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 445kg and car B 480kg .Car A approaches at 4.50 m/sand car B is moving at3.70

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 445kg and car B 480kg .Car A approaches at 4.50 m/s and car B is moving at3.70 m/s Determine the velocity of car A after the collision. Express your answer to three significant figures and include the appropriate units. Determine the velocity of car B after the collision. Express your answer to three significant figures and include the appropriate units. Calculate the change in momentum of car A after the collision. Express your answer to three significant figures and include the appropriate units. Calculate the change in momentum of car B after the collision. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

From the law of conservation of mechanical energy:

m?v?(i) + m?v?(i) = m?v?(f) + m?v?(f)
(450kg)(4.50m/s) + (550kg)(3.70m/s) = (450kg)v?(f) + (550kg)v?(f)
4060kg?m/s = (450kg)v?(f) + (550kg)v?(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:

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

Plugging (2) into (1) eliminates v?(f) allowing you to find v?(f):

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

Plugging this into (2):

v?(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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