Question

A light string passing over a smooth light pulley connects two blocks of masses \( m_1 \) and \( m_2 \) (where \( m_2>m_1 \)). If the acceleration of the system is \( \frac{g}{\sqrt{2}} \), then the ratio of the masses \( \frac{m_1}{m_2} \) is:

• A. \( \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \)
• B. \( \frac{1 + \sqrt{5}}{\sqrt{5} - 1} \)
• C. \( \frac{1 + \sqrt{5}}{\sqrt{2} - 1} \)
• D. \( \frac{\sqrt{3} + 1}{\sqrt{2} - 1} \)

Hint:

For Atwood’s machine, the acceleration is given by: \[ a = \frac{(m_2 - m_1)}{m_1 + m_2} g \] This equation helps determine the ratio of masses when acceleration is known.

Answer 1

The Correct Option is A

Step 1: {Equation of motion for the system}
The acceleration of the system is given by: \[ a = \frac{(m_2 - m_1)}{m_1 + m_2} g \] Step 2: {Equating given acceleration}
\[ \frac{g}{\sqrt{2}} = \frac{(m_2 - m_1)}{m_1 + m_2} g \] Step 3: {Solve for \( \frac{m_1}{m_2} \)}
\[ \sqrt{2} (m_2 - m_1) = m_1 + m_2 \] Rearranging: \[ \frac{m_1}{m_2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \] Thus, the correct answer is (A) \( \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \).

Answer 2

To determine the ratio of the masses \( \frac{m_1}{m_2} \), we start by analyzing the forces acting on the two-block system connected by a string over a frictionless pulley. The system's acceleration is \( a = \frac{g}{\sqrt{2}} \). By applying Newton's second law, we consider the forces on each block:

For \( m_1 \):

\( T - m_1g = m_1a \)
\( T = m_1(g + a) \)

For \( m_2 \):

\( m_2g - T = m_2a \)
\( T = m_2(g - a) \)

Equating the expressions for \( T \):

\( m_1(g + a) = m_2(g - a) \)

Substituting \( a = \frac{g}{\sqrt{2}} \):

\( m_1\left(g + \frac{g}{\sqrt{2}}\right) = m_2\left(g - \frac{g}{\sqrt{2}}\right) \)

Simplifying gives:

\( m_1\left(1 + \frac{1}{\sqrt{2}}\right) = m_2\left(1 - \frac{1}{\sqrt{2}}\right) \)

Dividing both sides by \( g \), we obtain:

\( m_1\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right) = m_2\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \)

The ratio \( \frac{m_1}{m_2} \) becomes:

\( \frac{m_1}{m_2} = \frac{\sqrt{2}-1}{\sqrt{2}+1} \)

This matches the provided correct answer: \( \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \).