A 876 kg sports car collides into the rear end of a 2169 kg SUV stopped at a red

Question

A 876 kg sports car collides into the rear end of a 2169 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 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. What was that speed? A 876 kg sports car collides into the rear end of a 2169 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 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. What was that speed?

Explanation / Answer

momentum before collision Pi = m1*v1

momentum after collison Pf = (m1+m2)*Vf

from momentum conservation

Pf = Pi

vf = m1*v1/(m1+m2)

vf = (876*v1)/(876+2169) = 0.287*v1

after collison the cars comes to rest

work by friction = change in KE

uk*(m1+m2)*g*x = (1/2)*(m1+m2)*vf^2

0.8*9.8*2.8 = (1/2)*(0.287*v1)^2

v = 23.1 m <<<<-----------answer

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