A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travel independently. The ball hits the ground at a distance 30 m and the bullet at a distance 120 m from the foot of the post. The value of initial velocity of the bullet will be: [g = 10 m s⁻²]
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Question

A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travel independently. The ball hits the ground at a distance 30 m and the bullet at a distance 120 m from the foot of the post. The value of initial velocity of the bullet will be: [g = 10 m s⁻²]

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G1 Physics · Accepted Answer

Both objects fall from the same height, so their time of flight is t = √(2h/g)=2 s. Thus ball horizontal speed after collision is 30/2 = 15 m s⁻¹ and bullet speed after collision is 120/2 = 60 m s⁻¹. Momentum conservation: 0.01u = 0.20×15 + 0.01×60 = 3.6, so u = 360 m s⁻¹.