Question

Two trucks of mass $M$ each are moving in opposite direction on adjacent parallel tracks with same velocity $u$. One is carrying potatoes and other is carrying onions, bag of potatoes has a mass $m_1$ and bag of onions has a mass $m_2$ (included in the mass of truck $M$). When trucks get close to each other while passing the drivers exchange a bag with the other one by throwing towards the other one. Find the final velocities of the truck after exchange of the bags.

• A. The final velocities of both trucks are $0$ m/s.
• B. The final velocity of the truck carrying potatoes is $u$ and the truck carrying onions is $-u$.
• C. The final velocity of the truck carrying potatoes is $-u$ and the truck carrying onions is $u$.
• D. The final velocities depend on the masses $m_1$ and $m_2$ and cannot be determined without further information.

Answer 1

Answer: (A)

The final velocities of both trucks are $0$ m/s.

Answer 2

The problem can be solved using the principle of conservation of linear momentum.

1. Initial State: Let's assume the initial direction of the truck carrying potatoes is the positive direction.

• Truck 1 (carrying potatoes): Mass $M$, initial velocity $v_{1i} = +u$.
• Truck 2 (carrying onions): Mass $M$, initial velocity $v_{2i} = -u$.

The total initial momentum of the system (two trucks and their loads) is: $P_i = M v_{1i} + M v_{2i}$ $P_i = M(u) + M(-u) = Mu - Mu = 0$

2. Final State: After the exchange of bags:

• Truck 1 (now carrying onions) has a mass of $(M - m_1) + m_2$. Let its final velocity be $v_{1f}$.
• Truck 2 (now carrying potatoes) has a mass of $(M - m_2) + m_1$. Let its final velocity be $v_{2f}$.

The total final momentum of the system is: $P_f = ((M - m_1) + m_2)v_{1f} + ((M - m_2) + m_1)v_{2f}$ $P_f = (M - m_1 + m_2)v_{1f} + (M - m_2 + m_1)v_{2f}$

3. Conservation of Momentum: Since there are no external horizontal forces acting on the system during the exchange, the total linear momentum is conserved: $P_i = P_f$ $0 = (M - m_1 + m_2)v_{1f} + (M - m_2 + m_1)v_{2f}$

This equation has two unknowns ($v_{1f}$ and $v_{2f}$). However, the problem implies a symmetric exchange. In such problems, especially when the initial objects are identical and moving with equal and opposite velocities, a common implicit assumption that leads to a solvable unique answer is that the final velocities of the primary objects (the trucks) are equal.

Let's assume $v_{1f} = v_{2f} = v_f$. Substituting this into the conservation of momentum equation: $0 = (M - m_1 + m_2)v_f + (M - m_2 + m_1)v_f$ $0 = [(M - m_1 + m_2) + (M - m_2 + m_1)]v_f$ $0 = [M - m_1 + m_2 + M - m_2 + m_1]v_f$ $0 = [2M]v_f$

For this equation to hold true, $v_f$ must be $0$. Therefore, $v_{1f} = 0$ and $v_{2f} = 0$.

The final velocities of both trucks are $0$ m/s.