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 Ib and was traveling westward at 45.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 22.5 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.) Number mph

Explanation / Answer

mA = 1515 lb = 1515(0.45 Kg) = 681.75 Kg

mB = 1125 lb = 1125(0.45 Kg) = 506.25 Kg

assuming the east direction as positive

vB = -45 mph = - 45(0.447) m/s = - 20.12 m/s

Since the collision is perfecly ineleastic after collision both cars travel together with a common velocity v

friction force

F = uN

F = u(mA + mB )g

As the two cars stop after travlling 22.5 ft due to friction (d = 22.5 ft = 22.5(0.3048) = 6.86 m)

Work done by friction = kinetic energy of cars after collision

W = (1/2)(mA + mB )v2

Fd = (1/2)(mA + mB )v2

u((mA + mB )g)d = (1/2)(mA + mB )v2

ugd = (1/2)v2

v2 = 2ugd

v2 = 2(0.75)(9.81)(6.86)

v = 10.05 m/s

Applying conservation of momentum we get

mAvA + mBvB = (mA + mB )v

681.75vA + 506.25 (- 20.12) = ( 681.75 + 506.25) (10.05)

681.75vA = 2.21 X 104

vA = 32.45 m/s

vA = 32.45 (2.237) mph

vA = 72.6 mph

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