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In the arrangement shown in figure, mA=2...

In the arrangement shown in figure, `m_A=2kg` and `m_B=1kg`. String is light and inextensible. Find the acceleration of centre of mass of both the blocks. Neglect friction everywhere.

Text Solution

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Net pulling force on the system is `(m_(A) - m_(B))g` or `(2-1)g=g`
Total mass being pulled is `m_(A) + m_(B)` or 3 kg

`therefore a = ("Net pulling force")/("Total mass") =g/3`= acceleration of each block.
Now, `veca_(CM) =(m_(A)veca_(A) + m_(B)veca_(B))/(m_(A) + m_(B)) =((2)(a) -(1)(a))/(1+2) =a/3 = g/9` downwards.
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