Collision Lab Classroom Activity, 40 Minutes

Run a 40 minute lesson in four blocks: a 5 minute hook on equal and opposite forces, 25 minutes of guided exploration on the collision lab where students predict a reading before every change to mass, velocity, or elasticity, a 5 minute check for understanding, and a 5 minute exit ticket.

AP Physics: Unit 4 (topics 4.3 Conservation of Linear Momentum, 4.4 Elastic and Inelastic Collisions). Built for AP Physics 1 Unit 4, Topics 4.3 and 4.4, which together carry 10 to 15 percent of the multiple-choice section. The predict-then-reveal structure targets EK 4.3.B.2, that total momentum is constant when the net external force on the system is zero, and EK 4.4.A.3, that inelastic collisions lose total kinetic energy while conserving momentum.

Lesson at a glance

Essential question: two carts push on each other with forces that are always equal in size and opposite in direction. So why does one cart's velocity change so much more than the other's?

This lesson targets Topics 4.3 and 4.4, Conservation of Linear Momentum and Elastic and Inelastic Collisions, in Unit 4 of AP Physics 1, worth 10 to 15 percent of the multiple-choice section. Students should already know momentum, p = mv, and the impulse momentum theorem before this lesson; if that is shaky, run impulse momentum theorem first.

Block	Minutes	Purpose
Hook	5	Surface the equal force, unequal effect misconception before naming it
Guided exploration	25	Predict, then reveal, on the collision lab
Check for understanding	5	Fast formative check, whole class
Exit ticket	5	Individual, collected

By the end, students should be able to state that total momentum is conserved in every collision on the lab's frictionless track regardless of how bouncy it is, that total kinetic energy is only conserved when the collision is elastic, and that an equal and opposite force pair still produces very different velocity changes when the two masses are different.

Materials

A projector or one device per pair, open to the collision lab
Scratch paper or a mini whiteboard per student, for writing a prediction before each reveal
The exit ticket questions below, printed as a half sheet or posted for students to answer on paper
For the no-tech variant: two identical coins, one much larger coin or a marble against a heavier object such as a golf ball, two small lumps of modeling clay or blue tack, and a ruler or meter stick on a smooth tabletop

No sign-up, login, or install is required. The simulator runs in the browser and every reading it shows, including momentum, kinetic energy, and the two final velocities, is computed live from the same mass, velocity, and elasticity values you set on screen.

Hook: does the bigger cart hit harder? (5 minutes)

Roll two identical coins toward each other at the same speed so they meet in the middle and bounce apart. Then swap one coin for a much bigger one, a large washer or a stack of coins against a single small coin, and flick the heavy one into the light one at a similar speed.

Ask the room to vote with a show of hands on one question: when the heavy coin hits the light one, is the force the heavy coin exerts on the light one bigger than, smaller than, or the same size as the force the light one exerts back on the heavy one? Do not resolve the vote yet. Write the two positions on the board exactly as students phrase them, typically something like "bigger, because it is heavier" against "the same, forces always come in pairs."

Tell the class the collision lab draws each cart's momentum as a bar, before and after, and that it will settle the vote in the next 25 minutes. Do not give the answer here. The unresolved vote is what makes the light cart, heavy cart move in the guided exploration land.

Guided exploration on the collision lab (25 minutes)

Open the collision lab on the projector at its starting values: cart 1 at 2.0 kg moving 3.0 m/s, cart 2 at 1.0 kg at rest, elasticity e = 1. Each move below has a prediction step before you touch a slider. Cold-call for the prediction, then make the change and read the numbers out loud.

Baseline reveal. Before touching anything, ask: cart 1 is twice the mass of cart 2 and hits it head on. Does cart 1 bounce backward, keep moving forward slower, or stop dead? Take two or three guesses, then point out that elasticity already reads e = 1 and read the after row: cart 1 slows to 1.00 m/s, cart 2 takes off at 4.00 m/s, and total momentum reads 6.00 both before and after. Point out the elastic badge and "no kinetic energy lost."

Make the masses equal. Predict: if I drag cart 2's mass slider up from 1.0 kg to 2.0 kg, so both carts match, what happens to the two final velocities? Set cart 2's mass to 2.0 kg. Reveal: cart 1 drops to 0.00 m/s and cart 2 leaves at 3.00 m/s, cart 1's exact starting speed. Name the pattern: equal masses in an elastic collision trade velocities completely.

Change the velocities, not the masses. Keep both masses at 2.0 kg. Predict: if cart 2 is not sitting still but moving toward cart 1 at 2.0 m/s, do the two carts still swap velocities the same way? Drag cart 2's velocity slider to negative 2.0 m/s. Reveal: cart 1 ends at negative 2.00 m/s and cart 2 ends at 3.00 m/s, the exact incoming values swapped between carts. Total momentum reads 2.00 before and after. The swap rule from move 2 holds even when both carts were moving.

Break the elastic case. Keep the same masses and velocities from move 3. Predict: if I drag the elasticity slider down from 1 to 0.5, does the total momentum reading change, does the total kinetic energy reading change, or both? Set e to 0.5. Reveal: total momentum still reads 2.00 before and after, but kinetic energy drops from 13.00 J to about 3.63 J, a loss of about 9.38 J. Momentum did not notice the change; kinetic energy did.

Go fully inelastic. Keep the same masses and velocities. Predict: at e = 0, do the two carts still end up at different speeds, or do they move off together? Drag elasticity to 0. Reveal: the carts stick, and both read 0.50 m/s afterward, the shared velocity total momentum divided by total mass. Kinetic energy falls to 0.50 J, down from 13.00 J, the largest possible loss for this pair of speeds.

Lose every joule. Elasticity stays at 0. Predict: if I lower cart 1's velocity from 3.0 down to 2.0 m/s, so it now approaches at exactly the same speed cart 2 is approaching from the other side, will any kinetic energy survive the crash? Drag cart 1's velocity to 2.0 m/s. Reveal: both carts read 0.00 m/s afterward and kinetic energy drops from 8.00 J to 0.00 J, a complete loss. Ask why the carts could not end up moving at all: total momentum was exactly 0.00 before the crash, so the only way to conserve it afterward is for the stuck-together mass to sit still.

Resolve the hook. Reset elasticity to 1. Now set cart 1's mass down to 0.2 kg at 3.0 m/s and cart 2's mass up to 5.0 kg at rest, matching the hook's heavy-versus-light setup. Predict: does the light cart bounce backward, keep going forward, or stop? Make the change. Reveal: cart 1 rebounds to about negative 2.77 m/s, almost its original speed reversed, while cart 2 barely moves at about 0.23 m/s. Total momentum still reads 0.60 both before and after. The force pair between them was equal and opposite the entire time; the light cart simply accelerates far more for the same force, because its mass is so much smaller.

Try to force a collision that cannot happen. Reset cart 1 to 2.0 kg and cart 2 to 1.0 kg. Predict: if I set cart 1's velocity to 1.0 m/s and cart 2's velocity to 3.0 m/s, both moving in the positive direction, will they still collide? Make the change. Reveal: the badge reads "They never meet" and the animation shows no contact, because cart 1 is drawn on the left and can only reach cart 2 by closing the gap; a slower cart behind a faster one never catches up. The before and after readouts are identical, since nothing happened.

Check for understanding (5 minutes)

Fast, whole class, thumbs up or down or a one-word shout, no writing required:

True or false: in an elastic collision, each individual cart keeps the same kinetic energy it started with. (False. The total kinetic energy is preserved, but the two carts can and usually do trade energy between them.)
Two carts collide and stick together. Is total momentum conserved? Is total kinetic energy conserved? (Momentum, yes, always. Kinetic energy, no, some of it becomes heat, sound, and deformation.)
In the hook, is the force the heavy coin exerts on the light one bigger, smaller, or the same size as the force the light coin exerts back? (The same size, by Newton's third law. The light coin just accelerates far more because its mass is so much smaller.)

If more than a few thumbs go the wrong way on question 3, rerun move 7 before moving on. That one move carries the whole lesson.

Exit ticket (5 minutes)

Individual, on paper, collected at the door:

Two carts of equal mass collide elastically. One is moving, the other is at rest. Using only the pattern from move 2, predict what happens to each cart's velocity without redoing any algebra.
A 2 kg cart moving right at 2 m/s collides with a 2 kg cart moving left at 2 m/s and they stick together. What is their combined velocity right after the crash, and how do you know this without calculating any kinetic energy?
Two carts always push on each other with equal and opposite forces during a collision. Explain in one or two sentences why the lighter cart can still end up with a much bigger change in velocity than the heavier one.

Score questions 1 and 2 as understood or not yet; question 3 as a correct force-versus-mass explanation or not. A class where most students get question 2 but miss question 3 is ready for practice problems; a class that misses question 2 needs move 6 repeated before independent work.

No-tech variant

Run the same four blocks with coins and clay instead of the screen, useful for a room with no projector or when devices are the distraction rather than the tool.

Pairs get two identical coins on a smooth tabletop or lab bench. One partner flicks a coin into an identical, stationary one along a ruled line. Predict first: does the moving coin keep going, stop, or bounce back? Because the masses match, the moving coin should nearly stop while the resting coin takes off at close to the flicking speed, the coin version of move 2's velocity swap. Repeat a few times and compare how consistently the swap happens.

For the light-versus-heavy move, swap one coin for something much heavier, a stack of coins, a large washer, or a golf ball against a small coin or marble. Predict first, then flick the light object into the heavy one and watch it rebound while the heavy object barely shifts, matching move 7 without any screen at all.

For the fully inelastic case, press two small, equally sized lumps of modeling clay or blue tack together so they stick on contact. Push them toward each other by hand at what feels like matching speed and watch the combined lump land close to the impact point rather than skidding off, the hands-on version of move 6's total loss. If a stopwatch and meter stick are available, timing how far the combined lump travels in a fixed time after a gentler, one-sided push gives a rough speed to compare against a prediction, the same distance-over-time idea the on-screen tool skips by reading velocity directly.

Common misconceptions

Momentum conservation is not kinetic energy conservation. The most common error is assuming that if momentum is conserved, energy must be too, or that a collision that loses kinetic energy must also lose momentum. They are separate claims. Total momentum is conserved in every collision on the lab's frictionless track, elastic or not, because the two carts always exert equal and opposite forces on each other for the same length of time, and those impulses cancel in the total no matter how bouncy the collision is. Kinetic energy is different: it is one category of energy, not a protected quantity on its own, and a collision is always free to move energy out of that category into heat, sound, and permanent deformation. Moves 4, 5, and 6 hold the masses and velocities fixed and change only elasticity specifically so students watch the momentum readout stay still while the kinetic energy readout falls.

Equal and opposite forces still produce unequal accelerations. Newton's third law guarantees that the force cart 1 exerts on cart 2 matches the force cart 2 exerts on cart 1 in size, at every instant, for the entire time they are in contact. Students often skip straight from "equal forces" to "equal effects," expecting both carts to change speed by the same amount. Acceleration also depends on mass, a = F divided by m, so the same force pair produces a much bigger velocity change on the smaller mass. Move 7 makes this concrete with numbers: the 0.2 kg cart's velocity swings by about 5.77 m/s while the 5 kg cart's velocity shifts by only about 0.23 m/s, from the identical force pair acting for the identical time.

Perfectly inelastic does not usually mean zero final kinetic energy. Students who have just seen a sticking collision often generalize that e = 0 always means the wreck ends up motionless. Move 5 shows a perfectly inelastic collision where the combined cart still moves off at 0.50 m/s, because the total momentum going in was not zero. Move 6 is the deliberately built exception: only when the incoming momentum is exactly zero, meaning the two carts approach with equal and opposite momenta, does a perfectly inelastic collision leave nothing moving at all.

The baseline reveal: 2 kg at 3 m/s hits 1 kg at rest, elastically

Move 1 of the guided exploration opens the collision lab at its starting values: cart 1 at $m_1 = 2$ kg and $v_1 = 3$ m/s, cart 2 at $m_2 = 1$ kg and $v_2 = 0$ , elasticity $e = 1$ . Before reading the readouts, find both final velocities by hand using conservation of momentum and conservation of kinetic energy.

Conserve momentum: $m_1 v_1 + m_2 v_2 = m_1 v_{1f} + m_2 v_{2f}$ , so $2(3) + 1(0) = 2 v_{1f} + v_{2f}$ , giving $6 = 2 v_{1f} + v_{2f}$ .
Conserve kinetic energy, since $e = 1$ : $\tfrac{1}{2} m_1 v_1^2 + \tfrac{1}{2} m_2 v_2^2 = \tfrac{1}{2} m_1 v_{1f}^2 + \tfrac{1}{2} m_2 v_{2f}^2$ , so $9 = v_{1f}^2 + \tfrac{1}{2} v_{2f}^2$ .
Use the standard one dimensional elastic collision result for two objects, which solves that pair of equations once: $v_{1f} = \dfrac{(m_1 - m_2)v_1 + 2 m_2 v_2}{m_1 + m_2} = \dfrac{(2-1)(3) + 2(1)(0)}{3} = \dfrac{3}{3} = 1$ m/s.
Solve the matching expression for cart 2: $v_{2f} = \dfrac{(m_2 - m_1)v_2 + 2 m_1 v_1}{m_1 + m_2} = \dfrac{(1-2)(0) + 2(2)(3)}{3} = \dfrac{12}{3} = 4$ m/s.
Check both conservation laws: momentum after is $2(1) + 1(4) = 6$ , matching the value before. Kinetic energy after is $\tfrac{1}{2}(2)(1)^2 + \tfrac{1}{2}(1)(4)^2 = 1 + 8 = 9$ J, matching the 9 J before.

Cart 1 finishes at 1.00 m/s, cart 2 finishes at 4.00 m/s, matching the collision lab's own readouts from move 1's baseline reveal. Total momentum holds at 6.00 kg times m/s and total kinetic energy holds at 9.00 J, both before and after, since the collision is elastic.

Frequently asked questions

How long does this collision lab activity take?

40 minutes as written: a 5 minute physical hook, 25 minutes of guided exploration on the collision lab, a 5 minute check for understanding, and a 5 minute exit ticket. Trim a guided-exploration move or two to fit a 30 minute block, or add a ninth move testing three different elasticity values on the same masses and velocities to stretch it toward a full period.

What do students need to know before this lesson?

Momentum as p = mv and the impulse momentum theorem. If either is shaky, run impulse momentum theorem first, since this lesson applies conservation of momentum to two objects at once rather than reteaching the single-object case.

Does the collision lab simulator include friction?

No. The track is frictionless, so the only forces during a collision are the equal and opposite forces the two carts exert on each other, which is exactly what keeps the total momentum reading fixed in every move. That is also why total kinetic energy never comes back once elasticity drops below 1; there is no friction to blame the loss on, only the collision itself.

How do I run this without a projector or student devices?

Use the no-tech variant: two identical coins for the equal-mass swap, a heavy object against a light one for the mass-asymmetry move, and two lumps of modeling clay or blue tack for the sticking case. It keeps the same predict-then-reveal structure as the on-screen version without needing a screen at all.

What is the single most important move if I only have time for one?

Move 7, the light cart bouncing off the heavy one. Watching a 0.2 kg cart's velocity swing by nearly 6 m/s while a 5 kg cart's velocity barely moves, from an equal and opposite force pair, is what actually resolves the hook's vote about whether the bigger object hits harder.