Two blocks A and B of masses 1, kg and 2, kg | vedclass

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(B) Since there are no external horizontal forces acting on the system,the linear momentum of the system is conserved.Initially,the system is at rest,

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$2$

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(B) Since there are no external horizontal forces acting on the system,the linear momentum of the system is conserved.Initially,the system is at rest,so the total momentum is $0$.When the spring is released,the blocks move in opposite directions with momenta $p_A$ and $p_B$ such that $p_A = p_B = p$.The kinetic energy $(KE)$ of a body is given by $KE = \frac{p^2}{2m}$.Therefore,the ratio of the kinetic energies is $\frac{KE_A}{KE_B} = \frac{p^2 / (2m_A)}{p^2 / (2m_B)} = \frac{m_B}{m_A}$.Given $m_A = 1\, kg$ and $m_B = 2\, kg$,we have $\frac{KE_A}{KE_B} = \frac{2}{1} = 2$.

Two blocks $A$ and $B$ of masses $1\, kg$ and $2\, kg$ are connected together by a spring and are resting on a horizontal surface. The blocks are pulled apart so as to stretch the spring and then released. The ratio of $K.E.$ of block $A$ to the $K.E.$ of block $B$ is:

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