A spherical ball A with mass 3 kg is sent with velocity 10 m

A spherical ball A with mass 3 kg is sent with velocity 10 m/s inside a smooth
tube, where two other balls, B with mass 4 kg and C with mass 3 kg, are at rest.
The ball A collides with B (coefficient of restitution 2/3
), which in turn collides
with C (coefficient of restitution 1/2
).
(a) Find the speed of C.
(b) Which side(s) of the tube do A and B come out, compared to C?

Solution

Given: spherical ball A with mass 3 kg is sent with velocity 10 m/s inside a smooth tube,
            where two other balls, B with mass 4 kg and C with mass 3 kg, are at rest i,e(zero velocity)
             The ball A collides with B (coefficient of restitution 2/3),

            which in turn collides with C (coefficient of restitution 1/2).
(a) the speed of C is:

                      mAuA+mBuB=mAvA+mBvB

                      3(10)+0=3vA+4vB

                      3vA+4vB=30-----------(1)

          and (coefficient of restitution of A,B is 2/3)

             vB-vA/uB-uA=2/3   [here uB=0 uA=10]

              3vB-3vA=-20----------(2)

solve 1,2

       vB=10/7

      vA=170/21

after  mBuB+mCuC=mBvB+mCvC

    4uB+0=4(10/7)+3mC

        4uB+3mC=-40/7------------(3)

       (coefficient of restitution is for B,C is 1/2)

         vC-vB/uC-uB=1/2                [here vB=10/7 ;   uC=0]

         2vC+uB=20/7-------------(4)

        solve (3),(4)

           we get vC=40/77

   or speed of C is:40/77 m/s

   so vA,vB,vC=170/21,10/7,40/77

b) towords c side(s) B come out,

    opposite to C side(s) A come out.

    by compareing speeds of vA,vB,vC=170/21,10/7,40/77