A particle of mass m=1 kg collides with the end with velocity v_(0)=6 ms^(-1) of a spinning rod of mass 2m and length l=1 m at the end of he rod. If the coefficient of restitution of collision e=(2//3),find the a. velocity of the particle b. angular velocity of the rod just after the impact.

A particle of mass m=1 kg collides with the end with velocity v_(0)=6

1.	A particle of mass m=1 kg collides with the end with velocity v_(0)=6 ms^(-1) of a spinning rod of mass 2m and length l=1 m at the end of he rod. If the coefficient of restitution of collision e=(2//3),find the a. velocity of the particle b. angular velocity of the rod just after the impact.
Answer» Solution :If we take rod and particle as system we FIND no EXTERNAL impulse acting on the sytem. Hence, we can use conservation of linear moemntum just before collision just after collision. Let the velocities of particle and rod just after collision is `v_(1)` and `v_(2)` respectively. `mv_(1)+2mg_(2)=mv_(0)` `v_(1)+2v_(2)=6`..........i No applying Newton's restitution equation at the point of collision. `e=2/2=((v_(2)+omegal)=v_(1))/(v_(0)-(-omega_(0)l/2))` `2.(6+3)=3v_(2)+(3omega)/2-3v_(1)` `36=6v_(2)3omega-6v_(1)`.........ii Conservation of angular momentum about point of collision `(-2ml^(2))/12 .omega_(0)=(2ml^(2))/12.omega-2mv_(2).l/2` `-omega_(0)l=omegal-6v_(2)impliesomega=6v_(2)-6`..........iii PUT the value of `omega` in ii , `36=6v_(2)+18v_(2)-18-6v_(1)` `v_(1)-4v_(2)=9`.......iv From i and iii `v_(1)=1ms^(-1), v_(2)=2.5ms^(-1)` and `omega=9rads^(-1)`