Question

A truck with a mass of 1800 kg and moving with a speed of 15.0 m/s rear-ends a 591 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.

Answer 1

First, conserve momentum: initial p = final p
1800kg * 15m/s + 0 = 1800kg * u + 591kg * v
where u, v are the post-collision velocities of the truck, car respectively
27000 = 1800u + 591v

For an elastic collision, we know (from conservation of energy) that the
relative velocity of approach = relative velocity of separation, or
15 m/s = v - u, so
v = u + 15
Plug that into the momentum equation:
27000 = 1800u + 591(u + 15) = 2391u + 8865
u = 7.6 m/s ← truck ....Answer.
v =7.6m/s + 15m/s = 22.6 m/s ← car.....Answer.

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