A 980-kg sports car travelling with a speed of 21 m/s collides into the rear end of a 2300-kg SUV that is stationary at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward a certain distance "d" before stopping (called stopping distance). The police officer estimates the coefficient of kinetic friction between tires and road to be 0.80. Calculate this stopping distance.

Explanation:

After the collision , both the car will have common velocity , which according to conservation of momentum will be as follows

v = 980 x 21 / (980 + 2300)

= 20580 / 3280

= 6.274 m /s

The kinetic energy of both the cars after collision  

= 1/2 x (980+2300) x 6.274²

=  64555.44 J .

frictional force = μ mg where μ is coefficient of friction , mg is weight of both the cars

= .8 x (980+2300) x 9.8

= 25715.2 N

work done by friction will be equal to kinetic energy of car

25715.2 x d = 64555.44 ; where d is displacement of both the cars

d = 2.5  m

the answer is d. 4

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