2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Momentum and velocity are different things. The velocity of individual objects is not conserved if there is energy lost during a collision, but total momentum of the system is. Your question asks about velocity conservation. Can you edit your question to clarify? $\endgroup$ – Paul T. Nov 28 '20 at 14:26
  • 1
    $\begingroup$ No I know that the total momentum is conserved but how so? In an isolated system, if we have two objects A and B moving with a velocity vA and vB towards each other they collide, I know that the total momentum before and after collision is equal but how does that occur if there will be energy losses due to the two surfaces touching with each other for a split second, because this energy loss isn’t an external force it’s within the system itself. $\endgroup$ – Hamza Eyad Nov 28 '20 at 14:37
  • $\begingroup$ I finished editing it 👍🏻 $\endgroup$ – Hamza Eyad Nov 28 '20 at 14:44
  • $\begingroup$ @HamzaEyad Actually i can think of a case when momentum conservation for bodies with internal force fails. If you take a baloon and flick it keeping your fingers after flicking to be straight. $\endgroup$ – Anonymous Nov 29 '20 at 12:54
4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

While formulating the theory, we assume that the particles obey Newton's laws (third one) and thus, the forces exerted by them on each other gets cancelled out. This is sometimes called as the weak law of action and reaction.

Mathematically speaking, $$\mathbf{F} = m\mathbf{a}$$

$$\frac{d^2}{dt^2} \sum_i{m_i \mathbf{r}_i} = \sum_i{\mathbf {F}^{(ext)}_{i}} + \sum_{i, j,\ i \neq j}{\mathbf {F}_{ji}}$$

Where in the RHS the second term depicts the force that the internal particles exert on each other. Assuming the weak principle that second term vanishes. Since, $\mathbf{F}_{ij} + \mathbf{F}_{ji}= 0$.

$\endgroup$
1
$\begingroup$

I think the key is for you to take account that energy is a scalar quantity and momentum is a vector quantity. So let's say two particles are approaching each other and their momenta are 7 to the right and 5 to the left (omitting units here), so net momentum to the right of magnitude 2. If they have an inelastic collision, after the collision let's say their momenta are 2 to the left and 4 to the right (just making up numbers here). Both have lost energy, the total energy is less but the net momentum is still to the right with magnitude 2.

$\endgroup$
0
$\begingroup$

Okay when two bodies collide (say $A$ and $B$), both of them exert forces on each other and these form the action - reaction pairs and hence equal in magnitude and opposite in direction.

So , if we consider the motion of the block $A$ or block $B$ only, we cannot say that it's momentum is conserved since they experienced an external impulsive force during the collison individually.

But if we take both of them at a time (i.e. both are considered as a part of teh system) , then the net change in the momentum of the system's center of mass will be zero because both exerted equal and opposite forces on each other. And these forces are called internal forces since they are exerted by internal bodies of the system and not from the surrounding.


So the key point here is that during collision, the individual momenta of the interacting objects do change (since they experience force from their interacting partners) but if we consider the net change in momentum of the system then it will be zero because both the blocks experienced same force and thus same change in momentum for same time interval but in opposite directions.

This same logic cannot be applied for kinetic energy conservation (as others have said) since during collision bodies may deform and thus these inelastic deformations take away some of the initial kinetic energies of the system.

Summary :

During collision, kinetic energy of individual objects may change and hence their momentum also changes but when we talk of both as a system , change in momentum (being a vector) cancels out while energy being a scalar adds up. If we take the magnitude of the change in momentum of both the objects only then it will not be zero but as it is a vector , the change finally becomes zero.

Hope it helps 🙂.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.