Momentum and energy problem?

Question

A careless driver rear-ends a car that is halted at a stop sign. Just before the impact, the driver slams on his brakes, locking the wheels. The driver of the struck car also has his foot on the brake pedal, locking his wheels. THe struck car has a mass of 900 kg, that of the initially moving car is 1200kg. On collision, the bumpers of the 2 cars stick together. The sliding cars leave marks 0.760m long. Tests show that the coefficient of kinetic friction btw tires and pavement is 0.920. What is the velocity of the moving car as he approached the intersection?

Answer 1

using energy
.5*m*v^2=m*g*0.920*0.760
solve for v
3.7 m/s

The combined mass of the two vehicles is 2100 kg

using conservation of momentum
1200*vi=2100*3.7
solve for vi
6.475 m/s
j

Answer 2

µ=0.920
delta(KE+PE)=Wnc
M=big car; m=little car
U=intitial velocity of big car; u=intial velocity of little car(0) V=velocity of the system after the collision
Wnc=(.5(m+M)V²-.5MU²) +(0--no PE)
-µ(m+M)gd=(.5(m+M)V²-.5MU²) +(0--no PE)
-.92(2100)(9.8)(.76)=-.5(1200)(U)²
14,389.536=600U²
U=4.897m/s