Question

In the system shown in figure $M₁ > M₂$ and pulley and threads are ideal. System is held at rest by thread BC. Just after thread BC is burnt.

• A. Acceleration $M₁$ and $M₂$ will be upward
• B. Magnitude of acceleration of both masses will be $\fracM₁-M₂M₁+M₂g$
• C. Acceleration of $M₁$ and $M₂$ will be equal to zero
• D. Acceleration of $M₁$ will be equal to zero, while that of $M₂$ will be $\fracM₁-M₂M₂g$ upward

Hint:

For an Atwood machine, $a = \fracM₁-M₂M₁+M₂g$.

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the forces and motions involved immediately after the thread BC is burnt. Since we have an ideal pulley and thread, the system behaves as an Atwood machine. 

Given: \(M_1 > M_2\)

1. Free Body Diagram Analysis:

• \(M_1\) will have a gravitational force pulling it downward: \(M_1g\)
• \(M_2\) will have a gravitational force pulling it downward: \(M_2g\)
• The tension in the string is the same on both sides due to the ideal pulley.

2. Equations of Motion:

• For \(M_1\): \(T - M_1g = M_1 a\) (Equation 1)
• For \(M_2\): \(M_2g - T = M_2 a\) (Equation 2)

3. Solving the Equations:

Add Equation 1 and Equation 2 to eliminate tension (T):

\(M_2g - M_1g = (M_1 + M_2)a\)

Rearrange to solve for acceleration (a):

\(a = \frac{M_1 - M_2}{M_1 + M_2}g\)

This shows that the magnitude of the acceleration for both masses is \(\frac{M_1 - M_2}{M_1 + M_2}g\), which matches the given correct answer.

The direction of acceleration:

• \(M_1\) moves downward because it is heavier.
• \(M_2\) moves upward.

Hence, the correct answer is that the magnitude of acceleration of both masses will be \(\frac{M_1 - M_2}{M_1 + M_2}g\).