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(B) The acceleration $a$ of a system of two masses $m_1$ and $m_2$ connected by a string over a pulley is given by the formula: $a = \left( \frac{m_2

Vedclass · Accepted Answer

$9 : 7$

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(B) The acceleration $a$ of a system of two masses $m_1$ and $m_2$ connected by a string over a pulley is given by the formula: $a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$,assuming $m_2 > m_1$.Given that the acceleration $a = \frac{g}{8}$,we substitute this into the equation:$\frac{g}{8} = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$Canceling $g$ from both sides:$\frac{1}{8} = \frac{m_2 - m_1}{m_1 + m_2}$Cross-multiplying gives:$m_1 + m_2 = 8(m_2 - m_1)$$m_1 + m_2 = 8m_2 - 8m_1$Rearranging the terms to group $m_1$ and $m_2$:$m_1 + 8m_1 = 8m_2 - m_2$$9m_1 = 7m_2$Therefore,the ratio of the masses is:$\frac{m_2}{m_1} = \frac{9}{7}$

$A$ light string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$ (vertically). If the acceleration of the system is $\frac{g}{8}$,then the ratio of masses $m_2 : m_1$ is:

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