Two bodies of masses $m_{1}$ and $m_{2}$ have equal momenta. Their kinetic energies $E_{1}$ and $E_{2}$ are in the ratio.

\sqrt{m_{1}} : \sqrt{m_{2}}

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m_{1} : m_{2}

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m_{2} : m_{1}

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m_{1}^{2} : m_{2}^{2}

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Solution

The correct option is C $m_{2} : m_{1}$
$E = \frac{1}{2} m v^{2} = \frac{p^{2}}{2 m}$

Now, given that-

$p_{1} = p_{2} = p (s a y)$

$⟹ E \propto \frac{1}{m}$

$⟹ \frac{E_{1}}{E_{2}} = \frac{m_{2}}{m_{1}}$

$E_{1} : E_{2} = m_{2} : m_{1}$

Answer-(C)

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Two bodies of masses m1 and m2 have equal momenta. Their kinetic energies E1 and E2 are in the ratio.

The correct option is C m2:m1E=12mv2=p22mNow, given that-p1=p2=p(say)⟹E∝1m⟹E1E2=m2m1E1:E2=m2:m1Answer-(C)

Two bodies of masses $m_{1}$ and $m_{2}$ have equal momenta. Their kinetic energies $E_{1}$ and $E_{2}$ are in the ratio.

The correct option is C $m_{2} : m_{1}$
$E = \frac{1}{2} m v^{2} = \frac{p^{2}}{2 m}$

Now, given that-

$p_{1} = p_{2} = p (s a y)$

$⟹ E \propto \frac{1}{m}$

$⟹ \frac{E_{1}}{E_{2}} = \frac{m_{2}}{m_{1}}$

$E_{1} : E_{2} = m_{2} : m_{1}$

Answer-(C)