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

Answer

D

Explanation

`R=usqrt((2u)/g`
`implies 20=V_(1)sqrt((2xx5)/10)` and `100 =V_(2)sqrt((2xx5)/10)`
`implies V_(1)=20 m//s, V_(2)=100 m//s`
Applying momentum conservation just before and just after the collision `(0.01)(V)=(0.2)(20)+(0.01)(100)`
`V=500 m//s`