# Is vector sum of momentum constant if there is friction?

In a head-on collision of 2 objects with the same mass and speed in opposite directions, the vector sum of momenta before and after the collision stays the same and is equal to zero. I've read that if we assume that there's friction, then momentum in the system of two objects will not be conserved, because some momenta will be lost in the ground. But shouldn't friction affect both objects equally, since they have the same mass, and the force of friction depends on mass? So wouldn't both momenta vectors decrease, but equally for both objects, so their vector sum would be constant? Doesn't it mean that momentum in this scenario is still conserved even though there's friction?

• You should specify, where the friction occurs. Do you think about friction of each individual mass with a third entity, like a table, or friction of each mass in air? Mar 12, 2019 at 16:01
• While writing my question I mostly meant the ground or a table, but would it matter what's the source of friction if we only want to calculate the difference of it between two objects? Because both objects are the same and have the same speed, so they should both experience the same friction, which would mean that the difference (vector sum) is constant Mar 12, 2019 at 16:39

Well as per codename47 suggested you it's all correct.

See momentum is conserved in the system even if you take the two particles as the system and friction is the external force.

See friction has the same effect in opposite directions for the two bodies and when added vectorially the net friction on the system would cancel out.So the net external force on the system is zero and as per the definition of linear momentum conservation linear momentum of the whole system is conserved.

• I have give answer by assuming that friction is operating between the objects and the ground.
– user213933
Mar 12, 2019 at 16:09
• Momentum conservation is not necessarily true for individual objects, only for the total system. If a billard ball in a frictionless system collides elastically with another, it changes its momentum even though overall momentum conservation is true. Plus, in his case, if you add all the changes in momentum it will be zero at all times. Mar 12, 2019 at 16:09
• Well I haven't said that momentum is conserved individually and as the friction is the external force its sure that linear momentum of the system will not be conserved.
– user213933
Mar 12, 2019 at 16:12
• The linear momentum is generally not conserved when friction applies, but if the sum of frictional forces cancels, so does the changes in momentum, see my answer. Mar 12, 2019 at 16:25
• Well as per your answer you must consider earth and both the particles as the system.
– user213933
Mar 12, 2019 at 16:26

In the specific case you mention, both the momentum and the change in momentum cancels $$p_1(t) = -p_2(t), \qquad d p_1(t)/dt = -d p_2(t)/dt.$$ This is because the total external frictional force $$F$$ acting on the objects cancels, and $$F = dp/dt$$, so $$\sum_i F_i = 0 \Rightarrow \sum_i d p_i/dt = 0.$$ This means that technically it is conserved, but that is not true for the general situation. This is why we say that conservation of momentum is not true in general when friction applies, instead of saying that it is impossible to find a way to conserve momentum if friction applies.

Think about it like this: When conservation of momentum applies, we must force all momenta changes to cancel out to get the right result. In your case momentum changes just happen to cancel out, which is sort of the reverse situation.

If the two objects are considered the system, then as others have said when the net external friction force is zero momentum is conserved.

What is not conserved, however, is the total system energy. Since the friction forces are non-conservative (dissipative) some of the kinetic energy lost by the system goes to the surroundings (to the surface on which the objects slide). That makes this scenario different from that involving an inelastic collision between two objects. There the kinetic energy that is lost is converted to deformation (internal) energy of the colliding objects and therefore stays within the system, provided it is isolated (no heat transfer to the surroundings). Total system energy is conserved.

Hope this helps.