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 45.0 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

Let car A mass be mA and velocity be VA and let car B mass be mB and velocity be -VB( - minus is because moving in opposite)

Acc. to the conservation of momentum, Initial and final momentum is equal

mA* VA+ mB* VB= (mA+mB) * Vtot ( Vtot is the combined speed)

On subsitution, 1515 * VA - 1125 * 42 = (1515+1125) * Vtot ----------------------- eq 1

From the other track lets calculate Vtot

Work done will be change in kinetic energy, W = K.E2 - K.E1 ( K.E2 is zero since the cars stop)

Also, W = force * disp = Coeff. of friction * Normal force * disp

W = 0.75 * (mA+mB) * g * 21.5 { Normal force is the combined weight}

Using the relation, W= - K.E1

  0.75 * (mA+mB) * g * 21.5 = (0.5)* (mA+mB) *Vtot2 (g= 78999.64 m/h2) (21.5 ft = 0.004 mile)

with proper unit conversions, 0.75 * 78999.64 * 0.004 = 0.5 * Vtot2

Vtot= 21.77 m/h.

Substituin Vtot in eq 1 , 1515 * VA - 1125 * 42 = (1515+1125) * 21.77

Solving it, VA= 69.12 m/h.

Hope it helps you

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