Question

A ball of mass 0.2 kg rests on a vertical post of height 5m. 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 20m and the bullet at a distance of 100m from the foot of the post. The initial velocity of the bullet is

And is 500m/s.

Please give detailed explanation as I won't be able to understand concise answers.

Answer 1

The question is about conservation of momentum

Assume the ball and the bullet after collision travelled their respective distances in 1sec, which makes there speed 20m/s and 100m/s.(since they travel 20 and 100 m respectively)

You can calculate the momentum for each of them using the formule p=mv

0.2*20=4kgm/s

0.01*100=1kgm/s

Add the momentums (4+1=5) and that is the momentum transferred by the bullet.

And now you can find out the velocity because you have the mass and momentum of the bullet at the start.

p=mv

v=5/0.01 (p=5 and m=0.01)

v=500m/s

Hope this helps :)