The phase space trajectory of a single particle falling freely from height is?

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Phase space is a plot between momentum and position, and since kinetic energy increases the momentum must increase with position, so option "2" must be correct, but the answer key shows that answer in option "4". Please tell me the correct method to plot it if my observation is wrong


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  • 1
    $\begingroup$ That would depend completely on what $z$ is representing. $\endgroup$ – JMac May 9 at 19:10
  • $\begingroup$ You mean to say, it depends on where we fix our coordinate system $\endgroup$ – sawan kumawat May 9 at 19:13
  • 2
    $\begingroup$ I mean it depends on if $z$ represents current height above the ground, or displacement from the point it was released. $\endgroup$ – JMac May 9 at 19:15

At t = 0, p = 0 and z = H. So the first point should be on the horizontal axis. At the final position the object is at z = 0 and its p is max, hence it is on the vertical axis. You can use the relation

v^2 = 2g delta(z) to figure out these points. I think that supports the answer.

Though I do not like the apparent hump in the plot.

  • 1
    $\begingroup$ Okk I got it, the mistake I was doing is that, I was fixing my coordinate system at the ground and was considering velocity positive instead of negative $\endgroup$ – sawan kumawat May 9 at 18:29
  • 1
    $\begingroup$ You can choose different coordinates and it should still work. But from the kinematics equations the momentum goes as the square root of z and option B looks more like z^2. it gets tricky but phase space is the place to be. $\endgroup$ – ggcg May 9 at 20:40

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