In an elastic collision, a 475 kg bumper car collides directly from behind with a second bumper car that is traveling in the same direction. The mass of the leading bumper car is 30.0% greater than that of the trailing bumper car. The initial speed of the leading bumper car is 3.70 m/s, and that of the trailing car is 7.50 m/s. Assuming that the mass of the drivers is much, much less than that of the bumper cars, what are their final speeds? (Enter your answers in m/s. Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

Explanation:

mass of bumper car, m1 = 475 kg

mass of trailing car, m2 = m = 365.4 kg

mass of bumper car = 30% greater than the trailing car

475 = 130 x m / 100

m = 365.4 kg

initial velocity of bumper car, u1 = 3.7 m/s

initial velocity of trailing car, u2 = 7.5 m/s

Let the velocity of the bumper car and the trailing car after collision is v1 and v2 respectively.

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

475 x 3.7 + 365.4 x 7.5 = 475 v1 + 365.4 v2

4498 = 475 v1 + 365.4 v2 .... (1)

As the collision is elastic so the coefficient of restitution is 1.

v2 - v1 = 3.7 - 7.5

v1 - v2 = 3.8 ... (2)

from equation (1)

v1 - (4498 - 475 v1) / 365.4 = 3.8

365.4 v1 - 4498 + 475 v1 = 1388.52

840.4 v1 = 5886.52

v1 = 7 m/s

Put in equation (2)

7 - v2 = 3.8

v2 = 3.2 m/s  

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