Conservation of momentum on an inclined plane

Question

Please look at the image given.

The force in the x-direction is N" which is a component of the normal reaction force.Now the body starts sliding down the slope, even if I consider kinetic friction opposite to N" it is not equal to N" so Fnet is not equal to zero and momentum is NOT conserved.However many books still say that momentum is conserved.Can anyone explain how?

NOTE:The inclined plane is fixed

Can you provide specific references to some books that say that? — robphy, Feb 09 '22 at 05:29
well I don't have that book but this is what I remember from that page.This was the diagram — AJknight, Feb 09 '22 at 05:38
I'm hoping to see the context of some of the "many books" on the statement that momentum is conserved in this case. — robphy, Feb 09 '22 at 05:42
if Fnet$\ne 0$ then that block is accelerating, meaning $p=mv$ can't be constant. — Cotton Headed Ninnymuggins, Feb 09 '22 at 05:48
The book said that considering there is no friction the momentum is conserved in y direction but what I want to know is that if there is kinetic friction when the block slides can there be a scenario where the N" is equal to horizontal component of friction? — AJknight, Feb 09 '22 at 06:42
well I don't have that book but this is what I remember from that page. Are you sure that the example in the book was not for a wedge which could slide freely? Here is a recently posted example How is conservation of momentum applied even if there is component of mg acting as external force?. — Farcher, Feb 09 '22 at 09:15

score 1 · Answer 1 · answered Feb 09 '22 at 06:00

1

How could you say component of kinetic friction (kN) opposite to horizontal component of N let's say (kN.cosx) is not equal to horizontal component of N i.e. (N.sinx)?
Where 'x' is angle of inclined plane with horizontal. And if the x and coefficient of friction (k) is such that
kN.cosx = N.sinx
&
N.cosx + kN.sinx = Mg
then the net force acting on the block would be zero and it will slide down with constant velocity and hence momentum would be conserved. But if there exists a net force such that the block accelerates then no momentum is conserved.

answered Feb 09 '22 at 06:00

Apoorv Mishra

319

Yes I understood what you are trying to say but if you could also use some values to show this it will be crystal clear – AJknight Feb 09 '22 at 06:17
take k=1 and x=45degrees and put it in both the relations – Apoorv Mishra Feb 09 '22 at 07:11

score 1 · Answer 2 · answered Feb 09 '22 at 06:28

1

No books, or more specifically, no correct books, would say that momentum is conserved in this situation.

Momentum is conserved when the change in momentum is 0. When the mass of a system is constant, as it is here, the change in momentum is equal to the net force. If the net force is not 0 (that is to say, the mass is accelerating), the change in momentum is also not 0, and thus momentum is not conserved.

So is the momentum of your system conserved?

answered Feb 09 '22 at 06:28

bingus

11

The book said that considering there is no friction the momentum is conserved in y direction but what I want to know is that if there is kinetic friction when the block slides can there be a scenario where the N" is equal to horizontal component of friction? – AJknight Feb 09 '22 at 06:42
1

@AJknight If the book flatly stated "momentum is conserved in the $y$ direction" with no additional context, that is absolutely incorrect. However, what it is often done (and may be the case here) when dealing with situations involving an inclined plane, is to redefine coordinates so that the $x$-axis is parallel to the slope of the incline, and the $y$-axis is perpendicular to it. In this case, where the $y$-axis is redefined so that it is parallel to the normal force, momentum in the $y$ direction is conserved, but only if such a redefinition of coordinates is done. – bingus Feb 09 '22 at 19:55

score 0 · Answer 3 · answered Feb 09 '22 at 07:09

You need to take care when you apply the principle of conservation of momentum, as it always holds true only for an entire system of interacting bodies, and not for a subset of them. If a body accelerates down a ramp fixed to the Earth, its own momentum is clearly not fixed; however, its acceleration imparts an infinitesimal recoil to the ramp and the Earth, so the overall momentum is conserved.

You also need to take account of friction. At a macroscopic level, if a body is decelerated by frictional forces such as air resistance, the macroscopic momentum of the body is not conserved; however, were you able to observe and measure the effects of the countless collisions between the body and the individual molecules of air, you would find that the momentum lost by the body had been transferred to the air molecules, so momentum overall was conserved.

Conservation of momentum on an inclined plane

3 Answers3