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 is the algebra-based course, so no calculus is needed. Use g = 9.8 m/s² where gravity appears. Round final answers to 2 significant figures unless told otherwise, and keep unrounded values in your calculator until the end. All data are invented for practice.
Question 1 (multiple choice · foundation)
Two balls collide on a smooth table. Before the collision, ball A has 8.0 J of kinetic energy and ball B has 4.0 J. After the collision, ball A has 3.0 J and ball B has 9.0 J. Which statement is correct?
- (A) The collision is elastic, because the total kinetic energy is 12 J before and after.
- (B) The collision is inelastic, because ball A lost kinetic energy.
- (C) The collision is perfectly inelastic, because energy was transferred from A to B.
- (D) The collision is inelastic, because the total momentum must have decreased.
Answer and explanation
Answer: (A). Classify with the system total: 8.0 + 4.0 = 12 J before and 3.0 + 9.0 = 12 J after. It is unchanged, so the collision is elastic.
- (B) judges by one object. In an elastic collision individual kinetic energies can change.
- (C) Perfectly inelastic means the balls stick together and kinetic energy decreases. Neither is stated or true here.
- (D) Momentum is conserved in every collision of an isolated system, elastic or not.
Question 2 (multiple choice · core)
Take +x to the right. A 2.0 kg trolley moving at +3.0 m/s collides with a 1.0 kg trolley moving at −3.0 m/s. They lock together. What is their velocity just after the collision?
- (A) +1.0 m/s
- (B) +3.0 m/s
- (C) 0
- (D) +2.0 m/s
Answer and explanation
Answer: (A). Momentum before = (2.0)(+3.0) + (1.0)(−3.0) = +3.0 kg·m/s. After, the combined 3.0 kg moves at v_f: v_f = 3.0 ÷ 3.0 = +1.0 m/s.
- (B) adds the momenta without signs, (6.0 + 3.0) ÷ 3.0. The trolleys move in opposite directions.
- (C) assumes equal speeds cancel. The momenta are not equal, because the masses differ.
- (D) uses only the 2.0 kg trolley’s momentum, 6.0 ÷ 3.0, ignoring the second trolley’s.
Question 3 (multiple choice · core)
A block of mass m slides at speed v on a frictionless surface and sticks to a block of mass 3m at rest. What fraction of the original kinetic energy is transformed into other forms?
- (A) 1/4
- (B) 3/4
- (C) 1/3
- (D) 0, because momentum is conserved
Answer and explanation
Answer: (B). Momentum gives v_f = mv / 4m = v/4. K_after = ½(4m)(v/4)² = ⅛mv², which is ¼ of K_before = ½mv². So ¾ of the kinetic energy is transformed. The shortcut m₂ / (m₁ + m₂) = 3m / 4m gives the same.
- (A) is the fraction that remains as kinetic energy.
- (C) is the mass ratio m / 3m; kinetic energy depends on mass and the square of speed.
- (D) confuses the two conservation laws. Momentum is conserved, but kinetic energy is not in a sticky collision.
Question 4 (calculation · core)
Take +x to the right. Cart A (0.60 kg) moves at +1.5 m/s and cart B (0.40 kg) moves at −0.50 m/s. They collide head-on. Just afterwards, A moves at +0.10 m/s.
(a) Find B’s velocity just after the collision. (b) Classify the collision, with calculations. (c) In a second trial with the same starting velocities, the carts stick together. Find their common velocity and compare the kinetic energy transformed in the two trials.
Worked solution and suggested Marlbridge rubric
(a) Momentum before = (0.60)(1.5) + (0.40)(−0.50) = 0.90 − 0.20 = 0.70 kg·m/s. After: 0.70 = (0.60)(0.10) + (0.40)v_B, so 0.40v_B = 0.64 and v_B = +1.6 m/s.
(b) K before = ½(0.60)(1.5)² + ½(0.40)(0.50)² = 0.675 + 0.050 = 0.725 J. K after = ½(0.60)(0.10)² + ½(0.40)(1.6)² = 0.003 + 0.512 = 0.515 J. K decreased by 0.21 J, so the collision is inelastic (but not perfectly inelastic: the carts separate).
(c) v_f = 0.70 ÷ 1.00 = +0.70 m/s. K after = ½(1.00)(0.70)² = 0.245 J, so 0.725 − 0.245 = 0.48 J is transformed. That is more than twice the 0.21 J of the first trial, as expected: sticking together transforms the most kinetic energy for a given momentum.
| Point | What earns it |
|---|---|
| 1 | Momentum before with signs (0.70 kg·m/s) |
| 1 | v_B = +1.6 m/s |
| 1 | Both kinetic energy totals |
| 1 | “Inelastic” because the total decreased |
| 1 | v_f = +0.70 m/s and 0.48 J transformed, with the comparison stated |
Question 5 (graph · core)
Take +x to the right. Motion sensors record two carts on a level track. The velocity–time graph shows:
- Cart A (0.50 kg): +0.80 m/s until t = 1.00 s, then a steady +0.20 m/s from t = 1.05 s.
- Cart B (0.30 kg): −0.20 m/s until t = 1.00 s, then a steady new value from t = 1.05 s.
(a) Find cart B’s velocity after the collision. (b) Is the collision elastic? Show your reasoning. (c) Find the average force exerted on cart B during the collision. (d) A student adds a third line to the graph for the velocity of the centre of mass. Describe that line.
Worked solution and suggested Marlbridge rubric
(a) Momentum before = (0.50)(0.80) + (0.30)(−0.20) = 0.40 − 0.06 = 0.34 kg·m/s. After: 0.34 = (0.50)(0.20) + (0.30)v_B, so v_B = 0.24 ÷ 0.30 = +0.80 m/s.
(b) K before = ½(0.50)(0.80)² + ½(0.30)(0.20)² = 0.160 + 0.006 = 0.166 J. K after = ½(0.50)(0.20)² + ½(0.30)(0.80)² = 0.010 + 0.096 = 0.106 J. The total decreased by 0.060 J, so the collision is not elastic (it is inelastic).
(c) Contact time 0.05 s. Δp_B = (0.30)(0.80 − (−0.20)) = 0.30 kg·m/s. F_avg = 0.30 ÷ 0.05 = 6.0 N, to the right. (Cart A’s Δp is −0.30 kg·m/s, so it feels 6.0 N to the left.)
(d) A horizontal line at v_cm = 0.34 ÷ 0.80 = +0.43 m/s, unchanged through the collision, because there is no net external force on the two-cart system.
| Point | What earns it |
|---|---|
| 1 | v_B = +0.80 m/s with correct signs |
| 1 | Kinetic energy totals compared; “inelastic” |
| 1 | Δp_B = 0.30 kg·m/s using the change from −0.20 to +0.80 |
| 1 | 6.0 N to the right |
| 1 | Horizontal line at about 0.43 m/s, with a reason |
Question 6 (experimental · stretch)
Two identical 0.50 kg carts have hook-and-loop pads. Cart A is launched at different speeds into cart B, which is at rest, and they stick together. Take +x in A’s direction. The results are:
| Trial | v_A before (m/s) | v after, together (m/s) |
|---|---|---|
| 1 | 0.40 | 0.19 |
| 2 | 0.60 | 0.30 |
| 3 | 0.80 | 0.41 |
| 4 | 1.00 | 0.49 |
| 5 | 1.20 | 0.61 |
(a) Derive the predicted value of K_after / K_before for these carts. (b) Calculate K_before and K_after for each trial. (c) Describe the graph you would plot to test your prediction, and find its slope. (d) Do the data support the prediction? Give one check you would make on the data before trusting them.
Worked solution and suggested Marlbridge rubric
(a) Momentum: m v = 2m v_f, so v_f = v/2. K_after / K_before = [½(2m)(v/2)²] / [½mv²] = ½. (In general, m₁ / (m₁ + m₂).)
(b)
| Trial | K_before (J) | K_after (J) |
|---|---|---|
| 1 | 0.040 | 0.018 |
| 2 | 0.090 | 0.045 |
| 3 | 0.160 | 0.084 |
| 4 | 0.250 | 0.120 |
| 5 | 0.360 | 0.186 |
(c) Plot K_after (vertical) against K_before (horizontal). The prediction is a straight line through the origin with slope 0.50. A best-fit line through the origin has slope ≈ 0.51.
(d) Yes. The slope 0.51 agrees with the predicted 0.50 within the scatter of the points (individual ratios range from about 0.45 to 0.53). A useful check is momentum: m v_A should equal 2m v_f. For example, trial 4 gives 0.50 kg·m/s before and 0.49 kg·m/s after. Momentum agrees within about 5% in every trial, so the velocity measurements are trustworthy.
| Point | What earns it |
|---|---|
| 1 | Uses momentum to get v_f = v/2 |
| 1 | Ratio ½ derived |
| 1 | Correct kinetic energies |
| 1 | K_after against K_before; straight line through origin, slope 0.50 expected |
| 1 | Slope ≈ 0.51 from a best-fit line |
| 1 | Supports the prediction, with a sensible data check (momentum, repeats, track level) |
Alternative method. Plotting v_after against v_before (expected slope 0.50) tests the momentum prediction rather than the energy one. It earns the slope point only if the student then explains why a speed ratio of ½ gives an energy ratio of ½ here (mass doubles, speed squared quarters).
Question 7 (constructed response · stretch)
Take +x along the track. A 1.5 kg cart rolls at 3.0 m/s along a level track and sticks to a 0.50 kg cart at rest. The pair then rolls up a frictionless ramp.
(a) Derive an expression for the greatest height h the pair reaches, in terms of m₁, m₂, v and g, and calculate it. (b) A student says: “Energy is conserved, so the height is v²/2g = 0.46 m.” Explain the error. (c) How much kinetic energy was transformed in the collision?
Model answer and suggested Marlbridge rubric
(a) Stage 1, the collision (momentum): m₁v = (m₁ + m₂)v_f, so v_f = m₁v / (m₁ + m₂) = (1.5)(3.0) ÷ 2.0 = 2.25 m/s. Stage 2, up the ramp (energy, no friction): ½(m₁ + m₂)v_f² = (m₁ + m₂)gh, so
h = v_f² / 2g = [m₁v / (m₁ + m₂)]² / 2g = (2.25)² ÷ (2 × 9.8) = 0.26 m.
(b) The student used kinetic-energy conservation through the collision. A sticky collision is perfectly inelastic, so kinetic energy is not conserved there. Only momentum can link the speeds before and after the collision. Energy conservation is valid only after the collision, on the frictionless ramp. The student’s answer is too large by a factor of 1 / (0.75)² ≈ 1.8.
(c) K before = ½(1.5)(3.0)² = 6.75 J. K after = ½(2.0)(2.25)² = 5.06 J. Transformed: 1.7 J (a quarter, which matches m₂ / (m₁ + m₂) = 0.50 ÷ 2.0).
| Point | What earns it |
|---|---|
| 1 | Momentum used for the collision: v_f = 2.25 m/s |
| 1 | Energy used for the ramp only |
| 1 | Symbolic expression for h |
| 1 | h ≈ 0.26 m |
| 1 | Explains that kinetic energy is not conserved in the sticky collision |
| 1 | 1.7 J transformed |
How did you do?
- Q1 or Q3 wrong: re-read “Two conservation questions for every collision”, Worked example 2 and “Comparing scenarios with the expressions” in the study guide.
- Q2 or Q4 wrong: revisit Worked examples 1 and 3. Signed velocities for momentum; squared speeds for kinetic energy.
- Q5 incomplete: compare with Figure 1, and recall that v_cm stays constant (Topic 4.3).
- Q6 or Q7 incomplete: see “Testing whether a collision is elastic” and the misconception about energy conservation across a sticky collision.
Then tick off the topic checklist.