Two particles of masses `m_(1)` and `m_(2)` in projectile motion have velocities `vec(v)_(1)` and `vec(v)_(2)` , respectively , at time `t = 0`. They collide at time `t_(0)`. Their velocities become `vec(v')_(1)` and `vec(v')_(2)` at time ` 2 t_(0)` while still moving in air. The value of `|(m_(1) vec(v')_(1) + m_(2) vec(v')_(2)) - (m_(1) vec(v)_(1) + m_(2) vec(v)_(2))|`

Options

A. Zero
B. `(m_(1)+m_(2))g t_(0)`
C. `2(m_(1)+m_(2))g t_(0)`
D. `(1)/(2) (m_(1)+m_(2))g t_(0)`

Answer

C

Explanation

The momentum of the two-particle system, at t = 0 is
`vec(P)_(i) = m_(1)vec(v)_(1) + m_(2)vec(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 = 2t_(0)` is `(m_(1) + m_(2))g xx 2t_(0)`
`therefore |` Change in momentum in this interval.
`= |m_(1)vec(v.)_(1) + m_(2)vec(v.)_(2) - (m_(1)vec(v)_(1) + m_(2)vec(v)_(2))| = 2(m_(1) + m_(2)) g t_(0)`