This review is for the algebra-based Physics 1 course, Unit 4 (Linear Momentum). Do the unit diagnostic first.
Big ideas of the unit
- Momentum p = mv is a vector. In one dimension it takes the sign of v; in two, use components (Topic 4.1).
- Momentum and kinetic energy scale differently: K = p² ÷ (2m), so equal momenta can carry different kinetic energies (Topic 4.1).
- Collisions and explosions are models. Forces between the objects dwarf the net external force, so only the states just before and after matter (Topic 4.1).
- Impulse J = F_avg Δt equals Δp. It is the area under a force–time graph; force is the slope of a momentum–time graph. The same Δp over a longer time needs a smaller force (Topic 4.2).
- Internal impulses cancel by the third law, so without a net external force p_sys and v_cm stay constant (Topic 4.3).
- Your system decides whether momentum stays inside it. Any change in p_sys equals the external impulse (Topic 4.3).
- In two dimensions, conserve each component separately (Topic 4.3).
- Kinetic energy sorts collisions: unchanged (elastic), decreased (inelastic), or decreased with one shared velocity (perfectly inelastic) (Topic 4.4).
Key relationships and methods
| Idea | Relationship or method | When it applies |
|---|---|---|
| Momentum | p = mv; p_x = m v_x, p_y = m v_y | any object; sign or direction of v |
| Momentum and kinetic energy | K = ½mv² = p² ÷ (2m) | comparing objects or scenarios |
| Impulse | J = F_avg Δt = area under F_net–t graph | any time interval |
| Impulse–momentum theorem | J = Δp = p − p₀; F_net = slope of p–t graph | net external force on the chosen system |
| System momentum | p_sys = Σ m v; v_cm = p_sys ÷ M_total | add signed values or components |
| Conservation | Σp before = Σp after, per component | no net external impulse, or a very short interaction |
| Third law in momentum form | J_on B = −J_on A, so Δp_B = −Δp_A | the two objects of any interaction |
| Collision types | elastic: K_sys same; inelastic: K_sys falls; stuck: one shared velocity | after checking momentum |
Practice questions
These are original Marlbridge practice questions, not past exam questions. The rubric tables are a suggested Marlbridge rubric, not official scoring. Use g = 9.8 m/s², ignore air resistance, and treat tracks, air tables and ice as frictionless unless told otherwise.
Question 1 (multiple choice · mixed)
Two carts are at rest on a level, frictionless track, with a compressed spring between them. Cart A has mass 2m and cart B has mass m. The spring is released and the carts move apart. Which statement is correct?
- (A) The impulses are equal in size, and the carts end with equal kinetic energies.
- (B) The impulses are equal in size, and B ends with twice A’s kinetic energy.
- (C) A receives the larger impulse, so the carts end with equal speeds.
- (D) B receives the larger impulse because it ends up moving faster.
Answer and explanation
Answer: (B). Third-law forces of equal size act for the same time, so the impulses, and the final momenta, are equal in size. K = p² ÷ (2m) then gives K_B ÷ K_A = 2.
- (A) gets the impulses right but assumes equal momenta mean equal kinetic energies.
- (C) breaks the third law; equal speeds would also give the system non-zero momentum.
- (D) mistakes B’s larger velocity change, caused by its smaller mass, for a larger impulse.
Question 2 (multiple choice · mixed)
Take +x to the right. Cart A (0.40 kg) moves at +1.5 m/s towards cart B (0.60 kg), which is at rest. A force sensor on B records a symmetric triangular force–time graph: from 0 up to a peak of 12 N and back to 0, over 0.080 s in total. Which is correct just after the collision?
- (A) B moves at +0.80 m/s, and the collision is elastic because momentum is conserved.
- (B) B moves at +1.6 m/s, and A rebounds at −0.90 m/s.
- (C) B moves at +0.80 m/s, and the carts stick together.
- (D) B moves at +0.80 m/s, and the collision is inelastic.
Answer and explanation
Answer: (D). Impulse on B = triangle area = ½ × 0.080 s × 12 N = 0.48 N·s, so v_B = 0.48 ÷ 0.60 = +0.80 m/s. By the third law A receives −0.48 N·s, so p_A = 0.60 − 0.48 = 0.12 kg·m/s and v_A = +0.30 m/s. K before = ½(0.40)(1.5)² = 0.45 J. K after = 0.018 + 0.192 = 0.21 J. The total fell, so the collision is inelastic.
- (A) uses momentum to classify. Momentum is conserved in every collision; only kinetic energy decides the type.
- (B) uses the peak force for the whole 0.080 s (0.96 N·s). The kinetic energy would rise to 0.93 J.
- (C) Stuck carts would share +0.60 m/s. Here A and B move at different velocities.
Question 3 (multiple choice · mixed)
Take +y up. A rubber ball falls onto a concrete floor and bounces back up. Which statement about the short time the ball is in contact with the floor is correct?
- (A) The ball’s momentum is constant, because a bounce is a collision.
- (B) The momentum of the ball–Earth system changes, because the floor exerts a very large force on the ball.
- (C) The ball’s momentum changes, but the momentum of the ball–Earth system is constant: Earth’s momentum changes by the same amount in the opposite direction.
- (D) Momentum is not conserved, because the ball bounces back more slowly than it arrived.
Answer and explanation
Answer: (C). For the ball alone, the floor’s push is external, so the ball’s momentum reverses. For the ball–Earth system, that push and its third-law partner are internal, as is gravity. Earth gains an equal and opposite momentum, with an unnoticeable velocity change.
- (A) misuses the collision model: momentum is constant only for both interacting objects together.
- (B) The floor’s force is large, but it is internal to the ball–Earth system.
- (D) A slower rebound shows kinetic energy being transformed, not momentum being lost.
Question 4 (constructed response · mixed)
Take +x to the right. On a level, frictionless track, cart A (0.60 kg) moves at speed v = 2.0 m/s towards cart B (0.40 kg), which is at rest. A light spring with k = 250 N/m is fixed to the front of B. A compresses the spring, and then the carts separate.
(a) Explain why the carts have the same velocity at the instant of maximum compression, and find that velocity. (b) Derive an expression for the maximum compression x in terms of m_A, m_B, v and k, and evaluate it. (c) After separating, A moves at +0.40 m/s. Find B’s velocity, and classify the collision with calculations. (d) A student says: “At maximum compression the carts have lost kinetic energy, so this must be an inelastic collision.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) While A is faster than B the gap closes; once B is faster it opens. So maximum compression is the instant the velocities are equal. The spring force is internal, so momentum is constant: (0.60)(2.0) = (1.00)v_c, so v_c = +1.2 m/s.
(b) The kinetic energy missing at that instant is stored in the spring: ½m_A v² − ½(m_A + m_B)v_c² = ½kx², with v_c = m_A v ÷ (m_A + m_B). This simplifies to ½kx² = ½ [m_A m_B ÷ (m_A + m_B)] v², so x = v √[m_A m_B ÷ ((m_A + m_B)k)]. Numbers: 1.2 J − 0.72 J = 0.48 J, and x = √(2 × 0.48 ÷ 250) = 0.062 m (6.2 cm).
(c) 1.2 = (0.60)(0.40) + (0.40)v_B, so v_B = +2.4 m/s. K after = ½(0.60)(0.40)² + ½(0.40)(2.4)² = 0.048 + 1.152 = 1.2 J, equal to K before. The collision is elastic.
(d) Incorrect. The type compares the system’s kinetic energy before and after the whole collision. The 0.48 J stored at maximum compression is all returned as the carts separate.
| Point | What earns it |
|---|---|
| 1 | Equal velocities at maximum compression, with a reason |
| 1 | v_c = +1.2 m/s from momentum conservation |
| 1 | Energy statement: missing kinetic energy equals spring energy |
| 1 | Correct expression for x, and 0.062 m |
| 1 | v_B = +2.4 m/s |
| 1 | Both kinetic energy totals (1.2 J) and “elastic” |
| 1 | Rejects the claim because the type compares before and after, and the spring returns its energy |
Total: 7 points.
Question 5 (constructed response · mixed)
To find the mass m_A of a cart, a student places it at rest, back to back with cart B of known mass, on a level, low-friction track. A compressed spring between them is released, and motion sensors measure each cart’s speed just after they separate. She repeats this with different masses for B.
| m_B (kg) | 0.20 | 0.40 | 0.60 | 0.80 | 1.00 |
|---|---|---|---|---|---|
| speed of A, v_A (m/s) | 0.62 | 0.86 | 1.07 | 1.16 | 1.27 |
| speed of B, v_B (m/s) | 1.20 | 0.88 | 0.70 | 0.59 | 0.51 |
(a) Explain why the carts move off with momenta of equal size in opposite directions. (b) Show that v_A ÷ v_B = m_B ÷ m_A. State what to plot to get a straight line through the origin, and what its slope means. (c) Use the data to find m_A. (d) Using your value of m_A, find the impulse the spring gives each cart in the third trial, and comment on the result.
Worked solution and suggested Marlbridge rubric
(a) System: both carts, starting with p_sys = 0. The spring forces are internal and friction is negligible, so p_sys stays zero: m_A v_A − m_B v_B = 0, with the carts moving in opposite directions.
(b) m_A v_A = m_B v_B, so v_A ÷ v_B = m_B ÷ m_A = (1 ÷ m_A) × m_B. Plot v_A ÷ v_B (vertical) against m_B (horizontal). It should be a straight line through the origin with slope 1 ÷ m_A.
(c) Ratios: 0.52, 0.98, 1.53, 1.97, 2.49. A best-fit line through the origin has slope about 2.5 kg⁻¹ (2.49 kg⁻¹), so m_A = 1 ÷ 2.49 = 0.40 kg.
(d) Trial 3: impulse on B = m_B v_B = (0.60)(0.70) = 0.42 N·s; on A = (0.40)(1.07) = 0.43 N·s, in the opposite direction. They agree within about 2%, as the third law predicts.
| Point | What earns it |
|---|---|
| 1 | Zero initial momentum and internal spring forces, so the total stays zero |
| 1 | Derives v_A ÷ v_B = m_B ÷ m_A |
| 1 | Plots v_A ÷ v_B against m_B (or an equivalent straight-line choice) with the slope identified |
| 1 | Slope about 2.5 kg⁻¹ from a best-fit line |
| 1 | m_A ≈ 0.40 kg |
| 1 | Both impulses about 0.42 N·s, linked to equal and opposite forces for equal times |
Total: 6 points.
Question 6 (constructed response · mixed)
Take +x east and +y north, looking down on smooth, level ice. A 0.20 kg puck slides east at 3.0 m/s. A stick strikes it for 0.010 s and gives it an impulse of 0.80 N·s due north.
(a) Find the puck’s momentum just after the strike: its components, its size and its direction. (b) Find the average force of the stick. Explain why the puck’s weight does not need to appear in your answer. (c) The puck then hits a 0.40 kg puck at rest. Just afterwards, the first puck moves due east. Write the x and y momentum equations, then use one of them to find the second puck’s velocity component v₂y. (d) Predict v₂y if the second puck had mass 0.80 kg and the first puck again left due east. State the y-component of the impulse on the first puck in the collision.
Worked solution and suggested Marlbridge rubric
(a) The impulse changes only the y-component: p_x = +0.60 kg·m/s, p_y = +0.80 kg·m/s. Size √(0.60² + 0.80²) = 1.0 kg·m/s (speed 5.0 m/s) at tan⁻¹(0.80 ÷ 0.60) = 53° north of east.
(b) F_avg = J ÷ Δt = 0.80 ÷ 0.010 = 80 N, north. The weight is vertical and balanced by the normal force, so neither changes the horizontal momentum.
(c) With v₁′ for the first puck and (v₂x, v₂y) for the second: x: 0.60 = (0.20)v₁′ + (0.40)v₂x y: 0.80 = 0 + (0.40)v₂y The y-equation has one unknown: v₂y = +2.0 m/s (north).
(d) The second puck must still carry 0.80 kg·m/s of y-momentum. With twice the mass, its northward component halves to 1.0 m/s. The impulse on the first puck has y-component −0.80 N·s (south).
| Point | What earns it |
|---|---|
| 1 | Components +0.60 and +0.80 kg·m/s |
| 1 | 1.0 kg·m/s at 53° north of east |
| 1 | 80 N north, with the weight balanced by the normal force |
| 1 | Both component equations set up correctly |
| 1 | v₂y = +2.0 m/s from the y-equation |
| 1 | Halves to 1.0 m/s for the heavier puck; impulse −0.80 N·s in y |
Total: 6 points.
Question 7 (constructed response · mixed)
Take +x to the right. Cart A (0.60 kg) and cart B (0.30 kg) collide on a level, frictionless track. Motion sensors give this momentum–time data:
| Time interval | Cart A, p_x (kg·m/s) | Cart B, p_x (kg·m/s) |
|---|---|---|
| 0 to 0.40 s | +1.2 (constant) | −0.30 (constant) |
| 0.40 s to 0.50 s | falls steadily to +0.30 | rises steadily to +0.60 |
| after 0.50 s | +0.30 (constant) | +0.60 (constant) |
(a) Show that the data are consistent with conservation of momentum. (b) Find the average force on each cart during the collision, and explain how the two forces are related. (c) Find each cart’s velocity before and after, and classify the collision. (d) A student says: “The center of mass moved at 0.50 m/s before the collision, the average of 2.0 m/s and −1.0 m/s.” Evaluate this claim and find the correct value.
Worked solution and suggested Marlbridge rubric
(a) Before: 1.2 + (−0.30) = +0.90 kg·m/s. After: 0.30 + 0.60 = +0.90 kg·m/s. The total is unchanged.
(b) Force = slope of the momentum–time graph. A: (0.30 − 1.2) ÷ 0.10 = −9.0 N. B: (0.60 − (−0.30)) ÷ 0.10 = +9.0 N. Equal and opposite: a third-law pair acting for the same 0.10 s.
(c) v = p ÷ m. A: +2.0 m/s → +0.50 m/s. B: −1.0 m/s → +2.0 m/s. K before = ½(0.60)(2.0)² + ½(0.30)(1.0)² = 1.2 + 0.15 = 1.35 J. K after = ½(0.60)(0.50)² + ½(0.30)(2.0)² = 0.075 + 0.60 = 0.675 J. Half the kinetic energy is transformed and the carts separate: inelastic, but not perfectly inelastic.
(d) Incorrect. The center-of-mass velocity is a mass-weighted average: v_cm = p_sys ÷ M_total = 0.90 ÷ 0.90 = +1.0 m/s. The plain average ignores that A has twice B’s mass.
| Point | What earns it |
|---|---|
| 1 | Both totals +0.90 kg·m/s |
| 1 | −9.0 N and +9.0 N from slopes |
| 1 | Links the equal and opposite forces to the third law |
| 1 | All four velocities with signs |
| 1 | Both kinetic energy totals and “inelastic” |
| 1 | Rejects the claim; v_cm = +1.0 m/s from p_sys ÷ M_total |
Total: 6 points.
How did you do?
Question 4 is worth 7 points; Questions 5–7 are worth 6 each. Use the points you lost to choose what to study.
- Questions 1, 6 or 7 (momentum, K = p² ÷ 2m): Topic 4.1 checklist.
- Questions 2, 4, 6 or 7 (impulse, areas and slopes): Topic 4.2 checklist.
- Questions 3, 5, 6 or 7 (systems, conservation, two dimensions): Topic 4.3 checklist.
- Questions 2, 4 or 7 (elastic or inelastic): Topic 4.4 checklist.
- Quick check of every topic: retake the unit diagnostic.