A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is :
(take g = 9.8 ms⁻²)

Solution

Physics sir · 4 steps

Split it in two: momentum carries you through the collision, energy carries you up the swing.

1. Swing upward. After the bullet is embedded, only gravity does work on the pair, so mechanical energy is conserved. Their kinetic energy at the bottom becomes potential energy at height h, giving V = √2gh.
2. Speed after impact. With h = 0.098 m, V = √(2 × 9.8 × 0.098) = √1.9208 ≈ 1.386 m s⁻¹.
3. The collision. The collision is perfectly inelastic, so kinetic energy is lost but momentum is conserved. The bullet's momentum becomes the momentum of the combined mass M + m = 6.00 kg.

   mv = (M+m)V ⇒ v = (M+m)/m√2gh = 6.00/0.010 × 1.386

4. So. v = 600 × 1.386 ≈ 831.6 m s⁻¹. The closest option is 831.4 m/s, which is option (a).

The idea to keep. In a ballistic pendulum, never use energy conservation across the collision itself. Use momentum for the impact, then energy for the swing.

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