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 velocity V of the bullet is A. `250 m//s`B. `250sqrt(2)m//s`C. `400m//s`D. `500m//s`

A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet

1.	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 velocity V of the bullet is A. `250 m//s`B. `250sqrt(2)m//s`C. `400m//s`D. `500m//s`
Answer» Correct Answer - D `R=usqrt((2u)/g` `implies 20=V_(1)sqrt((2xx5)/10)` and `100 =V_(2)sqrt((2xx5)/10)` `implies V_(1)=20 m//s, V_(2)=100 m//s` Applying momentum conservation just before and just after the collision `(0.01)(V)=(0.2)(20)+(0.01)(100)` `V=500 m//s`

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