Easy MCQ +4 / -1 PYQ · JEE Mains 2022

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

  1. A 0.018 s
  2. B 0.036 s Correct answer
  3. C 0.009 s
  4. D 0.072 s

Solution

<p>F = 100 N</p> <p>$\Delta$P = 2 $\times$ 0.15 $\times$ 12</p> <p>= 3.6</p> <p>$\Rightarrow$ t = ${{3.6} \over {100}}$ = 0.036 s</p>

About this question

Subject: Physics · Chapter: Rotational Motion · Topic: Impulse and Momentum

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