Question

Two small bodies of mass 10 kg and 20 kg are placed at a distance of 1m apart and released. Assuming that only mutual gravitational forces are acting, find the speeds of the particles when the separation decreases to 0.5 m.

So I attempted to conserve the energy... But I don't know whether I should conserve the system's
(The center of mass cannot develop a velocity, right? as the forces are internal?) Or the individual bodies'.... This should include which terms?

Answers: 4.2E-5 m/s and 2.1E-5

Answer 1

First you need to determine the equation of motion. Note that there are two forces acting on each particle. Write Newton's second equation and later try to solve for the speed of each particle after the time each has moved 0.5 m horizontally.

Answer 2

please, note that, since the system of the two masses is an isolated system, then also the total momentum has to be conserved, so we can write:
\[m _{1}v _{1}=m _{2}v _{2}\]
and since \[m _{2}=2m _{1}\]
we get:
\[v _{1}=2v _{2}\]
Please, add that equation to that of conservation of total energy,
(namely kinetic + potential ) and you will get your answer

Answer 3

please, @Saeeddiscover note that, the problem of @decidedlypretentious is another question, namely the sistuation described in the question of @decidedlypretentious can be reassumed as below:

Answer 4

That's it. Thanks, so I solved it.
\[- \frac{ GMm }{ 1 }=-\frac{ GMm}{ 0.5 }+\frac{ M(V)^{2} }{ 2 }+\frac{ m(2V)^{2} }{ 2 }\]

Where the capital letters are for for the heavier object. Substitue, and Voila!