Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 
Sep 15th 2011, 04:27 PM

#1  Junior Member
Join Date: Sep 2011
Posts: 1
 How to calculate resultant force\velocity vector after collision in 3d space?
I basically need to calculate the resultant force\velocity vector caused by hitting a racket with a ball, and it is in 3d space. Basically i need to implement the effect of motion in a game that i am developing in unity3d, and the built in Physics engine calculates the effect of ball bouncing off a static surface or if it is moving with a velocity, but it doesn't calculate the hitting force if the racket is being moved directly by mouse. So i have to write the code which does it when the ball impacts the racket, and changes the velocity of the ball accordingly.
I have the velocity of the ball, velocity of the racket, and at the time of collision, i can get the normal of the surface it collided with. With this information, is it better to calculate the resultant force of the impact (i also have the mass of the racket and the ball and the engine implements real world physics with great precision, and the mass i assign to the ball, the friction of the surfaces, the bounciness of the surface etc, all matters) i just need to do the calculation correctly and i can add force, add velocity or assign a completely new velocity to the ball) or should i calculate the resultant velocity? I think velocity is more easy but again i am not sure.
What can be more accurately calculated, and how should i calculate (formula)?
Velocity, normal and Force are in (x, y, z) components, all vectors are in this form, and i am also attaching a snapshot of the engine to give a better idea. Coordinate system is the same as the one used in Physics books all over the world.
Also the engine allows me to calculate angles (i give two vectors and it returns the angle between them, but i didn't find it very useful, because it needs three angles to calculate the resultant force or velocity in 3d space, i can get individual angles by setting one component of a vector to zero).
And if need arises, i can completely override the internal Physics engine and make it calculate things the way i want, but that may not be necessary in this case.
Last edited by SpeedBirdNine; Sep 15th 2011 at 04:31 PM.

 
Sep 16th 2011, 12:31 AM

#2  Physics Team
Join Date: Jun 2010 Location: Mauritius
Posts: 609

Hm.. I can suggest making use of momentum, which is basically:
MU + mu = MV + mv
M is the mass of the first object,
U is the initial velocity of the first object,
m is the mass of the second object,
u is the initial velocity of the second object,
V is the final velocity of the first object (after collision),
v is the final velocity of the second object.
And from there, you can work out the force, using:
F = d(mv) * t
d(mv) is the change in momentum of either object and does not depend on the object, as according to Newton's third law, both the racket and the ball will exert the same force on each other.
EDIT: Change "F = d(mv) * t" to "F = d(mv) / t"
__________________
Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet.
No one can go back and change a bad beginning; but anyone can start now and create a successful ending.
If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
Last edited by Unknown008; Sep 16th 2011 at 09:23 AM.
Reason: Typo

 
Sep 16th 2011, 08:04 AM

#3  Senior Member
Join Date: Dec 2009
Posts: 209
 Yes, except ...
Originally Posted by Unknown008 Hm.. I can suggest making use of momentum, which is basically:
MU + mu = MV + mv
M is the mass of the first object,
U is the initial velocity of the first object,
m is the mass of the second object,
u is the initial velocity of the second object,
V is the final velocity of the first object (after collision),
v is the final velocity of the second object.
And from there, you can work out the force, using:
F = d(mv) * t
d(mv) is the change in momentum of either object and does not depend on the object, as according to Newton's third law, both the racket and the ball will exert the same force on each other. 
Hi there,
I have read many of your replies to other people's post's and I always find them useful. I think you gave a good answer here also. I think maybe though you may have a "typo".
Impulse and momentum are related and in fact a change in one yields a change in the other*, so:
F = ma
F = mdV/dt
Fdt = mdV, or put in non calculus form:
Fdelta(t) = mdelta(V).
So where you have, "F = d(mv) * t", I think you meant to put:
Fdt = mdV, or F = d(mV)/dt.
I like Fdt = mdV because one can integrate each side and then solve for the force applied over some period of time that caused the yielded velocity.
Your answer of setting the initial momentum of everything equal to the final momentum of everything, and relating it to impulse, I like because of course momentum is conserved.
Yes, no?
Many Smiles,
Craig
* The exception is when an impulsive force is not "enough" to cause a change in momentum. Like waking over to a wall and putting your hand against it. You may apply say 5 Netwons for 2 seconds, but the change in momentum is zero. An impulsive force will always change a mass's momemtum when that mass is free to move; perhaps not by a "detectable" amount for a small force over a short time upon a large mass, but of course it does.
Last edited by clombard1973; Sep 16th 2011 at 08:29 AM.

 
Sep 16th 2011, 09:23 AM

#4  Physics Team
Join Date: Jun 2010 Location: Mauritius
Posts: 609

Oops, you're right, I was thinking of too many things at the same time, as always
That said, I also forgot to mention that you will have to find the overall velocity SpeedBirdNine. And I think that's it
__________________
Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet.
No one can go back and change a bad beginning; but anyone can start now and create a successful ending.
If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying? 
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