Question

Two identical particles collide in the air in an perfectly inelastic manner. Before collision, one moves horizontally and the other moves in vertical direction with equal speed. After collision, the fractional loss in kinetic energy of the system of particles is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $0$

Answer 1

The correct option is A $\frac{1}{2}$


Let mass of each ball is $M$, velocity of first ball is $V\hat{i}$, velocity of the second ball is $V\hat{j}$ and let $\overrightarrow{V^{\prime }}$ be the final velocity after the collision.
Applying conservation of momentum,
$MV\hat{i}+MV\hat{j}=2M\overrightarrow{V^{\prime }}$
$|\overrightarrow{V^{\prime }}|=\frac{V}{2}\times \sqrt{1^2+1^2}=\frac{V}{\sqrt{2}}$
Initial $KE=\frac{1}{2}M{V}^2+\frac{1}{2}M{V}^2=M{V}^2$
Final $KE=\frac{1}{2}\left(2M\right)(V^{\prime }{)}^2$
$=\frac{1}{2}\left(2M\right)\frac{(V{)}^2}{2}$
$=\frac{1}{2}M{V}^2$
Therefore, fractional change in KE $=\frac{\text{Initial KE}-\text{Final KE}}{\text{Initial KE}}$
$=\frac{M{V}^2-\frac{1}{2}M{V}^2}{M{V}^2}$
$=\frac{1}{2}$
option A is correct.