The problems are on the impulse momentum worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Momentum is the product of mass and velocity, p = m v, measured in kg m/s. These worked answers correct the 20 problems of the Physics 11-12 worksheet on impulse momentum, from key relationships to elastic collision proof: each solution shows the formula, the substitution and the final answer with its unit.
Each answer states the principle, then the substitution and the result with its unit. Check the signs of velocities carefully, because a lost minus sign is the most common error in momentum problems. Keep extra digits until the final step.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB A.2
Problem 1 Key relationships
1. p = m v, in kg m/s.2. Impulse = F Δt = Δp = m v − m u, in N s.
3. No external resultant force acts on the system.
4. Kinetic energy is conserved in an elastic collision and not in an inelastic one; momentum is conserved in both.
p = m v; J = F Δt = Δp.
Problem 2 Motorbike momentum
1. 54 km/h = 54 000 ÷ 3600 = 15 m/s.2. p = m v = 210 × 15 = 3150 kg m/s.
3. Momentum is proportional to speed, so the value doubles: 2 × 3150 = 6300 kg m/s.
p = 3150 kg m/s; 6300 kg m/s.
Problem 3 Pushing a cart
1. J = F Δt = 70 × 6.0 = 420 N s.2. Δp = J = 420 kg m/s.
3. v = Δp ÷ m = 420 ÷ 85 = 4.9 m/s.
J = 420 N s; v = 4.9 m/s.
Problem 4 Same unit, two names
1. 1 N = 1 kg m/s², so 1 N s = 1 kg m/s² × s = 1 kg m/s.2. m = 2500 kg and v = 10 m/s, so p = 2500 × 10 = 25 000 kg m/s.
p = 2.5 × 10⁴ kg m/s.
Problem 5 Catching a volleyball
1. Δp = 0 − 0.27 × 6.0 = −1.62 kg m/s, a loss of 1.62 kg m/s.2. F = Δp ÷ Δt = 1.62 ÷ 0.15 = 10.8 N.
Δp = 1.62 kg m/s; F = 11 N.
Problem 6 Jumping off a raft
1. p = 60 × 2.5 = 150 kg m/s.2. Total momentum stays zero, so the raft has 150 kg m/s the opposite way: v = 150 ÷ 90 = 1.7 m/s.
v = 1.7 m/s backwards.
Problem 7 Railway wagons
1. 12 000 × 2.0 = 20 000 v, so v = 24 000 ÷ 20 000 = 1.2 m/s.2. Before: ½ × 12 000 × 2.0² = 24 000 J. After: ½ × 20 000 × 1.2² = 14 400 J.
3. Lost = 24 000 − 14 400 = 9600 J.
v = 1.2 m/s; 9.6 kJ lost.
Problem 8 Basketball bounce
1. Area = ½ × 0.020 × 260 = 2.6 N s.2. F = J ÷ Δt = 2.6 ÷ 0.020 = 130 N.
3. J = m v − m u: 2.6 = 0.40 v − 0.40 × (−5.0), so 0.40 v = 0.60 and v = 1.5 m/s upwards.
J = 2.6 N s; average F = 130 N; v = 1.5 m/s.
Problem 9 Cycling into soft sand
1. F = Δp ÷ Δt = 75 × 6.0 ÷ 0.80 = 562.5 N.2. F = 450 ÷ 0.050 = 9000 N.
3. The change in momentum is identical, so a longer stopping time gives a smaller average force and less injury.
560 N; 9.0 × 10³ N.
Problem 10 Rain on a shed
1. Volume per hour = 0.025 × 40 = 1.0 m³, so mass = 1000 kg per 3600 s = 0.278 kg/s.2. F = (Δm/Δt) v = 0.278 × 12 = 3.3 N.
0.28 kg/s; F = 3.3 N.
Problem 11 Gliders on an air track
1. v₁ = (0.60 − 0.20) × 5.0 ÷ 0.80 = 2.5 m/s and v₂ = 2 × 0.60 × 5.0 ÷ 0.80 = 7.5 m/s.2. Before: 0.60 × 5.0 = 3.0 kg m/s. After: 0.60 × 2.5 + 0.20 × 7.5 = 1.5 + 1.5 = 3.0 kg m/s.
3. Approach 5.0 m/s; separation 7.5 − 2.5 = 5.0 m/s.
v₁ = 2.5 m/s; v₂ = 7.5 m/s.
Problem 12 Model rocket
1. p = 0.050 × 120 = 6.0 kg m/s downwards.2. The rocket and its remaining fuel (0.75 kg) gain 6.0 kg m/s upwards: v = 6.0 ÷ 0.75 = 8.0 m/s.
v = 8.0 m/s upwards.
Problem 13 Bullet and ball
1. Bullet: 0.0040 × 500 = 2.0 kg m/s. Ball: 0.50 × 4.0 = 2.0 kg m/s.2. Bullet: ½ × 0.0040 × 500² = 500 J. Ball: ½ × 0.50 × 4.0² = 4.0 J.
3. Both need the same impulse (2.0 N s) to stop, but the bullet carries 125 times the kinetic energy, so it does more damage.
Equal momentum 2.0 kg m/s; KE 500 J and 4.0 J.
Problem 14 Checking glider data
1. Before: 0.30 × 0.80 + 0.20 × (−0.60) = 0.24 − 0.12 = 0.12 kg m/s. After: 0.30 × (−0.32) + 0.20 × 1.08 = −0.096 + 0.216 = 0.12 kg m/s.2. Before: ½ × 0.30 × 0.64 + ½ × 0.20 × 0.36 = 0.096 + 0.036 = 0.132 J. After: ½ × 0.30 × 0.1024 + ½ × 0.20 × 1.1664 = 0.01536 + 0.11664 = 0.132 J.
3. Both are conserved, so the collision was elastic.
Problem 15 Catching a fast ball
1. F = 0.16 × 25 ÷ 0.010 = 400 N.2. F = 4.0 ÷ 0.060 = 66.7 N.
3. The impulse (4.0 N s) is the same, so six times the time gives one sixth of the force.
400 N and 67 N.
Problem 16 Crossroads collision
1. Car: 1000 × 20 = 20 000 kg m/s east. Van: 2000 × 10 = 20 000 kg m/s north.2. Total momentum = √(20 000² + 20 000²) = 28 284 kg m/s. v = 28 284 ÷ 3000 = 9.4 m/s.
3. tan θ = 20 000 ÷ 20 000 = 1, so θ = 45° north of east.
4. Before: ½ × 1000 × 400 + ½ × 2000 × 100 = 300 000 J. After: ½ × 3000 × 9.43² = 133 333 J. Lost = 1.7 × 10⁵ J.
v = 9.4 m/s at 45° north of east; 1.7 × 10⁵ J lost.
Problem 17 Ballistic pendulum
1. After the impact, ½ M V² = M g h, so V = √(2 × 9.81 × 0.20) = 1.98 m/s.2. Momentum: 0.015 v = 2.5 × 1.98, so v = 4.95 ÷ 0.015 = 330 m/s.
3. KE before = ½ × 0.015 × 330² = 817 J. KE after = ½ × 2.5 × 1.98² = 4.9 J. Fraction lost = (817 − 4.9) ÷ 817 = 0.994.
V = 1.98 m/s; v = 330 m/s; 99 % lost.
Problem 18 Grain on a conveyor
1. Δp per second = 40 × 2.0 = 80 kg m/s per second.2. F = 80 N.
3. P = F v = 80 × 2.0 = 160 W.
4. KE per second = ½ × 40 × 2.0² = 80 W. The other 80 W is lost as thermal energy because the grain slips on the belt before it reaches belt speed.
F = 80 N; P = 160 W; 80 W wasted.
Problem 19 Force readings in a kick
1. Each 2.0 ms strip has area ½ (F₁ + F₂) × 0.0020: (250 + 750 + 1250 + 1250 + 750 + 250) × 0.0020 = 4500 × 0.0020 = 9.0 N s.2. v = J ÷ m = 9.0 ÷ 0.45 = 20 m/s.
3. F average = 9.0 ÷ 0.012 = 750 N.
4. The peak force of 1500 N is twice the average of 750 N.
J = 9.0 N s; v = 20 m/s; average F = 750 N.
Problem 20 Elastic collision proof
1. Momentum: m₁(u₁ − v₁) = m₂(v₂ − u₂). Energy: m₁(u₁² − v₁²) = m₂(v₂² − u₂²). Dividing the second by the first gives u₁ + v₁ = v₂ + u₂, so u₁ − u₂ = v₂ − v₁.2. v₂ = v₁ + 8.0. Momentum: 0.20 × 6.0 + 1.0 × (−2.0) = −0.80 = 0.20 v₁ + 1.0 (v₁ + 8.0), so 1.2 v₁ = −8.8 and v₁ = −7.33 m/s, v₂ = 0.67 m/s.
3. Before: ½ × 0.20 × 36 + ½ × 1.0 × 4 = 3.6 + 2.0 = 5.6 J. After: ½ × 0.20 × 53.8 + ½ × 1.0 × 0.444 = 5.38 + 0.22 = 5.6 J.
v₁ = −7.3 m/s; v₂ = +0.67 m/s.
