Four point masses $m, 2 m, M_{1}$ and $M_{2}$ are placed on a plane as shown in figure. If the co-ordinates of the centre of mass of the system be $(\frac{6}{5}, \frac{7}{10})$ then the values of $M_{1}$ and $M_{2}$ are respectively.

3 m

and

4 m

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4 m

and

3 m

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\frac{29}{4} m

and

\frac{43}{4} m

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\frac{29}{4} m

and

6 m

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Solution

The correct option is B $4 m$ and $3 m$
Given
$m_{1} = m$
$m_{2} = 2 m$
$m_{3} = M_{1}$
$m_{4} = M_{2}$

The center of mass of the system is ,

${\to R}_{c m} = \frac{m_{1} {\to r}_{1} + m_{2} {\to r}_{2} + m_{3} {\to r}_{3} + m_{4} {\to r}_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$ . . . . .(1)

The coordinate of the center of mass of the system $(\frac{6}{7}, \frac{7}{10})$

From the diagram given above.
Along x-direction, put all value in equation (1), we get

$X_{c m} = \frac{4 m + 2 M_{1}}{3 m + M_{1} + M_{2}} = \frac{6}{7}$

$2 M_{1} = 4 m + 3 M_{2}$ . . . . . .(2)

Along y_axis,

$Y_{c m} = \frac{M_{1} + M_{2}}{3 m + M_{1} + M_{2}} = \frac{7}{10}$

$M_{1} + M_{2} = 7 m$

$M_{1} = 7 m - M_{2}$ . . . .(3)

Put equation (3) in (2), we get,

$M_{2} = 2 m$

Now, from equation (3)

$M_{1} = 5 m$

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