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Potential Energy

Find the cumu...

Question

Find the cumulative kinetic energy $E$ of these particles in the frame $K^{'}$ , where it's value is minimum. (Given $μ = \frac{m_{1} m_{2}}{(m_{1} + m_{2})}$ )

E = μ \frac{{(_{1} -_{2})}^{2}}{2}

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E = μ \frac{{(_{1} -_{2})}^{2}}{4}

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E = μ \frac{{(_{1} -_{2})}^{2}}{8}

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E = μ \frac{{(_{1} -_{2})}^{2}}{9}

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Solution

The correct option is A $E = μ \frac{{(_{1} -_{2})}^{2}}{2}$
Here, two particles travel along the $x$ axis in the reference frame $K$ , one of mass $m_{1}$ with velocity $_{1}$ , and the other of mass $m_{2}$ with velocity $_{2}$ .

Let, $\to v$ be the velocity of the reference frame K' and $E_{1}$ and $E_{2}$ be the kinetic energies for particles in this frame.

$E_{1} = \frac{1}{2} m_{1} (\to v -_{1})^{2}$ and $E_{2} = \frac{1}{2} m_{2} (\to v -_{2})^{2}$

Now, $E = E_{1} + E_{2}$

$E = \frac{1}{2} m_{1} (\to v -_{1})^{2} + \frac{1}{2} m_{2} (\to v -_{2})^{2}$ ...(1)

$E = \frac{1}{2} [m_{1}^{2} - 2 m_{1} v_{1} + m_{1} {_{1}}^{2} + m_{2} {\to v}^{2} - 2 m_{2} v_{2} + m_{2} {_{2}}^{2}]$

$E = \frac{1}{2} [^{2} (m_{1} + m_{2}) - 2 \to v (m_{1}_{1} + m_{2}_{2}) + m_{1} {_{1}}^{2} + m_{2} {_{2}}^{2}$

Differentiating with respect to $v$ we get

$\frac{d E}{d v} = \frac{1}{2} [2 \to v (m_{1} + m_{2}) - 2 (m_{1}_{1} + m_{2}_{2})]$

Since, reference frame $K^{'}$ is moving with respect to frame $K$ and the cumulative kinetic energy of two particles is minimum there hence,
$\frac{d E}{d v} = 0$

$\frac{1}{2} [2 \to v (m_{1} + m_{2}) - 2 (m_{1}_{1} + m_{2}_{2})] = 0$

$\to v (m_{1} + m_{2}) = m_{1}_{1} + m_{2}_{2}$

$\to v = \frac{m_{1}_{1} + m_{2}_{2}}{m_{1} + m_{2}}$ ....(2)

Using equation (2) in equation (1), we get

$E = \frac{1}{2} m_{1} (\frac{m_{1}_{1} + m_{2}_{2}}{m_{1} + m_{2}} -_{1})^{2} + \frac{1}{2} m_{2} (\frac{m_{1}_{1} + m_{2}_{2}}{m_{1} + m_{2}} -_{2})^{2}$

$E = \frac{1}{2} m_{1} (\frac{m_{1}_{1} + m_{2}_{2} - m_{1}_{1} - m_{2}_{1}}{m_{1} + m_{2}})^{2} + \frac{1}{2} m_{2} (\frac{m_{1}_{1} + m_{2}_{2} - m_{1}_{2} - m_{2}_{2}}{m_{1} + m_{2}})^{2}$

$E = \frac{1}{2} m_{1} m_{2}^{2} (\frac{(_{2} -_{1})^{2}}{(m_{1} + m_{2})^{2}}) + \frac{1}{2} m_{2} m_{1}^{2} (\frac{(_{1} -_{2})^{2}}{(m_{1} + m_{2})^{2}}$

$E = \frac{1}{2} \frac{(_{1} -_{2})^{2}}{(m_{1} + m_{2})^{2}} m_{1} m_{2} (m_{1} + m_{2})$

$E = \frac{1}{2} \frac{(_{1} -_{2})^{2}}{(m_{1} + m_{2})} m_{1} m_{2}$

$E = μ \frac{(_{1} -_{2})^{2}}{2}$

where, $μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}$

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μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}

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