A particle of mass $\mathit{\text{m}}$ 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

  $62\%$
• B

  $44\%$
• C

  $50\%$
• Correct answer
  D

  $56\%$

Solution

The initial momentum of system is $\overrightarrow{P_i}=m\left(2V\right)\widehat{i}+\left(2m\right)V \widehat{j}$

According to question as



On perfectly inelastic collision the particles stick to each other.

$\overrightarrow{P_f}=3m \overrightarrow{V_f}$

By conservation of linear momentum

$\overrightarrow{P_f}=\overrightarrow{P_i} \Rightarrow 3m \overrightarrow{V_f}=m2V \widehat{i}+2mV\widehat{j}$

$\Rightarrow \overrightarrow{V_f}=\frac{2V}{3}\left(\widehat{i}+\widehat{j}\right)\Rightarrow \left|V_f\right|=\frac{2\sqrt{2}}{3}V$

$\therefore$ loss in KE. of system $=K_{initial}-K_{final}$

$=\frac{1}{2}m{\left(2V\right)}^2+\frac{1}{2}\left(2m\right)V^2-\frac{1}{2}\left(3m\right){\left(\frac{2\sqrt{2}V}{3}\right)}^2$

$=2m{V}^2+m{V}^2-\frac{4}{3}m{V}^2=3m{V}^2-\frac{4m{V}^2}{3}=\frac{5}{3}m{V}^2$

% Loss in KE $=100\times \frac{∆K}{K_i}=\frac{\frac{5}{3}m{v}^2}{3m{V}^2}=\frac{5}{9}\times 100=56\%$

Hence, the correct answer is option: 'D'.