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In my book ''Ncert Class 12" It is written

We begin by considering the collision of one molecule with one of the walls of the container, oriented with a unit normal vector pointing out of the container in the positive $ˆi$-direction. Suppose the molecule has mass m and is moving with velocity $v=vxˆi+vyˆj+vzˆk$. Because the collision with the wall is elastic, the $y$-and $z$-components of the velocity of the molecule remain constant and the $x$-component of the velocity changes sign (Figure 29.2), resulting in a change of momentum of the gas molecule.

I can't understand why this thing will get due to elastic collision. How should I use conservation of kinetic energy and conservation of momentum?

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It is because the only force acting on the molecule is acting perpendicular to the wall, i.e. in the $x$ direction.

The force is acting perpendicular to the wall because we assume that there is no friction between particle and wall (they are both perfectly smooth and rigid). In the absence of friction, there can be no force not perpendicular to the wall. This is of course an idealization, but it works pretty well because the particle is much, much smaller than the wall so we can consider it to be point-like.

Since energy is conserved during an elastic collision the velocity of the particle must have the same magnitude before and after the collision, so since the other two components ($v_y$ and $v_z$) are unchanged,$v_x$ must change its sign.

PS: Notice that momentum is not conserved because we are neglecting the motion of the wall itself. This is reasonable because we assume that its mass is far greater than the mass of the particle.

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  • $\begingroup$ So how did you find that force acts is perpendicular to wall? $\endgroup$ – Vaibhav Patel May 9 '16 at 10:19
  • $\begingroup$ If by change in momentum in unit time than it means u preasumed that only vx is changing $\endgroup$ – Vaibhav Patel May 9 '16 at 10:20
  • $\begingroup$ @VaibhavPatel What would be the source of a force that's applied parallel to the wall? If the wall is smooth, and the ball is smooth, there's nothing to provide such a force. To clarify the P.S: momentum is conserved. Work out the details for yourself. The wall is attached to the earth, so you have to use the earth as the second object in the collision. $\endgroup$ – garyp May 9 '16 at 10:43
  • $\begingroup$ @garyp Of course momentum is conserved, it just isn't conserved in our schematization of the problem. If the velocity of the particle goes from $v_x$ to $-v_x$, its mass unchanged, momentum is not conserved. This is our schematization. If it was conserved, final velocity would be very slightly different. And who said the wall is "attached to the Earth?" the box could be on a frictionless surface, nothing would change. You don't need the Earth. Anyway, I updated my answer to clarify why the force is perpendicular. $\endgroup$ – valerio May 9 '16 at 10:55
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If we described the collision as a physics problem describing a particle on a background (which includes the wall), the theory is translationally symmetric with respect to translations in the $y,z$ directions. It's ultimately because the wall looks the same if one shifts it in the $y,z$ directions. (It's true for a flat wall, and similarly it holds locally for any shape, and that's what matters.)

By Emmy Noether's theorem, the symmetry of the laws of physics with respect to the translations along $y,z$-axes is equivalent to the conservation of the $y,z$-components of the momentum. The theory unavoidably has these two conservation laws for $p_y$, $p_z$.

This fact may be shown in many ways. In the basic school physics, the conservation is right because the wall only acts by forces that are perpendicular to the walls. So the force from the wall has no component in the parallel directions.

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  • $\begingroup$ What an unnecessarily complicated answer... $\endgroup$ – valerio May 9 '16 at 9:59
  • $\begingroup$ It is only complicated for those who are not thinking in terms of fundamental ideas. $\endgroup$ – Luboš Motl May 9 '16 at 10:10
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Why only x component changes is : The definition of elastic collision (https://en.wikipedia.org/wiki/Elastic_collision) states exactly that. You can see it as ping pong ball that hits the table. All the speed in the direction of the table will make it bounce away from it : if you hit the ball down, it will come up after the bounce on the table. But the other components of the speed remain unchanged by the wall so if you hit the ball down and forward, it will go up and still forward after the bounce.

How to use conservation of energy : An elastic collision doesn't change the amount of energy of the moving object, only the direction of its speed.

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  • $\begingroup$ Ok now consider the conservation of momentum of system consisting of a particle in a wall.. Then initial momentum of balls is equals to final momentum of ball after collision. Which implies( vector p initial) is equal to( vector p final) resulting in equal velocity vectors a contradiction,isn't it. $\endgroup$ – Vaibhav Patel May 9 '16 at 9:55
  • $\begingroup$ This answer is not responsive to the question. $\endgroup$ – garyp May 9 '16 at 10:46
  • $\begingroup$ I am not sure I understood your comment but in your example, what is equal is the norms of the vector p, not the vectors themselves. $\endgroup$ – Jean Mercat May 10 '16 at 12:27

protected by Qmechanic May 9 '16 at 14:23

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