#### Archived

This topic is now archived and is closed to further replies.

# Collision plus Impulse Force

This topic is 5857 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I need to determine the interaction between a volleyball player and a volleyball. If I assume that the volleyball player is standing motionless with his arms extended I can use the law of conservation of linear momentum to determine the ball''s velocity after it collides with the player. If instead of standing motionless the player applies an impulse force to the ball (affecting the ball''s momentum) I can not determine the ball''s velocity with the same formula as when the impulse force is not present. How do I determine the total effect of the collision and the impulse force on the ball''s velocity? I''m unable to figure out what the math should look like.

##### Share on other sites
I suppose you could make some assumptions about the amount of impulse the player is able to impart on the ball. This would be a function of the speed of the players hands or feet and the mass of the player. You could treat the impulse as a force being applied over one frame, or just calculate the acceleration that would be caused by a true instantaneous impulse.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

##### Share on other sites
The way I see it there are three important quantities to consider. There is the momentum of the ball at the time of the collision, the momentum of the player and his extremities at the time of the collision, and the impulse force provided by the player's muscles for a time interval (t0,t). Can I really just ignore the impulse force during (t0,t) and rely solely on the speeds of the colliding objects?

Imagine these three examples:

A) Player stands motionless with arms extended in a bumping stance. Player waits for the ball to hit his "rigid" arms.
B) Player assumes the same stance as before but instead of waiting for the ball to hit him he jumps into the air and collides with the ball while keeping his arms locked in place.
C) Player keeps his feet on the ground but instead of standing motionless he moves his arms to hits the ball, thus applying an impulse force to the ball for a time period (t0,t) during which the ball's momentum changes by AvgForce * (t - t0).

Can example C really be solved in terms of mass and velocity at the time of collision? Isn't the impulse force provided by the player's muscle external to the system so that it's not enough to rely on the conservation of momentum as in examples A and B?

If I use the law of conservation of momentum to determine the change in the ball's momentum I1 = m(v - v0) due to the collision, calculate AvgForce * (t - t0) to obtain another vector quantity I2 (the impulse provided by the player's muscles), and add the two vectors together, would this allow me to determine the ball's actual momentum after the collision?

I don't know that much about physics. Am I going about it the wrong way?

[edited by - chronos on November 5, 2002 2:43:54 AM]

##### Share on other sites
Am I making the wrong assumptions about the physics of striking a moving volleyball? Could somebody at least point me to a resource that might help me understand the physics of a moving object being struck by another that is currently accelerating?

Let''s say we have two spheres. The first sphere has a mass M1 and initial velocity 0, the second sphere has a mass M2 and initial negative velocity V2. The first sphere has an engine which is currently producing a positive force F1. The two spheres have a head-on (1D) elastic collision. How do I approach such a problem?

I''ve read about collisions in my physics textbook and am currently reading Chris Hecker''s Game Developer articles on dynamics. I''m still as confused as ever. Any help is appreciated, even if it''s just pointing me to additional resources.

##### Share on other sites
Okay, we just studied this in physics. So since the ball and the player basically constitue a closed system, momentum will be pretty much conserved. If the ball is object 1, and the player is object 2, m1v1 + m2v2 = m1v1'' + m2v2''. The '' means "prime" and is the velocity after the collision. For an accelerating object, you would just need to find the velocity at the time of the collision. The interaction time between the objects doesn''t affect the end results. So if the person (or just their arms) isn''t moving at the time of the collision, set v2 equal to zero. Now, since you don''t know either v1'' or v2'', you need a second equation, for kinetic energy: .5m1(v1)^2 + .5m2(v2)^2 = .5m1(v1'')^2 + .5m2(v2'')^2. That''s for a completely elastic collision, which this isn''t so you would need to adjust it a little by including a factor for the elasticity multiplied by the total initial kinetic energy. That''s gives you a linear system of two equations, solve how you prefer. Hopefully that was clear enough.

tj963

##### Share on other sites
I see what you're saying, but I still can't quite understand why the acceleration can be ignored. For example:

- An object with constant velocity and momentum M1V1 collides with an object with constant velocity and momentum M2V2. The objects stick together for some time period (T1, T1') and then go their separate ways.
- An object with constant acceleration and momentum M1V1 at the time of collision collides with an object with constant velocity and momentum M2V2. The objects stick together for some time period (T2, T2') and then go their separate ways.

Are the two situations equivalent? I don't understand why.

[edited by - chronos on November 6, 2002 9:56:46 PM]

##### Share on other sites
Actually, you have to put yourself in an instantenous moment to really understand why both situation are basically similar.

The only difference is that at the moment of impact, the velocity will be different. (supposing that 2 objects with the same velocity at the beginning arrive at the same point)

Acceleration is similar to an outside force which is applied on the objects. When you calculate the energy, you only need the velocity. The same principle apply to the conservation of momentum.

Nick

##### Share on other sites
But what if the momenta of the objects in the first example (M1V1 and M2V2) are the same as in the second example (M1V1 and M2V2) at the time of collision, the only difference being the acceleration of the first object in the second example? The momenta just before the collision are the same in both examples, but is the end result the same?

##### Share on other sites
quote:
Original post by chronos
But what if the momenta of the objects in the first example (M1V1 and M2V2) are the same as in the second example (M1V1 and M2V2) at the time of collision, the only difference being the acceleration of the first object in the second example? The momenta just before the collision are the same in both examples, but is the end result the same?

The situation is not the same, but for your purposes the difference will probably be negligible. Don't fret about your simulation being 100% accurate (impossible), as long as it looks/feels right.

If you care, here's why the situations aren't the same. If the volleyball is accelerating before the collision then it must be experiencing a net force, which we'll call F1. That means the net force during the collision will be the sum of F1 and the force of the player's hands, which we'll call F2. So the net force is F1+F2 if the ball is accelerating, but F2 if it isn't. F1+F2 != F2

I say the difference will be negligible because in your simulation F1 will be gravity, which will be much smaller than F2, the force exerted by the player's hands. So F1+F2 = (sort of) F2

edit: I suck at teh HTML

[edited by - Dobbs on November 7, 2002 1:44:59 AM]

##### Share on other sites
The accelerating sphere in my example is supposed to represent the player''s swing, to account for "follow through" impulse forces on the ball. I don''t intend to take the ball''s acceleration due to gravity into account at the time of collision exactly for the reasons you suggest. So it''s the swing''s acceleration that I''m trying to account for, though perhaps I''m wasting my time?

1. 1
Rutin
49
2. 2
3. 3
4. 4
5. 5

• 10
• 28
• 20
• 9
• 20
• ### Forum Statistics

• Total Topics
633409
• Total Posts
3011722
• ### Who's Online (See full list)

There are no registered users currently online

×