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(B) The system consists of two blocks $A$ and $B$ with masses $m_A = 2\,kg$ and $m_B = 3\,kg$ connected by a spring on a smooth horizontal floor.Since

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(B) The system consists of two blocks $A$ and $B$ with masses $m_A = 2\,kg$ and $m_B = 3\,kg$ connected by a spring on a smooth horizontal floor.Since the floor is smooth,there is no external horizontal force acting on the system.According to the law of conservation of linear momentum,the velocity of the centre of mass $(v_{cm})$ remains constant if the net external force on the system is zero.The initial velocity of block $A$ is $v_A = 10\,m/s$ and block $B$ is at rest,$v_B = 0\,m/s$.The velocity of the centre of mass is given by:$v_{cm} = \frac{m_A v_A + m_B v_B}{m_A + m_B}$$v_{cm} = \frac{2 	imes 10 + 3 	imes 0}{2 + 3} = \frac{20}{5} = 4\,m/s$Since the net external force is zero,the velocity of the centre of mass will remain $4\,m/s$ at any time $t$,including at $t = 10\,s$.

In the given figure,if the speed given to block $A$ is $10\,m/s$ and block $B$ is initially at rest,then the speed of the centre of mass after $10\,s$ will be ....... $m/s$.

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