A ball is released from height `80 m`. When the ball is at height `60 m`, it explodes in two parts of mass ratio `1 : 2`. One of them moves horizontally with speed `20 m//s`. The distance between the two pecies on the ground is

Options

A. `20 m`
B. `40 m`
C. `60 m`
D. `80 m`

Answer

C

Explanation

The center of mass will follow the vertical straight line. The time taken by ball to fall `80 m, t_(1) = sqrt((2 xx 80)/(10)) = 4 s` and the time taken to fall by `20 m , t_(2) = sqrt((2 xx 20)/(10)) = 2 s` , hence the time taken by the ball (or two pieces) from `B` to the ground , `t_(0) = t_(1) - t_(2) = 2 s`.
At `B` , the momentum conservation in horizontal direction :
`0 = m xx 20 - 2 m v_(2)`
`rArr v_(2) = 10 m//s , x_(1) = 20 t_(0) = 40 m , x_(2) = v_(2) t_(0) = 20 m`
Separation between blocks `= x_(1) + x_(2) = 60 m`.