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, sticking to it. The rod is pivoted at its upper end O, hangs vertically, and the block strikes its lower end. After the collision the rod swings through an angle θ befo…

Question

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, sticking to it. The rod is pivoted at its upper end O, hangs vertically, and the block strikes its lower end. After the collision the rod swings through an 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)

Accepted Answer

(1) 63°. The collision and the swing are two separate stages, and different things are conserved in each. Stage 1 — the collision. The pivot exerts a force, so linear momentum is not conserved, but angular momentum about O is. Before: L=m v l=(1)(6)(1)=6 kg m²/s. After, the rod and the stuck block turn together: I=(Ml²)/3+ml²=2/3+1=5/3 kg m². ω=L/I=6/(5/3)=3.6 rad/s. Stage 2 — the swing. The collision is over, so now energy is conserved. Kinetic energy just after impact: 1/2 Iω²=1/2(5/3)(3.6)²=10.8 J. Rising through θ, the rod's centre climbs l/2(1-cos θ) and the block climbs l(1-cos θ). Δ U=Mgl/2(1-cos θ)+mgl(1-cos θ)=(10+10)(1-cos θ). Set 20(1-cos θ)=10.8, so 1-cos θ=0.54 and cos θ=0.46. θ≈63°.

Step-by-step solution

Why the other options are wrong