Momentum in accidents?

Question

An accident involves a car of mass 2000 kg which approachs a stationary car of mass 970.0 kg. The driver of car A applies his brakes 15.6 meters before he crashes into car B. After the collision, car A slides 15.6 meters while car B slides 27.9 meters. The coefficient of kinetic friction between the locked wheels and the road for both cars is 0.530. How fast was the driver of car A going before he hit the brakes?

Answer 1

First, let's calculate the energy lost to friction for each car after the crash and use that to find their speeds just after impact

Car A
.5*ma*va^2=ma*g*µk*15.6
solve for va
va=12.7 m/s
Car B
.5*mb*vb^2=mb*g*µk*27.9
solve for vb
vb=17.0 m/s

Now let's calculate the speed of car A just before impact using conservation of momentum
2000*va2=2000*12.7+970*17
solve for va2
20.9 m/s

Now let's calculate the energy lost to friction just before impact and use it to calculate the initial v of car A

.5*ma*vai^2=ma*g*µk*15.6+.5*ma*va2^2
solve for vai
24.5 m/s

j