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 680.0 kg and was traveling eastward. Car B weighs 510.0 kg and was traveling westward at 72.0 km/h. The cars locked bumpers and slid eastward with their wheels locked for 6.00 m before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750. 1) How fast (in kilometer per hour) was car A traveling just before the collision? (Express your answer to three significant figures.)

Explanation / Answer

F=ma, so a=F/m

F_friction = normal force times coefficient of friction

= 0.75*(680+510)*g and m =(680 + 510). 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)=9.4 m/s,

therefore total momentum is mass times velocity is 9.4*(510+680) = 11186

you know that momentum is conserved. 72 km/h is 20 m/s initial momentum of car B is 510*20.

But they were going in opposite directions. so momentum of car A - momentum of car B = 11186

so (11186 -510*72000)

= momentum car A.

velocity car A = (11186 +510*20)/680

= 31.45 m/s is like 113.222 km/h

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