Problem Coordinate System

Begin with a diagram of the car and barge in their initial state. Place the origin of the rectangular coordinate system at the initial position of the car. The velocity of the barge when the car reaches the end of the barge is

     v_B = -0.885i m/s

Because the horizontal force of the water on the barge is neglected, the total linear momentum in the horizontal direction is conserved.

Since the car and barge are initially stationary, the total linear momentum is initially zero.

     m_Cv_C + m_Bv_B = constant = 0

The velocity of the car at the end of the barge can now be solved as

     v_C = -m_B/m_C v_B

          = -(2,500/500)(-0.885)i m/s

          = 4.425i m/s

Position of Barge and Car
after motion starts

Barge and Car Position
when Car Leaves Barge

The combined CG of the car and the barge is initially stationary, so it must remain stationary. Knowing the position of the combined CG, the position of the car and the barge relative to their original positions can be determined. The diagram at the left shows the position after motion starts. The initial combined r_CG is equal to zero since that is the location of the coordinate system.

     0 = m_C x_C - m_B x_B

There are still two unknowns, x_C and X_B. However, when the car leaves the barge, the combined distance x_B and x_C must be half the barge distance,

     x_C + x_B = 0.5 L

Combining with the previous equation gives,

     0 = m_C x_C - m_B (0.5L - x_C)

     0 = (500 + 2,500)x_C - 2,500 [0.5 (42)]

     x_c = 17.5 m

The distance from the dock is

     x_dock = 21 - 17.5

     x_dock = 3.5 m