The problems are on the momentum worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Momentum is the product of an object's mass and its velocity, p = mv, measured in kg m/s. These worked answers correct the 20 problems of the Physics 9-10 worksheet on momentum, from units and direction to spot the error: each solution shows the formula, the substitution and the final answer with its unit.
Check your working against these answers: the formula, the substitution and the result with its unit. If a sign or a unit differs from yours, find the step where it changed, because that is usually where the mistake began.
Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-PS2-2, HS-PS2-3
Problem 1 Units and direction
1. The unit is kg m/s (equivalent to N s).
2. Momentum is a vector, because it has size and direction.
3. It points in the same direction as the velocity.
Problem 2 Momentum of two movers
1. p = mv = 65 × 4.0 = 260 kg m/s.
2. p = 0.045 × 40 = 1.8 kg m/s.
Problem 3 Rearranging p = mv
1. v = p ÷ m = 9000 ÷ 1500 = 6.0 m/s.
2. m = p ÷ v = 36 ÷ 12 = 3.0 kg.
Problem 4 Convert then calculate
1. 250 g = 0.250 kg and 72 km/h ÷ 3.6 = 20 m/s. p = 0.250 × 20 = 5.0 kg m/s.
2. 3.0 tonnes = 3000 kg. p = 3000 × 10 = 30 000 kg m/s (3.0 × 10⁴ kg m/s).
Problem 5 Light and fast or heavy and slow
1. Ball: 0.15 × 40 = 6.0 kg m/s. Person: 60 × 0.50 = 30 kg m/s.
2. The person has 5.0 times more momentum (30 ÷ 6.0 = 5.0).
Problem 6 Force for a time
1. Impulse = Ft = 20 × 0.50 = 10 N s.
2. Δp equals the impulse: 10 kg m/s.
3. v = Δp ÷ m = 10 ÷ 2.0 = 5.0 m/s.
Problem 7 Coupling trolleys
1. 1.5 × 4.0 + 0 = 6.0 kg m/s.
2. 6.0 = 2.0 × v, so v = 3.0 m/s.
3. Before: ½ × 1.5 × 4.0² = 12 J. After: ½ × 2.0 × 3.0² = 9.0 J. Lost: 3.0 J.
Problem 8 Kicking a football
1. Δp = 0.43 × 22 − 0 = 9.46, so 9.5 kg m/s.
2. F = Δp ÷ t = 9.46 ÷ 0.010 = 946, so 950 N.
Problem 9 Throwing from skates
1. p = 3.0 × 8.0 = 24 kg m/s forward.
2. Total momentum stays zero, so 48 × v = 24 and v = 0.50 m/s.
3. She moves backwards, opposite to the ball.
Problem 10 Bouncing off a wall
1. +0.40 × 6.0 = +2.4 kg m/s.
2. 0.40 × (−4.0) = −1.6 kg m/s.
3. Δp = −1.6 − 2.4 = −4.0 kg m/s, so 4.0 kg m/s away from the wall.
4. F = 4.0 ÷ 0.050 = 80 N away from the wall.
Problem 11 Seat belts
1. Δp = 70 × 15 = 1050 kg m/s.
2. F = 1050 ÷ 0.10 = 10 500 N.
3. F = 1050 ÷ 0.50 = 2100 N.
4. The momentum change is the same, but the longer time makes the force five times smaller, so there is less risk of injury.
Problem 12 Area under a force-time graph
1. The peak force is 600 N.
2. Area = ½ × 0.020 × 600 = 6.0 N s.
3. v = 6.0 ÷ 0.30 = 20 m/s.
4. F = 6.0 ÷ 0.020 = 300 N, half of the peak.
Problem 13 Spring-loaded trolleys
1. It equals the total before: zero.
2. 3.0 × v = 2.0 × 6.0 = 12, so v = 4.0 m/s to the right.
3. ½ × 2.0 × 6.0² + ½ × 3.0 × 4.0² = 36 + 24 = 60 J, supplied by the spring.
Problem 14 Truck and car
1. Truck: 4000 × 5.0 = 20 000 kg m/s. Car: 1000 × 20 = 20 000 kg m/s. They are equal.
2. Truck: ½ × 4000 × 5.0² = 50 000 J. Car: ½ × 1000 × 20² = 200 000 J.
3. Equal momentum change over equal time gives equal forces.
4. The car has four times more kinetic energy, so it needs four times more work from the brakes and is harder to stop in a short distance.
Problem 15 Checking a collision record
1. 0.80 × 1.50 + 0 = 1.20 kg m/s.
2. 0.80 × 0.50 + 0.40 × 1.9 = 0.40 + 0.76 = 1.16 kg m/s.
3. (1.20 − 1.16) ÷ 1.20 × 100 = 3.3 %.
4. The totals agree within normal experimental error such as timing and friction, so the data support conservation of momentum.
Problem 16 Head-on carts
1. 3.0 × 2.0 + 1.0 × (−4.0) = 6.0 − 4.0 = +2.0 kg m/s.
2. v = 2.0 ÷ 4.0 = +0.50 m/s, in the direction of the heavier cart.
3. Before: ½ × 3.0 × 2.0² + ½ × 1.0 × 4.0² = 6.0 + 8.0 = 14 J. After: ½ × 4.0 × 0.50² = 0.50 J. Lost: 13.5 J.
Problem 17 Find the missing mass
1. p = 4.0 × 1.5 = 6.0 kg m/s to the left.
2. The right-hand cart has equal momentum: m × 2.0 = 6.0, so m = 3.0 kg.
3. ½ × 4.0 × 1.5² + ½ × 3.0 × 2.0² = 4.5 + 6.0 = 10.5 J.
Problem 18 Does the energy match?
1. 0.60 × 5.0 = 0.60 × 1.0 + 0.40 × v, so 3.0 = 0.60 + 0.40v and v = 6.0 m/s.
2. Before: ½ × 0.60 × 5.0² = 7.5 J. After: ½ × 0.60 × 1.0² + ½ × 0.40 × 6.0² = 0.30 + 7.2 = 7.5 J.
3. The kinetic energy is unchanged, so the collision is elastic.
Problem 19 Landing from a vault
1. v² = 2gh = 2 × 9.8 × 1.8 = 35.28, so v = 5.9 m/s.
2. p = 45 × 5.94 = 270 kg m/s (267 before rounding).
3. F = 267 ÷ 0.050 = 5300 N.
4. F = 267 ÷ 0.40 = 670 N, eight times smaller.
Problem 20 Spot the error
1. The student treated both momenta as positive, but the trolleys move in opposite directions, so one momentum must be negative.
2. (2.0 × 3.0 − 1.0 × 3.0) ÷ 3.0 = 3.0 ÷ 3.0 = 1.0 m/s in the direction of the 2.0 kg trolley.
3. The momenta partly cancel while the mass moved doubles up, so the speed drops: the result is less than 3.0 m/s.
