Q.172. Work, energy, law of conservation of linear momentum.

Question 172.
A 930 kg sports car collides into the rear end of a 2300 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 3.1 m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.80, calculates the speed of the sports car at impact.

Answer 172.
Frictional force
= (930 + 2300) * (9.81) * 0.80 N
Work done by the frictional force
= Force x displacement
= (930 + 2300) * (9.81) * 0.80 * 3.1 J

If w = velocity of both the cars after the collision,
kinetic energy of the two cars = work done by friction
=> (1/2) (930 + 2300) w^2 = (930 + 2300) * (9.81) * 0.80 * 3.1
=> w^2 = 2 * (9.81) * 0.80 * 3.1 = 48.6576 m/s
=> w = 6.975 m/s

Let v = speed of the sports car at impact.
By the law of conservation of momentum,
(930) v = (930 + 2300) * (6.975)
=> v
= 24.227 m/s
= 24.227 * 3.6 km/h
= 87.2 km/h.

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