Q: 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 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 initial velocity V of the bullet is

(a) 250 m/s

(b) 250√2 m/s

(c) 400 m/s

(d) 500 m/s

Ans: (d)

Sol: Applying momentum conservation,

0.01 × V = 0.01 v_{1} + 0.2 v_{2} …(i)

Time taken by ball & bullet to hits the ground is

$ \displaystyle t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times 5}{10}}$ = 1 sec

A/c to question,

v_{1} t = 100 …(ii)

v_{2} t = 20 …(iii)

v_{1}= 100 m/sec

v_{2}= 20 m/sec

on putting the value of v_{1} and v_{2} we get

V = 500 m/s