Normally, in questions of collisions, we tend to apply conservation of momentum (and in cases of elastic collisions, conservation of energy as well) and we usually ignore the frictional force acting. But what actually happens, when friction is to be accounted for, theoretically, and mathematically. (Say ù is coefficient of friction between the two colliding bodies.)

  • $\begingroup$ Friction is often not directly relevant. Collisions are perpendicular interactions. Friction is a parallel force. For instance, think of a billiard ball hitting another ball on the pool table - it may come in at an angle, but the momentum transfer during collision happens perpendicular to the contact plane. Do you have a specific scenario we can look at? $\endgroup$
    – Steeven
    Commented Nov 12, 2019 at 9:25
  • $\begingroup$ A ball strikes a plank(at rest) at an angle with some velocity, there is friction between the plank and the ball, how would the final velocities be affected, taking friction into account. $\endgroup$ Commented Nov 12, 2019 at 9:28
  • $\begingroup$ @Qmechanic♦ I don't think friction between two colliding bodies have any effect on energy conservation as friction will not do any work $\endgroup$ Commented Nov 12, 2019 at 9:42
  • $\begingroup$ @Steeven When they come at an angle, friction would play a role as the relative velocity between the points of contact of the balls would have a tangential component--so the friction would act to reduce the magnitude of that relative velocity. So, there would be a momentum transfer along the tangential direction. And, if the coefficient of kinetic friction is close to unity, this would be an effect of the same order as the effect of the normal forces. Correct me if I am mistaken. $\endgroup$
    – user87745
    Commented Nov 12, 2019 at 9:48
  • $\begingroup$ Are you restricting it to friction between the colliding bodies or can it include kinetic friction between the bodies and their environment before and after the collision? $\endgroup$
    – Bob D
    Commented Nov 12, 2019 at 9:56

2 Answers 2


Normally the friction will not do any work during a quick collision of two hard bodies, like billiard balls. The reason is that work is equal to force times displacement: if the surfaces of the balls which are touching momentarily do not slip relative to each other then the work is precisely zero! Nevertheless, it will have observable consequences: there will be a transfer of angular momentum between the balls leading to rotation. You might even be able to calculate exactly the collision of two finite-sized non-slippery perfectly-elastic billiard balls but it is not an easy problem.


With friction operating in the system between the two bodies while impact(elastic),mechanical energy conservation law still holds good because in elastic impact friction doesn't do work as it is instantaneous(somewhat impulsive).

Linear momentum may be conserved if there is no external force on two bodies but as friction(between colliding bodies) will be an internal force in the system so it will not change the total momentum of the system.

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    $\begingroup$ You’re talking about friction between the colliding bodies, but taking the billiards example, what about kinetic friction between the balls and surface? $\endgroup$
    – Bob D
    Commented Nov 12, 2019 at 9:51
  • $\begingroup$ @Bob D the OP has mentioned that he is talking about the friction between colliding bodies $\endgroup$ Commented Nov 12, 2019 at 9:57
  • $\begingroup$ Yes but it almost seemed as a “for example” or parenthetical comment. Anyway I asked for clarification $\endgroup$
    – Bob D
    Commented Nov 12, 2019 at 10:09

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