Abumper car with mass m1 = 103 kg is moving to the right with a velocity of v1 = 5 m/s. a second bumper car with mass m2 = 81 kg is moving to the left with a velocity of v2 = -3.3 m/s. the two cars have an elastic collision. assume the surface is frictionless. what is the final velocity of car 1 in the ground (original) reference frame?

The final velocity of car 1 in the center-of-mass reference frame is- 2.308 m/s

Explanation:

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)

The velocity of the center of mass = [m1v1 + m2v2] / [m1 + m2] = [103*5 + 81(-3.3)] / 184 = 1.346 m/s (to the right)

The conservation law of  momentum says that the centre of mass velocity remains the same after impact.

If we assume that Right is the positive direction , then the initial velocity of car 1 in the center-of-mass reference frame is:

= 5 - 1.346 = 3.654 m/s

this velocity would appear to be in the positive direction

The final velocity of car 1 in the center-of-mass reference frame is the initial velocityo of car 1, reversed.

= - 3.654 m/s )

The final velocity of car 1 in the center-of-mass reference frame: is then :

-3.654 + 1.346 = -2.308 m/s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west