A light body of momentum P_L and a heavy body | vedclass

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(B) The kinetic energy $K$ of a body with momentum $P$ and mass $m$ is given by the relation $K = \frac{P^2}{2m}$.Given that both bodies have the same

Vedclass · Accepted Answer

$P_H > P_L$

Vedclass · Answer

(B) The kinetic energy $K$ of a body with momentum $P$ and mass $m$ is given by the relation $K = \frac{P^2}{2m}$.Given that both bodies have the same kinetic energy,we have $K_H = K_L$.Substituting the formula,we get $\frac{P_H^2}{2 m_H} = \frac{P_L^2}{2 m_L}$.Rearranging the terms,we get $\frac{P_H^2}{P_L^2} = \frac{m_H}{m_L}$,which implies $\frac{P_H}{P_L} = \sqrt{\frac{m_H}{m_L}}$.Since the body is heavy,$m_H > m_L$,which means $\frac{m_H}{m_L} > 1$.Therefore,$\frac{P_H}{P_L} > 1$,which leads to $P_H > P_L$.

$A$ light body of momentum $P_L$ and a heavy body of momentum $P_H$,both have the same kinetic energy,then:

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