Answer:
70.6 mph
Explanation:
Car A mass= 1515 lb
Car B mass=1125 lb
Speed of car B is 46 miles/h
Distance before locking, d=19.5 ft
Coefficient of kinetic friction is 0.75
Initial momentum of car B=mv where m is mass and v is velocity in ft/s
46 mph*1.46667=67.4666668 ft/s
[tex]Momentum_B=1125*67.4666668 ft/s[/tex]
Initial momentum of car A is given by
[tex]Momentum_A=1515v_a[/tex] where [tex]v_a[/tex] is velocity of A
Taking East as positive and west as negative then the sum of initial momentum is
[tex]1515v_a-(1125*67.4666668 ft/s)[/tex]
The common velocity is represented as [tex]v_c[/tex] hence after collision, the final momentum is
[tex]Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c[/tex]
From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence
[tex]1515v_a-(1125*67.4666668 ft/s)= 2640v_c[/tex]
The acceleration of two cars [tex]a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}[/tex]
From kinematic equation
[tex]v^{2}=u^{2}+2as[/tex] hence
[tex]v^{2}-u^{2}=2as[/tex]
[tex]0^{2}-(v_c)^{2}=2*-24.1275*19.5[/tex]
[tex]v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s[/tex]
Substituting the value of [tex]v_c[/tex] in equation [tex]1515v_a-(1125*67.4666668 ft/s)= 2640v_c[/tex]
[tex]1515v_a-(1125*67.4666668 ft/s)= 2640*30.67[/tex]
[tex]1515v_a=(1125*67.4666668 ft/s)+2640*30.67[/tex]
[tex]v_a=\frac {156868.8}{1515}=103.5438 ft/s[/tex]
[tex]\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph[/tex]
