Two small smooth particles, A and B of mass m are attached to the two ends of an inelastic string of length l and are at rest on a smooth horizontal plane . The system now moves with speed u in the direction AB with the string taut. The particle B, after some time, collides with a small smooth elastic particle C of mass M which is at rest on the plane. If e is the coefficient of restitution between the particles B and C,
show that the particle A collides with the particle B after a time
`(( m + M)l) / (M(1 +e)u)` from the moment of collision between B and C.
The first part is answered at http://www.enotes.com/homework-help/two-small-smooth-particles-b-mass-m-attached-two-448239
From the previous part it has shown that after collision B has a velocity of ;
`V_B = (u(m-eM))/(M+m)`
Let us take the velocity of A relative to the particle B considering E as Earth. This means B is stationary and A is moving.
`V_(A,E) = u`
`V_(B,E) = (u(m-eM))/(M+m)`
`V(A,B) = V_(A,E)-V_(E,B)`
`V(A,B) = u-(u(m-eM))/(M+m)`
`V_(A,B) = (u(M+m-m+eM))/(M+m)`
Relative to B, particle A should travel a distance of l to collide with B.
Time to collide `= l/V_(A,B)`
Time to collide `= l/(Mu(1+e))/(m+M) =( (M+m)l)/(Mu(1+e))`
So the particle A will collide with particle B, after;
`( (M+m)l)/(Mu(1+e))` time where B collide with C.