A bomb of mass `4 m`, while moving on a parabolic path, explodes at highest point of its path. If breaks into two parts of mass ratio `1 : 3`, smaller part coming to rest. The range of this projectile was `60 m` in the absence of explosion. The distance of the second part from the point of projection when it strikes the ground is

Options

A. `60 m`
B. `70 m`
C. `80 m`
D. `90 m`

Answer

B

Explanation

At the highest point , the bomb has horizontal velocity . The smaller part comes to rest, hence the larger part will move with higher horizontal velocity (due to the momentum conservation in horizontal direction)
`4m v_(0) = m xx 0 + 3 mv rArr v gt v_(0)`
In the vertical direction , both parts have no velocity , hence both will strike the ground simultaneously.
As the internal forces do not affect the motion of the center of mass , the center of mass hits the ground at the position where the original bomb (projectile) would have landed.
`x_(c.m.) = R = (m xx (R )/(2) + 3m x_(0))/(m + 3m) rArr x_(0) = 70 m`