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. ` 25 sqrt 2 m//s`
C. `400 m//s`
D. ` 500 m//s`

Answer

D

Explanation

Refer to Fig. 2 (CF) . 5 9, taking wertical downared motion of ball or bullet , we have
.
time of flight, ` T= sqrt (2 h)/g = sqrt ( 2 xx 5 m)/910 m//s^@) = 1 s`
After collision the velocity of ball,
` V_(ball) = 20 m//s.
the velocity of bullet , ` V_(bullet ) = 100 /1 = 100 m//s`. ,
According to law of conservation of linera moventume, we get
` m_(ball) xx 0 + m_(bullet) xx V=m_(ball)`
` + m_(bullet ) xx V_(bullet)`
` 0 + 0.01 xx V=0.2 xx 0.2 xx 20 + 0.01 ` or `V= 500 m//s`.