Two blocks of masses in the ratio m:n are con | vedclass

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(B) Let the masses of the two blocks be $M_1 = m$ and $M_2 = n$. Assume $m > n$.The acceleration $a$ of the blocks in an Atwood machine is given by $a

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$\left(\frac{m-n}{m+n}\right)^2 g$

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(B) Let the masses of the two blocks be $M_1 = m$ and $M_2 = n$. Assume $m > n$.The acceleration $a$ of the blocks in an Atwood machine is given by $a = \frac{|M_1 - M_2|}{M_1 + M_2} g = \frac{m-n}{m+n} g$.The acceleration of block $1$ is $a_1 = a$ (downwards) and the acceleration of block $2$ is $a_2 = a$ (upwards).Taking the downward direction as positive,the acceleration of the centre of mass $a_{cm}$ is given by:$a_{cm} = \frac{M_1 a_1 + M_2 a_2}{M_1 + M_2} = \frac{m(a) + n(-a)}{m+n} = \frac{(m-n)a}{m+n}$.Substituting the value of $a$:$a_{cm} = \frac{(m-n)}{m+n} \cdot \left( \frac{m-n}{m+n} g \right) = \left( \frac{m-n}{m+n} \right)^2 g$.

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).

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