A ball of mass 0.2 kg rests on a vertical post of height 5m. 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 ullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100m from the foot of the post. the initial velocity v of the bullet is

Options

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

Answer

D

Explanation

`R = u sqrt((2h)/(g)) implies 20 = V_1sqrt((2 xx 5)/(10)) and 100 = V_2 sqrt((2xx5)/(10)) implies V_1 = 20m// s` Applying momentum conservation just before and just after the collision `(0.01) (V) = (0.2) (20) + (0.01) (100)`
`V = 500 m//s`.