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 `72//55 v//l`
B. the centre of the rod will travel a distance it `pil//3` in the time in which it makes half rotation
C. impulse of the impact force is `24m V//55`
D. none of these

Answer

A, B, C

Explanation

The ball has `V'` component of its velocity perpendicular to the length of the rod immediately after the collison. `U` is the velocity of `CM` of the rod and `omega` is angular velocity of the rod just after collision. The ball strikes the rod with a speed of `vcos53^@` in the perpendicular direction and its component along the length of the rod after the colision is unchanged.

Using for the point of collision
Velocity of separation `k=` Velocity of approach
`(3V)/5=((omegal)/4+u)+V'`.....i
Conserving linear momentum of (rod `=` particle) in the direction perpendicular to the rod
`mV3/5="mu"-mV`' ............ii
Conserving angular momentum about point `D` as shown in the figure
`0=0+["mu"l/4-(ml^(2))/12 omega]impliesu=(omegal)/3`...........iii
`implies u=(24V)/55, W=(72V)/(55l)`
Time taken to rotate by `pi` angle `t=pi/omega`

In the same tme, distance travelled `=u_(2)t=(pil)/3`
Using angular impulse angular momentum equation
`intNdtl/4=(ml^(2))/4.(72V)/(55l)omega {because'intNdt l/4=(24mV)/55}`
[Using impulse momentum equation on the rod `intNdt="mu"=(24mv)/55`]