Answer:
The velocity imparted to barge 2 just after the car has come to rest on the barge = 0.096 km/h
Explanation:
Mass of the two barges, [tex]m_{b_1} = m_{b_2} = 570 Mg[/tex] = 570000 kg
Mass of the car, [tex]m_c = 1870 kg[/tex]
[tex]v_0 = 30 km/h[/tex]
[tex]\theta = 15^0[/tex]
The horizontal component of car velocity:
[tex]v_x = v_0 cos \theta\\v_x = 30 cos 15\\v_x = 30 * 0.97\\v_x = 29.1 km/h[/tex]
The driver brings his car to rest relative to barge 2 at B:
[tex]v_{x_2} = 0 m/s[/tex]
[tex]v_{B_1} = 0 m/s[/tex]
Applying the principle of conservation of momentum:
[tex]m_cv_x + m_{B_1}v_{B_1} = m_cv_{x_2} +m_{B_2}v_{B_2}\\\\(1870*29.1) + (570000*0) = (1870*0) + (570000*v_{B_2})\\\\(1870*29.1) = (570000*v_{B_2})\\\\v_{B_2} = \frac{54417}{570000} \\\\v_{B_2} = 0.096 km/h[/tex]
