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Two bodies of masses m1 and m2( < m1) a...

Two bodies of masses `m_1 and m_2( < m_1)` are connected to the ends of a massless cord and allowed to move as shown in figure. The pulley is both massless and frictionless. The acceleration of the centre of mass is `((m_1 - m_2)^n)/((m_1 + m_2)^2) g`. Find the value of n.

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The correct Answer is:
2
If `veca` is the acceleration of`m_1` then `-veca` is the acceleration of `m_2`.
We know that `(m_1 + m_2) vecA cm. m_1veca_1 + m_2 veca_2`
or `(m_1 + m_2) vecA cm = m_1 veca - m_2veca " or "vecacm. = (m_1 - m_2)/(m_1 + m_2)veca`
But, `veca = (m -1 - m_2)/(m_1 + m_2) vecg` [connected - body problem] `:. " " vecAcm . = ((m_1 - m_2)^2)/((m_1 + m_2)^2) vecg`.
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