Question

Two bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear. Car A has a mass of 450 kg and car B 550 kg, owing todifferences in passenger mass. If car A approaches at 4.5 m/s and car B is moving at 3.7 m/s calculate (a) their velocities after the collision, and (b) the change inmomentum of each.

Answer 1

(a) From the law of conservation of mechanical energy:

m1v1(i) + m2v2(i) = m1v1(f) + m2v2(f)
(450kg)(4.50m/s) + (550kg)(3.70m/s) = (450kg)v1(f) + (550kg)v2(f)
4060kg·m/s = (450kg)v1(f) + (550kg)v2(f)----------------->(1)

In your text, you should find the derivation of an equation relating the velocity of approach to the velocity of recession of two colliding objects, theequation is:

v1(i) - v2(i) = -[v1(f) - v2(f)]
4.50m/s - 3.70m/s = v2(f) - v1(f)
v1(f) = v2(f) - 0.800m/s----------------->(2)

Plugging (2) into (1) eliminates v1(f) allowing you to find v2(f):

4060kg·m/s = (450kg)[v2(f) - 0.800m/s] + (550kg)v2(f)
4060kg·m/s = (450kg)v2(f) - 360kg·m/s + (550kg)v2(f)
(1000kg)v2(f) = 4420kg·m/s
v2(f) = 4.42m/s

Plugging this into (2):

v1(f) = 4.42m/s - 0.800m/s
= 3.62m/s

(b) From (1) we know that the initial momentum is 4060kg·m/s. Now that we have the final velocities, the final momentum is:

(450kg)(3.62m/s) + (550kg)(4.42m/s) = 4060kg·m/s

So there is no change in momentum, this is because momentum is always conserved in all types of collisions.