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In my dynamics class, we learned that the individual momenta of two colliding particles is conserved in the direction tangential to the collision. Why would this be the case?

Consider the diagram above. The direction tangential to (i.e parallel to the surface of) the collision is labeled t, and the direction normal to the collision is labeled n. The blue ball has a higher speed than the red ball. During the time that they are colliding (deformation and restoration) they would move horizontally (or 'slip') relative to eachother. There would be a certain normal force, and therefore a friction force between the two, given by

$f_f = \mu_kN$,

where $f_f$ is friction force, $\mu_k$ is coefficient of kinetic friction, and $N$ is normal force.

Let's assume $\mu_k$ is on the order of 0.5 - 1.0 (which it often is), and therefore the friction force is significant.

Given that the integral of forces in both directions taken with respect to time gives change in momentum in each direction, the change in momentum of each particle in the t-direction would be $\mu_k$ times the change of momentum in the n-direction.

E.g. for the blue ball:

$\Delta P_n = \int_{t_0}^{t_1} \sum{F_{n}} \space dt = \int_{t_0}^{t_1} N \space dt$

$\Delta P_t = \int_{t_0}^{t_1} \sum{F_{t}} \space dt = \int_{t_0}^{t_1} -\mu_kN \space dt = -\mu_k (\int_{t_0}^{t_1} N \space dt) = -\mu_k \Delta P_n$

So why is it valid to assume change in each individual particle's momentum in the t-direction is negligible? Wouldn't the blue ball cause an increase in the horizontal speed of the red ball, and the red ball cause a decrease in the horizontal speed of the blue ball?

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  • $\begingroup$ Were the collisions specified to be elastic? $\endgroup$ – Mark H Oct 5 '19 at 12:52
  • $\begingroup$ This wasn't actually a textbook scenario, just one that I'd made up, but let's assume it is elastic for simplicity $\endgroup$ – Adrian Oct 5 '19 at 14:42
  • $\begingroup$ If the collisions are elastic, then there can be no frictional forces. Elastic means that kinetic energy is conserved, so there can be no dissipative forces like friction involved. In a non-elastic collision, only total momentum is conserved, not individual momentum nor any individual component of momentum. $\endgroup$ – Mark H Oct 5 '19 at 16:04
  • $\begingroup$ What direction is tangential to a collision in 2D? $\endgroup$ – R.W. Bird Oct 5 '19 at 17:59
  • $\begingroup$ @MarkH Oh right, of course. I guess this would be a non-elastic problem then, as there is the possibility of frictional effects. $\endgroup$ – Adrian Oct 6 '19 at 0:12
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In my dynamics class, we learned that the individual momenta of two colliding particles is conserved in the direction tangential to the collision. Why would this be the case?

In a collision, any forces between the objects are internal forces for the system of the two objects. Therefore, total momentum is conserved.

Since momentum is a vector quantity, this means that you can also say momentum conservation holds for components of the total momentum as well. Therefore, momentum is conserved in the direction tangential to the collision.

You have found the magnitudes of the change in momentum components of the blue ball. For the red ball these must be equal and opposite, by Newton's third law. Also note that these are vector components, so getting an expression that relates their magnitudes isn't of importance here.

So why is it valid to assume change in momentum in the t-direction is negligible?

Where is this assumption made? Overall this doesn't change at all, and for each individual ball their momentum can change in this direction with overall momentum still being conserved. In either case nothing is being assumed to be negligible.

Wouldn't the blue ball cause an increase in the horizontal speed of the red ball, and the red ball cause a decrease in the horizontal speed of the blue ball?

Yes. This is what you need to have happen for momentum in the horizontal direction to be conserved. If both horizontal speeds increased or decreased, then momentum in this direction has not been conserved.

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  • $\begingroup$ I may not have been clear enough... We learned that the momentum of each individual particle, as opposed to system momentum, is conserved along the t-direction in the absence of external forces (whereas along the n-direction this is not the case; the momentum of each individual particle can change along this direction). That is, the speed of the blue ball in the t-direction will not change, same for the red ball. But you have answered my question nonetheless: the assumption of each individual particle's change in momentum being negligible along the t-direction is invalid. Thank you! $\endgroup$ – Adrian Oct 5 '19 at 14:08
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    $\begingroup$ @Adrian That could be true, but it doesn't have to be true in general $\endgroup$ – BioPhysicist Oct 5 '19 at 14:20
  • $\begingroup$ Is the significance of the coeff. of friction the only determining factor in whether or not to neglect change in each particle's t-direction momentum? $\endgroup$ – Adrian Oct 5 '19 at 14:40
  • $\begingroup$ @Adrian if $F_t\approx0$ then $\Delta P_t\approx0$ (for a sufficiently small time over which the force acts). In general $\Delta P_t=\int F_t\,\text dt$. So whatever makes the integral negligible will work. $\endgroup$ – BioPhysicist Oct 5 '19 at 14:52
  • $\begingroup$ So either friction force is small, or the time of collision is small. But only the former (and therefore the insignificance of $\mu_k$) would make $\Delta P_t$ insignificant when compared to $\Delta P_n$, right? If time of collision is small, $\Delta P_t = -\mu_k \Delta P_n$ still holds, and $\Delta P_t$ is no less significant than for a longer collision. When I say 'significant', I mean significant in the context of the problem. I.e. if you're considering normal force, and friction force is comparable to this normal force, friction force should also be considered. $\endgroup$ – Adrian Oct 5 '19 at 15:05

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