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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.

Answer :

Answer:

[tex]113.17\frac{km}{h}[/tex]

Explanation:

In order to know the velocity of the cars after the collision, we must calculate the acceleration after the collision. According to Newton's second law:

[tex]\sum F_x:f_k=ma\\\sum F_y=N-mg=0\\N=mg\\f_k=\mu_k N=\mu_k mg\\a=\frac{f_k}{m}\\a=\frac{\mu_k mg}{m}\\a=0.75*9.8\frac{m}{s^2}\\a=7.35\frac{m}{s^2}[/tex]

Using the following kinematics equation, we calculate the initial velocity which is the same velocity after collision, so, the final velocity is zero since the cars stopped after 6 meters:

[tex]v_f^2=v_0^2-2ax\\v_0=\sqrt{v_f^2+2ax}\\v_0=\sqrt{0^2+2(7.35\frac{m}{s^2})(6m)}\\v_0=9.39\frac{m}{s}*\frac{3600s}{1h}\frac{1km}{1000m}=33.81\frac{km}{h}[/tex]

According to the law of conservation of momentum and solving for [tex]v_1[/tex]:

[tex]\Delta p=0\\p_i=p_f\\m_1v_1-m_2v_2=(m_1+m_2)v_0\\\\v_1=\frac{m_2v_2+(m_1+m_2)v_0}{m_1}\\v_1=\frac{510kg(72\frac{km}{h})+(680kg+510kg)33.81\frac{km}{h}}{680kg}\\v_1=113.17\frac{km}{h}[/tex]

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