A uniform thin rod of length 2L and mass m lies on a horizontal table. A horizontal impulse J is given to the rod at one red. There is no friction. The total kinetic energy of the rod just after the impulse will be

Options

A. `J^2/(2m)`
B. `J^2/(m)`
C. `(2J^2)/(m)`
D. `(6J^2)/(m)`

Answer

C

Explanation

`J = mv" "...(i)`
When v is the velocity of centre of mass after impulse rod get angular velocity `omega`
Angular impulse `=I omega`
`implies J xx L = (m(2L)^2)/12 xx omega`
`impliesJ=(mLomega)/3 implies omega =(3J)/(mL)`
From equation (i) `v=J/m`
Kinetic energy = KE = `1/2 mv^2 +1/2 Iomega^2`
`implies KE=1/2 m J^2/m^2 +1/2 xx(mxx4L^2)/(12) xx(9J^2)/(m^2L^2)`
`implies KE = (2J^2)/m`