A thin, uniform rod of mass M and length L is at rest on a smooth horizontal surface. A particle of same mass M collides with the rod at one end perpendicular to its length with velocity `V_(@)` and sticks to it. C is the middle point of the rod and D is a point at a distance of L/4 from end A. the time taken by the rod turn through `90^(@)` after collisions.

Options

A. `(L)/(V_(@))`
B. `(piL)/(V_(@))`
C. `(3piL)/(5V_(@))`
D. `(5piL)/(12V_(@))`

Answer

D

Explanation

By COAM about point D (ie about new centre of mass)
`MV_(@)(L)/(4)=[(ML^(2))/(12)+(ML^(2))/(16)+(ML^(2))/(16)]omega`
or `w=(6v_(@))/(5L)`
Now, `t=(theta)/(w)=((pi//2))/(w)=(5piL)/(12v_(@))`