Question

A uniform rod of length `l` and mass 2 m rests on a smooth horizontal table. A point mass `m` moving horizontally at right angles to the rod with velocity `v` collides with one end of the rod and sticks it. Then
A. `1/6mv^(2)`
B. `1/3mv^(2)`
C. `mv^(2)`
D. `(mv^(2))/5`

Answer 1

Correct Answer - A
Loss in kinetic energy of the system as a result of collision
`=K_("Initial")-K_("Final")=1/2mv^(2)-{1/2(m+2m)v_(CM)^(2)=1/2I_(CM)omega^(2)}`
`=1/2mv^(2)-{1/2(3m)(v/3)^(2)+1/2((ml^(2))/3)(v/l)}`
For the composite `I_(CM)=(ml^(2))/3=1/2mv^(2)-1/3m v^(2)=1/6mv^(2)`