A wooden block of mass M hangs from a rigid rod of lengthl having negligible mas

Question

A wooden block of mass M hangs from a rigid rod of lengthl having negligible mass. The rod is pivoted at its upper end. A bullet of mass mtraveling horizontally and normal to the rod with speed v hitsthe block and gets embedded in it.
What is the angular momentum L of the block-bullet system, with respect to the pivot point immediately after the collision?

1. L = M v l 2. L = mv l 3. L = (M m) v l 5. L = (m +M) v l
What is the fraction KF/KI(the final kinetic energycompared to the initial kinetic energy) in the collision?

A wooden block of mass M hangs from a rigid rod of length l having negligible mass. The rod is pivoted at its upper end. A bullet of mass m traveling horizontally and normal to the rod with speed v hits the block and gets embedded in it. What is the angular momentum L of the block-bullet system, with respect to the pivot point immediately after the collision? What is the fraction KF/KI(the final kinetic energy compared to the initial kinetic energy) in the collision?

Explanation / Answer

mv+M*0=(M+m)v1 v1=mv/(M+m) The angular momentum is L=(M+m)*mv/(M+m) L=mv

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