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Question

Two masses m and 2m are at rest initially. Find the velocity of mass 2m, when their separation reduces to d2 due to gravitational force of attraction.

A
Gm2d
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B
Gmd
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C
2Gm3d
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D
Gm3d
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Solution

The correct option is C 2Gm3d

Let v1 and v2 are the velocities of m & 2m respectively when the separation between both decreases to d2.


Since, there is no external force on the given system, therefore momentum will be conserved

pi=pf

0=mv1(2m)v2

v1=2v2 ..........(1)


Initial potential energy of the system

Ui=Gm(2m)d=2Gm2d

Final potential energy of the system

Uf=Gm(2m)d2=4Gm2d

Initial and final kinetic energy of the system will be

Ki=0;

Kf=12mv21+12(2m)v22

From eq. (1),

Kf=12m(2v2)2+12(2m)v22=3mv22

Energy of the system is also conserved because only gravitation force is present, which is conservative force.

Ei=Ef

Ui+Ki=Uf+Kf

2Gm2d+0=4Gm2d+3mv22

2Gm2d=3mv22

v2=2Gm3d

Hence, option (c) is correct.

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