Question

A smooth sphere of mass $m=5\text{kg}$ is moving on a horizontal plane with a velocity $(5\hat{i}+\hat{j})\text{m/s}$ when it collides with a vertical wall which is parallel to the vector $\hat{j}.$ If the coefficient of restitution between the sphere and the wall is $\frac{1}{2}$, find the loss in kinetic energy caused by the impact.

• A. $46.875\text{J}$
• B. $0\text{J}$
• C. $32.5\text{J}$
• D. $49.375\text{J}$

Answer 1

The correct option is A $46.875\text{J}$

We must separate the velocity components parallel and perpendicular to the wall. The component of velocity parallel to the wall remains unchanged, while component perpendicular to the wall changed according to.
$e=\frac{\text{speed of separation}}{\text{speed of approach}}$
Let $\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$ be the velocity after collision.
Then, $v_x=-e{u}_x=-\frac{5}{2}\text{m/s}$
and $v_y=1\text{m/s}$

So, $\overrightarrow{v}=-\frac{5}{2}\hat{i}+\hat{j}$ ...(i)



Hence the velocity of sphere after the impact is $\overrightarrow{v}=\frac{-5}{2}\hat{i}+\hat{j}$

Loss in kinetic Energy $=$ Initial Kinetic Energy - Final Kinetic Energy
$=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$
$=\frac{1}{2}\times 5\times (5^2+1^2)-\frac{1}{2}\times 5[{\left(\frac{5}{2}\right)}^2+1^2]$
$△K.E.=\frac{375}{8}=46.875\text{J}$