Two blocks of mass 1 kg and 2 kg are connecte | vedclass

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(A) The velocity of the centre of mass $(v_{cm})$ is given by: $v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{1 	imes 0 + 2 	imes 6}{1 + 2} = \

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(A) The velocity of the centre of mass $(v_{cm})$ is given by: $v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{1 	imes 0 + 2 	imes 6}{1 + 2} = \frac{12}{3} = 4\ m/s$.In the centre of mass frame,the total energy is conserved. The reduced mass is $\mu = \frac{m_1m_2}{m_1 + m_2} = \frac{1 	imes 2}{1 + 2} = \frac{2}{3}\ kg$.The kinetic energy in the centre of mass frame is $K_{cm} = \frac{1}{2} \mu v_{rel}^2$,where $v_{rel} = v_2 - v_1 = 6 - 0 = 6\ m/s$.$K_{cm} = \frac{1}{2} 	imes \frac{2}{3} 	imes (6)^2 = 12\ J$.At the maximum velocity of the $1\ kg$ block,the spring is at its natural length again,and the relative velocity is reversed $(v_{rel} = -6\ m/s)$.The velocity of the $1\ kg$ block in the ground frame is $v_1 = v_{cm} + v_{1,cm}$.In the centre of mass frame,$m_1v_{1,cm} + m_2v_{2,cm} = 0$,so $v_{1,cm} = -\frac{m_2}{m_1+m_2} v_{rel} = -\frac{2}{3} 	imes (-6) = 4\ m/s$.Thus,$v_{1,max} = v_{cm} + v_{1,cm} = 4 + 4 = 8\ m/s$.

Two blocks of mass $1\ kg$ and $2\ kg$ are connected with each other by a spring of spring constant $60\ N/m$. If the velocity given to the block of mass $2\ kg$ at the natural length is $6\ m/s$,find the maximum velocity of the $1\ kg$ block in the subsequent motion (in $m/s$).

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