These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. This set is for the calculus-based Physics C: Mechanics course. Masses are in kg, velocities in m/s, momenta in kg·m/s and energies in J. In every question, u is a velocity just before the collision and v a velocity just after. Tracks and air tables are level and frictionless unless stated. Round final answers to 2 significant figures unless told otherwise.
Question 1 (multiple choice · foundation)
Take +x to the right. A 1.0 kg cart moving at +4.0 m/s hits a 3.0 kg cart at rest. Just after, the 1.0 kg cart moves at −2.0 m/s and the 3.0 kg cart at +2.0 m/s. Which statement is correct?
- (A) The collision is elastic: the total kinetic energy is 8.0 J before and after.
- (B) The collision is inelastic, because the 1.0 kg cart’s kinetic energy fell from 8.0 J to 2.0 J.
- (C) The collision is perfectly inelastic, because the carts move off with the same speed.
- (D) The collision is impossible, because the total momentum after is 8.0 kg·m/s, double the momentum before.
Answer and explanation
Answer: (A). Momentum: 1.0 × 4.0 = 4.0 before; 1.0 × (−2.0) + 3.0 × 2.0 = 4.0 after, so it is conserved. Kinetic energy: ½(1.0)(4.0)² = 8.0 J before; ½(1.0)(2.0)² + ½(3.0)(2.0)² = 2.0 + 6.0 = 8.0 J after. The total is unchanged, so the collision is elastic.
- (B) looks at one object only. The 6.0 J the first cart lost went to the second cart; the system total is what matters.
- (C) confuses equal speeds with a common velocity. The carts move in opposite directions, so they have not stuck together.
- (D) adds speeds, ignoring the minus sign on the 1.0 kg cart’s velocity.
Question 2 (multiple choice · core)
Take +x to the right. A 2.0 kg cart moving at +3.0 m/s hits a 4.0 kg cart at rest, and they stick together. What fraction of the initial kinetic energy is transformed into other forms?
- (A) 0
- (B) 1/3
- (C) 2/3
- (D) 8/9
Answer and explanation
Answer: (C). Common velocity v = (2.0 × 3.0) ÷ 6.0 = 1.0 m/s. K_before = ½(2.0)(3.0)² = 9.0 J; K_after = ½(6.0)(1.0)² = 3.0 J. Transformed: 6.0 J ÷ 9.0 J = 2/3.
- (A) assumes that because momentum is conserved, kinetic energy is too. Sticking together is always inelastic.
- (B) is the fraction that remains as kinetic energy (3.0 J of 9.0 J), not the fraction transformed.
- (D) uses only the 2.0 kg mass after the collision: ½(2.0)(1.0)² = 1.0 J. Both carts move at 1.0 m/s, so use the combined 6.0 kg.
Question 3 (multiple choice · core)
Two carts with spring bumpers collide head-on on a track. The collision is elastic. At the instant the bumpers are most compressed, which statement is correct?
- (A) Both carts have the same velocity, equal to v_cm, and the system’s kinetic energy is at its minimum.
- (B) The system’s kinetic energy equals its initial value, because the collision is elastic.
- (C) The system’s total momentum is zero for an instant.
- (D) Both carts are momentarily at rest.
Answer and explanation
Answer: (A). The bumpers stop compressing when the carts stop approaching each other, which is when their velocities are equal. Momentum conservation then makes that shared velocity v_cm. Kinetic energy is lowest here, because the most energy is stored in the springs.
- (B) applies “elastic” to the middle of the collision. Elastic means total kinetic energy after equals total before; during contact some is stored as elastic potential energy.
- (C) The total momentum stays at its initial value throughout. It is zero only if it was zero to start with.
- (D) is true only when v_cm = 0. In general the carts share a nonzero velocity.
Question 4 (calculation · core)
Take +x to the right. Ball 1 (0.20 kg) moves at +5.0 m/s and ball 2 (0.30 kg) moves at −1.0 m/s. They collide head-on, and the collision is elastic. Find both velocities just after, and check your answer with kinetic energy.
Worked solution
- Momentum: 0.20(5.0) + 0.30(−1.0) = 0.70 = 0.20v₁ + 0.30v₂.
- Elastic, one dimension: v₂ − v₁ = −(u₂ − u₁) = −(−1.0 − 5.0) = 6.0.
- Substitute v₂ = v₁ + 6.0: 0.20v₁ + 0.30v₁ + 1.8 = 0.70, so 0.50v₁ = −1.1 and v₁ = −2.2 m/s.
- v₂ = −2.2 + 6.0 = +3.8 m/s.
- Check: K_before = ½(0.20)(25) + ½(0.30)(1.0) = 2.5 + 0.15 = 2.65 J. K_after = ½(0.20)(4.84) + ½(0.30)(14.44) ≈ 0.484 + 2.166 = 2.65 J. ✓
Suggested mark points (3): 1 for the momentum equation with signs; 1 for the second equation (relative velocity reversal or kinetic energy); 1 for both velocities with directions.
Common error: writing u₂ = +1.0 m/s. Ball 2 moves left, so its velocity is negative.
Question 5 (constructed response · core)
Take +x along the track. Students test three bumper types. In each trial cart A (0.40 kg) moves at +0.50 m/s towards an identical cart B (0.40 kg) at rest. Velocities just after the collision are measured to ±0.01 m/s:
| Trial | Bumpers | v_A (m/s) | v_B (m/s) |
|---|---|---|---|
| 1 | repelling magnets | +0.01 | +0.49 |
| 2 | rubber pads | +0.10 | +0.40 |
| 3 | hook-and-loop strips | +0.25 | +0.25 |
(a) Show that momentum is conserved in every trial. (b) Calculate the fraction of the initial kinetic energy that remains after each trial. (c) Classify each collision, using the data and their uncertainty as evidence. (d) In trial 2, say what happened to the “missing” kinetic energy. (e) Explain why no trial could ever keep less kinetic energy than trial 3.
Worked solution and suggested Marlbridge rubric
(a) P_before = 0.40 × 0.50 = 0.20 kg·m/s. After: 0.40(0.01 + 0.49) = 0.40(0.10 + 0.40) = 0.40(0.25 + 0.25) = 0.20 kg·m/s in each trial.
(b) K_before = ½(0.40)(0.50)² = 0.050 J. After: trial 1, 0.048 J (96%); trial 2, 0.034 J (68%); trial 3, 0.025 J (50%).
(c) Trial 1: the ideal elastic result for equal masses is a swap (0 and 0.50 m/s), which lies within ±0.01 m/s of the readings. The 4% shortfall is within what the uncertainty allows, so it is elastic within experimental uncertainty. Trial 2: 68% is far below anything the uncertainty can explain, and the carts separate, so it is inelastic. Trial 3: the carts share one velocity, so it is perfectly inelastic.
(d) The rubber pads deformed and did not fully spring back. Nonconservative forces in the rubber transformed about 0.016 J into internal (thermal) energy and sound.
(e) Momentum stays 0.20 kg·m/s, so the center of mass keeps moving at 0.25 m/s. Kinetic energy of center-of-mass motion, ½(0.80)(0.25)² = 0.025 J, cannot be removed by any collision. Trial 3 already has only this energy, so it is the minimum.
| Point | What earns it |
|---|---|
| 1 | (a) Momentum before and after for all three trials |
| 1 | (b) All three fractions correct |
| 1 | (c) Trial 1 called elastic with the uncertainty argument |
| 1 | (c) Trials 2 and 3 classified correctly with reasons |
| 1 | (d) Names a nonconservative process and at least one destination for the energy |
| 1 | (e) Links the minimum to the fixed center-of-mass velocity (or K_min = P²/2M) |
Alternative for (c). Comparing separation speed with approach speed (0.48, 0.30 and 0 m/s against 0.50 m/s) earns the points if the same conclusions are reached.
Question 6 (constructed response · stretch)
Take +x along the motion. Object 1, mass m₁, moves at u₁ and collides elastically and head-on with object 2, mass m₂, at rest.
(a) Starting from momentum and kinetic energy conservation, derive v₂ in terms of m₁, m₂ and u₁. (b) Show that the fraction of object 1’s kinetic energy transferred to object 2 is f = 4m₁m₂ / (m₁ + m₂)². (c) Evaluate f for m₂ = m₁, m₂ = 4m₁ and m₂ = m₁/4. (d) The target is replaced by one four times as massive as the first object, instead of an equal one. By what factor does v₂ change?
Worked solution and suggested Marlbridge rubric
(a) Momentum: m₁(u₁ − v₁) = m₂v₂. Kinetic energy: m₁(u₁² − v₁²) = m₂v₂². Divide the second by the first: u₁ + v₁ = v₂, so v₁ = v₂ − u₁. Substitute into the momentum equation: m₁(2u₁ − v₂) = m₂v₂, giving v₂ = 2m₁u₁ / (m₁ + m₂).
(b) f = (½m₂v₂²) ÷ (½m₁u₁²) = (m₂/m₁) × 4m₁² / (m₁ + m₂)² = 4m₁m₂ / (m₁ + m₂)².
(c) m₂ = m₁: f = 4m₁² / 4m₁² = 1 (all the energy moves). m₂ = 4m₁: f = 16/25 = 0.64. m₂ = m₁/4: f = 1/(25/16) = 0.64. The formula is symmetric in the two masses, so a mismatch either way lowers the transfer by the same amount.
(d) Equal masses: v₂ = u₁. With m₂ = 4m₁: v₂ = 2u₁/5 = 0.40u₁. So v₂ falls by a factor of 0.40 (to 2/5 of its value).
| Point | What earns it |
|---|---|
| 1 | (a) Both conservation equations written correctly |
| 1 | (a) Algebra to v₂ = 2m₁u₁/(m₁ + m₂), via the relative-velocity step or substitution |
| 1 | (b) Forms the ratio of kinetic energies and simplifies |
| 1 | (c) All three values, with a comment on the symmetry or on the maximum at equal masses |
| 1 | (d) Factor 0.40 (or 2/5), from the formula, not from a numerical guess |
Question 7 (explanation · stretch)
A student says: “In a perfectly inelastic collision all of the kinetic energy is lost, because the objects stop bouncing.”
Take +x to the right. Test the claim with (i) a 0.40 kg cart at +2.0 m/s that sticks to a 0.20 kg cart moving at −4.0 m/s, and (ii) the same 0.40 kg cart at +2.0 m/s sticking to the 0.20 kg cart at rest. Then state the general rule.
Model answer and suggested Marlbridge rubric
(i) P = 0.40(2.0) + 0.20(−4.0) = 0. The joined carts stop. K_before = 0.80 + 1.6 = 2.4 J, and all 2.4 J is transformed. Here the claim happens to be true.
(ii) P = 0.80 kg·m/s, so v = 0.80 ÷ 0.60 ≈ 1.33 m/s. K_before = 0.80 J; K_after = ½(0.60)(1.33)² ≈ 0.53 J. Two-thirds of the kinetic energy remains, so the claim is false here.
General rule. After sticking, all the mass moves at v_cm, so K_after = P² / (2M). This is zero only when the total momentum is zero. Otherwise, momentum conservation forces the joined object to keep moving, and it must carry kinetic energy. What is lost is the kinetic energy of motion relative to the center of mass, which is the most any collision can remove.
| Point | What earns it |
|---|---|
| 1 | (i) Shows P = 0, so the carts stop and all 2.4 J is transformed |
| 1 | (ii) Common velocity and K_after ≈ 0.53 J, so the claim fails |
| 1 | States that K_after = P²/(2M) (or ½Mv_cm²) remains |
| 1 | Explains that only the energy relative to the center of mass can be removed |
How did you do?
- Q1 or Q3 wrong: re-read “Two checks for every collision” and “Elastic collisions” in the study guide. Judge the system total, before and after.
- Q2 or Q7 wrong: revisit “Inelastic and perfectly inelastic collisions” and Figure 1.
- Q4 or Q6 incomplete: work through “Solving a one-dimensional elastic collision” and check signs on every velocity.
- Q5 incomplete: a classification needs evidence: give the kinetic energy fraction and say whether the uncertainty could explain it.
Then tick off the topic checklist.