JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Conservation of Momentum

A ball of mass 0.2 kg rests on a vertical post of height 5 m . A bullet of mass 0.01 kg, traveling with a velocity

Vm / 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

(A)250 m/s
(B) $2502 m / s$
(C)400 m/s
(D)500 m/s✓ Correct

Answer: (D)

Solution

Time to fall from height 5 m:

5 = \frac{1}{2} (10) t^{2} \Rightarrow t = 1

s. Horizontal velocities after collision: ball

v_{ba l l} = 20/1 = 20

m/s, bullet

v_{b u l l e t} = 100/1 = 100

m/s. By conservation of momentum:

0.01 V = 0.01 (100) + 0.2 (20) = 1 + 4 = 5

, so

V = 500

m/s.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option

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JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Conservation of Momentum

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