Question

You’re an accident investigator at a scene where a drunk driver in a 2300 kg car has plowed into a 2000 kg parked car with its brake set. You measure skid marks showing that the combined wreckage moved 27 m before stopping, and you determine a frictional coefficient of 0.74. What do you report for the drunk driver’s speed just before the collision?

Answer 1

a drunk driver in a 2300 kg car has plowed into a 2000 kg parked car with its brake set."
Classic inelastic collision. Momentum is conserved. Kinetic energy is not. Whatever we find is the momentum of the two cars, that is the momentum the 2300 kg car had before the collision.

"the combined wreckage moved 27 m before stopping"
Frictional force F * d = (1/2)mv^2 with d = 27 m, where (1/2)mv^2 is the kinetic energy of the COMBINED masses right AFTER the collision.

"you determine a frictional coefficient of 0.74."
F = mu * mg = 0.74 * (2300 + 2000) * 9.8 = 31183.6 N
m = 2300 + 2000 = 4300 kg.

I prefer to solve in symbols before plugging in numbers.
F * d = (1/2)mv^2
2Fd = mv^2
v^2 = 2Fd/m
Now plug in the numbers.
v = sqrt(2Fd/m) = sqrt(2 * 31183.6 N * 27m/4300 kg) = 19.789 m/s

Now apply conservation of momentum. Right after the collision, the two cars have momentum 4300 * 19.789 m/s. This was all equal to mv of the 2300 kg car before the collision.

2300 * v = 4300 * 19.789
v = 4300 * 19.789 / 2300 = 36.996 m/s.