Question
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.
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
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.
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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