Q. Two particles of masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${\overrightarrow{v}}_{1}$ and ${\overrightarrow{v}}_{2}$ respectively at time $t = 0$. They collide at time $t_{0}$. Their velocities becomes ${\overrightarrow{v}}_{1}$’ and ${\overrightarrow{v}}_{2}$’ at time $2t_{0}$ while still moving in air. The value of $\left| \left( m_{1}\overrightarrow{v_{1}^{\ }}' + m_{2}\overrightarrow{v_{2}}' \right) \right| - \left( m_{1}\overrightarrow{v_{1}} + m_{2}\overrightarrow{v_{2}} \right)$ is
A) Zero
B) $\left( m_{1} + m_{2} \right)gt_{0}$
C) $2\left( m_{1} + m_{2} \right)gt_{0}$
D) $\frac{1}{2}\left( m_{1} + m_{2} \right)gt_{0}$
Explanation
(c) The momentum of the two-particle system, at $t = 0$ is
${\overrightarrow{P}}_{i} = m_{1}{\overrightarrow{v}}_{1} + m_{2}{\overrightarrow{v}}_{2}$
Collision between the two does not affect the total momentum of the system
A constant external force $\left( m_{1} + m_{2} \right)g$ acts on the system
The impulse given by this force, in time $t = 0$ to $t = 2t_{0}$
is $\left( m_{1} + m_{2} \right)g \times 2t_{0}$
$\therefore\ $Change in momentum in this interval
$= \left| m_{1}{\overrightarrow{v}'}_{1} + m_{2}{\overrightarrow{v}'}_{2} - (m_{1}{\overrightarrow{v}}_{1} + m_{2}{\overrightarrow{v}}_{2}) \right| = 2\left( m_{1} + m_{2} \right)gt_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.