A ball of mass 200 g rests on a vertical post of height 20 m . A bullet of mass 10 g , travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if g = 10 m s - 2 ) :

Step-by-step solution

Let initial velocity of bullet be u . Given, mass of bullet m = 10   g and mass of ball M = 200   g , Initially ball is at height 5   m and at rest, the only acceleration acting is due to gravity. Time of flight of each ball and bullet is t = 2 h g . Now, velocity of ball and bullet as v 1 = 30 2 h g   and   v 2 = 120 2 h g As the collision is elastic, so applying conservation of linear momentum. m u + M 0 = M v 1 + m v 2 ⇒ 0 . 01 u = 0 . 2 30 g 2 h + 0 . 01 120 g 2 h ⇒ u = 300