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 initial velocity V of the bullet is :
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m1 = 0.01 kg m2 = 0.2 kg
Let v1 & v2 be velocity of bullet & ball respectively just after collision.
v2 × 1 = 20 ⇒ v2 = 20
& v1 = 100
From conservation of momentum
0.01 × v = (0.01 × 100) + (0.2 × 20)
0.01 v = 1 + 4 = 5