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

External force acting on the system `= (m_(1) + m_(2))g`.
We know `F Delta t = Delta p`
From `t = 0 to t = 2t_(0)`,
`Delta p = (m_(1) + m_(2))g xx = 2(m_(1) + m_(2))g t_(0)`