Question

Two barges, each with a displacement (mass) of 570 Mg, are loosely moored in calm water. A stunt driver starts his 1870-kg car from rest at A, drives along the deck, and leaves the end of the 15° ramp at a speed of 30 km/h relative to the barge and ramp. The driver successfully jumps the gap and brings his car to rest relative to barge 2 at B. Calculate the velocity v2 imparted to barge 2 just after the car has come to rest on the barge.

Answer 1

The velocity imparted to barge 2 just after the car has come to rest on the barge = 0.096 km/h

Step-by-step explanation:

Mass of the two barges,
`m_(b_1) = m_(b_2) = 570 Mg` = 570000 kg

Mass of the car,
`m_c = 1870 kg`

`v_0 = 30 km/h`

`\theta = 15^0`

The horizontal component of car velocity:

`v_x = v_0 cos \theta\\v_x = 30 cos 15\\v_x = 30 * 0.97\\v_x = 29.1 km/h`

The driver brings his car to rest relative to barge 2 at B:

`v_(x_2) = 0 m/s`

`v_(B_1) = 0 m/s`

Applying the principle of conservation of momentum:

`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) = (54417)/(570000) \\\\v_(B_2) = 0.096 km/h`