System of Particles and Rotational Motion NEET PYQ

(2)

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Masses, $m_{1} = 10 kg$ and $m_{2} = 20 kg$

Let the distance between the two masses be $d = 10 m$ .

Now, centre of masses for two particle system is written as,

$X_{COM} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$ ...(i)

Considering point A as origin, so $B = (d, 0)$

$X_{COM} = (\frac{m_{1} \times 0 + m_{2} d}{m_{1} + m_{2}}) = \frac{m_{2} d}{m_{1} + m_{2}}$

$X_{COM} = \frac{20 kg \times 10 m}{10 kg + 20 kg} = \frac{20}{3} m$

(3)

Given : $m_{1} = 5 kg,$ $m_{2} = 10 kg$ and $L = 1 m$

Here centre of mass, $X_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$

$= \frac{5 \times 0 + 10 \times 1}{15} = \frac{10}{15} = \frac{2}{3} = 0.666 m; 0.67 m = 67 cm$

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