A drunken driver crashes his car into a parked car that has its brakes set. The

Question

A drunken driver crashes his car into a parked car that has its brakes set. The two cars move off together (perfectly inelastic collision), with the locked wheels of the parked car leaving skid marks 4.0 m in length.

If the mass of the moving car is 2310kg and the mass of the parked car is 1300 kg how fast was the first car travelling when it hit the parked car?
Assume that the driver of the moving car made no attempt to brake and that the coefficient of friction between the tires and the road is 0.44.

Explanation / Answer

Deaccelaration due to friction = g*u = 9.8*0.44 = 4.312 m/s^2

m1*v1 = m2*v2 (v2 is the initial velocity of both cars combined and v1 is the velocity of the 1st car when it collided)

2310*v1 = (2310+1300)*v2 = 3610*v2

v2 = (2310/3610)*v1 = 0.64*v1

So for the combination of two cars,

v = 0

s = 4 m

u = v2

a = 4.312

v^2 = u^2 - 2*a*s

2*4.312*4 = 0.64*0.64*v1^2

v1 = sqrt(34.496/0.4096) = 9.177 m/s = 33.012 km/hr

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