The carts are placed at the position as indicated in the figure above. They move on a friction less surface. When  m2  collides with m1, both will move together and let’s say their combined speed is V.  In accordance with the principle of conservation of momentum, the following equation will hold good

               (0.42)(0.68) + (0.84) (0) = ( 0.84 +0.42) V

 

Accordingly , V=    =    m/s…Answer to part (a)

Again, according to the principle of conservation of mechanical energy, the total mechanical energy of a system remains constant, provided it is not acted upon  by any external force, other than gravity.  Hence, at the time of collision,

 

 (0.42) (0.68)2 =   (0.42) ()2  +(0.84)( )2 + k x2

 

KE of m2                   KE of m2                   KE of m1            EPE of Spring Bumper

before colliding        after colliding          after colliding      (Energy stored in the bumper)

 

Energy stored in the bumper = (0.42) (0.68)2 -   (0.42) ()2  -(0.84)( )2

 

                                              =

 

                                              =

 

                                              = (0.14)(0.68)2 Joules .. Answer to part (b)

 

 

After the collision, stored in the bumper will get transferred to m1. The KE of m1 will then be  = (0.14)(0.68)2+(0.84)( )2  =(0.68)2 (0.14+) = (0.68)2

 If at this point of time speed of  m1  is v, then  = (0.68)2

 

Solving for v ,  we will have v (speed of m1)= (0.68) m/s

The speed of m2 will continue to  remain    m/s…Answer to part (c)

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