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JEE Main 202226 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
26 July
Shift
Evening Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2022 · 26 July · Evening Shift · Physics · Center of Mass, Momentum and Collision

A ball of mass 0 . 15kg hits the wall with its initial speed of 12m...

A ball of mass 0.15kg0 . 15kg hits the wall with its initial speed of 12ms−112m{s}^{- 1} and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100N100N. calculate the time duration of the contact of ball with the wall.

Options

  1. A0.018s0 . 018s
  2. B0.036s0 . 036s
  3. C0.009s0 . 009s
  4. D0.072s0 . 072s

Correct answer

B. 0.036s0 . 036s

Step-by-step solution

Given: m=1.5kg and u=12ms−1m = 1 . 5kg \text{ and } u = 12m{s}^{- 1}

Change in the momentum of the ball after the collision=2mv= 2mv

=2×1.5×12= 2 \times 1 . 5 \times 12

=36Ns= 36Ns

As the force applied during collision is equal to 100N100N and if tt is the duration of collision, so

100×t=\Deltap100 \times t = \Deltap

⇒t=36100s⇒ t = \dfrac{36}{100}s

t=36×10−2s=0.036st = 36 \times {10}^{- 2}s = 0 . 036s

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