A drunken driver crashes his car into a parked car that has its brakes set. The two cars move off together (perfectly inelastic collision) for 6.0 m.
If the mass of the moving car is 1030 kg and the mass of the parked car is 1410 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 0.02.
Conservation of momentum: MaVa + MbVb = MaVa' + MbVb'
Conservation of momentum of KE: (1/2)MaVa2+ (1/2)MbVb2 = (1/2)MaVa'2 + (1/2)MbVb'2
Frictional force = μFn
The Attempt at a Solution
I calculated the frictional force acting on the car at rest: 276.4 N
And also the frictional force acting on the whole system: 479 N
Because I'm not sure which one to use.
Since the second car is at rest and they have the same velocity following the collision, the conservation of momentum can be simplified to: MaVa = V' (Ma + Mb)
I don't know how to go about this problem, and any help would be greatly appreciated. Thank you!