A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity 100 m/s. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v = ____________ m/s so that the pendulum describes a circle. (Assume the string to be inextensible and g = 10 m/s²)

Solution

Physics sir · 4 steps

First find the speed the bob needs to loop, then use momentum with the bullet's recoil sign taken as negative.

1. Speed to loop. With a string, the bob must just keep the string taut at the top, where tension is zero and gravity alone provides the centripetal force. That needs a speed of √5gL at the lowest point.
2. Put in numbers. uₘᵢₙ = √(5 × 10 × 0.5) = √25 = 5 m s⁻¹. This is the bob's speed just after the collision.
3. Conserve momentum. Take the bullet's initial direction as positive. The bullet recoils, so its final velocity is -100 m s⁻¹. The bullet has mass 0.01 kg and the bob has mass 1 kg.

   0.01 v = 0.01 × (-100) + 1 × 5

4. So. 0.01 v = -1 + 5 = 4, which gives v = 400 m s⁻¹.

The common slip: 600. Taking the recoil velocity as +100 instead of -100 gives 0.01 v = 1 + 5, so v = 600, not 400.

The idea to keep. For a bob on a string to complete a vertical circle, it needs √5gL at the bottom. Get that speed first, then use momentum conservation, and keep the signs of rebounding bodies careful.

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