91Ó°ÊÓ

Two identical railroad cars sit on a horizontal track, with a distance \(D\) between their two centers of mass. By means of a cable between them, a winch on one is used to pull the two together. (a) Describe their relative motion. ( \(b\) ) Repeat the analysis if the mass of one car is now three times that of the other. Keep in mind that the velocity of the center of mass of a system can only be changed by an external force. Here the forces due to the cable acting on the two cars are internal to the two-car system. The net external force on the system is zero, and so its center of mass does not move, even though each car travels toward the other. Taking the origin of coordinates at the center of mass, $$ x_{\mathrm{cm}}=0=\frac{\sum m_{i} x_{i}}{\sum m_{i}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}} $$ where \(x_{1}\) and \(x_{2}\) are the positions of the centers of mass of the two cars. (a) If \(m_{1}=m_{2}\), this equation becomes $$ 0=\frac{x_{1}+x_{2}}{2} \quad \text { or } \quad x_{1}=-x_{2} $$ The two cars approach the center of mass, which is originally midway between the two cars (that is, \(D / 2\) from each), in such a way that their centers of mass are always equidistant from it. (b) If \(m_{1}=3 m_{2}\), then we have $$ 0=\frac{3 m_{2} x_{1}+m_{2} x_{2}}{3 m_{2}+m_{2}}=\frac{3 x_{1}+x_{2}}{4} $$ from which \(x_{1}=-x_{2} / 3\). Since \(m_{1}>m_{2}\), it must be that \(x_{1}

Step by step solution

01

Understanding the Problem

Two identical railroad cars are on a horizontal track separated by a distance \(D\). You will use a winch to pull them together. There are no external forces, so the center of mass of the system remains stationary. We need to analyze the motion of the cars both when their masses are equal and when one car is heavier.

02

Determine the Center of Mass Equation

The center of mass of the two-car system is given by \( x_{\mathrm{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \). This will remain zero, as there are no external forces. The origin of coordinates is at the system's center of mass.

03

Case (a) Identical Masses \(m_1 = m_2\)

Substitute \(m_1 = m_2\) into the center of mass equation: \(0 = \frac{x_1 + x_2}{2}\). This simplifies to \(x_1 = -x_2\). The two cars move toward each other symmetrically, always equidistant from the center of mass, originally \(D/2\) from each.

04

Case (b) Different Masses \(m_1 = 3m_2\)

Substitute \(m_1 = 3m_2\) into the equation: \(0 = \frac{3m_2 x_1 + m_2 x_2}{3m_2 + m_2} = \frac{3x_1 + x_2}{4}\). Solving it gives \(x_1 = -\frac{x_2}{3}\). The heavier car \(m_1\) moves proportionally less compared to the lighter car \(m_2\).

05

Position Analysis for Different Masses

Given \( |x_1| + |x_2| = D\), and substituting \(x_1 = -\frac{x_2}{3}\), we get \(\frac{x_2}{3} + x_2 = D\). Solving for \(x_2\) gives \(x_2 = \frac{3D}{4}\). Therefore, the lighter car was originally \(\frac{3D}{4}\) from the center of mass, and the heavier car, \(\frac{D}{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Internal Forces

In a system like these railroad cars, internal forces play a key role in determining motion. Internal forces are those forces that act between the components of a system rather than coming from outside influences. In this problem, the pulling force exerted by the winch through the cable is an internal force because it is applied by one railcar onto the other.

These forces do not affect the movement of the system's center of mass. This is crucial because while parts of the system (the cars) move, the overall system center of mass does not move since the internal forces produce equal and opposite reactions, canceling each other out. This means our system's center of mass remains stationary.

Relative Motion

Relative motion is about how objects move concerning each other. In this exercise, when the cars are pulled towards one another, each moves relatively to the other, and to their common center of mass.

Initially, when both cars are identical in mass, they approach each other symmetrically. This means that their relative speeds are the same, and they cover the same distance in the same amount of time. When the mass distribution changes, however, the heavier car will move less than the lighter one. Thus, the cars no longer approach symmetrically, and the heavier car moves one-third of the distance covered by the lighter car.

Mass Distribution

Mass distribution determines the position of the center of mass and affects the movement of individual parts in a system. In the exercise, when both cars had equal mass, the center of mass was halfway (\(D/2\)) between them.

Altering mass distribution by making one car heavier shifts the center of mass closer to the heavier car, but not beyond it. With one car weighing three times as much as the other, the center of mass is located one-third of the total distance away from the heavier car, and two-thirds from the lighter one. Thus, mass distribution heavily influences how each car moves to maintain the center of mass at rest.

Conservation of Momentum

The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces impact it. Since the forces between the cars are internal, the momentum before and after they start moving must be conserved.

For the two cars initially at rest, their combined momentum is zero. As the cable pulls them, each car gains momentum, but in opposite directions. These opposite and equal momenta keep the center of mass fixed, showing the momentum is conserved. The movement of each car relative to the other respects this conservation, emphasizing the role and importance of internal forces within the system.