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Two particles of masses m(1) and m(2) in...

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))|`

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)`

Text Solution

Verified by Experts

The correct Answer is:
C
`F_("net")= (m_(1)+ m_(2))g`
Since, `F_("net")= (Delta P)/(Delta t)`
`rArr Delta P= F_("net") xx Delta t=` Change is momentum
`= (m_(1) + m_(2))g xx 2t_(0)`
`=2 (m_(1) + m_(2))g t_(0)`
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