Question

A particle of mass $\text{m}$ 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. $50\%$
• D. $56\%$

Answer 1

Answer: (D)

The initial momentum of system is $\overrightarrow{P_i}=m\left(2V\right)\hat{i}+\left(2m\right)V\hat{j}$
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
$\overrightarrow{P_f}=\overrightarrow{P_i}\Rightarrow 3m\overrightarrow{V_f}=m2V\hat{i}+2mV\hat{j}$
$\Rightarrow \overrightarrow{V_f}=\frac{2V}{3}\left(\hat{i}+\hat{j}\right)\Rightarrow |V_f|=\frac{2\sqrt{2}}{3}V$
$\therefore$ 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 $=100\times \frac{\Delta K}{K_i}=\frac{\frac{5}{3}m{v}^2}{3m{V}^2}=\frac{5}{9}\times 100=56\%$