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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 different kinetic energies are $E_1$ and $E_2$ respectively. If $m_1 > m_2$ then
$E_1 < E_2$
$\frac{{{E_1}}}{{{E_2}}}\, = \,\frac{{{m_1}}}{{{m_2}}}$
$E_1 > E_2$
$E_1 = E_2$
Solution
Kinetic energy is given by
$E=\frac{1}{2} m v^{2}=\frac{1}{2 m}(m v)^{2}$
but $\mathrm{mv}=$ momentum of the particle $=\mathrm{p}$
$\mathrm{E}=\frac{\mathrm{p}^{2}}{2 \mathrm{m}} \quad$ or $\quad \mathrm{p}=\sqrt{2 \mathrm{mE}}$
Therefore, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{m}_{1} \mathrm{E}_{1}}{\mathrm{m}_{2} \mathrm{E}_{2}}}$
but it is given that $\mathrm{p}_{1}=\mathrm{p}_{2}$
$\mathrm{m}_{1} \mathrm{E}_{1}=\mathrm{m}_{2} \mathrm{E}_{2}$
or $\frac{E_{1}}{E_{2}}=\frac{m_{2}}{m_{1}}$ $…(i)$
Now $\mathrm{m}_{1}>\mathrm{m}_{2}$
$\text { or } \frac{\mathrm{m}_{1}}{\mathrm{m}_{2}}>1$ $…(ii)$
Thus, Eqs. $(i)$ and $(ii)$ give
$\frac{E_{1}}{E_{2}}<1$
or $\quad \mathrm{E}_{1}<\mathrm{E}_{2}$
