A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass 10⁻² kg moving with a speed of 2 × 10² ms⁻¹. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g=10 m / s² )

Solution

Physics sir · 4 steps

Momentum carries you through the collision, then energy carries the bob up. Never mix the two stages.

1. Collision stage. The bullet sticks in the bob, so kinetic energy is lost as heat and sound. Only momentum is conserved during the impact.

   mu = (m + M)v ⇒ v = (10⁻² × 200)/(1 + 10⁻²) = 2/1.01 ≈ 1.98 m s⁻¹

2. Swing stage. After the impact only gravity does work, so mechanical energy is conserved. Then 1/2(m+M)v² = (m+M)gh, and the mass cancels, giving h = v²/2g.
3. Put in numbers. h = (1.98)²/(2 × 10) = 3.92/20 ≈ 0.196 m. This is less than the 1 m string length, so the bob swings back as expected.
4. So. The bob rises about 0.20 m, which is option (a).

The idea to keep. In a ballistic pendulum, use momentum conservation for the instant of impact and energy conservation for the rise afterwards. Kinetic energy is not conserved in the sticking collision itself.

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