Question

A man throws a ball of mass $m$ on a vertical wall with speed $v$. If the ball bounces back with the same speed, find out the magnitude of impulse on the ball, imparted by normal reaction from the wall.

• A. $mv$
• B. $\sqrt{2}mv$
• C. $2mv$
• D. $0$

Answer 1

The correct option is C $2mv$

Assuming leftward direction (in the figure) as $-vex$ axis.
Initially, the ball is moving along $-vex-$ axis.
Initial velocity of the ball $\overrightarrow{v_1}=-v\hat{i}$
After collision with the wall, the ball reverses its direction
$\Rightarrow$ Final velocity of ball $\overrightarrow{v_2}=v\hat{i}$

Impulse imparted by normal reaction on the ball:
$J=m\Delta v=m[v_f-v_i]$
$\Rightarrow J=m\left[\right(+v)-(-v\left)\right]=+2mv$

Hence direction of impulse is along the $+vex-$ axis direction.