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 bum

Question

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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Khánh Gia 3 years 2021-08-29T11:06:27+00:00 1 Answers 45 views 0

Answers ( )

  1. Sigrid
    0
    2021-08-29T11:08:04+00:00
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    Answer:

    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

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