Ball bounce angle on inclined surface

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The discussion centers on calculating the velocity and direction of a ball after it strikes an inclined plane at a vertical velocity, focusing on elastic collisions. Participants explore various methods, including breaking down the velocity into components along and perpendicular to the inclined plane, which yields a horizontal velocity of v. There is a desire for alternative approaches, particularly those that do not rely on collision mechanics, such as conservation of energy. However, it is noted that while energy conservation can apply, it does not provide direction, making the collision method more straightforward for this scenario. The conversation reflects a search for diverse problem-solving techniques in physics, emphasizing the challenge of recalling previously used methods.
Faris Shajahan
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Homework Statement


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Ball strikes inclined plane of infinite mass with velocity v vertically. Elastic collisions. Velocity and direction after collision?

One way of solving is take components along and perpendicular to inclined plane and then solve easily.

Is there any way to solve is using energy conservation or some other way?

Homework Equations


Irrelevant. ;)

The Attempt at a Solution


The collision attempt (rather the solution):
Component ##\frac{v}{\sqrt{2}}## and ##\frac{v}{\sqrt{2}}##.
One component reversed, other the same.
Hence answer velocity v horizontally.

P.S.: This is NOT a homework question but for some reason the site admins don't accept that!
 
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What other methods did you have in mind?
Are you looking for a more general method that will work for different angles, inelastic collisions, etc.?
Conservation of energy is in effect in your method since ##\| v_{in}\| = \| v_{out}\|##
 
RUber said:
What other methods did you have in mind?
Are you looking for a more general method that will work for different angles, inelastic collisions, etc.?

I don't care. Just any other method other than the one I used. And I am kind of looking for a "not collision" method.
eg. Body thrown with velocity ##v## from surface of earth. Find final height ##h##. (assume ##g## constant)
Method 1: Newton's laws of motion and hence ##v = \sqrt{2gh}##
Method 2: Conservation of energy and hence ##\frac{1}{2}mv^2 = mgh##
So this is an example of what I am looking for.

RUber said:
Conservation of energy is in effect in your method since ##\| v_{in}\| = \| v_{out}\|##
We can't obtain direction by conservation of energy. Right?
 
You can break the vertical velocity of the ball before the bounce into a component parallel to the inclined plane and vertical to it. Since the plane is at 45 degrees to both horizontal and vertical, that easy. The two components must be equal- calling that equal length "x", we have x^2+ x^2= 2x^2= v so that x= \frac{v}{\sqrt{2}}
 
No matter what method you use, the answer will be the same. You will just be dancing around it differently.
In the problem you are given, using infinite mass for the inclined plane and velocity only for the ball, it lends itself to a collision method to solve.
Any other approach I can think of is unnecessarily complicated.
 
Reason I asked this is because I feel like I used to do it another way before. It just bugs me. I can't do anything else basically. It is an issue I have, if I forget something I can't do anything for the next few days!

So if anyone's got any other solution please post it if you don't mind. I don't mind how unnecessarily complicated it is. All you've got to do is hint me to your solution if it terribly long.

TIA
 
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