1
$\begingroup$

Classical mechanics says that if I throw a ball with velocity perpendicular to the wall and it collides elastically with the wall with a velocity $v_0$, then it bounces back with the same velocity $v_0$.

However, if I shoot a beam of water perpendicular to the wall, in most cases it will not deflect back perpendicular to the wall instead it gains velocity perpendicular to the initial velocity and continues to move on the surface. Isn't this a violation of conservation of momentum since for any small molecule inside the beam of water we had no momentum in the perpendicular direction to get started with?

$\endgroup$
  • $\begingroup$ do you have a video for this assertion? $\endgroup$ – anna v May 23 at 4:28
3
$\begingroup$

If a tennis ball hits a wall, it bounces normally, but with several balls in a continuous jet, the outcome is different. Many bouncing balls will hit incoming ones, and the result will be a scattering pattern. The advantage of the direction parallel to the wall is that it is a free path from new coming balls.

I think that the process with water jet is similar.

By the way, in a elastic collision against a wall, the moment is not conserved if we forget to include the Earth recoil in the system. The point of elastic collisions is that the ball kinetic energy is conserved.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You are right in saying that the result will be a scattering pattern, however it's worth mentioning that this is only true if the balls aren't undergoing head-on collisions. $\endgroup$ – user258881 May 25 at 14:09
1
$\begingroup$

Momentum conservation does not work like that. You can conserve momentum only when there is no external force acting in the direction. Actually, you don't have to take an example as complex as this to understand thhat momentum need not be conserved always.

Imagine that you have a ball in your hand. You leave it at rest, with obviously $0$ velocity in the vertical direction. If you conserve momentum, then the ball should never fall (since its initial momentum is zero), but it does. This is because momentum is not conserved when an external force comes into the play.

The simple reason as to why momentum is not conserved when a force acts can be obtained from the very definition of Force, which is the rate of change of momentum. A change in momentum points to a force and presence of a force (net force, to be more precise) implies a change in momentum.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ The wall doesn't apply a force on the water molecules in a direction parallel to the wall. So, this doesn't answer OP's question. $\endgroup$ – Dvij D.C. May 23 at 5:12
  • 1
    $\begingroup$ @DvijD.C., but, sir, gravity does act, doesn't it? I can consider the water+wall as a system, so, my guess would be that any force exerted by the wall should have no effect on the system's vertical momentum, and hence the water jet's momentum can be assumed to be constant (in absence of external force, assuming the wall does not move much) $\endgroup$ – Krishna May 23 at 5:18
  • $\begingroup$ @DvijD.C. Well actually , it does . When water molecules tend to return perpendicularly from the wall , the wall attracts the polar water molecules in a direction opposing the flow of reflected water. $\endgroup$ – Noah J. Standerson May 23 at 6:23
  • 1
    $\begingroup$ @NoahJ.Standerson I am not sure how the polarity works here but the point is that the momentum of water is still zero parallel to the wall because it is democratically distributed among all directions in the plane of the wall, no? Correct me if I am missing something. $\endgroup$ – Dvij D.C. May 23 at 6:29
  • 1
    $\begingroup$ @DvijD.C. okay , I get your point now. Initially the momentum of water parallel to the surface was zero. But that scenario is only possible if you consider a place without gravity. If gravity is present , the water molecules surely will have a momentum parallel to the surface (before striking it) . On the other hand , in a gravity free environment , I imagine that the molecules would just splash out unilaterally in all directions and stay stationary after striking the surface. $\endgroup$ – Noah J. Standerson May 23 at 6:42
1
$\begingroup$

The Fact that your first statement is wrong symbolizes why you were unable to understand the situation.
It is true that ideally if a ball is thrown to a wall with coefficient of restitution=1 (elastically as you said)it will return with the same velocity (horizontally) but in the case of water that is not the case.... coefficient of restitution of water and wall system is very less+ you should also take into consideration the fact that it is not liable to return in the same line of impact for gravity is acting on it, this causes it to move along the surface and adhesion forces+ consider other forces like air drag and what not
Ideal situations aren't real(as implied by the name).....amalgamation of both can cause problems- like this question here.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Conservation of momentum is valid only for systems with no external forces.

In the case of water, the molecules are polar so they get attracted by the surface you are throwing it on. Hence the external forces on the system is not zero.

Also since water molecules are ejected as a stream of particles, Even if some molecules are ejected perpendicular to the surface , you wont be able to see it. Its just like opening a water pipe inside a bucket of water. You wouldn't notice whether the pipe is working or not.

If you imagine an experiment where you have a completely non polar fluid , then you would observe the following things if you ejected the fluid perpendicular to the surface:

1)You would notice that the Beam of fluid attains almost 0 velocity near the surface. The fluid will thereby fall from there to the ground (without sticking to the surface) due to gravity.

This happens because the fluid particles after striking the wall come back with the same velocity and collide with the incoming beam of fluids and thus nullifying the part of the beam near the surface.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Momentum along the direction perpendicular to the wall is not conserved in both the cases. However, the momentum along the parallel direction is conserved (if we neglect gravity or any other force acting parallel to the wall) in both the cases. Here's how:

  • In the first case, it's easy to see that momentum is conserved in the direction parallel to the wall, because the inital and final momentum along the wall are, both, zero.

  • In the second case, however, the water particles end up having a velocity along the wall, which they did not have initially. But, if we calculate the total final momentum of these water particles along the direction parallel to the wall, we'll see that it comes out to be equal to zero. The final momenta (in the direction parallel to the wall) of all the particles cancel out perfectly.

Why so? Because of symmetry. The second situation possesses an axis of symmetry along the direction perpendicular to the wall. In other words, if we rotate the wall about an axis perpendicular to the wall (and in line with the water stream), the scenario/situation remains unchanged. Due to this symmetry, there is no special direction parallel to the wall, and all such directions are equivalent. Thus, there is no specific and special direction (parallel to the wall) along which the final angular momentum can point, and thus, the fnal angular momentum (parallel to the wall) of all the particles cancels out exactly to preserve the symmetry of this situation.

| cite | improve this answer | |
$\endgroup$

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.