Question

A ball of mass 0.2 kg rests on a vertical 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. They hit 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

Answer 1

Answer:

Let the mass of bullet be m and mass of ball be M

initially ball is at height 5m and t rest , the only acceleration is due to gravity

therefore applying equation of motion

S=ut+

2

1

gt

2

5=0+

2

1

(10)t

2

therefore t = 1 sec

so V(ball)=20m/sec

V(bullet) = 100m/sec

So by collision

M×V(ball:final)+m×V(bullet:final)= M×V(boll:initial)+m×V(bullet:initial)

0.01×V(bulletfinal)=0.01×100+0.20×20

V=500m/sec