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. `250m//s`
B. `250sqrt2m//s`
C. `400m//s`
D. `500m//s`

Answer

d

Explanation

Here, `h = 5 m, m_(1) =0.2kg, m_(2) = 0.01kg`
`d_(1) = 20m , d_(2) = 100m`
As `h = ut + (1)/(2)gt^(2) = 0xx t+(1)/(2) xx 10xxt^(2)`
`5 = 5 t^(2) or t = 1s`
Further as `upsilon_(1) xx t = 20 m or upsilon_(1) xx 1=20`
`upsilon_(1) = 20m//s`
`upsilon_(2) xx t = 100 m` or `upsilon_(2) xx 1 =100`
`upsilon_(2) = 100m//s`
Applying law of conservation of linear momentum
`(0.01) upsilon = (0.01) + (0.2) (20)`
`0.01 upsilon = 1+ 4` or `upsilon = (5)/(0.01) = 500m//s` .