A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity `V m//s` in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is

Options

A. `250 m//s`
B. `250 sqrt(2) m//s`
C. `400 m//s`
D. `500 m//s`

Answer

D

Explanation

By the momentum conservation
`0.01 v = 0.01 v_(1) + 0.2 v_(2)` (i)
Now the projectile motion of bullet and ball
Using equation of trajectory
`y = (gx^(2))/(2 v_(1)^(2)) rArr 5 = (10(100)^(2))/(2 v_(1)^(2)) rArr v_(1) = 100 m//s`
`5 = (10 (20)^(2))/(2v_(2)^(2))`
`v_(2) = 20 m//s`
Putting values of `v_(1)` and `v_(2)` in (i)
`0.01 v = 0.01 xx 100 + 0.2 xx 20 = 1 + 4 rArr v = 500 m//s`