# Plotting expression as function of another expression [closed]

How can I plot $\frac{m}{(M+m)}$ as function of $\frac{m}{M}$? I've tried the regular Plot[] but it didn't work.

This is a function of the loss of kinetic energy during plastic collision in physics. So, when $\frac{m}{M}$ is zero, it means that the mass $M$ is a way bigger than $m$, so the expression $\frac{m}{M+m}$ would be approximately zero as well (we can neglect $m$ if $M$ is huge). That's what I mean. It should not be very rigorous, but this graph should be quite similar to the real life experiment.

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You've seen ParametricPlot[]? –  Ｊ. Ｍ. Apr 2 '13 at 16:54
@J.M. - Yes, I tried it, but unfortunately with no success. I'm not that big expert on Mathematica, if you could just show me one example. –  ffyh3h3 Apr 2 '13 at 16:56
@belisarius - thanks for the reply, but no. I added some explanation. –  ffyh3h3 Apr 2 '13 at 17:12
So you want Plot[1/(1 + 1/x), {x, 0, 3}, PlotRange -> All] instead ... –  belisarius Apr 2 '13 at 17:23
You can do all the algebra and the plotting in one go as Plot[y /. First@Solve[Eliminate[{y == m/(m + M) && x == m/M}, {m, M}], y], {x, 0, 10}]. –  whuber Apr 2 '13 at 17:51

## closed as too localized by Jens, Leonid Shifrin, Artes, Simon Woods, m_goldbergApr 2 '13 at 23:40

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$$\frac{m}{(M+m)} = \frac{ \frac{m}{m} }{(\frac{M}{m}+1)} = \frac{1}{1+\frac{1}{x}}$$
where $$x = \frac{m}{M}$$
Plot[1/(1 + 1/x), {x, 0, 3}, PlotRange -> All]