Two blocks of masses $2 k g$ and $1 k g$ respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in $2 s$ is :
$(g = 10 m / s^{2})$

1.42 m

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2.22 m

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3.12 m

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3.33 m

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Solution

The correct option is B $2.22 m$
The masses of two blocks are $2 k g$ and $1 k g$ , the time is $2 sec$ .

The acceleration of masses is given as,

$a = \frac{(m_{2} - m_{1}) g}{(m_{2} + m_{1})}$

$= \frac{(2 - 1) g}{(2 + 1)}$

$= \frac{g}{3}$

The acceleration of the centre of mass is given as,

$a_{c m} = \frac{(m_{2} - m_{1}) \times \frac{g}{3}}{(m_{2} + m_{1})}$

$= \frac{(2 - 1) \times \frac{g}{3}}{(2 + 1)}$

$= \frac{g}{9}$

The distance travelled by centre of mass is given as,

$s = \frac{1}{2} a t^{2}$

$= \frac{1}{2} \times \frac{g}{9} \times 4$

$= 2.22 m$

Thus, the distance travelled by the centre of mass is $2.22 m$ .

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Two blocks of masses 2kg and 1kg respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 2s is :(g=10m/s2)

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$(g = 10 m / s^{2})$