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.

Question

Scottie

Physics

1 October, 00:52

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.

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Brenna · Answer 1

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