description before you sit it, because the numbers move: the College Board has announced changes to the multiple-choice count and the section timings effective with the May 2027 administration. What does not move is the kind of thinking each question type rewards, and that is what this lesson drills.

Ninety-five minutes over four questions is about 24 minutes each. The last of the four types is the one students most often lose points on while feeling that they answered it, so we will take one apart.

A student answer that looks fine and is not

Here is a prompt of the qualitative/quantitative translation kind.

A solid steel ball and a hollow plastic ball of the same radius are released from rest at the top of a ramp and roll without slipping. A student claims that both balls reach the bottom at the same time, because objects of different mass fall together. In a clear, coherent paragraph-length response that may also contain equations and drawings, evaluate the student's claim.

And here is an answer that a capable student writes in three minutes and feels good about.

The student is partly right. Mass does not matter because of gravity, so the balls should tie. But the hollow one is different because it has more rotational inertia, which means it is harder to get going. So it will be slower and the student is wrong.

Every physical instinct in that paragraph is sound. It would earn very little. Work out why before reading on, because the gap between that answer and a full one is the difference between a 3 and a 5 on this exam.

Debugging it, line by line

• "Mass does not matter because of gravity." This is a restatement, not a reason. Gravity is why things roll down at all; it is not why mass cancels. The reason is that the energy available, mgh, and the energy needed, one half m v squared times (1 + c), both contain m, so it divides out. No principle is named and no cancellation is shown, so no reasoning point is available.
• "It has more rotational inertia." More than what? A steel ball of the same radius has vastly more rotational inertia in kg m² than a plastic shell, because steel is far denser. The quantity that decides the race is the dimensionless coefficient c in I = cMR², which is 0.400 for the solid ball and 0.667 for the shell. Writing "more rotational inertia" is not merely vague; taken literally it is false, and a grader must mark what is written.
• "Harder to get going, so it will be slower." Plausible, unsupported, and it skips the step that matters: the shell must put a larger fraction of the same gravitational energy into spin, leaving less for forward motion.
• No equation anywhere. This question type explicitly invites equations as part of the reasoning. An equation with a sentence attached is worth far more than either alone.
• The claim is never squarely evaluated. The student's claim had two parts: the balls tie, and the reason is that mass does not matter. The correct verdict is that the conclusion is wrong and the reasoning is only half right, and the answer never says that in those terms.

Bottom line: a paragraph-length response is graded on the visible chain from principle to conclusion. Correct instincts with the chain missing score like a guess, because from the page they are indistinguishable from one.

The rebuilt answer

The student's conclusion is wrong, although part of the reasoning is right. Apply conservation of energy from release to the bottom: mgh = (1/2)mv² + (1/2)I omega², and since the balls roll without slipping, omega = v/R. Writing I = cMR² and substituting gives mgh = (1/2)mv²(1 + c), so v = sqrt(2gh/(1 + c)). Both m and R cancel, so the student is correct that mass alone does not decide the outcome, and two solid balls of different masses really would tie. However, the balls here have different shapes. For a solid sphere c = 0.400 and for a hollow shell c = 0.667, so the shell must divert a larger fraction of the same gravitational energy into rotation and therefore arrives with a smaller speed and at a later time. The solid steel ball wins, and it would still win if the shell were made of steel and the solid ball of plastic.

That is roughly 150 words and takes about six minutes. Look at what it does structurally, because the structure is reusable for any prompt of this type.

1. Verdict first. The opening sentence answers the question asked. Graders read for it.
2. Name the principle. "Conservation of energy" is written out. "Using the formula" is not a principle.
3. Show the equation and the substitution. The cancellation of m and R is the heart of the answer, so it appears as algebra, not as a claim.
4. Connect with because and therefore. Those words are where reasoning points live. If you can delete every connective in your paragraph without changing it, you have written a list of facts.
5. Name the actual objects. Not "the object" but "the hollow shell", with its number, 0.667.
6. Grant what is right in the claim. Evaluating means saying which part holds and which fails, not just declaring a winner.
7. Close on the comparison asked for. The final sentence rules out the mass explanation by changing the materials, which shows the distinction is understood rather than remembered.

The multiple choice, where the calculator mostly stays down

A calculator is allowed in Section I, and you will rarely want it. The questions are built so that reasoning is quicker than arithmetic, and reaching for the keypad is usually a sign you have missed the intended route. Four habits do most of the work.

Ratio reasoning. Almost every relation in this course is a power law, so ask how the answer scales. A car's speed rises from 20 m/s to 60 m/s: its kinetic energy goes up by 3² = 9 and so does its stopping distance. A satellite's orbital radius is multiplied by 4: since T² is proportional to r³, the period is multiplied by 4^1.5 = 8. No numbers were needed.

Limiting cases. Push a variable to zero or to infinity and see whether a candidate answer stays sensible. In an elastic collision of m₁ with a stationary m₂, an answer of (m1 - m2)/(m1 + m2) passes the test: with m₂ enormous it gives a bounce straight back, and with m₂ tiny it gives no change. An option reading (m1 + m2)/(m1 - m2) blows up when the masses are equal, which is physically absurd, and can be eliminated on sight.

Units. An option offering a period in metres per second is gone. Less obviously, if you have derived sqrt(L/g) and the options contain sqrt(g/L), note that the second has units of 1/s and cannot be a time.

Estimate. A question asking for the centripetal acceleration of a car at 25 m/s on a 100 m bend wants 625/100, and you can see it is between 6 and 7 without touching anything. If three options sit near 6.25 and one reads 0.0625, the question is testing whether you divide or multiply, not whether you can do long division.

Two further points of exam mechanics. There is no penalty for a wrong answer, so leave nothing blank. And a multi-part question set built on one stimulus often gives its later parts away: a graph you have already interpreted usually answers the next question too.

The other three free-response types, briefly

Module 1 Lesson 3 of this course worked the experimental type in full, by taking a flawed procedure for measuring g and repairing it. If that lesson is more than a few weeks behind you, reread it beside this one: those two question types are half of Section II, and neither is answered by recalling a formula.

Common misconceptions

• "Writing more earns more." A paragraph that names conservation of energy once and uses it scores above a page that describes the situation at length. Graders look for specific reasoning steps, and padding buries them.
• "Equations alone are enough on the paragraph question." This question type asks for coherent prose. An unexplained string of algebra may earn the calculation and lose the reasoning, and the reasoning is most of the marks.
• "If my final answer is right, the working does not matter." On the free response, most points sit on the steps. A correct number with no visible route is worth less than a wrong number reached by sound physics from a stated principle.
• "The multiple choice is about speed of calculation." It is about recognising which relation applies. Ratio reasoning and limiting cases answer a large fraction of the section faster than any arithmetic.
• "Hedging is safe." Writing both that the shell is faster and that it is slower, in the hope one lands, loses the point. A contradiction inside an answer cancels the part that was right.

What you now know

The exam is two equal halves, currently 42 multiple-choice questions in 85 minutes and four free-response questions in 95 minutes, with a calculator and an equation sheet allowed and with the format under revision for 2027, so check the current Course and Exam Description. The four free-response types are fixed and public, and the one that costs most marks is qualitative/quantitative translation, because it is graded on a visible chain of reasoning. A paragraph that says the right thing without naming a principle, showing the cancellation and joining the steps with because and therefore reads, from the page, exactly like a guess. Build every such answer the same way: verdict first, principle named, equation substituted, connectives doing real work, the actual objects named with their numbers, and a closing sentence that answers the comparison that was asked. In Section I, keep the calculator down and reach instead for ratios, limiting cases, units and rough estimates, and answer every question, because nothing is deducted for being wrong.

Sources

1. College Board. (n.d.). AP Physics 1: Algebra-Based exam, exam format and question types. AP Central. apcentral.collegeboard.org
2. College Board. AP Physics 1: Algebra-Based course and exam description. apcentral.collegeboard.org
3. OpenStax. (2022). 10.4 Rotational kinetic energy: work and energy revisited. In College Physics 2e. openstax.org
4. Knight, R. D., Jones, B., and Field, S. (2019). College Physics: A Strategic Approach (4th ed.), ch. 7. Pearson.