Question

Two masses m1 and m2 are connected by a spring of spring constant k and are placed on a smooth horizontal surface. Initially the spring is stretched through a distance 'd' when the system is released from rest. Find the distance moved by the two masses when spring is compressed by a distance 'd'.

Answer 1

Distance moved by m1 = $\frac{2d m_2}{m_1 + m_2}$, Distance moved by m2 = $\frac{2d m_1}{m_1 + m_2}$

Answer 2

Let's solve this problem step-by-step using the principles of conservation of momentum and conservation of the center of mass.

1. Identify the System and Forces: The system consists of two masses ($m_1$ and $m_2$) and a spring. The surface is smooth, meaning there are no external horizontal forces like friction. The spring force is an internal force within the system.

2. Conservation of Linear Momentum: Since there are no external horizontal forces acting on the system, the total linear momentum of the system must be conserved. Initially, the system is released from rest, so the initial velocity of both masses is zero. Initial momentum $P_i = m_1(0) + m_2(0) = 0$. Let $v_{1f}$ and $v_{2f}$ be the velocities of $m_1$ and $m_2$ respectively when the spring is compressed by distance 'd'. Final momentum $P_f = m_1 v_{1f} + m_2 v_{2f}$. By conservation of momentum: $P_i = P_f \implies 0 = m_1 v_{1f} + m_2 v_{2f}$.

3. Conservation of Center of Mass Position: Since the total linear momentum is conserved and is equal to zero, the velocity of the center of mass ($V_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$) is always zero. If the velocity of the center of mass is always zero, its position ($X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$) must remain constant. Let $x_{1i}$ and $x_{2i}$ be the initial positions of $m_1$ and $m_2$. Let $x_{1f}$ and $x_{2f}$ be the final positions of $m_1$ and $m_2$. The initial position of the center of mass is $X_{CM,i} = \frac{m_1 x_{1i} + m_2 x_{2i}}{m_1 + m_2}$. The final position of the center of mass is $X_{CM,f} = \frac{m_1 x_{1f} + m_2 x_{2f}}{m_1 + m_2}$. Since $X_{CM,i} = X_{CM,f}$, we have $\frac{m_1 x_{1i} + m_2 x_{2i}}{m_1 + m_2} = \frac{m_1 x_{1f} + m_2 x_{2f}}{m_1 + m_2}$. $m_1 x_{1i} + m_2 x_{2i} = m_1 x_{1f} + m_2 x_{2f}$. Rearranging this, we get $m_1 (x_{1f} - x_{1i}) + m_2 (x_{2f} - x_{2i}) = 0$. Let $\Delta x_1 = x_{1f} - x_{1i}$ be the displacement of $m_1$ and $\Delta x_2 = x_{2f} - x_{2i}$ be the displacement of $m_2$. So, $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$. This is our first equation relating the displacements.

4. Relation between Displacements and Spring Extension/Compression: Let the natural length of the spring be $L_0$. Initially, the spring is stretched by a distance 'd'. The distance between the masses is $x_{2i} - x_{1i} = L_0 + d$. Finally, the spring is compressed by a distance 'd'. The distance between the masses is $x_{2f} - x_{1f} = L_0 - d$. Consider the change in the distance between the masses: $(x_{2f} - x_{1f}) - (x_{2i} - x_{1i}) = (L_0 - d) - (L_0 + d)$ $(x_{2f} - x_{2i}) - (x_{1f} - x_{1i}) = -2d$ $\Delta x_2 - \Delta x_1 = -2d$. This is our second equation relating the displacements.

5. Solve the System of Equations: We have two linear equations for $\Delta x_1$ and $\Delta x_2$:

1. $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$
2. $\Delta x_2 - \Delta x_1 = -2d$

From equation (1), $\Delta x_2 = -\frac{m_1}{m_2} \Delta x_1$. Substitute this into equation (2): $-\frac{m_1}{m_2} \Delta x_1 - \Delta x_1 = -2d$ $-\Delta x_1 \left(\frac{m_1}{m_2} + 1\right) = -2d$ $-\Delta x_1 \left(\frac{m_1 + m_2}{m_2}\right) = -2d$ $\Delta x_1 = \frac{2d m_2}{m_1 + m_2}$.

Now find $\Delta x_2$ using $\Delta x_2 = -\frac{m_1}{m_2} \Delta x_1$: $\Delta x_2 = -\frac{m_1}{m_2} \left(\frac{2d m_2}{m_1 + m_2}\right)$ $\Delta x_2 = -\frac{2d m_1}{m_1 + m_2}$.

The distance moved by the masses are the magnitudes of their displacements. Distance moved by $m_1$ is $|\Delta x_1| = \frac{2d m_2}{m_1 + m_2}$. Distance moved by $m_2$ is $|\Delta x_2| = \frac{2d m_1}{m_1 + m_2}$.

Note that the displacements are in opposite directions, as indicated by the signs, which is consistent with the conservation of momentum (or center of mass being stationary).

Explanation: The system starts from rest, so its center of mass is initially at rest. Since there are no external forces, the center of mass remains at rest throughout the motion. The displacement of the masses relative to the center of mass are related by $m_1 \Delta x'_1 + m_2 \Delta x'_2 = 0$, and since the CM is fixed in the inertial frame, the displacements in the inertial frame are also related by $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$. The change in the relative position of the masses is the change in the spring length. The spring goes from an extension of 'd' to a compression of 'd'. If we consider $x_2 - x_1$ as the relative position, it changes from $L_0+d$ to $L_0-d$. The change in relative position is $(L_0-d) - (L_0+d) = -2d$. This change is also equal to $\Delta x_2 - \Delta x_1$. Solving the two equations $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$ and $\Delta x_2 - \Delta x_1 = -2d$ simultaneously gives the displacements $\Delta x_1 = \frac{2d m_2}{m_1 + m_2}$ and $\Delta x_2 = -\frac{2d m_1}{m_1 + m_2}$. The distances moved are the magnitudes of these displacements.