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Standard XII

Physics

Applying NLM along 3 Axis

Two blocks of...

Question

Two blocks of masses $2 k g$ and $1 k g$ respectively are tied to the ends of a strings 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 centre of mass in $2$ seconds 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 $2.22 m$
Given,

Two blocks having masses $2 k g, 1 k g, t = 2 s$

So, $a = \frac{(2 - 1) m g}{(2 + 1) m} = \frac{g}{3} a_{C M} = \frac{2 \times \frac{g}{3} - 1 \frac{g}{3}}{2 + 1} = \frac{g}{9}$

So, $S = u t + \frac{1}{2} a t^{2} = \frac{1}{2} \times \frac{g}{2} \times 4 = \frac{20}{9} = 2.22 m$

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Two blocks of masses 2kg and 1kg respectively are tied to the ends of a strings 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 centre of mass in 2 seconds is:(g=10m/s2)

The correct option is B 2.22mGiven,Two blocks having masses 2kg,1kg,t=2sSo, a=(2−1)mg(2+1)m=g3aCM=2×g3−1g32+1=g9So, S=ut+12at2=12×g2×4=209=2.22m

Two blocks of masses $2 k g$ and $1 k g$ respectively are tied to the ends of a strings 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 centre of mass in $2$ seconds is:( $g = 10 m / s^{2}$ )