Collisions worksheet answers for AP Physics 1 (free PDF download)

Problem 1 Sorting collisions

1. The clay lumps move off together: perfectly inelastic.
2. No kinetic energy is lost: elastic.
3. Kinetic energy is lost but the van still separates from the barrier: inelastic.
The quantity conserved in all three is the total momentum of the system, provided external forces are negligible.

Problem 2 Balancing point of two masses

x_cm = Σmx ÷ Σm.
1. (3.0 × 0 + 1.0 × 0.80) ÷ 4.0 = 0.20 m.
2. (2.0 × (−1.0) + 6.0 × 3.0) ÷ 8.0 = 16 ÷ 8.0 = 2.0 m.

Problem 3 Carts that stick

1. p = 0.80 × 1.5 = 1.2 kg m/s.
2. Momentum is conserved: 1.2 = (0.80 + 0.40) v, so v = 1.2 ÷ 1.20 = 1.0 m/s.

Problem 4 Choosing the system

1. Truck B exerts a force on A during the contact, and that force is external to A, so the momentum of A changes.
2. The forces between the trucks are equal and opposite (Newton’s third law) and are internal to the pair. With no net external force, the total momentum of the pair is conserved.

Problem 5 Energy lost when carts stick

1. Before: ½ × 0.80 × 1.5² = 0.90 J. After: ½ × 1.20 × 1.0² = 0.60 J.
2. Loss = 0.90 − 0.60 = 0.30 J, which is 0.30 ÷ 0.90 = 33 %.

Problem 6 Centre of mass velocity

1. v_cm = (2.0 × 4.0 + 3.0 × (−2.0)) ÷ 5.0 = 2.0 ÷ 5.0 = +0.40 m/s.
2. It does not change: +0.40 m/s, because the net external force is zero.

Problem 7 Curling takeout

1. m₁ = m₂, so v₁ = 0 × 2.0 ÷ 38 = 0 and v₂ = 2 × 19 × 2.0 ÷ 38 = 2.0 m/s. The moving stone stops and the other takes all the motion.
2. Before: 19 × 2.0 ÷ 38 = 1.0 m/s. After: 19 × 2.0 ÷ 38 = 1.0 m/s, unchanged.

Problem 8 Crash test at the proving ground

1. 1200 × 20 = 2000 v, so v = 24 000 ÷ 2000 = 12 m/s.
2. Before: ½ × 1200 × 20² = 2.4 × 10⁵ J. After: ½ × 2000 × 12² = 1.44 × 10⁵ J. Loss = 9.6 × 10⁴ J.
3. It becomes work deforming the metal, thermal energy and sound.

Problem 9 Find the unknown cart

1. 0.50 × 2.4 = (0.50 + m) × 1.5.
2. 1.2 = 0.75 + 1.5 m, so m = 0.45 ÷ 1.5 = 0.30 kg.

Problem 10 Reading a velocity graph

1. Cart A: 5.0 m/s before, 1.0 m/s after. Cart B: 0 before, 2.0 m/s after.
2. Momentum: 2.0 × 5.0 = 2.0 × 1.0 + m × 2.0, so m = 8.0 ÷ 2.0 = 4.0 kg.
3. Before: ½ × 2.0 × 5.0² = 25 J. After: ½ × 2.0 × 1.0² + ½ × 4.0 × 2.0² = 1 + 8 = 9 J. Energy is lost, so the collision is inelastic.

Problem 11 Heavy hits light

1. v₁ = (2.0 − 1.0) × 3.0 ÷ 3.0 = 1.0 m/s; v₂ = 2 × 2.0 × 3.0 ÷ 3.0 = 4.0 m/s. Check: 6.0 = 2.0 + 4.0.
2. Approach: 3.0 − 0 = 3.0 m/s. Separation: 4.0 − 1.0 = 3.0 m/s. They are equal.

Problem 12 Head-on trolleys stop dead

1. Take the first direction as positive: 20 × 1.5 + 30 × (−1.0) = 30 − 30 = 0, so v = 0.
2. Before: ½ × 20 × 1.5² + ½ × 30 × 1.0² = 22.5 + 15 = 37.5 J. After: 0. Loss = 37.5 J, all of it.

Problem 13 Equal and opposite impulses

1. Larger cart: 0.80 × (1.0 − 1.5) = −0.40 kg m/s. Smaller cart: 0.40 × (1.0 − 0) = +0.40 kg m/s. They are equal and opposite.
2. F = Δp ÷ Δt = 0.40 ÷ 0.050 = 8.0 N on each cart, in opposite directions.

Problem 14 Three masses on a light rod

1. x_cm = (1.0 × 0 + 2.0 × 0.30 + 3.0 × 0.90) ÷ 6.0 = 3.3 ÷ 6.0 = 0.55 m.
2. Place the pivot at the centre of mass, 0.55 m from the 1.0 kg end.

Problem 15 Checking recorded data

1. Before: 0.50 × 0.80 = 0.400 kg m/s. After: 0.50 × 0.27 + 0.25 × 1.06 = 0.135 + 0.265 = 0.400 kg m/s.
2. Before: ½ × 0.50 × 0.80² = 0.160 J. After: ½ × 0.50 × 0.27² + ½ × 0.25 × 1.06² = 0.0182 + 0.1405 = 0.159 J.
3. The energy differs by less than 1 %, so the collision is elastic within the measurement error.

Problem 16 Sticking to a heavy cart

1. m₁u = (m₁ + m₂) v, so v = m₁u ÷ (m₁ + m₂).
2. KE after ÷ KE before = [½ (m₁ + m₂) v²] ÷ [½ m₁u²] = (m₁ + m₂) m₁²u² ÷ [(m₁ + m₂)² m₁u²] = m₁ ÷ (m₁ + m₂).
3. With m₂ = 9 m₁: v = 5.0 ÷ 10 = 0.50 m/s. Retained fraction = 1 ÷ 10, so 90 % of the kinetic energy is lost.

Problem 17 Designing a test of momentum conservation

1. Measure the mass of each cart. Fit a card of known length to each cart; speed = card length ÷ time to pass the photogate. Measure the speed of the moving cart before and the joined carts after. Repeat with different masses.
2. Calculate p_before = m₁u and p_after = (m₁ + m₂) v. Plot p_after against p_before: a straight line of gradient 1 through the origin within uncertainty supports conservation.
3. Friction on the track removes momentum, so p_after is slightly low. Tilt the track very slightly or use an air track, and take the speed just after the collision.

Problem 18 Timeline of a point collision

1. The carts approach at 4.0 m/s over a gap of 0.80 m: t = 0.80 ÷ 4.0 = 0.20 s. Cart 1 is then at 0.20 + 2.0 × 0.20 = 0.60 m.
2. v₁ = [(1.5 − 0.5) × 2.0 + 2 × 0.5 × (−2.0)] ÷ 2.0 = 0; v₂ = [(0.5 − 1.5) × (−2.0) + 2 × 1.5 × 2.0] ÷ 2.0 = +4.0 m/s. Check: momentum 2.0 kg m/s both times.
3. Method one: v_cm = 2.0 ÷ 2.0 = 1.0 m/s from x_cm = (1.5 × 0.20 + 0.50 × 1.00) ÷ 2.0 = 0.40 m, so x_cm = 0.40 + 1.0 × 0.50 = 0.90 m. Method two: cart 1 at 0.60 m; cart 2 at 0.60 + 4.0 × 0.30 = 1.80 m; x_cm = (1.5 × 0.60 + 0.50 × 1.80) ÷ 2.0 = 0.90 m.

Problem 19 Partly inelastic bounce

1. Momentum before: 6.0 − 1.5 = 4.5 kg m/s. After: 2.0 × 0.50 + 1.0 × v = 4.5, so v = +3.5 m/s.
2. Before: 9.0 + 1.125 = 10.125 J. After: 0.25 + 6.125 = 6.375 J. Loss = 3.75 J.
3. Approach 4.5 m/s, separation 3.5 − 0.50 = 3.0 m/s. Ratio = 0.67. A ratio of 1 means elastic and 0 means perfectly inelastic, so this collision is partly inelastic.

Problem 20 Evaluating a claim

1. Taking one direction as positive: 1500 × 10 − 1500 × 10 = 0 before and 0 after. Kinetic energy before = 2 × ½ × 1500 × 10² = 1.5 × 10⁵ J, all transformed.
2. The claim is wrong. For the system of both cars the total momentum was zero before and is zero after, so it is conserved. Each car lost its own momentum because the other car exerted a force on it, but the momentum of one car is the opposite of the other, so the total is unchanged. What disappeared was kinetic energy, which became deformation work, sound and thermal energy.