When an object collides with another object in an elastic collision, the first object exerts a force on the second object (and vice versa); how do you calculate the force that the object experiences? (and I am just thinking in terms of super basic point-like particles).

All I have been able to find on this has to do with calculating the resultant velocities of the collision, not the resultant forces. The next closest thing that I've come across has to do with impulses (http://en.wikipedia.org/wiki/Impulse_(physics)), but I'm new to all of this so I am not sure how to apply it (would an elastic collision of two point-like particles have an infinitesimal duration?).

Getting the force just seems like it would be pretty simple, so I must be missing something. I'm a beginner so any help would be great! Thanks!

  • $\begingroup$ Is the duration infinitesimal (i.e. extremely short)? Infinitesimally long makes no sense. Also, it is not clear what you ask: you say that you want to calculate the force, but also say that getting the force ... it would be pretty simple. $\endgroup$ – Sofia Jan 23 '15 at 1:22
  • $\begingroup$ Yea, I just mean extremely short. I was trying to talk in Calculus jargon. $\endgroup$ – M Schu Jan 23 '15 at 1:25

The force for a realistic collision will not be constant over the course of the collision. There is no one way to figure out what the force on each body will be over time as the collision progresses. The one thing we can say for sure is that

$$\Delta \vec{p} = \int_{t_0}^{t_1}\vec{F}(t)dt$$

where $\Delta \vec{p}$ is the change in momentum of one of the colliding bodies, $\vec{F}(t)$ is the force on that body as a function of time, $t_0$ is the time the bodies first start to interact, and $t_1$ is the time they stop interacting.

If you want to study a simplified collision where the force is constant ($\vec{F}(t) = \vec{F}$), this can be simplified:

$$\Delta \vec{p} = \int_{t_0}^{t_1}\vec{F}dt$$ $$\Delta \vec{p} = \vec{F} \int_{t_0}^{t_1}dt$$ $$\Delta \vec{p} = \vec{F} (t_1 - t_0)$$ $$\Delta \vec{p} = \vec{F} \Delta t$$

in which case,

$$\vec{F} = \frac{\Delta \vec{p}}{\Delta t}$$

But as stated above, $\vec{F}(t) = \vec{F}$ is not a very realistic interaction force. If you examine that last equation, in this case, (as will also be true for any interaction), as the time interval of the interaction approaches zero, the force on the objects approaches infinity.


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