Two blocks of masses m and 2m are connected b | vedclass

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(A) The acceleration $a$ of the system is given by:$a = \left( \frac{m_2 - m_1}{m_1 + m_2} ight) g = \left( \frac{2m - m}{2m + m} ight) 	imes 10

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$6$

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(A) The acceleration $a$ of the system is given by:$a = \left( \frac{m_2 - m_1}{m_1 + m_2} ight) g = \left( \frac{2m - m}{2m + m} ight) 	imes 10 = \frac{10}{3} \ ms^{-2}$After time $t = 5.4 \ s$,the speed of each block is:$v = at = \frac{10}{3} 	imes 5.4 = 18 \ ms^{-1}$The block of mass $m$ moves upwards with velocity $v_1 = 18 \ ms^{-1}$ (let this be positive direction),and the block of mass $2m$ moves downwards with velocity $v_2 = -18 \ ms^{-1}$.The velocity of the centre of mass is:$v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{m(18) + 2m(-18)}{m + 2m} = \frac{18m - 36m}{3m} = \frac{-18m}{3m} = -6 \ ms^{-1}$The speed is the magnitude of the velocity,which is $6 \ ms^{-1}$.

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