Q64 / 90 · JEE Main 2022 (25 Jul Shift 2) · Physics

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?

  • A
    100 N
  • Correct answer
    B
    200 N
  • C
    300 N
  • D
    400 N

Solution

After collision, the balls change their directions of motion without changing the magnitudes of their velocity.

Figure

Change in momentum of any one ball

Δp→=2×0.05×10

⇒Δp→=1 kg m s-1

 The force exerted on the ball is F→av =Δp→Δt=10.005

Fav =200 N

Hence, the correct answer is option: 'B'.

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