Question

Two point masses m and 2m are kept at a distance
a. Their relative velicity of approach when separation becomes a/2

Answer 1

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Explanation:

Reduced mass μ=m.2mm+2m=2m3

Changining in potential energy is

U1−U2=−Gm.2ma−(−Gm.2ma/2)

=2Gm2a

12μv21/2=2Gm2a⇒12=6Gma−−−−−√

v1/2=v1+v2=6Gma−−−−−√ ......(i)

mv1=2mv2

v1=2v2

v1+v2=3v2=6Gma−−−−−√⇒v2=136Gma−−−−−√=2Gm3a−−−−−√

Answer 2

The relative velocity is $2\sqrt{\dfrac{3Gm}{a}}$.

Explanation:

Given that,

Mass of first point = m

Mass of second point = 2m

Distance = a

We need to calculate the reduces mass

Using formula of reduce mass

$\mu=\dfrac{m_{1}m_{2}}{m_{1}+m_{2}}$

put the value into the formula

$\mu=\dfrac{m\times2m}{m+2m}$

$\mu=\dfrac{2m^2}{3m}$

$\mu=\dfrac{2m}{3}$

We need to calculate the relative velocity

From conservation of mechanical energy

Decrease in potential energy is equal to increase in kinetic energy

$\dfrac{Gm_{1}m_{2}}{r}=\dfrac{1}{2}\mu v_{\frac{1}{2}}^2$

$v_{\frac{a}{2}}=\sqrt{\dfrac{2Gm_{1}m_{2}}{\mu r}}$

Put the value into the formula

$v_{\frac{a}{2}}=\sqrt{\dfrac{2G\times m\times 2m}{\dfrac{2m}{3}\times\dfrac{a}{2}}}$

$v_{\frac{a}{2}}=2\sqrt{\dfrac{3Gm}{a}}$

Hence, The relative velocity is $2\sqrt{\dfrac{3Gm}{a}}$.

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Topic : relative velocity

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