Question

A plank of length L and mass M hangs from an axle passing through one end. The other end is allowed to hang down so that it just brushes a workbench. You place a box of height h and mass m so that it just touches the plank when sitting on the bench. You then raise the plank up, turning around the axle, so that its movable end is a height H above the bench. You let the plank go, it pivots around the axle, and hits the ball elastically. How fast does the box move after the collision if it slides without friction?

Answer 1

Answer:

v = ((M(√2gH)/3m)

Explanation:

Initial Moment of Inertia= Moment of Inertia of Rod

I = (ML²)/3

Linear Velocity of moveable end of the rod, just before collision is given by = v = (√2gH)/L

Initial Angular Momentum, about the point of the suspension:

Li= Iw = {(ML²)/3} . {(√2gH)/L} = {ML(√2gH)}/3

Final Angular Momentum = Li = mvl, where 'v' is the speed of the mass 'm' after the collision

Since the collision is elastic, all momentum will be conserved, which means

Initial Angular Momemtum = Final Angular Momentum

{ML(√2gH)}/3 = mvL

solving for v = {(M)/3m} . {(√2gH)/L}