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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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