Question

The figure below is the TOP-VIEW of two equal masses 'm' that can slide over frictionless rails on a horizontal floor. The rails are separated by distance $l$ as shown. The masses are connected with a massless spring of natural length $l$ and spring constant $K$. At the initial moment of time the masses are at rest.

• A. The maximum magnitude of relative velocity between the two masses is given by $l(\frac{1-cos\theta}{cos\theta})\sqrt{\frac{2K}{m}}$
• B. The maximum magnitude of relative velocity between the two masses is given by $l(\frac{1-cos\theta}{sin\theta})\sqrt{\frac{K}{2m}}$
• C. As the spring comes to its natural length it is cut suddenly from the center. The speed of lower block as shown in figure just after the spring is cut will be $\frac{l}{4}\sqrt{\frac{K}{2m}}$ (Take $\theta=37^\circ$)

Answer 1

Answer: (A), (C)

A, C

Answer 2

The problem describes two equal masses 'm' on frictionless horizontal rails, separated by a vertical distance 'l'. They are connected by a massless spring of natural length 'l' and spring constant 'K'. The system is released from rest from an initial configuration where the spring makes an angle $\theta$ with the vertical.

1. Initial State Analysis:

• From the figure, the vertical separation between the rails is $l$.
• Let the initial length of the spring be $L_{initial}$. The angle $\theta$ is between the spring and the vertical line.
• Using trigonometry, $\cos\theta = \frac{\text{vertical separation}}{\text{spring length}} = \frac{l}{L_{initial}}$.
• Therefore, the initial length of the spring is $L_{initial} = \frac{l}{\cos\theta}$.
• The natural length of the spring is $l_0 = l$.
• The initial extension of the spring is $\Delta L_{initial} = L_{initial} - l_0 = \frac{l}{\cos\theta} - l = l \left(\frac{1}{\cos\theta} - 1\right) = l \left(\frac{1 - \cos\theta}{\cos\theta}\right)$.
• Since the masses are initially at rest, the initial kinetic energy $K_{initial} = 0$.
• The initial potential energy stored in the spring is $U_{initial} = \frac{1}{2} K (\Delta L_{initial})^2 = \frac{1}{2} K \left[ l \left(\frac{1 - \cos\theta}{\cos\theta}\right) \right]^2$.

2. State of Maximum Speed (Minimum Potential Energy):

• The masses will move towards each other. Due to the absence of external horizontal forces and equal masses, the center of mass remains stationary in the horizontal direction. This implies the masses will move with equal speeds in opposite directions.
• The speeds of the masses will be maximum when the potential energy stored in the spring is minimum. This occurs when the spring is at its natural length ($l_0 = l$).
• At this point, the horizontal separation between the masses becomes zero, and the spring is vertical.
• The final potential energy $U_{final} = 0$.
• Let the maximum speed of each mass be $v_{max}$.
• The total kinetic energy at this point is $K_{final} = \frac{1}{2} m v_{max}^2 + \frac{1}{2} m v_{max}^2 = m v_{max}^2$.

3. Conservation of Mechanical Energy:

• Applying the principle of conservation of mechanical energy ($U_{initial} + K_{initial} = U_{final} + K_{final}$): $\frac{1}{2} K \left[ l \left(\frac{1 - \cos\theta}{\cos\theta}\right) \right]^2 + 0 = 0 + m v_{max}^2$ $v_{max}^2 = \frac{K}{2m} l^2 \left(\frac{1 - \cos\theta}{\cos\theta}\right)^2$ $v_{max} = l \left(\frac{1 - \cos\theta}{\cos\theta}\right) \sqrt{\frac{K}{2m}}$

4. Evaluating Options (A) and (B) - Maximum Relative Velocity:

• The relative velocity between the two masses is $v_{rel} = v_{upper} - v_{lower}$. If the upper block moves left ($+v_{max}$) and the lower block moves right ($-v_{max}$), then the magnitude of relative velocity is $|v_{max} - (-v_{max})| = 2v_{max}$.
• The maximum magnitude of relative velocity is: $(v_{rel})_{max} = 2 v_{max} = 2 \cdot l \left(\frac{1 - \cos\theta}{\cos\theta}\right) \sqrt{\frac{K}{2m}}$ $(v_{rel})_{max} = l \left(\frac{1 - \cos\theta}{\cos\theta}\right) \sqrt{\frac{4K}{2m}}$ $(v_{rel})_{max} = l \left(\frac{1 - \cos\theta}{\cos\theta}\right) \sqrt{\frac{2K}{m}}$
• Comparing this with the given options:
  • (A) $l(\frac{1-cos\theta}{cos\theta})\sqrt{\frac{2K}{m}}$ - This matches our derived expression.
  • (B) $l(\frac{1-cos\theta}{sin\theta})\sqrt{\frac{K}{2m}}$ - This does not match.

5. Evaluating Option (C) - Speed of lower block after spring is cut:

• The problem states that the spring is cut suddenly from the center "As the spring comes to its natural length". This means the cutting occurs at the instant when the blocks have their maximum speeds ($v_{max}$).
• Cutting the spring instantaneously removes the spring force but does not change the velocities of the blocks instantaneously.
• Therefore, the speed of the lower block just after the spring is cut will be $v_{max}$.
• We need to calculate this speed for $\theta = 37^\circ$.
• We know $\cos(37^\circ) \approx 0.8$ (or $4/5$).
• Speed of lower block $= v_{max} = l \left(\frac{1 - \cos(37^\circ)}{\cos(37^\circ)}\right) \sqrt{\frac{K}{2m}}$ $v_{max} = l \left(\frac{1 - 0.8}{0.8}\right) \sqrt{\frac{K}{2m}}$ $v_{max} = l \left(\frac{0.2}{0.8}\right) \sqrt{\frac{K}{2m}}$ $v_{max} = l \left(\frac{1}{4}\right) \sqrt{\frac{K}{2m}}$ $v_{max} = \frac{l}{4} \sqrt{\frac{K}{2m}}$
• Comparing this with option (C): $\frac{l}{4}\sqrt{\frac{K}{2m}}$ - This matches our derived expression.

Both options (A) and (C) are correct statements based on the given information and standard physics principles.