Two bumper cars in an amusement park ride collide elastically as one approaches

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 to differences 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 in momentum of each.

Explanation / Answer

(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, the equation 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.

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