Two blocks of masses 2 text{ kg} and 1 text{ | vedclass

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(C) Let the masses be $m_1 = 2 	ext{ kg}$ and $m_2 = 1 	ext{ kg}$.The acceleration of the system is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \

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$2.22$

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(C) Let the masses be $m_1 = 2 	ext{ kg}$ and $m_2 = 1 	ext{ kg}$.The acceleration of the system is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(2 - 1) 	imes 10}{2 + 1} = \frac{10}{3} 	ext{ m/s}^2$.The $2 	ext{ kg}$ block moves downwards with acceleration $a_1 = 10/3 	ext{ m/s}^2$ and the $1 	ext{ kg}$ block moves upwards with acceleration $a_2 = 10/3 	ext{ m/s}^2$.Taking the downward direction as positive, the acceleration of the center of mass is $a_{cm} = \frac{m_1 a_1 - m_2 a_2}{m_1 + m_2} = \frac{2(10/3) - 1(10/3)}{2 + 1} = \frac{10/3}{3} = 10/9 	ext{ m/s}^2$.The distance traversed by the center of mass in $t = 2 	ext{ s}$ is $d = \frac{1}{2} a_{cm} t^2 = \frac{1}{2} 	imes \frac{10}{9} 	imes (2)^2 = \frac{1}{2} 	imes \frac{10}{9} 	imes 4 = \frac{20}{9} \approx 2.22 	ext{ m}$.

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