JEE Main202226 Jul 2022Evening ShiftPhysicsCenter of Mass, Momentum and CollisionActual
A ball of mass 0 . 15 kg hits the wall with its initial speed of 12 m s - 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.
Options
- A0 . 018   s
- B0 . 036   s
- C0 . 009   s
- D0 . 072   s
Correct answer
B. 0 . 036   s
Step-by-step solution
Given: m = 1 . 5   kg   &   u = 12   m   s - 1 Change in the momentum of the ball after the collision = 2 m v = 2 × 1 . 5 × 12 = 36   N   s As the force applied during collision is equal to 100   N and if t is the duration of collision, so 100 × t = Δ p ⇒ t = 36 100   s t = 36 × 10 - 2   s = 0 . 036   s