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 automobile remain locked together, and they skid along the pavement with all their wheels locked. An investigation of this accident establishes that the length of this skid marks made by the automobiles after the impact was 19 m. The mass of the moving automobile was 2360 kg, and that of the parked automobile was 1120 kg. The coefficient of sliding friction between the wheels and the pavement is 0.95.

(a) What was the speed of the two automobiles immediately after the impact? m/s

(b) What was the speed of the moving automobile before the impact? m/s

Explanation / Answer

a) The kinetic friction coefficient is given here mu = 0.95

The initial velocity = v(suppose)

the acceleration a = -mu*g = -0.95*9.81 = -9.34 m/s^2

Final velocity vf = 0

distance travelled d = 19m

hence applying equation of motion, vf^2 = vi^2-2as

0 = v^2-2*9.34*19

v = sqrt(2*9.34*19) = 18.84 m/s

b) Suppose the automobile is moving with a velocity = v_a

then from conservation of momentum,

2360*v_a = (2360+1120)*v = (2360+1120)*18.84

hence v_a = (2360+1120)*18.84/2360 = 27.78 m/s

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