A bumper car with mass m 1 = 115 kg is moving to the right with a velocity of v

Question

A bumper car with mass m1 = 115 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2 = 96 kg is moving to the left with a velocity of v2 = -3.8 m/s. The two cars have an elastic collision. Assume the surface is frictionless.

1) What is the velocity of the center of mass of the system?    .94 m/s

2)What is the initial velocity of car 1 in the center-of-mass reference frame?   3.90 m/s

I need help with:

What is the final velocity of car 1 in the center-of-mass reference frame? and What is the final velocity of car 1 in the ground (original) reference frame?

Explanation / Answer

The basic concept is that for any collision, conservation of momentum holds.

For elastic collisions, the relative velocity of each mass in relation to the center of mass will be equal in magnitude but opposite in direction for each mass. (relative to the center of mass)

1. What is the velocity of the center of mass of the system?
If we assume that Right is the positive direction
v' = 115(4.9) + 96(-3.8) / (115 + 96)  = 0.94 m/s (to the right)

2. What is the initial velocity of car 1 in the center-of-mass reference frame?
as the center of mass is moving to the right, someone riding the center of mass would see the first car approaching from behind at
4.9 - 0.94 = 3.96 m/s this velocity would appear to be in the positive direction

3. What is the final velocity of car 1 in the center-of-mass reference frame?
the initial velocity would appear to be reversed v = - 3.8 m/s

4. What is the final velocity of car 1 in the ground (original) reference frame?
v = -3.8 + 0.94 = - 2.86 m.s

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