Question

A particle of mass $m$ moving in the x direction with speed $2v$ is hit by another particle of mass $2m$ moving in the y direction with speed $v$. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

• A. 50%
• B. 56%
• C. 62%
• D. 44%

Answer 1

The correct option is B 56%

As the collision is inelastic the masses will move together
Assuming the speed of block A and B becomes $v_1\hat{i}+v_2\hat{j}$
Writing momentum equation in x direction
$m\left(2v\right)+2m\left(0\right)=3m{v}_1\Rightarrow v_1=\frac{2v}{3}$
Writing momentum equation in Y direction
$m\left(0\right)+2m\left(v\right)=3m{v}_2\Rightarrow v_2=\frac{2v}{3}$
The velocity of both blocks will be $\frac{2v}{3}\hat{i}+\frac{2v}{3}\hat{j}$

So,
$E_{initial}=\frac{1}{2}m(2v{)}^2+\frac{1}{2}2m(v{)}^2=3m{v}^2$
$E_{final}=\frac{1}{2}3m(\frac{4}{9}{v}^2+\frac{4}{9}{v}^2)=\frac{4}{3}m{v}^2$
Therefore the fractional loss is
$\frac{E_{initial}-E_{final}}{E_{initial}}\times 100=\frac{3-\frac{4}{3}}{3}\times 100$

$=\frac{5}{9}\times 100$
$=56\%$