A ball of mass 0.15 kg hits the wall with its initial speed of 12 ms⁻¹ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N, calculate the time duration of the contact of ball with the wall.

Solution

Physics sir · 3 steps

The ball reverses, so the momentum change is twice mv, not mv. Then divide it by the force.

1. Change in momentum. Take the direction away from the wall as positive. The ball arrives with -mv and leaves with +mv, so Δ p = mv - (-mv) = 2mv.
2. Put in numbers. Δ p = 2 × 0.15 × 12 = 3.6 kg m s⁻¹.

   F Δ t = Δ p ⇒ Δ t = 3.6/100 s

3. So. The contact lasts Δ t = 0.036 s, which is option (b).

The idea to keep. Whenever a body bounces back with the same speed, its momentum change is 2mv. Impulse FΔ t equals that change, so time of contact is 2mv/F.

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