A 920 kg sports car collides into the rear end of a 2300 kg SUV stopped at a

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tlk3rd

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red light. The bumpers lock, and? A 920 kg sports car collides into the rear end of a 2300 kg SUV stopped at a red light. The bumpers lock, and the two cars skid 3.10 meters before coming to rest. Knowing that the coefficient of kinetic friction between the tires and the road is 0.08, the police officer on the scene determines the speed of the sports car prior to impact. How fast was the sports car going?

I have the answer to this question but I cannot work it out. Please help and show your work so I can understand. thank you.
 
Let the velocity of the two cars after the collision is v m/s
By Work done = change in KE
=>F(friction) x s = 1/2mv^2
=>coefficient of kinetic friction x (m1+m2) x g x s = 1/2(m1+m2)v^2
=>2 x 0.08 x 9.8 x 3.10 = v^2
=>v = sqrt4.84 = 2.20 m/s
By the law of momentum conservation:-
m1u1+m2u2 = (m1+m2) x v
920 x u + 0 = (920+2300) x 2.20
u = 7.70 m/s
 
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