Two masses of 1 kg and 2 kg are hanged from | vedclass

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(A) The acceleration of the system is given by $a = \left(\frac{m_2 - m_1}{m_1 + m_2}ight)g = \left(\frac{2 - 1}{2 + 1}ight) 	imes 10 = \frac{10}

Vedclass · Accepted Answer

$\frac{20}{9} \ m$

Vedclass · Answer

(A) The acceleration of the system is given by $a = \left(\frac{m_2 - m_1}{m_1 + m_2}ight)g = \left(\frac{2 - 1}{2 + 1}ight) 	imes 10 = \frac{10}{3} \ m/s^2$.Let the upward direction be positive. The acceleration of the $1 \ kg$ mass is $a_1 = +\frac{10}{3} \ m/s^2$ and the acceleration of the $2 \ kg$ mass is $a_2 = -\frac{10}{3} \ m/s^2$.The acceleration of the centre of mass is $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$.Substituting the values,$a_{cm} = \frac{1(\frac{10}{3}) + 2(-\frac{10}{3})}{1 + 2} = \frac{\frac{10}{3} - \frac{20}{3}}{3} = \frac{-10/3}{3} = -\frac{10}{9} \ m/s^2$.The magnitude of acceleration of the centre of mass is $|a_{cm}| = \frac{10}{9} \ m/s^2$.The distance travelled by the centre of mass in $t = 2 \ s$ starting from rest is $S_{cm} = \frac{1}{2} |a_{cm}| t^2$.$S_{cm} = \frac{1}{2} 	imes \frac{10}{9} 	imes (2)^2 = \frac{1}{2} 	imes \frac{10}{9} 	imes 4 = \frac{20}{9} \ m$.

Two masses of $1 \ kg$ and $2 \ kg$ are hanged from a pulley as shown in the figure. If the system starts from rest,find the distance travelled by the centre of mass in the first two seconds.

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