Question

Two identical particles collide in air inelastically. One moves horizontally and the other moves vertically with equal speed just before collision. The fractional loss in kinetic energy of the system of particles is equal to :

• A. $1/\sqrt{2}$
• B. 1/2
• C. $\sqrt{3}/2$
• D. none of these

Answer 1

The correct option is B 1/2

The velocities of the respective particle before collision, $\overrightarrow{u_1}=v\hat{i}$ and $\overrightarrow{u_2}=v\hat{j}$

$\therefore$ $\overrightarrow{u_1}-\overrightarrow{u_2}=v(\hat{i}-\hat{j})$ $⟹|\overrightarrow{u_1}-\overrightarrow{u_2}|=v\sqrt{2}$

Loss in Kinetic energy $\Delta K.E=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}|\overrightarrow{u_1}-\overrightarrow{u_2}{|}^2(1-e^2)$

Given : $m_1=m_2=m$ $e=0$

$\therefore$ $\Delta K.E=\frac{1}{2}\frac{m\times m}{m+m}(v\sqrt{2}{)}^2$ $⟹\Delta K.E=\frac{m{v}^2}{2}$

Initial Kinetic energy of the system $K.E_i=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=m{v}^2$

Thus fraction loss in kinetic energy $\frac{\Delta K.E}{K.E}=\frac{\frac{m{v}^2}{2}}{m{v}^2}$

$⟹\frac{\Delta K.E}{K.E}=\frac{1}{2}$