Answer:
Part a)
[tex]v_a = 3.94 m/s[/tex]
Part b)
[tex]v_b = 3.35 m/s[/tex]
Part C)
[tex]v_b = 6.44 m/s [/tex]
Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Explanation:
Part a)
As we know that car A moves by distance 6.1 m after collision under the frictional force
so the deceleration due to friction is given as
[tex]a = -\frac{F_f}{m}[/tex]
[tex]a = -\frac{\mu mg}{m}[/tex]
[tex]a = - \mu g[/tex]
now we will have
[tex]v_f^2 - v_i^2 = 2ad[/tex]
[tex]0 - v_i^2 = 2(-\mu g)(6.1)[/tex]
[tex]v_a = \sqrt{(2(0.13)(9.81)(6.1)}[/tex]
[tex]v_a = 3.94 m/s[/tex]
Part b)
Similarly for car B the distance of stop is given as 4.4 m
so we will have
[tex]v_b = \sqrt{2(0.13)(9.81)(4.4)}[/tex]
[tex]v_b = 3.35 m/s[/tex]
Part C)
By momentum conservation we will have
[tex]m_1v_{1i} = m_1v_{1f} + m_2v_{2f}[/tex]
[tex]1400 v_b = 1100(3.94) + 1400(3.35)[/tex]
[tex]v_b = 6.44 m/s [/tex]
Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.