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Potential Energy

In the refere...

Question

In the reference frame $K$ two particles travel along the $x$ axis, one of mass $m_{1}$ with velocity $_{1}$ , and the other of mass $m_{2}$ with velocity $_{2}$ .Find the velocity $\to v$ of the reference frame $K^{'}$ in which the cumulative kinetic energy of these particles is minimum.

A

\to v = \frac{(2 m_{1}_{1} + 2 m_{2}_{2})}{(3 m_{1} + 3 m_{2})}

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B

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

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C

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

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D

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

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Solution

The correct option is B $\to v = \frac{(m_{1}_{1} + m_{2}_{2})}{(m_{1} + m_{2})}$
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, $E = E_{1} + E_{2}$

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

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

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

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,
$\frac{d E}{d v} = 0$

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

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

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

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