JEE Main 2023 · 31 January · morning · Single correct · +4 / −1
100 balls each of mass m moving with speed v simultaneously strike a wall normally and reflected back with same speed, in time t s. The total force exerted by the balls on the wall is
- (a) (200 m v)/t
- (b) (100 m v)/t
- (c) (m v)/(100 t)
- (d) 200 m v t
Answer and solution
Answer (a) (200 m v)/t
Solution
Physics sir · 3 steps
The ball turns right around, so its momentum changes by 2mv, not mv. That factor 2 is the whole question.
- One ball. Take the direction towards the wall as positive. Before the hit the ball has momentum +mv, after it has -mv, so its change is -mv - mv = -2mv, of size 2mv.
- All 100 balls. All the balls hit together, so the total change in momentum is 100 × 2mv = 200mv. This happens in time t.
F = (Δ p)/(Δ t) = (200 mv)/t
- So. The wall pushes the balls back with this force, and by Newton's third law the balls push the wall with the same force, 200mv/t. That is option (a).
The idea to keep. Before using F = Δ p/Δ t, always write the momentum before and after with signs. A bounce at the same speed gives 2mv, and a ball that just stops gives mv.
Answer from the NTA final answer key. Solution by PrepWiser.