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(B) For a system of two masses connected by a string over a pulley, the acceleration of each mass is given by $a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}

Vedclass · Accepted Answer

$\frac{(m_{1} - m_{2})^{2}}{m_{1} + m_{2}} g$

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(B) For a system of two masses connected by a string over a pulley, the acceleration of each mass is given by $a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} g$.Since the two masses move in opposite directions, the acceleration of the center of mass $(a_{CM})$ is calculated as:$a_{CM} = \frac{m_{1}a_{1} + m_{2}a_{2}}{m_{1} + m_{2}}$.Taking the direction of $m_{1}$ as positive, $a_{1} = a$ and $a_{2} = -a$.$a_{CM} = \frac{m_{1}a - m_{2}a}{m_{1} + m_{2}} = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}ight) a$.Substituting the value of $a$:$a_{CM} = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}ight) 	imes \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}ight) g = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}ight)^{2} g$.The total external force acting on the system is $F_{ext} = (m_{1} + m_{2}) a_{CM}$.$F_{ext} = (m_{1} + m_{2}) 	imes \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}ight)^{2} g = \frac{(m_{1} - m_{2})^{2}}{m_{1} + m_{2}} g$.

$A$ smooth massless string passes over a smooth fixed pulley. Two masses $m_{1}$ and $m_{2}$ $(m_{1} > m_{2})$ are tied at the two ends of the string. The masses are allowed to move under gravity starting from rest. The total external force acting on the two masses is

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