Two masses, \(M_1\) and \(M_2\), are connected through a massless spring of spring constant k, as shown in the figure below. The mass \(M_1\) is at rest against a rigid wall. Both \(M_1\) and \(M_2\) are on a frictionless surface. The mass \(M_2\) is pushed towards \(M_1\) by a distance x from its equilibrium position and then released. After \(M_1\) leaves the wall, the speed of the center of mass of the composite system is
Show Hint
In problems where a system's constraints change (like a mass leaving a wall), the velocity of the center of mass is conserved only *after* the external constraint force disappears. Calculate the state of the system (velocities of all parts) at the very instant the constraint is removed, and use that to find the subsequent constant velocity of the center of mass.
Step 1: Understanding the Concept:
This problem involves two stages. In the first stage, mass \(M_2\) oscillates while \(M_1\) is held stationary by the wall. In the second stage, after \(M_1\) leaves the wall, the two-mass system moves freely. The key is to identify the state of the system at the precise moment \(M_1\) leaves the wall. From that moment on, the total momentum of the \(M_1-M_2\) system is conserved, and the velocity of the center of mass becomes constant. Step 2: Key Formula or Approach:
1. Conservation of Energy for the first stage to find the velocity of \(M_2\).
2. Definition of Velocity of Center of Mass: \(v_{cm} = \frac{M_1v_1 + M_2v_2}{M_1 + M_2}\).
3. Condition for leaving the wall: Mass \(M_1\) will leave the wall when the spring is no longer pushing on it, i.e., when the spring reaches its natural, uncompressed length. Step 3: Detailed Explanation:
1. Find the state of the system when \(M_1\) leaves the wall:
The mass \(M_2\) is initially released from a position where the spring is compressed by a distance \(x\). The initial potential energy stored in the spring is \(U_i = \frac{1}{2}kx^2\).
\(M_2\) will accelerate to the right. Mass \(M_1\) remains against the wall as long as the spring is compressed, because the spring pushes on it, and the wall provides an opposing normal force.
\(M_1\) will leave the wall at the exact moment the spring reaches its natural length. At this instant, the force from the spring on \(M_1\) becomes zero.
At this moment, all the initial potential energy has been converted into the kinetic energy of mass \(M_2\) (since \(M_1\) is still momentarily at rest).
2. Calculate the velocity of \(M_2\) at this instant:
Using conservation of energy for the system of \(M_2\) and the spring:
\[ E_{\text{initial}} = E_{\text{final}} \]
\[ \frac{1}{2}kx^2 + \frac{1}{2}M_2(0)^2 = \frac{1}{2}k(0)^2 + \frac{1}{2}M_2v_2^2 \]
\[ \frac{1}{2}kx^2 = \frac{1}{2}M_2v_2^2 \]
\[ v_2 = \sqrt{\frac{k}{M_2}}x \]
At this same instant, the velocity of mass \(M_1\) is \(v_1 = 0\).
3. Calculate the velocity of the center of mass:
After this instant, there are no external horizontal forces acting on the \(M_1-M_2\) system (the force from the wall is now zero). Therefore, the velocity of the center of mass will remain constant from this point forward. We calculate this constant velocity at the moment \(M_1\) leaves the wall.
\[ v_{cm} = \frac{M_1v_1 + M_2v_2}{M_1 + M_2} \]
Substitute the velocities we found:
\[ v_{cm} = \frac{M_1(0) + M_2 \left(\sqrt{\frac{k}{M_2}}x\right)}{M_1 + M_2} \]
\[ v_{cm} = \frac{M_2 \sqrt{k/M_2}}{M_1 + M_2}x = \frac{\sqrt{M_2^2 \cdot k/M_2}}{M_1 + M_2}x \]
\[ v_{cm} = \frac{\sqrt{kM_2}}{M_1 + M_2}x \]
Step 4: Final Answer:
The speed of the center of mass of the composite system after \(M_1\) leaves the wall is \(\frac{\sqrt{kM_2}}{M_1 + M_2}x\). This matches option (C).
