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Standard XII

Physics

Distance

Two blocks of...

Question

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 frictioless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by 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 $2.22 m$
$20 - T = 2 a$
$T - 10 = 19 - ----------- -$
$10 = 3 a$
Distance travelled by both man in $2 s e c$
$\Rightarrow S = \frac{1}{2} a t^{2}$

$= \frac{1}{2} \times \frac{10}{3} \times 2 \times 2$

$= \frac{20}{3} m$
Distance a position of COM from initial $= \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m - 2}$
$= \frac{2 \times \frac{20}{3} + 1 \times (\frac{- 20}{3})}{3}$
$= \frac{20}{9} m$
$= 2.22 m$
Hence, the answer 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 frictioless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by centre of mass in 2s is (g=10m/s2)

The correct option is B 2.22m20−T=2aT−10=19–––––––––––––10=3aDistance travelled by both man in 2sec⇒S=12at2 =12×103×2×2 =203mDistance a position of COM from initial =m1x1+m2x2m1+m−2 =2×203+1×(−203)3 =209m =2.22mHence, the answer is 2.22m.

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 frictioless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by centre of mass in $2 s$ is $(g = 10 m / s^{2})$