A speeding automobile strikes the rear of a parked automobile. After the impact

Question

A speeding automobile strikes the rear of a parked automobile. After the impact the two automobiles remain locked together, and the skid along the pavement with all their wheels locked. An investigation of this accident establishes that the length of the skid marks made by the automobiles after the impact was 18 m; the mass of the moving automobile was 2200 kg and that of the parked automobile was 1400 kg, and the coefficient of sliding friction between the wheels and the pavement was 0.95.

a) What was the speed of the two automobiles immediately after impact?
b) What was the speed of the moving automobiles before impact?

Explanation / Answer

force of friction for the locked automobiles after collision = mg = .95*(1400+2200)*9.8=33516 N

so work done = force*distance = 333516*18 = 603288

so initial kinetic E after collision = work done

so (1/2)(1400+2200)*v2 = 603288

so, v = 18.31m/s

if initial velocity of moving one is v,

2200*v + 1400*0 = (2200+1400)* 18.31

v= 29.96 m/s

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