A ball of mass 0.2 ~kg rests on a vertical post 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. 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

Step-by-step solution

Time taken by the bullet and ball to strike the ground is t= 2 h g = 2 5 10 =1 ~s Let v₁ and v₂ are the velocities of ball and bullet after collision. Then applying We have, 20=v₁ 1 or v₁=20 ~ms ⁻¹ 100=v₂ 1 or v₂=100 ~m / s ⁻¹ Now, from conservation of linear momentum before and after collision we have, 0.01 v=(0.2 20)+(0.01 100) On solving, we get v=500 ~ms ⁻¹ Correct answer is (d). Analysis of Question Question is moderately lengthy from calculation point of view, otherwise it is simple.