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Question:

A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, traveling 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 initial velocity V of the bullet is?

500m/s

250m/s

250√2m/s

400m/s

Solution:

Let the mass of bullet be m and mass of ball be M.
Initially ball is at height 5m and at rest, the only acceleration is due to gravity.
Therefore applying equation of motion
S = ut + 1/2gt²
5 = 0 + 1/2(10)t²
Therefore t = 1 sec
so V(ball) = 20m/sec
V(bullet) = 100m/sec
So by collision
M × V(ball:final) + m × V(bullet:final) = M × V(ball:initial) + m × V(bullet:initial)
0.01 × V(bullet:final) = 0.01 × 100 + 0.20 × 20
V = 500m/sec