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 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?

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wagonbelleville wagonbelleville · Answer 1 · 2019-11-07T05:33:54+01:00

Answer

given,

Car A weight = 1515 lb traveling eastward

Car B weight = 1125 lb traveling westward at a speed of 45 mph

distance to taken to stop the vehicle = 22.5 ft

1 ft = 0.000189394 mile

22.5 ft = 0.000189394 x 22.5 = 0.0043 mile

coefficient of friction = 0.750

Energy after the collision

[tex]\dfrac{1}{2}(M_a +M_b)V^2 = \mu (M_a +M_b)g d[/tex]

[tex]\dfrac{1}{2}(1515 +1125)V^2 = 0.75\times (1515 +1125)\times 9.8 \times 0.

0043 [/tex]

[tex]V = \sqrt{0.06321}[/tex]

[tex]V =0.251\ mile/h[/tex]

using momentum conservation

[tex]m_1v_1 - m_2v_2 = (m_1+m_2)V[/tex]

[tex]1515\times v_1 - 1125 \times 45= (2640)\times 0.251[/tex]

[tex]1515\times v_1 = 51287.64[/tex]

[tex] v_1 =33.85\ mile/h[/tex]