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I'm trying to understand how mass factors into forces acting on particles in a simulation.

For the sake of argument, imagine a 1-dimensional simulation of a particle, whose position is 0 on the x-axis of this hypothetical 1D world.

If I implement a force/mass model, the particle's acceleration will be inversely proportionate to its mass, so if I give the particle a mass of 100 and push it with a force of 15, the resulting position on the x-axis after a single time step would be 0.15 (0 + 15 * (1/100)).

That all makes sense to me....heavier objects take more force to push.

However, what I'm failing to understand is what happens if my particle has a small mass. Imagine instead of a mass of 100, it has a mass of .00001. Using the same formula, its new position along the x-axis would be 1500000 (0 + 15 * (1/.00001)). That seems....wrong?

In real life, it makes sense that pushing a 2kg object takes twice the force of a 1kg object to go the same distance (and plugging those values into my formula yields that intuitive result)...but if I push a mote of dust with a tiny mass it doesn't soar halfway across the universe like the previous formula would suggest. I'm sure my error is obvious but I can't figure it out.

Any insights are appreciated!

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You just aren't using the right equations of motion.

With mass $m$, force $F$, initial position $x_0$ and position after some time interval is $x_1$, you are using the equation $$x_1=\frac Fm\left(x_0+\frac Fm\right)$$

Issues with the equation:

  • No reference to a time interval (you would expect a larger distance to be covered if the force is applied over a longer time interval)
  • The units don't match up (both between the left and right side as well as with the math on the right side)
  • The equation is just not how position evolves over time due to a constant force

What you really want is the kinematic equation taught in introductory physics:$^*$ $$x_1=v_0\Delta t+\frac12a\left(\Delta t\right)^2$$ Or, using Newton's second law $$x_1=x_0+v_0\Delta t+\frac12\frac Fm\left(\Delta t\right)^2$$ where $v_0$ is the initial velocity and $\Delta t$ is a time interval. This is the correct way to incorporate the mass into your analysis.


I was implying a time step of 1 for simplicity. Your solution still doesn't answer my question....tiny masses will result in huge distances traveled.

This is exactly correct. For the same force applied over the same time, the acceleration $a=\frac Fm$ is larger for smaller masses. Based on the equation above this means that the term $\frac12\frac Fm\left (\Delta t\right)^2$ is larger for smaller masses, which results in a larger distance traveled.

As for the time step being $1$, your original equation cannot be fixed by putting in a time interval.


$^*$Note that this is technically true for just the initial conditions for position $x_0$ and velocity $v_0$. If you are wanting to do things with time steps, then you can't keep adding this onto the position. In other words, $$x_2\neq x_0+2\left (v_0\Delta t+\frac12\frac Fm\left(\Delta t\right)^2\right)$$

You would either do $$x_2=x_0+v_0\left(2\Delta t\right)+\frac12\frac Fm\left(2\Delta t\right)^2$$ or you could do $$x_2=x_1+v_1\Delta t+\frac12\frac Fm\left(\Delta t\right)^2$$ where $v_1$ it's given by another kinematic equation $$v_1=v_0+\frac Fm\Delta t$$

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  • $\begingroup$ I was implying a time step of 1 for simplicity. Your solution still doesn't answer my question....tiny masses will result in huge distances traveled. But that seems incorrect, or at least counter intuitive? And I don't mean to imply that your equations are wrong, just moreso trying to understand why my intuitions are wrong. $\endgroup$ – Tyson Nov 5 '18 at 4:11

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