Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms⁻¹ collide and rebound with the same speed. If the time duration of contact is t = 0.005 s, then what is the force exerted on the ball due to each other?

Solution

Physics sir · 3 steps

The ball reverses direction, so its momentum changes by 2mv, not mv. Divide that by the contact time.

1. Change in momentum. Take one ball. It arrives at +10 m s⁻¹ and leaves at -10 m s⁻¹, so its velocity changes by 20 m s⁻¹.

   Δ p = m(v - (-v)) = 2mv = 2 × 0.05 × 10 = 1 kg m s⁻¹

2. Force from impulse. The average force equals the change in momentum divided by the contact time: F = (Δ p)/t = 1/0.005 = 200 N.
3. So. Each ball pushes the other with an average force of 200 N, option (b). By Newton's third law the two forces are equal and opposite.

The idea to keep. When a body rebounds with the same speed, its momentum change is 2mv. Force on one body comes from that body's own momentum change, never the sum for both bodies.

Answer from the NTA final answer key. Solution by PrepWiser.