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 v1 + 0.2 v2 …(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,
v1 t = 100 …(ii)
v2 t = 20 …(iii)
v1= 100 m/sec
v2= 20 m/sec
on putting the value of v1 and v2 we get
V = 500 m/s