Question

Two blocks of masses in the ratio \( m:n \) are connected by a light inextensible string passing over a frictionless fixed pulley. If the system of the blocks is released from rest, then the acceleration of the centre of mass of the system of the blocks is (g = acceleration due to gravity)

• A. \( \left(\frac{m+n}{m-n}\right)^2 g \)
• B. \( \left(\frac{m-n}{m+n}\right)^2 g \)
• C. \( \left(\frac{m+n}{m-n}\right) g \)
• D. \( \left(\frac{m-n}{m+n}\right) g \)

Hint:

Acceleration of block in Atwood machine: \( a = \frac{\Delta m}{\sum m} g \). Acceleration of Center of Mass: \( a_{cm} = \left( \frac{\Delta m}{\sum m} \right) a = \left( \frac{\Delta m}{\sum m} \right)^2 g \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We have an Atwood machine with masses \( M_1 \) and \( M_2 \) where \( M_1/M_2 = m/n \). We need to find the acceleration of the center of mass \( a_{cm} \). The individual acceleration of the blocks is \( a \). \( a_{cm} = \frac{M_1 a_1 + M_2 a_2}{M_1 + M_2} \).
Step 2: Key Formula or Approach:

1. Acceleration of blocks: \( a = \frac{|M_1 - M_2|}{M_1 + M_2} g \). 2. Acceleration of Center of Mass: Since one moves up and one down, \( a_1 = a \) and \( a_2 = -a \) (or vice versa). \( a_{cm} = \frac{M_1(a) + M_2(-a)}{M_1 + M_2} = a \frac{M_1 - M_2}{M_1 + M_2} \).
Step 3: Detailed Explanation:

Let the masses be \( m \) and \( n \) (proportionality constants cancel out in ratios). Acceleration of the blocks magnitude: \[ a = \frac{m - n}{m + n} g \] (Assuming \( m \textgreater n \), direction is towards \( m \)). Vector acceleration of Center of Mass: \[ \vec{a}_{cm} = \frac{m \vec{a}_m + n \vec{a}_n}{m + n} \] If \( m \) moves down, \( \vec{a}_m = a \hat{j} \) (down). \( n \) moves up, \( \vec{a}_n = -a \hat{j} \) (up). Wait, let's set down as positive. \( \vec{a}_m = a \). \( \vec{a}_n = -a \). \[ a_{cm} = \frac{m(a) + n(-a)}{m + n} = a \frac{m - n}{m + n} \] Substitute \( a = \frac{m - n}{m + n} g \): \[ a_{cm} = \left( \frac{m - n}{m + n} g \right) \left( \frac{m - n}{m + n} \right) \] \[ a_{cm} = \left( \frac{m - n}{m + n} \right)^2 g \]
Step 4: Final Answer:

The acceleration of the centre of mass is \( \left(\frac{m-n}{m+n}\right)^2 g \).