How is momentum conserved? I know that the condition is that no external resultant force should act on the interacting objects. But how is the momentum conserved if the objects surfaces touch and there’s friction between them. Even if for a split second
Newtonian Mechanics – Why Internal Forces Do Not Affect Momentum Conservation?
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Related Solutions
Objects are not damaged by momentum $\vec{p}$ they are damaged by force $\vec{F}$. When two objects collide their momenta changes because of forces they apply each other while being in contact. According to 2nd Newton's law the force can be calculated as follows: $$ \vec{F} = \frac{d\vec{p}}{dt} $$
So the force is determined by the time of interaction. When two objects contact their surfaces are flexed. The bigger is the flex the higher is the force which changes the momentum. At some moment of time the relative normal momentum (and velocity) becomes zero and the objects start move backward.
At this moment the force and the flex is maximal. If the surface (armor) is not strong enough for this flex the object is damaged. This can happen long before the relative normal velocity become zero. In that case the interaction between internal parts of the objects started.
Momentum
Hence damage is not determined only by momentum. It is determined by the force and the ability of the object resist it. The force is not constant during the collision and its maximal value depends on momentum and the time of interaction of the surfaces. The time of interaction depends on both flexibility and strength.
When a basketball hits the floor its momentum changes from $\vec{p}$ to almost $-\vec{p}$ so the total change is almost $2p$. When a glass hits the floor its momentum becomes zero so the change is $p$. The ball has higher flexibility and strength and is not destroyed even though the momentum change is two times higher.
Another example is a bullet that hits a door. It makes hole before the door opens. The force is huge but the momentum of the door is almost not changed because the time necessary to reach critical flex is too short. When one pushes the door with his finger the force is small and does not destroy the surface before the flex stop the finger. The door gets enough momentum to start move.
Energy
When the surface is flexed in irreversible way or damaged some part of kinetic energy of the objects turns into kinetic energy of their parts, heat, sound, light etc. This is called inelastic collision. Elastic collision means no damage. If you need a good model this should be taken into account.
You define a system which you are interested in.
If there is no net external force acting on the system then linear momentum is conserved.
You can identify internal forces as they must occur in equal in magnitude but opposite in direction pairs - Newton's third law.
So you find a force in the system $\:\mathbf{f}_{12}\:$ which is the force on part $1$ of the system due to part $2$ of the system which has its equal in magnitude opposite in direction twin, $\:\mathbf{f}_{21}\:$ force on part $2$ of the system due to part $1$ of the system.
There is no such pairing of forces within the system for external forces which are forces on the system due to something outside the system so their Newton's third law pair would be a force on something outside the system due to force produced by system.
Best Answer
As you say, momentum is conserved when there is no external resultant force acting on the system. This is a statement of Newton's 2nd law: when the net force acting on a system is zero, there is no change in momentum. $$ \vec{F}_\mathrm{net} = \frac{d\vec{p}}{dt} $$
an example
It might help to think about an example. Lets consider the head on collision of two objects. During the collision each object applies a force to the other. Object $A$ pushes on object $B$ causing $B$'s momentum to change: $$\Delta \vec{p}_B = \vec{F}_{A,B}\, \Delta t $$ And object $B$ pushes on object $A$ causing $A$'s momentum to change: $$\Delta \vec{p}_A = \vec{F}_{B,A}\, \Delta t $$
Newton's 3rd law tells us that $\vec{F}_{B,A} = -\vec{F}_{A,B}$, so $\Delta\vec{p}_A = - \Delta\vec{p}_B$. We need to recognize that each force is applied for the same time interval $\Delta t$, too. The forces are applied during the collision, and both objects start and stop touching at the same times.
When we consider the whole system of both objects, the total change in momentum is zero. $$\Delta\vec{p}_\mathrm{sys} = \Delta\vec{p}_A + \Delta\vec{p}_B = 0$$ Total momentum is conserved during the collision. Because the interaction forces ($\vec{F}_{A,B}$ and $\vec{F}_{B,A}$) are internal to the system the system doesn't have a net force.
Notice we did not care how big those forces were, nor did we care how long the collision lasted. What matters is the symmetry of the forces that comes from Newton's 3rd law.
energy
So what does this have to do with energy? The example didn't use energy at all.
Lets take our example particles to be equal mass and initially moving with equal speed in opposite directions. In order to conserve momentum the the particles must have equal and opposite velocities after the collision, but they don't necessarily have to have the same velocities as before.
If energy is lost, they will each be moving slower than before. An explosion during the collision could add energy to the system, causing them to go faster than before. Each case would still conserve momentum.