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There is an inclined plane on a frictionless surface. A ball strikes the inclined plane horizontally with velocity $v_o$ and moves vertically after collision with velocity $v$ (see figure) mass of ball=$m$, mass of inclined plane=$M$

figure

Now we have to find the velocity of the ball after collision. Here's what I did,

figure 2

Now, $$mv_o-J\sin(\theta) = 0$$ $$J\cos(\theta)=mv$$

Therefore, $$v=v_o \cot (\theta)$$

The question also asks for the velocity of the inclined plane after collision.

So I did, $$J\sin (\theta) = Mv'$$ $$v'=mvo/M$$

All my answers are apparently correct according to the book but the momentum along y-direction remains unconserved even though there is no external force on the system in the y-direction. Why is this so?

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    $\begingroup$ Does this have anything to do with the normal reaction between the floor and inclined plane? $\endgroup$
    – Jennie
    Sep 17, 2021 at 12:11
  • $\begingroup$ Shouldn't $v'=\frac{mv_o}{M}$? Because $mv_o=Mv'$. $\endgroup$
    – Manu
    Sep 17, 2021 at 14:06

5 Answers 5

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Sorry but what I see in the y-direction, the sum of all forces is not zero, so it is not conserved. $$F_M=N\hat y-Mg\hat y=0$$ $$F_m=-mg\hat y$$ where $$\frac{dp}{dt}=\sum F=F_M+F_m=-mg\hat y \neq0$$

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  • $\begingroup$ Thank you for correcting my misconception that there is no external force in y-direction. But according to your dP/dt equation, the change in momentum is towards the negative y-axis which implies that the system must have a momentum in negative y-direction, which isn't the case? $\endgroup$
    – Jennie
    Sep 17, 2021 at 13:39
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Yes, it's because of the normal force between the floor and the wedge, which increases by the $y$ component of the normal force between the ball and the plane during the collision.

If the wedge and the ball collide in free fall, the system is the ball and the plane. We conserve momentum by tracking the velocity of the ball and the plane.

If the wedge and the ball collide in contact with the planet, the system is the ball, the wedge, and the planet. We conserve momentum by tracking the velocity of the ball, the wedge, and the planet. There's no planet in the x direction (and no friction), so in the x direction we don't need the planet's mass or $\Delta v_x$ to solve the x part.

In the y direction, there is a planet, so in the y direction we would conserve momentum by

$P_{y_{ball}}+P_{y_{wedge}}+P_{y_{planet}} = 0$

That is, the ball's temporary vertical speed is paid for by the planet+wedge very, very, very slightly moving downward.

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You've ignored that the inclined plane is in contact with the ground. And this being an impulsive interaction would provide some kind of an impulsive force and thus conservation of momentum would not be valid along the y-axis

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    $\begingroup$ Can I suggest that you rephrase your second sentence? As written, it could be read as saying that conservation of momentum doesn't hold for impulsive forces, which isn't true. I don't think that's what you meant, but it could be clearer. $\endgroup$ Sep 17, 2021 at 12:39
  • $\begingroup$ I understand what you mean, the normal reaction on the plane by the ground acts as an impulsive force negating the impulse given to the plane by the ball. Like @g s has stated the system is the plane, the ball and the earth. $\endgroup$
    – Jennie
    Sep 17, 2021 at 13:45
  • $\begingroup$ Momentum is always conserved, but it's not always constant. Impulse is the "current" or flow of momentum. Conservation is about limiting the creation or destruction of momentum. If the impulse on a system is zero, the momentum is *constant and conserved." If the impulse is not zero, momentum is still conserved (neither created nor destroyed) but the system momentum is not constant. $\vec{p}_f = \vec{p}_i + \vec{J}$. $\endgroup$
    – Bill N
    Sep 17, 2021 at 15:33
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The "system" with no external forces (ignoring gravity) for which momentum is conserved is the ball, wedge, and planet. The momentum of the wedge/planet after the collision is very small but is not zero. See the answer by @gs.

The ball itself does experience an external force: the impulsive force of the wedge on the ball during the collision, so the momentum of the ball alone is not conserved. Note momentum is a vector; if the ball has initial momentum in the horizontal direction and final momentum in the vertical direction, you can see that the angular momentum of the ball alone is not conserved, it changes due to the impulsive force from the wedge during the collision.

The force of gravity can be ignored if the collision is of very short duration, since the impulsive force is very large compared to the force of gravity in this case.

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    $\begingroup$ Momentum is always conserved, but it is not always constant within an object or a system of objects. Impulse is the transfer of momentum so that it is never created nor destroyed, i.e., conserved. If the system momentum changes, it flows in or out, but is conserved. $\vec{p}_f=\vec{p}_i+\vec{J}$ for everything! The impulse you use depends on your system definition. $\endgroup$
    – Bill N
    Sep 17, 2021 at 15:40
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I get those equations:

$$mv_{{{\it my}}}={\it dp}\cos \left( \theta \right)\\ Mv_{{{\it Mx}}}=-{\it dp}\sin \left( \theta \right) $$

and the relative velocity towards the normal vector equal zero

$$\left( v_{{0}}-v_{{{\it Mx}}} \right) \sin \left( \theta \right) +v_{ {{\it my}}}\cos \left( \theta \right)=0$$

where $~v~$ is the velocity and dp is the impulse.

you obtain 3 equations with 3 unknowns $~v_{my}~,v_{Mx}~,dp$

the conservation of momentum is: $$m \left( v_{{{\it my}}}+v_{{mx}} \right) +Mv_{{{\it Mx}}}-mv_{{0}}=0$$

where $v_{mx}=v_0$

substitute the results $~v_{my}~,v_{Mx}~$ of the equations :

$$ {\frac {mv_{{0}}M\sin \left( \theta \right) \left( -\cos \left( \theta \right) +\sin \left( \theta \right) \right) }{ \left( \cos \left( \theta \right) \right) ^{2}M+m-m \left( \cos \left( \theta \right) \right) ^{2}}} $$

thus the momentum is only conserved for $~\theta=\pi/4$

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