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. 2.5s
B. 4s
C. 1s
D. `(10)/(sqrt2)s`

Answer

A

Explanation

Position after time .t. Position of A is `(0, v_(A) t)` Position of B is `(v_(B)t, 10)` Distance between them - `l^(2)= (0- v_(B) t)^(2) + (v_(A) t- 10)^(2)`
`rArr l^(2)= 4t^(2) + 4t^(2)+ 100- 40t = 8t^(2)- 40t + 100`
For maxima and minima
`(dl)/(dt)= 0 rArr 16t - 40= 0 rArr t= (40)/(16)= 2.5s`
Now, `(d^(2)l)/(dt^(2))= 16(+ve) gt 0`, t value gives minima