A particle of mass m_1 is moving with a veloc | vedclass

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(A) The kinetic energy $E$ of a particle with momentum $p$ and mass $m$ is given by the relation:$E = \frac{p^2}{2m}$Since both particles have the sam

Vedclass · Accepted Answer

$E_1 < E_2$

Vedclass · Answer

(A) The kinetic energy $E$ of a particle with momentum $p$ and mass $m$ is given by the relation:$E = \frac{p^2}{2m}$Since both particles have the same momentum $(p_1 = p_2 = p)$,we can write the kinetic energies as:$E_1 = \frac{p^2}{2m_1}$ and $E_2 = \frac{p^2}{2m_2}$Taking the ratio of the two energies:$\frac{E_1}{E_2} = \frac{p^2 / 2m_1}{p^2 / 2m_2} = \frac{m_2}{m_1}$Given that $m_1 > m_2$,it follows that $\frac{m_2}{m_1} Therefore,$\frac{E_1}{E_2} < 1$,which implies $E_1 < E_2$.

$A$ particle of mass $m_1$ is moving with a velocity $v_1$ and another particle of mass $m_2$ is moving with a velocity $v_2$. Both of them have the same momentum,but their kinetic energies are $E_1$ and $E_2$ respectively. If $m_1 > m_2$,then:

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