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

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(B) According to the law of conservation of linear momentum,since the initial momentum of the system is zero,the magnitudes of the momenta of the two

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$2$

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(B) According to the law of conservation of linear momentum,since the initial momentum of the system is zero,the magnitudes of the momenta of the two blocks must be equal after release: $p_A = p_B = p$.The kinetic energy $(KE)$ of a body is related to its momentum $(p)$ and mass $(m)$ by the formula: $KE = \frac{p^2}{2m}$.Therefore,the ratio of the kinetic energies of block $A$ and block $B$ 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 $K.E.s$ of both the blocks is

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