A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle θ before momentarily coming to rest. If the rod has mass M = 2 kg, and length l = 1 m, the value of θ is approximately :
(take g = 10 m/s²)

Solution

Physics sir · 4 steps

Take angular momentum about O through the sticky collision, then use energy for the swing up.

1. Collision about O. The pivot force acts at O, so it has no torque about O. The collision is brief, so angular momentum about O is conserved. Kinetic energy is not, because the block sticks. Treat the block as a point mass at distance l from O.

   mvl = (Ml²/3 + ml²)ω

2. Find omega. I = 2/3 + 1 = 5/3 kg m² and mvl = 6 kg m² s⁻¹. So ω = 6/(5/3) = 3.6 rad s⁻¹, and the kinetic energy just after impact is 1/2 × 5/3 × 3.6² = 10.8 J.
3. Energy for swing. After the collision only gravity does work. The rod's centre rises l/2(1-cos θ) and the block rises l(1-cos θ). The gain in potential energy is (Mgl/2 + mgl)(1-cos θ) = 20(1-cos θ) J.
4. So. 20(1-cos θ) = 10.8 gives cos θ = 0.46, so θ ≈ 63°. The answer is (a).

The idea to keep. In a sticky collision with a pivoted body, the pivot force spoils linear momentum and the sticking spoils energy. Conserve angular momentum about the pivot first, then use energy for the motion after.

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