A rod `AC` of length `l` and mass `m` is kept on a horizontal smooth plane. It is free to rotate and move. A particle of same mass `m` moving on the plane with velocity `v` strikes the rod at point `B` making angle `37^@` with the rod. The collision is elastic. After collision,

Options

A. The angular velocity of the rod will be `(72v)/(55l)`
B. The centre of the rod will travel a distance `(pi l)/(3)` in the time in which it makes half rotation
C. Impulse of the impact force is `(24mV)/(55)`
D. none of these

Answer

A, B, C

Explanation

The ball has `V'`, component of its velocity perpendicular to the length of rod immediately after the collision. `u` is velocity of `COM` of the rod and `omega ` is angular velocity of the rod, just after collision. The ball strikes the rod with speed `v cos 53^(@)` in perpendicular direction and its component along the length of the rod after the collision is unchanged.
Using for the point of collision.
Velocity of separation `=` Velocity of approcah

`rArr (3V)/(5)=((omegal)/(4)+u)+V' .....(1)`
Conserving linear momentum `(` of rod `+` particle `)`, in the direction `_|_` to the rod.
`mV.(3)/(5)=m u =mV' ....(2)`
Conserving angular moment about point `'D'` as shown in the figure
`0=0+[m u(l)/(4)-(ml^(2))/(12)omega]rArr u=(omegal)/(3) ....(3)`
By solving
`u=(24V)/(55), w=(72V)/(55l)`
Time taken to rotate by `pi` angle `t=(pi)/(omega)`

In the same time, distance travelled `=u_(2).t=(pil)/(3)`
Using angular impulse`-` angular momentum equation
`int N.dt(l)/(4)=(ml^(2))/(4).(72V)/(55l)rArrintN.dt=(25mV)/(55) }`
or `{("usingimpu lse-momentum equation on Rod"),(intNdt=m u=(24mv)/(55)):}`