A thin-walled hollow cylinder (l=mhr^2) of mass mh=4 kg starts starts from rest

Question

A thin-walled hollow cylinder (l=mhr^2) of mass mh=4 kg starts starts from rest on top of an incline at a vertical height ho=6m and rolls down the incline without slipping. All the heights are measured relative to an arbitrary chosen zero level that passes through the center of mass of a cylinder when it is at the bottom of the incline (see figure below). At the end of the incline the hollow cylinder collides head on with with the solid cylinder (l=l/2mhr^2) of mass m=10 kg, initially at rest. After the collision The solid Sundar starts to roll up a second incline without slipping. Assuming that the collision between the two objects is perfectly elastic and that friction and air resistance can be neglected, calculate the highest point reached by the solid cylinder after the collision.

Explanation / Answer

The moment of inertia of any rigid body about centre axis is given by

I = c * m*R^2

where is different for different objects

the velocity of any object at the bottom of an incline is

v = sqrt((2 * g * h)/(1+C))

for thin cylinder c = 1

v = sqrt (( 2 * 9.8 * 6) / (1+1))

v = 7.67 m/s

After collision we have

v = 7.67 m/s

c = 1/2

V = sqrt((2 * g * h)/(1+C))

7.67 = sqrt((2 * 9.8 * h)/(1+1/2))

h = 4.5 m

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