Question

The 3.6-lb rod AB is hanging in the vertical position. A 2.2-lb block, sliding on a smooth horizontal surface with a velocity of 12 ft/s, strikes the rod at its end B.

What is the velocity of the block immediately after the collision. The coefficient of restitution between the block and the rod at B is e = 0.85?

Answer 1

v = 4.17 m/s

Step-by-step explanation:

as we know that block is moving towards the end of the rod

so let say just after the collision the block is moving with speed v' and rod is rotating with speed "w"

so we have angular momentum conservation about the hinge point

`mvL = mv'L + I\omega`

`(2.2)(12)L = (2.2)v'(L) + ((3.6 L^2)/(3)) \omega`

`12 = v' + 0.545(\omega L)`

now by the equation of coefficient of restitution we can say

`0.85 = ((\omega L) - v')/(12)`

now we have

`\omega L - v' = 10.2`

now we have

`\omega L = 14.37`

`v' = 4.17 m/s`

so velocity of block just after collision is 4.17 m/s