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

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Question

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{

Vedclass · Accepted Answer

$E_1 < E_2$

Vedclass · Answer

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 $\mathrm{m}_{1}>\mathrm{m}_{2}$ $ ext { or } \frac{\mathrm{m}_{1}}{\mathrm{m}_{2}}>1$ $...(ii)$ Thus, Eqs. $(i)$ and $(ii)$ give $\frac{E_{1}}{E_{2}}<1$ or $\quad \mathrm{E}_{1}<\mathrm{E}_{2}$

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

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