Question

You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515

lb and was traveling eastward. Car B weighs 1125
lb and was traveling westward at 41.0
mph. The cars locked bumpers and slid eastward with their wheels locked for 18.5
ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750
.

What speed
(in miles per hour) was car A traveling just before the collision? (This problem uses English units because they would be used in a U.S. legal proceeding.)

Answer 1

Solve for force:

Ff = UFn

Ff = 0.75(Fn)

Ff = 0.75(1515 + 1225 * g)

Ff = 20550N

Solve for acceleration:

F= ma

20550N = (1515 + 1225) a

a = 7.5m/s^2

solve for time:

a = d / t^2 ---> 7.5m/s^2 = 18.5/ t^2 ----> t = 0.85s

solve for velocity final

Impulse = F * t = 20550N * 0.85s

mv^2 = Impulse = 17467.5

(1515 + 1125)v^2 = 17467.5

vf = 2.5m/s

Plug in stuff:

1515 * v1 + 1125 * (-18.3m/s) = (1515 + 1125) * 2.5m/s

v1 = 9.23

Note: I converted 41mph(v2) to 18.3m/s, which is negative because "westward" is in the negative direction.

Explanation: Inelastic collision

I'm not sure but my guess is we can solve for the force of friction using the coefficient of friction. With that, we can solve for the acceleration in F = ma, and use that to solve for the time the two cars slide. And using that we can solve for the impulse, which is just the Force of friction times that time, which is also our momentum. Since we know the momentum, we can solve for the velocity of the two objects after the collision. Using that velocity, we can use the equation( m1v1 + m2v2 = (m1+m2)vf ), plug in the known quantities and solve for v1.(Note: don't forget to convert mph to mps and 18.5ft to meters)

Extra: I'm guessing because the two cars slide, the only force acting on them is the force of friction(so it's our net force), hence the Fnet = ma.