A ball of mass $600g$ strikes a wall with a velocity of $5m{s}^{-1}$ at an angle ${30}^{\circ }$ with the walls and rebounds with the same speed at the same angle with the wall. The magnitude of change in momentum of the ball is. (in $kgm{s}^{-1}$)

Given that,

$m=600g=0.6Kg$

Velocity, $v=5m/s$

Since, the ball bounces with same velocity and with same angle.

So, $u\sin \theta =v\sin \theta$

So, change in momentum,

$\Delta p=mu\sin \theta -(-mv\sin \theta )$

$=2mv\sin \theta$

$=2\times 0.6\times 5\times \sin 30$

$\Delta p=3Kgm/s$

Change in momentum is $3Kgm/s$