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Jan 23rd, 2005, 12:07 PM
#1
Thread Starter
PowerPoster
Conservation of Momentum
So I slam into a brick wall going a Velocity of (5,0,5)
My weight is 140.
Momentum = Velocity * Mass
Now I am not sure how much momentum the wall absorbes (I would love to know how to calculate or at least a chart of basic ones like concrete, metals, wood).
I am also not sure on how to apply the rebound to the velocity.
Assuming the brick wall absorbes a scalar between 0..1 then scaling the vector down would be easy. Then I can just move back at the negetive of the velocity because Angle of Incidence == Angle of Refraction, so the angle I bounce should be the same as the angle I hit.
"From what was there, and was meant to be, but not of that was faded away." - - Steve Damm
"The polar opposite of nothingness is existance. When existance calls apon nothingness it shall return to nothingness." - - Steve Damm
"When you do things right, people won't be sure if you did anything at all." - - God from Futurama
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Jan 23rd, 2005, 12:59 PM
#2
Re: Conservation of Momentum
Depends how accurate you want to be. There are all kinds of problems with this kind of thing.
Your assumption of "Angle of Incidence == Angle of [Reflection]" is correct for rays of light and point particles but for real-world things like tennis balls.
Say a tennisball is heading for a wall at 45 degrees, the bal is not spinning in any axis. The angle of reflection is slightly different, because the ball loses some speed due to its rotational inertia (the wall would cause the ball to spin in the direction of rotation.
I think that in this case the AOR would be a little greater than the AOI (although that's in my brain and not necessarily true in reality )
I don't live here any more.
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Jan 23rd, 2005, 01:17 PM
#3
Thread Starter
PowerPoster
Re: Conservation of Momentum
No, I believe you are absolutely correct.
It is the idea of putting backspin on a pool ball to make that shot but not go into the hole.
Backspin also effects for example how far a golf ball flys and what happens when it hits.
Those little tiny super bouncy balls.
If you put a hard backspin on that bouncy ball it will hit the ground and bounce one way, then when it hits again it will bounce back the other way, when it hits again it will bounce the original way again..and this repeats back and forth until it stops bouncing.
What I need is a calculation that determines how much momentum or velocity is lost when I slam into a wall.
Assuming I hit the wall softly, there will be no rebound.
Assuming I slam into it going 10000km then there should be a somewhat large rebound...
"From what was there, and was meant to be, but not of that was faded away." - - Steve Damm
"The polar opposite of nothingness is existance. When existance calls apon nothingness it shall return to nothingness." - - Steve Damm
"When you do things right, people won't be sure if you did anything at all." - - God from Futurama
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Jan 24th, 2005, 08:35 AM
#4
Re: Conservation of Momentum
 Originally Posted by Halsafar
Assuming I hit the wall softly, there will be no rebound.
Assuming I slam into it going 10000km then there should be a somewhat large rebound...
Since no collision is 100% efficient you can simply use a multiplier in the range 0 to 1. Multiply the initial speed by this number (0.5 would indicate 50% efficiency) and you end up with your rebound speed. Then you can factor in other factors like aero-drag and whatever.
A hot squash-ball might have an efficiency of 65% (drop it from 100cm onto a concrete slab and it should bounce back up to a height of around 65cm.
I don't live here any more.
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Jan 31st, 2005, 07:27 PM
#5
Re: Conservation of Momentum
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