A 667 kg car stopped at an intersection is rear-ended by a 1600 kg truck moving

Question

A 667 kg car stopped at an intersection is rear-ended by a 1600 kg truck moving with a speed of 13.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision. m/s (car) m/s (truck)

Explanation / Answer

The law of conservation of momentum says that the momentum before the collision will equal the collision after the collision. For an elastic collision the equation is : m1v1(i) + m2v2(i) = m1v1(f) + m2v2(f) Since the car is stopped (call it m2), its initial momentum is zero,so the second term on the left side of the equation is zero and becomes : (1650kg)(13.5m/s) = (1650kg)v1(f) + (720kg)v2(f) 22,275kg•m/s = (1650kg)v1(f) + (720kg)v2(f)----> equation (1) We need one more equation as we have two unknowns. From the combination of the laws of conservation of enrgy and momentum : v1(i) - v2(i) = -[v1(f) - v2(f)] 13.5m/s = v2(f) - v1(f) v2(f) = 13.5m/s + v1(f)--------> equation (2) Plugging (2) into (1), eliminates v2(f) so that v1(f) can be found : 22,275kg•m/s = (1650kg)v1(f) + (720kg)[13.5m/s + v1(f)] v1(f) = 5.30m/s So the truck continues in the same direction at 5.30m/s. The car's final speed is may be found most easily from equation (2) : v2(f) = 13.5m/s + 5.30m/s = 18.8m/s Pretty fast, but it is less than half the mass of the truck, and was hit at a pretty high speed (about 30mph).

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