Two particles of masses m and $2$ m are kept at a distance a. Find their relative velocity of approach when separation becomes $a / 2$ .

2 \sqrt{\frac{3 a}{2 G M}}

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\sqrt{\frac{a}{2 G M}}

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2 \sqrt{\frac{2 G M}{3 a}}

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\sqrt{\frac{6 G M}{a}}

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Solution

The correct option is D $\sqrt{\frac{6 G M}{a}}$
In arrangements $(1)$ and $(2)$ ,
Momentum conservation,
$0 = m v_{1} - 2 m v_{2}$ ...........(i)
Energy conseravtion
$- \frac{G m \cdot 2 m}{a} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} 2 m v_{2}^{2} - \frac{G m \cdot 2 m}{a / 2}$ ..........(ii)
From Eq. (i), we get
$v_{1} = 2 v_{2}$
Putting this value $(v_{1} = 2 v_{2})$ in Eq. (ii), we get
$- \frac{2 G m^{2}}{a} = \frac{1}{2} m (2 v_{2})^{2} + \frac{1}{2} \cdot 2 m v_{2}^{2} - \frac{4 G m^{2}}{a}$
$3 m v_{2}^{2} = \frac{2 G m^{2}}{a}$
$v_{2} = \sqrt{\frac{2 G m}{3 a}}$
$v_{1 / 2} = v_{1} + v_{2}$
$= 3 \sqrt{\frac{2 G m}{3 a}}$
$= \sqrt{\frac{6 G m}{a}}$ .

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Two particles of masses m and 2m are kept at a distance a. Find their relative velocity of approach when separation becomes a/2.

Two particles of masses m and $2$ m are kept at a distance a. Find their relative velocity of approach when separation becomes $a / 2$ .