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(C) According to the law of conservation of linear momentum,since the initial momentum is zero,the magnitudes of the momenta of the two pieces must be

Vedclass · Accepted Answer

$\frac{3E}{5}$

Vedclass · Answer

(C) According to the law of conservation of linear momentum,since the initial momentum is zero,the magnitudes of the momenta of the two pieces must be equal: $p_1 = p_2 = p$.The kinetic energy $K$ of a body is related to its momentum $p$ and mass $m$ by the formula $K = \frac{p^2}{2m}$.Since $p$ is the same for both pieces,the kinetic energy is inversely proportional to the mass: $K \propto \frac{1}{m}$.Let $E_1$ be the kinetic energy of the piece with mass $m_1 = 2M$ and $E_2$ be the kinetic energy of the piece with mass $m_2 = 3M$.Then,$\frac{E_1}{E_2} = \frac{m_2}{m_1} = \frac{3M}{2M} = \frac{3}{2}$.Given the total kinetic energy $E = E_1 + E_2$,we can write:$E_1 = \left( \frac{m_2}{m_1 + m_2} \right) E = \left( \frac{3M}{2M + 3M} \right) E = \left( \frac{3M}{5M} \right) E = \frac{3E}{5}$.

$A$ body initially at rest breaks up into two pieces of masses $2 M$ and $3 M$ respectively,together having a total kinetic energy $E$. The piece of mass $2 M$,after breaking up,has a kinetic energy:

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