All surfaces shown in the figure are smooth. The system is released from rest. The x and y components of acceleration of COM are

(a_{c m})_{x} = \frac{m_{1} m_{2} g}{m_{1} + m_{2}}

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(a_{c m})_{x} = \frac{m_{1} m_{2} g}{(m_{1} + m_{2})^{2}}

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(a_{c m})_{y} = \frac{m_{2}^{2} g}{(m_{1} + m_{2})}

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(a_{c m})_{y} = (\frac{m_{2}}{m_{1} + m_{2}}) g

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Solution

The correct option is A $(a_{c m})_{x} = \frac{m_{1} m_{2} g}{(m_{1} + m_{2})^{2}}$
$m_{1} g$ acting on $m_{1}$ is balanced by the normal reaction $N_{1}$ from the table.
Net force acting on the system ( $m_{1}$ and $m_{2}$ together) = $m_{2} g$ ;

Net acceleration=a= $\frac{n e t f o r c e}{t o t a l m a s s} = \frac{m_{2} g}{(m_{1} + m_{2})}$ ;

Let acceleration of $m_{1}$ = $a_{1}$ ;
Let x and y components of $a_{1} = (a_{1 x}, a_{1 y})$ ;
Let x and y components of $a_{2} = (a_{2 x}, a_{2 y})$ ;
$a_{1 x} = a$ ; since $m_{1}$ moves in y direction.
$a_{1 y} = 0$ ; since $m_{1}$ does not move in y direction.
$a_{2 x} = 0$ ; since $m_{2}$ does not move in x direction.
$a_{2 y} = a$ ; since $m_{2}$ moves in y direction only.

$(a_{c m})_{x} = \frac{m_{1} a_{1 x} + m_{2} a_{2 x}}{m_{1} + m_{2}} = \frac{m_{1} a_{1 x}}{m_{1} + m_{2}} = \frac{m_{1} m_{2} g}{(m_{1} + m_{2})^{2}}$ ; since $a_{2 x} = 0$ ;

$(a_{c m})_{y} = \frac{m_{1} a_{1 x} y + m_{2} a_{2 y}}{m_{1} + m_{2}} = \frac{m_{2} a_{2 y}}{m_{1} + m_{2}} = \frac{m_{2} m_{2} g}{(m_{1} + m_{2})^{2}} = \frac{m_{2}^{2} g}{(m_{1} + m_{2})^{2}}$ ; since $a_{1 y} = 0$ ;

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