You have been called to testify as an expert witness in a trial involving a head

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.)

Explanation / Answer

Mass of Car A = Ma = 1515 lb = 687.2 kg

Mass of Car B = MB = 1125 lb = 510.3 kg

Velocity of Car A before Collision= Va         in east direction (+ve)

Velocity of Car B before collision = Vb = - 41 mph = -18.33 m/s     in west direction (-ve)

Velocity of car A + Car b after the collision = V

Consider the motion of cars locked :

Distance travelled after collision by cars locked = d = 21.5 ft = 6.55 m

initial velocity of cars locked = Vi = V

final velocity of cars locked = Vf = 0

acceleration due to friction = a = - uk g = - (0.750) (9.8 m/s2 ) = - 7.35 m/s2     (uk = Coefficient of friction = 0.750)

Using the kinematics equation

Vf2 = Vi2 + 2 a d

02 = V2 + 2 (- 7.35 m/s2 ) (6.55 m )

V = 9.81 m/s

using conservation of momentum equation for inelastic collision since cars got locked:

Ma Va + Mb Vb = (Ma + Mb ) V

(687.2 kg) Va+ (510.3 kg) (-18.33 m/s ) = (687.2 kg + 510.3 kg) (9.81 m/s)

(687.2 kg) Va - 9353.8 kgm/s =   11747.48 kgm/s

Va = 30.71 m/s = 68.7 mph

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