Question

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

A. `250 m//s`
B. `250sqrt2 m//s`
C. `400m//s`
D. `500m//s`

Answer 1

Correct Answer - D
(d) For vetical motion of buller or ball
`u=0, s=5m, t=?, a=10m//s^2`

`S=ut+1/2at^2 rArr 5=1/2xx10xxt^2`
`rArr t=1sec`
For horizontal motion of ball
`x_(ball)=V_(ball)t rArr 20=V_(ball)xx1=V_(ball)`
For horizontal motion of bullet
`x_(bullet)=V_(bullet)xxt rArr 100= V_(bullet)xx1=V_(bullet)`
Applying conservation of linear momentum during
collision, we get
`mV=mV_(bullet)+mV_(ball)`
`0.01V=0.01xx100+0.2xx20`
`:. V=5/0.01=500m//s`