Question

A particle of mass m moving in the direction $x$ 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. $44\%$
• B. $50\%$
• C. $56\%$
• D. $62\%$

Answer 1

Answer: (C)

$56\%$

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 $\frac{2v}{3}\hat{i}+\frac{2v}{3}\hat{j}$
The loss in kinetic energy
$\frac{1}{2}m(2v{)}^2+\frac{1}{2}2m(v{)}^2-\frac{1}{2}3m(\frac{\sqrt{2}2v}{3}{)}^2$
$3m{v}^2-\frac{4}{3}m{v}^2=\frac{5}{3}m{v}^2$
percentage loss$\frac{\frac{5}{3}m{v}^2}{3m{v}^2}\times 100=56$ %