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

Car A weighs 1515 lb = 687.1924kg

Car B weighs 1125 lb = 510.2914kg

21.5 ft = 6.5532 m

45.0 mph =   72.4205km/h

  F=ma, so a=F/m F_friction = normal force times coefficient of friction = 0.75*(687.1924+510.2914)*g

and m =(687.1924 + 510.2914). Therefore a= 0.75*g.

v^2=v_i^2 +2*a*x, so initial velocity is sqrt(2*a*x)=sqrt(2*0.75*g*6.5532)=9.819 m/s, therefore total momentum is mass times velocity is 9.819*(510.2914+687.1924) = 11758.09.

you know that momentum is conserved. 72.4205 km/h is 220.11680556 m/s initial momentum of car B is 510.2914*20.11680556.

But they were going in opposite directions. so momentum of car A - momentum of car B = 11758.09

so (11099 -500*72000)= momentum car A.

velocity car A = (11758.09 +510.2914*20.11680556)/687.1924= 32.048 m/s is like 115.3728 km/h = 71.689334288mph

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