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A particle of mass $200\;g$ is dropped from a height of $50\;m$ and another particle of mass $100\; g$ is simultaneously projected up from the ground along the same line with a speed of $100\; m/s$. The acceleration of the centre of mass after $1\; s$ is

\[\begin {array} {1 1} (a)\;10\;m/s & \quad (b)\;\frac{10}{3}\;m/s^2 \\ (c)\;0 & \quad  (d)\;none\;of\;these \end {array}\]

Can you answer this question?
 
 

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$\bar {a}_{cm}=\large\frac{m_1\bar{a}_1+m_2 \bar {a}_2}{m_1+m_2}$
$\qquad=\large\frac{(m_1+m_2)g}{(m_1+m_2)}$
$\qquad=g$
answered Nov 22, 2013 by meena.p
 

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