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Revision:Motion With Variable Mass

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Motion with Variable Mass

These notes are based on the requirements of the M5 A Level mathematics module.


Motion with variable mass

You only need to know the impulse-momentum principle:

\displaystyle \mathsf{Impulse} = \mathsf{Change\ in\ momentum} = \mathsf{Force} \times \mathsf{Time}.


From this you can set up a differential equation that describes the motion of the system.


e.g. A particle is moving upward against gravity and is losing mass at a rate k mass/sec. The lost mass is ejected vertically downwards with speed u relative to the body.


To start off, the particle is moving against gravity. So the force acting on it is its weight. Thus:

\displaystyle I = Ft = -(m + \delta m)g \delta t, where \delta t is the time interval through which this impulse occurs and δm is the increase in mass (see below).


Now we want to find the change in momentum.

(i) particle

\displaystyle (m + \delta m)(v + \delta v) - mv, supposing that the body gains \delta m of mass and \delta v of speed (the fact that it loses mass will be accounted for later)


(ii) ejected mass

-\delta m(v + \delta v-u) - 0 = -\delta m(v-u), we always ignore \delta m\delta v since it's small.


So the change in momentum of the whole system is:

\displaystyle (m + \delta m)(v + \delta v) - mv - \delta m(v + \delta v-u)


Hence by the impulse-momentum principle:

\displaystyle (m + \delta m)(v + \delta v) - mv - \delta m(v-u) = -(m + \delta m)g \delta t


Some rearrangement (remember, we ignore \delta m \delta v):

\displaystyle m\delta v + u\delta m = -(m + \delta m)g \delta t


Divide by \delta t and take limits:

\displaystyle m(\frac{dv}{dt}) + u(\frac{dm}{dt}) = -mg


This is the differential equation that describes this motion. Now we take into account one last piece of information:

The body loses mass at constant rate, i.e. \displaystyle \frac{dm}{dt} = -k.

This transforms the differential equation into:

\displaystyle m(\frac{dv}{dt}) -ku = -mg, which you can now solve easily.


Some things you should know when solving questions related to this chapter:

  • Limiting speed is achieved when \displaystyle \frac{dv}{dt} = 0.
  • When dealing with rocket problems, sometimes the time of burnout helps. e.g. suppose that a rocket has variable mass m and that the mass of the fuel is Mf and of the rocket is Mr. The time of burnout T is when m=Mr. This fact is simple (and obvious), but easily over-looked.
  • When dealing with rain-drop type problems, use mass=density*volume, i.e. m=\rho v. This is helpful because sometimes you're told that the rain-drop is spherical, and this helps you get enough information to find an expression for \displaystyle \frac{dm}{dt}. That's because you know that p is constant, \displaystyle v=(\frac{4}{3})\pi r^3, and you're usually given that \displaystyle \frac{dr}{dt} is constant. In other words: \displaystyle \frac{dm}{dt} = 4p\pi r^2 \frac{dr}{dt}.


Comments

Originally written by dvs on TSR forums.

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