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Impulsive Forces in Collision

Two particles...

Question

Two particles of masses $m_{1} a n d m_{2}$ in projectile motion have velocities ${\to v}_{1} a n d {\to v}_{2}$ respectively at time t = 0 and that is the moment when they collide. Their velocities become ${\to v}_{1}^{'} a n d {\to v}_{2}^{'}$ at time $2 t_{0}$ while still moving in air. The value of magnitude of change in total momentum would be

A

$Z e r o$

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B

$(m_{1} + m_{2}) g t_{0}$

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C

$2 (m_{1} + m_{2}) g t_{0}$

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D

$\frac{1}{2} (m_{1} + m_{2}) g t_{0}$

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Solution

The correct option is C
$2 (m_{1} + m_{2}) g t_{0}$

The momentum of the two-particle system, at t = 0 is

${\to p}_{i} = m_{1} {\to v}_{1} + m_{2} {\to v}_{2}$

Collision between the two does not affect the total momentum of the system.

A constant external force $(m_{1} + m_{2}) g$ acts on the system.

The impulse given by this force, in time t = 0 to $t = 2 t_{0} i s (m_{1} + m_{2}) g \times 2 t_{0}$

$∴$ Change in momentum in this interval $= | m_{1} {\to v}_{1} + m_{2} {\to v}_{2} - (m_{1} {\to v}_{1} + m_{2} {\to v}_{2}) | = 2 (m_{1} + m_{2}) g t_{0}$

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