# Questions on force, collisions

1. Nov 3, 2012

### batballbat

1. An object of unit mass moving with unit speed applies unit force to any object when the two collide. True or False?
How can this definition shown to be equivalent to the definition F=ma of newton. Here its different because we are considering how much a body applies a force not how much force is being applied on the object as is done in classical mechanics. Or the two definitions are exclusive?
And similar manipulation for objects accelerating.

2. Is the time of impact during collision dependent on the parameters: mass, velocity of the bodies colliding?

3. Are the answers of this question not subtle and easily deducible from classical mechanics?

2. Nov 3, 2012

### AJ Bentley

1/ False.

2/ No. A collision between protons for example begins and ends at infinity - it could take millions of years. The collision between neutrons OTOH probably takes less than a femtosecond.

3/Yes.

3. Nov 3, 2012

### batballbat

why false?
What does gaussian unit of force mean?

So when time of impact is constant for all collisions in classical mechanics?

4. Nov 3, 2012

### AJ Bentley

When a mass collides with another, the force is not simply determined by the speed and mass. It is not even necessary for the force to be constant during the collision. The statement is false.

There is no such thing as a 'Gaussian Unit of Force'

I don't understand your last question 'So when time of impact is constant for all collisions in classical mechanics?'

5. Nov 3, 2012

### batballbat

So how do we determine the force? What else is required?
And how do we determine the time of impact when two bodies collide? And what about the direction of their motion after the collision?

6. Nov 3, 2012

### AJ Bentley

Generally in physics we study particular types of collision. In some of them (such as colliding billiard balls) the forces are irrelevant. In those cases we just use momentum/energy calculations.

In others, such as electrostatic collisions between charged particles, we use the known forces to calculate the changes in momentum of the particles. The forces are already known from the field equations.

If you ask a specific question about a real collision, we can probably explain in more detail. But your original question is too vague.

7. Nov 3, 2012

### batballbat

and answer this and then i will quit studying physics:
Is the direction of the motion of bodies after collision determinable? If so, how?

8. Nov 3, 2012

### arildno

We do not need access to the rate of momentum transfer (that is, FORCE) as long as we know, or premise on good grounds, that 1) momentum is conserved, and 2) energy is conserved, or the energy amount lost.

9. Nov 3, 2012

### Staff: Mentor

If you know details about the collision process and the objects, yes. This includes knowledge about the forces during the interaction - you can get them if you know more details about the setup.

If you just know some initial parameters (position, velocity and mass of the objects), it is not possible.

10. Nov 3, 2012

### batballbat

what details mfb?
@arildno: Ok so the definition is not correct. Then how do we solve that puzzle? When we analyse the conservation of momentum we analyse it in terms of a force and using newton's third law. This force should be determinable.

11. Nov 3, 2012

### Staff: Mentor

To solve a problem like this, you need to consider the deformational mechanics of the two bodies during the collision (assuming the bodies are large enough to be considered macroscopic solids). You would use a combination of Hooke's law of elasticity in tensorial form together with Newton's second law, and the analysis would result in a set of partial differential equations to solve. Both objects would undergo deformation during the collision, and this deformation would be non-uniform and time dependent. So the objects would not be treated as rigid bodies during the collision.

12. Nov 4, 2012

### batballbat

@Chestermiller
thanks. any resources where i can see these calculations?

13. Nov 4, 2012

### Staff: Mentor

A book giving solutions to Theory of Elasticity problems like Roark. You might also check Timoshenko.

14. Nov 4, 2012

### Staff: Mentor

Shape, rigidity, friction, maybe other material constants.

15. Nov 4, 2012

### batballbat

so perfectly elastic collision then. How do we go about solving this?

16. Nov 4, 2012

### Staff: Mentor

I already explained how to solve it for an elastic collision. You would have to solve the partial differential equations for the transient elastic deformation (displacements as a function of position and time). It is relatively easy to solve the problem for an elastic head-on collision of two cylinders. Then the problem is 1D. For spheres, it is much more complicated, and you might want to use finite element. It is also possible to solve a problem like this for viscoelastic materials, and also for materials which exhibit yield (in which case there is energy dissipation). Other interesting problems of this type are golf balls deforming while being hit by golf clubs, and baseballs deforming while being hit by baseball bats. All these problems can be solved using finite element, provided you have adequately characterized the stress-strain (rheological) response of the material.

17. Nov 4, 2012

### batballbat

ok atleast i know that it is subtle. And now have reasons to curse some teachers and physics books.
one last stupid question:
When two perfectly moving bodies moving towards eachother collide. How do we determine the minimum force that will push the body in opposite direction?

18. Nov 4, 2012

### Staff: Mentor

What is a "perfect movement"?
There is no minimal force, unless you specify a maximal range of the interaction.

19. Nov 4, 2012

### Staff: Mentor

Maybe this will help. For two identical elastic cylinders that collide head on, each traveling with velocity v prior to collision, the solution to the elasticity problem is as follows:

The contact force F is constant during the course of the collision, and given by

F = ρvAVS where

ρ is the density of the metal, A is the cross sectional area, and VS is given by:

VS = √(E/ρ) where

E = Young's modulus of the metal. Also, VS is equal to the speed of sound in the metal.

The time of contact is given by:

tc = 2L/VS

The impulse of the force is equal to the integral of the force with respect to time, and is given by:

I = F tc = (ρAL) (2v) = 2 m v

which, as expected, is the change in momentum of each cylinder.

20. Nov 4, 2012

### batballbat

one final question
what happens during collision? The two objects collide and are at rest for a moment when they compress. And then when they expand a force is applied to them and they move? But why not move always in opposite directions? Why does one retain its direction and the other change direction? Why does one stop and the other move in opposite direction?

21. Nov 5, 2012

### Staff: Mentor

The force can be perpendicular to the initial direction, or point in some other direction - it depends on the point (better: area) where they touch each other.

22. Nov 5, 2012

### Staff: Mentor

Good question. They begin compressing as soon as they make initial contact, but the compression of each cylinder is not uniform. There is a compressed region that begins forming at the contact end of each cylinder, and this compression region grows in size with time. The boundary of the compression region travels down each cylinder at the speed of sound (in the metal). Beyond the compression region, no deformation has occurred yet, and the material continues traveling with the initial velocity of the cylinder, unaware that anything has happened at the contact end. Finally, the compression region occupies the entire length of the cylinder, and it is at this point that the entire cylinder is at a dead stop. This happens at the half-way time of the contact. After this, the compression is released, starting at the end of the cylinder away from the contact point, and working its way back toward the contact point. In the region where the compression has been released, the velocity is now -v, while in the region closer to the contact point, it is still zero. The region where the compression is released travels along the cylinder at the speed of sound, and eventually reaches the contact end of the cylinder. At this time, the entire cylinder is traveling at -v, and the two cylinders lose contact. The time for the compression phase is exactly equal to the time for the compression release phase, which is equal to the length of the cylinder divided by the speed of sound. During the entire time interval of contact, the force between the cylinders is constant.

Incidentally, in an elastic collision of identical cylinders, both cylinders reverse direction symmetrically.

23. Nov 5, 2012

### batballbat

ok thank you guys for help.
But after the collision why do some objects continue in their initial direction? There is rest instance during the impact.

24. Nov 5, 2012

### Staff: Mentor

Objects can continue in their original directions if they have greater momentum than the object they are colliding with. Only if the two objects are identical in geometry and momentum can there a moment in time in which no portion of either mass is moving.