A 50. 0 kg child is riding in a 120. 0 kg bumper car at an amusement park. The child drives the bumper car at 4. 5 m/s head-on into the wall of the ride and rebounds at the speed. What is the magnitude of the change in momentum of the child and the bumper car?.

Explanation:
Total mass of child and car = 50 + 120 = 170 kg .
initial velocity u = 4.5 m/s
initial momentum = mass x velocity = 170 x 4.5 = 765 kg m/s
final velocity = - 4.5 m/s [ - ve is due to reversal of direction ]
final momentum = - 765 kg m/s
change in momentum = final momentum - initial momentum
= - 765 - ( 765 )
= - 1530 kg m/s .

Not the answer but here is the graph. 

B.  earth orbits at a slower rate than the rate at which the moon orbits earth.