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 / s²) :

Solution

Physics sir · 4 steps

Both fall the same 20 m, so they share one flight time. Get the speeds from the ranges, then use momentum.

1. Time of fall. Just after the collision, both ball and bullet move horizontally, so each starts its fall with zero vertical velocity. From h = 1/2gt², t = √((2 × 20)/10) = 2 s for both.
2. Speeds after impact. Horizontal speed is range divided by time. Ball: v₁ = 30/2 = 15 m s⁻¹. Bullet: v₂ = 120/2 = 60 m s⁻¹.
3. Conserve momentum. No horizontal external force acts during the impact, so horizontal momentum is conserved. Use masses in kg: 0.01 kg for the bullet and 0.2 kg for the ball.

   0.01 u = 0.2 × 15 + 0.01 × 60 = 3 + 0.6 = 3.6

4. So. u = 3.6/0.01 = 360 m s⁻¹, which is option (b).

The idea to keep. When objects leave the same height horizontally, they all take the same time to land. Turn each range into a speed with that time, then balance the momentum.

Answer from the NTA final answer key. Solution by PrepWiser.