Press Next (or Play) to walk through each story one small step at a time. The numbers come last. Tap a round play button to hear a step read aloud. After every three stories there is one quick check.
1. Two friends push off
Read the story as text
Two friends stand face to face on an ice rink, both at rest. They put their palms together and push. Both slide away, in opposite directions. The smaller friend shoots off fast. The bigger friend drifts back slowly.
Nobody outside pushed them. Before the push, nothing was moving. After, both are moving. So did they create motion out of nothing? And why does the smaller one always end up faster?
2. Throw forward, roll back
3. Passing momentum along
Quick check. One question before the next stories.
Two skaters stand still, then push apart. Skater A (40 kg) moves left at 3 m/s. Skater B is 60 kg. What is B’s velocity?
Another example: Jumping off a wagon
4. Catching on skates
5. From ice onto carpet
6. The falling apple and the Earth
Quick check. One question before the next stories.
When does the total momentum of a system not stay the same?
Another example: Spring carts in the lab
7. Check yourself
Think of your answer first, then tap to see it.
a) A cannon sits still and fires a cannonball forward. Why does the cannon roll backward?
Show answer
Before firing, the total momentum is zero. The ball gets forward momentum, so the cannon must get the same amount backward to keep the total at zero. (The ball and cannon push on each other equally hard.)
b) A ball falls and speeds up. Is the ball's momentum conserved? Which system keeps its momentum constant?
Show answer
No: Earth's gravity is an outside force on the ball, so the ball's momentum grows. For the ball + Earth system the pull is an inside force: Earth gains an equal, tiny upward momentum, and the total stays the same.
c) Two carts sit still with a squeezed spring between them. The spring pushes them apart. The 1 kg cart moves right at 3 m/s. What is the velocity of the 3 kg cart?
Show answer
Total before = 0. So 0 = 1 × 3 + 3 × v, which gives v = −1 m/s: 1 m/s to the left.
Already know this?
Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.
A 70 kg astronaut floats at rest in space. She throws a 7 kg tool forward at 2 m/s. How does she move?
The total momentum of a system is conserved when...
A 1 kg cart moving at 4 m/s hits a 1 kg cart at rest, and they stick. How fast do they move together?
The idea
Picture two skaters standing still on smooth ice. They put their hands together and push.
Both glide away, in opposite directions. The lighter skater moves faster. Nobody outside pushed them. They only pushed on each other.
First, choose a system: the objects you are going to watch. Everything else is outside.
Forces between objects inside the system are internal forces. Newton's third law says they come in pairs. The two forces are equal in size and opposite in direction. They act for the same time, so their impulses cancel. Internal forces move momentum from one part to another. They cannot change the total.
Only an external force (from something outside the system) can change the system's total momentum.
If the net external force on a system is zero, ptotal, before = ptotal, after
For two objects on a line: m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′ (primes mean "after"). Another way to say it: Δp₁ = −Δp₂. Whatever momentum one object gains, the other loses.
Because ptotal = M vcm, a constant total momentum means the centre of mass keeps moving at a steady velocity, no matter how the parts bump, stick or fly apart.
Explosions and recoil. If the system starts at rest, the total momentum is zero and stays zero. The pieces fly apart with momenta that add to zero: equal sizes, opposite directions. The lighter piece is faster.
System choice matters. In a collision between cart 1 and cart 2, the momentum of cart 1 alone is not conserved, because cart 2 is outside that system and pushes on it. The momentum of cart 1 + cart 2 together is conserved (if the track is level and friction is small).
The same idea with numbers
Read the steps as text
A 50 kg skater and a 75 kg skater stand at rest on smooth ice and push off each other. The 50 kg skater moves west at 3.0 m/s. Find the velocity of the 75 kg skater. East is positive.
System: both skaters. External forces: gravity and the normal force (vertical, they cancel), ice friction (tiny). Why: conservation only works if the net external force on the system is zero, so we check first.
Before: both at rest, so pbefore = 0. Why: p = mv and v = 0 for both.
After: pafter = (50 kg)(−3.0 m/s) + (75 kg)v = −150 kg·m/s + (75 kg)v. Why: west is negative; v is the unknown with its sign.
Set equal: 0 = −150 + 75v, so v = +2.0 m/s (east). Why: total momentum is conserved. The plus sign tells us the direction: opposite to the small skater.
Check: each skater has 150 kg·m/s of momentum, in opposite directions. The 75 kg skater is 1.5 times heavier and 1.5 times slower (3.0 ÷ 2.0). Why: a quick ratio check catches a wrong mass or sign.
The push did create kinetic energy (from the skaters' muscles) but it did not create momentum. Total momentum was zero before and is still zero.
Trap: assuming kinetic energy is conserved whenever momentum is. Instead: momentum is conserved when no net outside force acts, but KE can grow (push-off, explosion) or shrink (crash): in Example 3 it drops from 9.0 J to 6.75 J.
Lab 1: explosion (two carts pushed apart)
Two carts sit at rest with a compressed spring between them. At t = 0.4 s the spring is released and pushes them apart. Choose the masses and how hard the spring pushes (the impulse it gives each cart). Predict which cart ends up faster, then play. The dashed total momentum line stays at zero, and the centre of mass marker does not move.
Try this: set m₁ = 1 kg and m₂ = 4 kg. The speeds come out in the ratio 4 : 1, the opposite of the mass ratio. Now make the masses equal: the carts leave with equal speeds. Change the spring push: both speeds change, but the total momentum is always zero.
Lab 2: a collision, seen from the inside
Cart 1 runs into cart 2. Pick the masses, velocities and the bounciness e (1 = perfectly elastic, 0 = they stick). The two force graphs show the push each cart gets from the other. They are mirror images: equal size, opposite sign, same time. So the shaded areas (the impulses) cancel, and the total momentum line stays flat for every e.
Try this: make cart 2 very heavy (5 kg) and cart 1 light (0.5 kg). The heavy cart barely changes velocity, but its change in momentum is exactly as big as the light cart's. Then set e = 0 and watch the carts leave together at vcm.
Examples
Example 1: rifle recoil basic
A 4.0 kg rifle, held loosely and at rest, fires a 0.010 kg bullet forward at 400 m/s. Find the recoil velocity of the rifle. Forward is positive.
Show solutionRead the steps as text
System: rifle + bullet. The gas forces are internal. Why: choosing both lets the big unknown forces cancel.
Before: everything at rest, p = 0. Why: nothing moves before firing.
0 = (0.010 kg)(400 m/s) + (4.0 kg)v → 0 = 4.0 kg·m/s + 4.0v → v = −1.0 m/s. Why: momentum conserved; the minus sign means the rifle moves backward.
Trap: dropping the minus sign and saying the rifle moves at 1.0 m/s forward. Instead: keep the sign: v = −1.0 m/s means the rifle recoils backward, opposite to the bullet.
Example 2: walking on a boat medium
A 60 kg person stands at the back of a 120 kg rowing boat. Both are at rest on calm water (ignore water drag). The person walks toward the front at 2.0 m/s, measured relative to the water. (a) What is the velocity of the boat? (b) While the person walks 2.0 m (relative to the water), how far does the boat move?
Show solutionRead the steps as text
System: person + boat. Friction between feet and boat is internal. Take "toward the front" as +. Why: the force that moves the person forward also pushes the boat backward. Inside the system it cancels.
(a) 0 = (60)(+2.0) + (120)v → v = −1.0 m/s. The boat moves backward at 1.0 m/s. Why: total momentum starts at zero and stays zero.
(b) Both move for the same time. The person takes 2.0 m ÷ 2.0 m/s = 1.0 s, so the boat moves (1.0 m/s)(1.0 s) = 1.0 m backward. Why: constant velocities, so distance = speed × time.
Check: the centre of mass must stay still. (60)(+2.0 m) + (120)(−1.0 m) = 0. Why: zero total momentum means vcm = 0, so the mass-weighted displacements add to zero.
Example 3: one cart or two? (system choice) AP
A 2.0 kg cart moving at +3.0 m/s hits a 1.0 kg cart at rest on a level, low-friction track. After the collision the 1.0 kg cart moves at +3.0 m/s. (a) Find the velocity of the 2.0 kg cart after. (b) Find the change in momentum of each cart. (c) Is the momentum of the 2.0 kg cart conserved? Explain.
Show solutionRead the steps as text
(a) System: both carts. Before: (2.0)(3.0) + (1.0)(0) = 6.0 kg·m/s. After: (2.0)v + (1.0)(3.0). So 6.0 = 2.0v + 3.0, v = +1.5 m/s. Why: no net external horizontal force on the two-cart system.
(b) Δp of 2.0 kg cart = (2.0)(1.5 − 3.0) = −3.0 kg·m/s. Δp of 1.0 kg cart = (1.0)(3.0 − 0) = +3.0 kg·m/s. Why: Δp = m(v′ − v). They are equal and opposite, as Newton's third law says.
(c) No. If the system is the 2.0 kg cart alone, the 1.0 kg cart is outside it and exerts an external force on it during the collision, so its momentum changes by −3.0 kg·m/s. Why: "conserved" always refers to a chosen system. Only the two-cart system has zero net external force.
Bonus check: KE before = ½(2.0)(3.0)² = 9.0 J. KE after = ½(2.0)(1.5)² + ½(1.0)(3.0)² = 2.25 + 4.5 = 6.75 J. Less than before, so this is a possible (inelastic) collision. Why: a collision can never create kinetic energy unless something like a spring releases stored energy.
Trap: saying the 2.0 kg cart's own momentum is conserved in the collision. Instead: its momentum changes (by −3.0 kg·m/s) because the other cart pushes on it; only the two-cart total is conserved.
Example 4: an exploding firework in 2D AP
A 3.0 kg firework shell is at rest at the top of its flight (ignore gravity during the very short explosion). It bursts into three 1.0 kg pieces. One flies east at 6.0 m/s, one flies north at 8.0 m/s. Find the velocity of the third piece.
Show solutionRead the steps as text
Total momentum before = 0, so after: p₁ + p₂ + p₃ = 0 in both x and y. Why: momentum is conserved as a vector, component by component.
x (east +): (1.0)(6.0) + 0 + p₃ₓ = 0 → p₃ₓ = −6.0 kg·m/s. Why: only the east piece has an x part.
y (north +): 0 + (1.0)(8.0) + p₃ᵧ = 0 → p₃ᵧ = −8.0 kg·m/s. Why: only the north piece has a y part.
Size: |p₃| = √(6.0² + 8.0²) = 10 kg·m/s, so speed = 10 ÷ 1.0 = 10 m/s. Direction: tan⁻¹(8.0/6.0) = 53° south of west. Why: the third piece must balance the other two, so it points opposite their vector sum.
Practice (AP style)
1. Cart A (mass 2m) and cart B (mass m) are at rest with a compressed spring between them. The spring is released. Which statement is true after they separate?
Show answer
Total momentum is zero before and after, so 2m·vA = m·vB in size: vB = 2vA. Momenta are equal in size and opposite in direction. (KEs are not equal: KB = ½m(2v)² = 2mv², KA = ½(2m)v² = mv², so B has twice the KE.)
2. A ball is dropped from rest and falls toward the ground (ignore air). For which system is momentum conserved while the ball falls?
Show answer
For the ball alone, gravity from Earth is an external force, so the ball's momentum grows. For ball + Earth, the gravity pair is internal: the ball pulls Earth up with an equal force. Earth gains an equal and opposite (tiny-velocity) momentum, and the total stays zero.
3. An 80 kg astronaut floats at rest in space. She throws a 2.0 kg wrench away from her at 8.0 m/s. What is her velocity afterward?
Show answer
0 = (2.0)(8.0) + (80)v → v = −16/80 = −0.20 m/s: 0.20 m/s in the direction opposite to the wrench.
4. A heavy truck and a small car collide head-on. During the collision, how does the size of the truck's change in momentum compare with the car's?
Show answer
The forces on each are a Newton's third law pair: equal size, opposite direction, same time. So the impulses, and the changes in momentum, are equal in size. The car's velocity changes more because its mass is smaller.
5. Two carts collide on a level track. A momentum-time graph shows cart 1 going from +4.0 kg·m/s to +1.0 kg·m/s during the collision. Cart 2 had 0 kg·m/s before. What is cart 2's momentum after?
6. Two cars collide at an icy intersection. Road friction acts on both cars. Why is it still reasonable to use conservation of momentum for the instant just before and just after the crash?
Show answer
Friction from the road is external, but its impulse F·Δt is small because Δt is a fraction of a second and the friction force is much smaller than the crash forces. So the total momentum barely changes across the collision. Over several seconds of skidding afterward, friction does change the momentum.
7. (Short answer) A 2.0 kg cart moves at +1.5 m/s toward a 1.0 kg cart moving at −1.5 m/s on a frictionless track. (a) Find the velocity of the centre of mass before they collide. (b) What is the velocity of the centre of mass after they collide, whether they stick or bounce? Explain.
(b) Still +0.50 m/s. The net external force is zero, so the total momentum (= M vcm) does not change, and the total mass does not change. How the carts bounce does not matter.
8. (Short answer, experimental) In a lab, two carts start at rest and are pushed apart by a spring. A student records: m₁ = 0.50 kg, v₁ = −1.20 m/s; m₂ = 1.00 kg, v₂ = +0.58 m/s. (a) Do the data support conservation of momentum? (b) Name one likely reason the total is not exactly zero.
Show answer
(a) p₁ = (0.50)(−1.20) = −0.60 kg·m/s. p₂ = (1.00)(0.58) = +0.58 kg·m/s. Total = −0.02 kg·m/s, which is about 3% of the size of either momentum. That is close to zero (the starting value), so yes, the data support conservation within reasonable measurement uncertainty.
(b) Friction in the wheels slowing the carts before the speeds are measured, a slightly tilted track, or timing error in the speed measurement. (Any one, with a sentence on why.)
9. (Short answer, system choice) A student pushes a box across a rough floor at constant velocity. She says "the box's velocity is constant, so momentum is conserved, so there is no net force." Then she says "the student + box system has constant momentum too." Evaluate both claims.
Show answer
First claim: right conclusion, wrong word. The box's momentum is constant, which means the net force on the box is zero (her push balances friction). It is not conserved in the AP sense: the push and friction are external forces on the box; they just balance. Second claim: not justified. For the student + box system the floor exerts external forces: friction on the box (backward) and friction on her shoes (forward). The student's momentum is also constant here (she walks at constant velocity), so the system momentum happens to be constant, but only because those external forces balance, not because the system is isolated. A good answer names the external forces.
10. (Short free response: translation between representations) On a level low-friction track, cart A (2.0 kg) moves at +3.0 m/s toward cart B (1.0 kg) moving at −2.0 m/s. After they collide, A moves at +0.50 m/s. (a) Give the values of the momentum bars before the collision (A, B and total). (b) Calculate B's velocity after. (c) Give the bar values after. (d) Describe the graph of total momentum against time through the collision.
Show answer
(a) pA = 2.0 × 3.0 = +6.0; pB = 1.0 × (−2.0) = −2.0; total +4.0 kg·m/s.
(c) After: A +1.0, B +3.0, total +4.0 kg·m/s, the same total bar.
(d) A horizontal line at 4.0 kg·m/s: during the collision A's momentum falls and B's rises by the same amount, so the total never changes. (KE check: 11 J before, 4.75 J after, so this is possible.)
Point guide (4 points):
1 point: before bars with signs.
1 point: vB′ = +3.0 m/s.
1 point: after bars with the same total.
1 point: total-p graph is a flat line.
11. (Short free response: qualitative/quantitative translation) A person of mass M stands at rest on frictionless ice and throws a ball of mass m forward at speed v. (a) Without equations: how does the person's recoil speed change if the ball is heavier (same v)? If the person is heavier? (b) Derive the recoil speed V in terms of m, v and M. (c) Calculate V for M = 60 kg, m = 0.50 kg, v = 12 m/s, and for a 1.0 kg ball. (d) Compare the kinetic energies of the ball and the person in the first case. Are they equal? Explain with momentum.
Show answer
(a) Heavier ball: more momentum given to it, so the person recoils faster. Heavier person: the same momentum gives a smaller speed.
(b) Total momentum starts at 0: 0 = mv − MV, so V = mv / M, backward. It grows with m and shrinks with M, as in (a).
(c) V = 0.50 × 12 / 60 = 0.10 m/s. With 1.0 kg: 0.20 m/s.
(d) Ball: ½(0.50)(12)² = 36 J. Person: ½(60)(0.10)² = 0.30 J. Not equal: the momenta are equal in size (6.0 kg·m/s each), but KE = p²/2m, so the lighter object gets far more KE.
Point guide (4 points):
1 point: correct trends for both changes.
1 point: V = mv/M from ptotal = 0.
1 point: 0.10 m/s and 0.20 m/s.
1 point: 36 J vs 0.30 J with the equal-momentum explanation.
12. (Short free response: mathematical routines) A 60 kg skater stands at rest on frictionless ice and catches a 4.0 kg ball moving horizontally at 15 m/s. (a) Calculate the speed of the skater + ball just after the catch. (b) Calculate the kinetic energy lost in the catch. (c) Instead, the ball bounces straight back off her glove at 15 m/s. Calculate the skater's speed.
Show answer
(a) 4.0 × 15 = 64 v, so v = 60 / 64 = 0.94 m/s.
(b) Kbefore = ½ × 4.0 × 15² = 450 J. Kafter = ½ × 64 × 0.9375² ≈ 28 J. Lost ≈ 422 J.
(c) 60 = 60v + 4.0 × (−15), so 60v = 120 and v = 2.0 m/s.
Point guide (3 points):
1 point: 0.94 m/s.
1 point: about 422 J lost.
1 point: 2.0 m/s using −15 m/s for the ball.
13. (Short free response: experimental design and analysis) Design an experiment to test whether momentum is conserved when a moving cart sticks to a cart at rest. You have two carts with Velcro bumpers, a level track, a balance and a motion sensor. (a) Describe your procedure and what you measure. (b) In one run, cart A (0.50 kg) moves at 0.80 m/s and sticks to cart B (0.30 kg) at rest. Together they move at 0.48 m/s. Calculate the momentum before and after, and the percent difference. (c) Do the data support conservation of momentum? Give one likely reason for the difference.
Show answer
(a) Measure both masses on the balance. Level the track. Use the motion sensor to measure A's velocity just before and the pair's velocity just after the collision. Repeat for several masses and speeds and compare total momentum before and after.
(c) Yes, within a few percent. The small loss is likely from track friction, an external force on the system during the measurement.
Point guide (4 points):
1 point: measures masses and before/after velocities.
1 point: repeats runs, varies conditions.
1 point: 0.40 and 0.384 kg·m/s, 4%.
1 point: conclusion with an external-force reason.
Common mistakes
The mistake: saying "momentum is conserved" for one object in a collision.
Why it is wrong: the other object pushes on it. That push is an external force for a one-object system, so that object's momentum changes.
How to spot it: always name the system first. Then list every force from outside it.
The mistake: thinking an explosion or push-off creates momentum.
Why it is wrong: it creates kinetic energy (from stored energy) but the momenta of the pieces add to zero.
How to spot it: if the system starts at rest, your "after" momenta must have opposite signs and add to zero.
The mistake: believing the heavier object gets the bigger force or the bigger change in momentum.
Why it is wrong: Newton's third law: the forces are equal and opposite for the same time, so Δp₁ = −Δp₂. The lighter object has the bigger change in velocity.
How to spot it: compare Δp, not Δv. They should match in size.
The mistake: dropping the minus sign on a recoil velocity, or adding speeds instead of signed momenta.
Why it is wrong: momentum is a vector. Pieces that fly apart move in opposite directions.
How to spot it: state the positive direction in the first line and check the sign of every answer against the picture.
The mistake: assuming kinetic energy is conserved because momentum is.
Why it is wrong: momentum is conserved in every collision of an isolated system. Kinetic energy is conserved only in elastic collisions (see 4.4). In an explosion KE goes up; in a sticky collision it goes down.
How to spot it: compute KE before and after separately. Never write "KE before = KE after" unless the problem says elastic.
The mistake: using conservation of momentum over a long time with a big external force (friction during a long skid, gravity on a thrown ball).
Why it is wrong: the external impulse F·Δt grows with time and changes the momentum.
How to spot it: use conservation only across a short event (a collision, a throw, an explosion), or for a system with no net external force.