Question

A ball of mass m = 1 kg, rolling without slipping, strikes a smooth vertical wall and comes back. Before the impact the velocity of the ball is $v_1$ = 10 cm/s and just after the impact $v_1$=8 cm/s. Find the loss in energy of ball during the impact.

• A. 3.9 x 10^-3 J.
• B. 4.6 x 10^-3 J.
• C. 2.5 x 10^-3 J.
• D. 1.8 x 10^-3 J.

Answer 1

The correct option is D

1.8 x 10^-3 J.

During impact, the normal impulsive force offered by the wall changes the linear momentum of the ball whereas the angular momentum of the ball remains practically constant because the same normal impulse force passes through the center of the sphere thus produces no torque.

$\Rightarrow$ Angular velocity of the sphere remains constant before and after the impact.

The change in kinectic energy =$\Delta K.E=\left|\frac{1}{2}m\right(v_2^2-v_1^2\left)\right|$

$\Rightarrow \Delta K.E=\frac{1}{2}\left[1\right(\frac{8}{100}{)}^2-\left(\frac{10}{100}{)}^2\right]$

$\Rightarrow Energyloss=|\Delta KE|=1.8\times {10}^{-3}j$.