Q. Two particles of masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}$ and ${\overrightarrow{\mathbf{v}}}_{\mathbf{2}}$ respectively at time $t = 0$. They collide at time $t = 0.$ their velocities become ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}$ and ${\overrightarrow{\mathbf{v}}}_{\mathbf{2}}$ at time $2t_{0}$, while still moving in air. The value of $\left| \left( m_{1}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{+}m_{2}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}} \right)\mathbf{- (}m_{1}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{+}m_{2}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{)} \right|$ 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) As is clear from the equation,
$\left| \left( m_{1}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{+}m_{2}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}} \right)\mathbf{-}\left( m_{1}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{+}m_{2}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}} \right) \right|$
=change in linear momentum of the two particles
=external force on the system $\times$time interval
$\mathbf{= \lbrack}\left( m_{1} + m_{2} \right)g\rbrack \times \left( 2t_{0} \right)2\left( m_{1} + m_{2} \right)gt_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.