As no horizontal reaction occur at pivot end hence their is no impulsle on the system horizontal direction. We can use conservation linear momentum just before collision and just after collision in horizontal directioin. Let the velocities of particle and rod just after collisionis `v_(1)` and `v_(2)` respectively.
`mv_(0)=mv_(1)+3mv_(2)`
`v_(0)=v_(1)+3v_(2)`.....i
Now applying conservation of angular momentum about pivot end just before collision and just after colision. Let angular velocity of the rod about pivot end is `omega`.
`mv_(0).b=mv_(1).b+(3ml^(2))/3omega`
`v_(0)=v_(1)+(omegal^(2))/b`.......ii
From i and ii `3v_(2)=(omegal^(2))/b`........iii
For no impulse at end `v_(A)=0`
`v_(A)=0=v_(2)-omegal/2`
`v_(2)=omegal/2`.....iv
`implies b=(2l)/3`
b. If the ball falls dead just after the collision `v_(1)=0` ltbr Hence from conservation of angular momentum about pivot gives
`mv_(0)b=0(Ml^(2))/3 omegaimplies mv_(0).b=(Ml^(2))/3omega` ...............i
Again applying Newton's restitution equation at the point of collision.
`e=1/2=(omegab-0)/(v_(0)-0)implies omega=(v_(0))/(2b)implies m/M=(l^(2))/(6b^(2))`
