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 21.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750. How fast (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

mass of A, ma = 1515 lb

mb = 1225

speed of ma = ua

speed of mb = ub = 41 mph

coefficient of kinetic friction, u = 0.75

distance moved before stopping,d = 21.5 ft = 21.5*1.89*10^-4 miles

Initial momentum of a, Pa = ma*ua

Pb = mb*ub

So, initial momentum fo a, Pi = Pa +Pb = ma*ua + mb*ub

After collision le the final speed after locking be v

So, final momentum, Pf = (ma+mb)*v

By conservation of momentum,

Pi = Pf

So, ma*ua + mb*ub = (ma+mb)*v

So, 1515*ua-1125*41 = (1515+1125)*v

So, v = (1515*ua-1125*41)/(2640)

Now,

Initial Kinetic Energy(K.E) after collision, K.Ei = 0.5*(ma+mb)*v^2

Now all this K.E will be spent on the work due to friction

Workd due to friction, W = u(ma+mb)*g*d

So, K.Ei = W

So, 0.5*(ma+mb)*v^2 = u(ma+mb)*gd

So, 0.5*(1515+1125)* ((1515*ua-1125*41)/(2640))^2 = 0.75*(1515+1125)*g*21.5

Now , g = 9.8 m/s2 = 9.8*3.28/(0.000278)^2 ft/hr^2 = 4.25*10^7 ft/hr^2

So, 0.5*(1515+1125)* ((1515*ua-1125*41*5280)/2640)^2 = 0.75*(1515+1125)*4.25*10^7*21.5 <---- converting 41 mph to 41*5280 ft/hr

So, us = 96239 ft/hr = 96239/5280 = 18.2 mph <-------------answer