Physics • Year 11 • Module 2 • Lesson 12

Conservation of Momentum

Apply conservation of momentum to real data, real scenarios, and calculations involving collisions, explosions, and kinetic energy changes.

Apply · Data & Reasoning

1. Apply the Vector Protocol — four conservation problems

For each problem: (i) define positive direction; (ii) identify the collision/explosion type; (iii) write the conservation equation; (iv) solve; (v) state the direction of your answer. Show all working. 12 marks (3 each)

1.1 A 3 kg trolley moving at +4 m/s collides with a stationary 2 kg trolley. They stick together after the collision. Find their common final velocity.

1.2 A 0.010 kg bullet is fired from a 5 kg rifle that is initially at rest. The bullet leaves the barrel at +450 m/s. Find the recoil velocity of the rifle.

1.3 A 700 kg car travelling at +15 m/s collides head-on with a 500 kg car travelling at −10 m/s. The cars lock together after the collision. Find the final velocity and state the direction of motion.

1.4 A 0.5 kg ball moving at +6 m/s strikes a stationary 1.5 kg ball. After the collision the 0.5 kg ball moves at +1 m/s. Find the velocity of the 1.5 kg ball after the collision.

Stuck? Revisit the Vector Protocol checklist and the Three Scenario Setups table in Card 2 of the lesson.

2. Interpret experimental data — trolley collision trials

A student conducted five trolley-collision trials on a frictionless air track and recorded the data below. 8 marks

Trial	m₁ (kg)	v₁ before (m/s)	m₂ (kg)	v₂ before (m/s)	v₁′ after (m/s)	v₂′ after (m/s)
1	2.0	+4.0	2.0	0	0	+4.0
2	3.0	+5.0	2.0	0	stuck together	+3.0
3	1.0	+8.0	3.0	−2.0	−2.0	+2.0
4	4.0	+3.0	4.0	0	stuck together	+1.5
5	2.0	+6.0	4.0	−1.0	−2.0	+3.0

Illustrative data. Positive direction = right.

2.1 Calculate p_before and p_after for each trial. State whether momentum is conserved (Yes / No / Approx) for each trial and complete the table. 5 marks (1 per trial)

2.2 Trial 2 uses the phrase “stuck together” for v₁′. Identify the collision type and explain what this means for the equation used to check conservation. 2 marks

2.3 A student claims Trial 1 shows an elastic collision. Using the data, confirm or refute this claim by calculating KE before and after. 1 mark

Stuck? Revisit the worked examples and the Elastic vs Inelastic table in Card 3 of the lesson.

3. Real-world scenario — crash investigation

A crash investigator is analysing a collision on a highway in NSW. A 1200 kg sedan travelling east at an unknown speed collides with a stationary 1600 kg truck. After the collision the two vehicles lock together and skid east, leaving marks 12 m long. Accident data show that the combined coefficient of friction (μ) between the tyres and road is 0.65, and g = 9.8 m/s². 7 marks

Hint: First use kinematics (v² = u² + 2as with a = −μg) to find the velocity of the combined wreck immediately after impact, then apply conservation of momentum to find the sedan’s initial speed.

3.1 Calculate the deceleration of the combined wreck during the skid. 1 mark

3.2 Use kinematics to find the velocity of the combined wreck immediately after impact. 2 marks

3.3 Apply conservation of momentum to find the sedan’s initial speed before the collision. 2 marks

3.4 The highway speed limit is 100 km/h (= 27.8 m/s). Was the sedan speeding? State your conclusion with reference to your calculated value. 1 mark

3.5 The investigator’s method assumes the road is horizontal and friction acts uniformly. State one limitation of this assumption and explain how it might affect the calculated initial speed of the sedan. 1 mark

Stuck? Use v² = u² + 2as with final v = 0 to find u (wreck speed after collision), then use perfectly inelastic momentum conservation.

4. Predict and justify — astronaut scenario

An astronaut of mass 75 kg is floating at rest in space (no gravity, no air resistance). The astronaut throws a 3 kg tool kit forward at +8 m/s relative to the space station.

5 marks

4.1 Predict the direction in which the astronaut will move after throwing the tool kit, and justify your prediction using the law of conservation of momentum. 2 marks

4.2 Calculate the astronaut’s recoil speed. Show full working using the Vector Protocol. 2 marks

4.3 The astronaut then throws a second identical 3 kg tool kit (now in their other hand) also at +8 m/s relative to the station. Predict, without full calculation, whether the astronaut’s speed after the second throw will be greater than, equal to, or less than after the first throw. Justify your answer. 1 mark

Stuck? Revisit the Explosion from Rest section and Worked Example 2 in the lesson.

Answers — Do not peek before attempting

Q1 — Vector Protocol problems

1.1 Positive direction: right (+). Perfectly inelastic. (3 + 2)v_f = 3 × 4 + 2 × 0 = 12. v_f = 12/5 = +2.4 m/s (forward).

1.2 Explosion from rest. 0 = 0.010 × 450 + 5 × v_rifle. v_rifle = −4.5/5 = −0.90 m/s (backward, recoil).

1.3 Perfectly inelastic. (700 + 500)v_f = 700 × 15 + 500 × (−10) = 10 500 − 5000 = 5500. v_f = 5500/1200 = +4.58 m/s (east, same direction as the 700 kg car).

1.4 General two-object. 0.5 × 6 + 1.5 × 0 = 0.5 × 1 + 1.5 × v₂′. 3 = 0.5 + 1.5v₂′. v₂′ = 2.5/1.5 = +1.67 m/s (forward).

Q2.1 — Data table

Trial 1: p_before = 2.0×4.0 + 2.0×0 = 8.0 N s; p_after = 2.0×0 + 2.0×4.0 = 8.0 N s. Conserved: Yes.
Trial 2: p_before = 3.0×5.0 + 0 = 15.0 N s; p_after = (3.0+2.0)×3.0 = 15.0 N s. Conserved: Yes.
Trial 3: p_before = 1.0×8.0 + 3.0×(−2.0) = 8 − 6 = 2.0 N s; p_after = 1.0×(−2.0) + 3.0×2.0 = −2 + 6 = 4.0 N s. Not conserved — error in data or non-closed system.
Trial 4: p_before = 4.0×3.0 = 12.0 N s; p_after = (4.0+4.0)×1.5 = 12.0 N s. Conserved: Yes.
Trial 5: p_before = 2.0×6.0 + 4.0×(−1.0) = 12 − 4 = 8.0 N s; p_after = 2.0×(−2.0) + 4.0×3.0 = −4 + 12 = 8.0 N s. Conserved: Yes.

Q2.2 — Trial 2 collision type

Perfectly inelastic. Since the objects stick together, they move with a single final velocity v_f. The equation becomes m₁v₁ + m₂v₂ = (m₁ + m₂)v_f. Both v₁′ and v₂′ in the table refer to this single shared velocity (+3.0 m/s).

Q2.3 — Trial 1 elastic check

KE_before = ½ × 2.0 × 4.0² = 16 J. KE_after = ½ × 2.0 × 0² + ½ × 2.0 × 4.0² = 0 + 16 = 16 J. KE is conserved. Confirmed: elastic collision. (Equal masses, complete momentum transfer — this is the special elastic result for identical masses.)

Q3 — Crash investigation

3.1 a = −μg = −0.65 × 9.8 = −6.37 m/s².

3.2 v² = u² + 2as → 0 = u² + 2 × (−6.37) × 12 → u² = 152.9 → u = 12.4 m/s (velocity of combined wreck immediately after impact, eastward).

3.3 Perfectly inelastic: m₁v₁ = (m₁ + m₂)v_f. 1200 × v_sedan = (1200 + 1600) × 12.4 = 34 720. v_sedan = 34 720/1200 = 28.9 m/s.

3.4 28.9 m/s > 27.8 m/s. Yes, the sedan was speeding (by approximately 1.1 m/s, or ~4 km/h).

3.5 If the road is not perfectly horizontal (e.g. has a slight downhill gradient in the direction of skid), friction will decelerate the wreck less, giving a higher post-impact speed. This would cause the calculated sedan speed to be an overestimate. Conversely, an uphill gradient would cause an underestimate. Accept other valid limitations: variable μ across the road surface; weight transfer affecting normal force during skid; tyres not skidding uniformly.

Q4 — Astronaut scenario

4.1 The astronaut moves backward (opposite to the direction the tool kit was thrown). By conservation of momentum, total p before = 0 (both at rest). After the throw, p_toolkit = +24 N s (forward), so the astronaut must have p = −24 N s (backward) to keep total p = 0.

4.2 0 = 3 × 8 + 75 × v_ast. v_ast = −24/75 = −0.32 m/s (backward at 0.32 m/s).

4.3 Greater than after the first throw. After the first throw the astronaut is moving at −0.32 m/s and the system (astronaut + second toolkit) has p = 75 × (−0.32) + 3 × (−0.32) = −25.0 N s (not zero). Applying conservation again, the astronaut gains additional backward momentum from the second throw and ends up moving faster than −0.32 m/s. (Full calculation gives v_ast,final = −0.64 m/s, i.e. twice the original recoil speed.)