Question

Set up a two particle, totally elastic collision. Make the two particles have different masses. Analyze the collision (determine initial and final velocities of the two particles) in three different frames. One frame must be the center of mass frame, one frame must be co-moving with one of the particles before the collision, and the third frame is up to you. In each frame calculate the change in kinetic energy for each particle and compare these values. That is, does ∆K1 = −∆K2 as expected from energy conservation?

Answer 1

Let mass of first particle be m and second particle be 2m.Let the first particle is moving right with speed 2v and second particle with v towards left.

Velocity of center of mass Vcm = ( m*2v-2m*v)/(m+2m) =0

Final velocities in ground frame are V1 and V2 towards right.

mV1 +2mv2 = m*2v - 2mv =0

V1 =-2v2

And rate of separation = rate of approach.

V2 - V1 = 2v+v

V2+2V2 = 3v

V2 = 3v/3 = v

V1 = - 2v

Case1) Center of mass frame.

delta K1 = 0.5*m*[(-2v-vcm) ^2 - (2v-Vcm )^2

= 0.5*m*(4v^2 - 4v^2) =0

Delta K2 = 0.5*2m*(v- vcm) ^2 - (-v-vcm) ^2)

= 0

Hence delta k1= - delta K2

case2: Let the referance frame is comoving with first particle with velocity 2v rightward.

delta K1 = 0.5*m[(-2v - 2v)^2 - (2v - 2v)^2 ] = 8mv^2

delta K2 = 0.5*2m[(v - 2v)^2 - (-v - 2v)^2 ] = -8mv^2

hence delta K1 = - delta k2

case3] frame of reference moves rightward with v.

delta K1 = 0.5*m[(-2v - v)^2 - (2v - v)^2 ] = 4mv^2

delta K2 = 0.5*2m[(v - v)^2 - (-v - v)^2 ] = -4mv^2

hence delta K1 = - delta k2

Yes, ∆K1 = −∆K2 as expected from energy conservation as it is elastic collision, so whatever KE is increased in first body has to be decreased from second body.