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Classical Mechanics Level 3

A ball of mass 0.2 k g 0.2 ~kg rests on a vertical post of height 5 m 5~m . A bullet of mass 0.01 k g 0.01~kg , travelling with a velocity V m s − 1 V~ms^{-1} in the 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 V of the bullet is :

Note : Take g = 10 m s − 2 g = 10~ms^{-2}

Source : JEE - 2011

400 m s − 1 400~ms^{-1} 500 m s − 1 500~ms^{-1} 250 m s − 1 250~ms^{-1} 250 2 m s − 1 250\sqrt{2}~ms^{-1}

Rishu Jaar by Rishu Jaar , 21, India

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1 solution

Mark Hennings
Nov 12, 2017

Both particles take 1 1 second to hit the ground after the collision, and hence the ball has horizontal speed 20 20 m/s, and the bullet horizontal speed 100 100 m/s, after the collision. If the bullet has mass m m conservation of momentum tells us that 0.01 V = 0.2 × 20 + 0.01 × 100 0.01V \; = \; 0.2 \times 20 + 0.01\times100 so that V = 500 V = \boxed{500} m/s.

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Nice solution(+1)!

Rishu Jaar - 3 years, 7 months ago

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