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 - 1 in a horizontal direction, hits the centre of the ball. After the col

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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 - 1 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

TestHub · Accepted Answer

Time of flight t f = 2 h g = 2 × 5 10 = 1 s Horizontal range (R) = horizontal velocity × time of flight ∴ Horizontal velocities of the bullet and of the ball after the collision respectively are v b u l l e t = 100 1 = 100 m s - 1 v b a l l = 20 1 = 20 m s - 1 From the conservation of momentum, Total initial momentum = total final momentum ⇒ m b u l l e t × V = m b u l l e t × v b u l l e t + m b a l l × v b a l l ⇒ 0.01 V = 0.01 × 100 + 0.2 × 20 ⇒ V = 500 m s - 1

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