Please show work! A bumper car with mass m1 = 104 kg is moving to the right with

Question

Please show work! A bumper car with mass m1 = 104 kg is moving to the right with a velocity of v1 = 4.2 m/s. A second bumper car with mass m2 = 94 kg is moving to the left with a velocity of v2 = -3.6 m/s. The two cars have an elastic collision. Assume the surface is frictionless. I found that the velocity of the center of mass of the system to be .496

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

2) What is the final velocity of car 1 in the ground (original) reference frame?

3) What is the final velocity of car 2 in the ground (original) reference frame?

4) In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide. What is the final speed of the two bumper cars after the collision?

Explanation / Answer

mass m1 = 104 kg   velocity v1 = 4.2 m/s

mass m2 = 94 kg velocity v2 = -3.6 m/s.

velocity of the center of mass of the system = 0.497 m/s

The two cars have an elastic collision

Momentum and Energy conservation

Let the velocity of m1 and m2 be U1 and U2 respectively.

Initial momentum = Final momentum

104 *4.2 - 94 *3.6 = 104U1 + 94U2     

98.4 = 104U1 + 94U2      equation1

Since elastic collision Hence

velocity of seperation = velocity of approach

U2 - U1 = 4.2 + 3.6

U2 - U1 = 7.8 equation 2

Solving  equation 1 and 2

198U2 = 909.6

U2 = 4,6 m/s

U1 = -3.2 m/s

1.)  the final velocity of car 1 in the center-of-mass reference frame = -0.497 - 3.2 = -3.697 m/s

2.)  the final velocity of car 1 in the ground (original) reference frame = -3.2 m/s

3.)  the final velocity of car 2 in the ground (original) reference frame = 4.6 m/s

4.) (104+94)V = 98.4

V =0.497 m/s

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