Cars colliding on a barge - part I

1. Conservation of momentum: A linear momentum of a body can be calculated as $p = m\cdot v,$ where $m$ is a mass and $v$ is a velocity of the body. In a closed system (one that does not exchange any matter with its surroundings and is not acted on by external forces) the total momentum is constant. This fact is known as $\textit{conservation of momentum}$ . Suppose that $n$ particles can interact (e.g. collide) with each other. The conservation of momentum of $n$ particles can be expressed as: $m_1u_1 + m_2u_2 + \cdots m_n u_n = m_1v_1 + m_2v_2 + \cdots m_n v_n,$ where $u_i$ denotes the velocity of an i-th particle before and $v_i$ denotes the velocity of an i-th particle after the interaction between particles.

2. Relative motion: When you are walking in a bus which moves with a certain velocity $v$ , your velocity relative to chairs $v_{rel}$ in the bus is different from the one with respect to passing trees outside the bus. Your velocity with respect to the trees is the sum of the velocity of the bus and your velocity with respect to the bus, which can be written as: $v_{trees} = v + v_{rel}$ .

The task

A barge of a mass $m_L$ transports two cars A and B of masses $m_A$ and $m_B$ respectively. They are placed at the distance $d$ away from each other. They drive off with accelerations $a_A$ and $a_B$ respectively. Each one accelerates until it reaches a constant velocity $v$ with respect to the barge. What is the velocity of the barge shortly before the impact of the two cars? Include the sign of the number.

Assumptions: The barge initially at rest. Any resistances are neglected. The cars are treated as mass points. Given: $m_L = 22500$ kg, $m_A = 2000$ kg, $m_B = 1500$ kg, $a_A = 2\; \frac{\mathrm m}{\mathrm s^2}$ , $a_B = 4 \frac{\mathrm m}{\mathrm s^2}$ , $v = 3 \frac{\mathrm m}{\mathrm s}$ , $d = 15$ m.