A rod `AB` of mass `M` and length `8l` lies on a smooth horizontal surface. A particle `A` of mass `m` and velocity `v_(0)` strike's the rod perpendicular to its length, as shown in Fig. As a result of the collision, the centre of mass of the rod attains a speed of `v_(0)//8` and the particle rebounds back with a (not to scale) speed of `v_(0)//4`. Find the following:
a. The ratio `M//m`.
b. The angular velocity of the rod about `O`.
c. The coefficient of restitution `'e'` for the collision.
d. The velocities of the ends `'A'` and `'B'` of the rod, namely, `v_(A)` and `v_(B)` respectively.

Options

A. `51/128`
B. `61/128`
C. `21/128`
D. `31/128`

Answer

A

Explanation

Conservation of linear momentum,
`mv_(0)=m(v_(0))/8-m(v_(0))/4, 8m=m-2m`
`implies 10m=M`
`implies M/m=10`
conservation of angular momentum about point of collision.
`0=(M(8l)^(2))/12 omega-M(v_(0))/8l`
`implies omega=(3v_(0))/(128l)`
using coefficient of restitution equation
`(v_(2)-v_(1))=e(u_(1)=u_(2))`
`[((v_(0))/8+omegal)-(-(v_(0))/4)]=I[v_(0)-0]`
`implies e=51/128`
Velocity at end `A`
`v_(A)=(v_(0))/8+omega4l=28/128v_(0)`
`impliesv_(A)=7/32v_(0)`
velocity at end `B`
`v_(B)=(v_(0))/8-omega4l=4/128v_(0)=1/32v_(0)`