Question

A particle of mass $\textit{m}$ 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 $\overset{ \rightarrow }{P_{i}}=m\left(2 V\right)\hat{i}+\left(2 m\right)V \, \hat{j}$
According to question as

On perfectly inelastic collision the particles stick to each other.
$\overset{ \rightarrow }{P_{f}}=3m \, \overset{ \rightarrow }{V_{f}}$
By conservation of linear momentum
$\overset{ \rightarrow }{P_{f}}=\overset{ \rightarrow }{P_{i}} \, \Rightarrow 3m \, \overset{ \rightarrow }{V_{f}}=m2V \, \hat{i}+2mV\hat{j}$
$\Rightarrow \overset{ \rightarrow }{V_{f}}=\frac{2 V}{3}\left(\hat{i} + \hat{j}\right)\Rightarrow \left|\right. V_{f} \left|\right.=\frac{2 \sqrt{2}}{3}V$
$\therefore $ loss in KE. of system $=K_{i n i t i a l}-K_{f i n a l}$
$=\frac{1}{2}m\left(2 V\right)^{2}+\frac{1}{2}\left(2 m\right)V^{2}-\frac{1}{2}\left(3 m\right)\left(\frac{2 \sqrt{2} V }{3}\right)^{2}$
$=2mV^{2}+mV^{2}-\frac{4}{3}mV^{2}=3mV^{2}-\frac{4 m V^{2}}{3}=\frac{5}{3}mV^{2}$
% Loss in KE $=100\times \frac{\Delta K}{K_{i}}=\frac{\frac{5}{3} m v^{2}}{3 m V^{2}}=\frac{5}{9}\times 100=56\%$