Elastic collision between two spheres

In summary: I don't think you need trigonometry to solve a collision. What you need is a vector calculus course.When the distance between the centres of two elastic spheres is equal to (or very slightly less than) the sum of their radii, the line between their centres is normal to the plane of contact at the point of reflection. You can solve that collision and the outcome with vectors in 2D or 3D.
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Tim667
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Suppose I have two spheres in 3 dimensions of equal mass. In cartesian coordinates, sphere A is traveling with velocity uAi, and sphere B travels with vBi. They will collide elastically.

I want to find the final velocities after the collision, ie uAf and vBf.

Am I correct in saying that elasticity means all kinetic energy from A will be transferred to B, and vice versa? In this case, uAf=vBi and vBf=uAi.

If this is not correct, is there a general expression for the final velocities of two colliding spheres, given two initial velocity vectors (in cartesian coords)?

Thank you
 
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  • #2
The final velocities depend on the variable scattering angle of the collision.
 
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  • #3
PeroK said:
The final velocities depend on the variable scattering angle of the collision.
Is there a way to do this with just the velocity vectors? Without the trigonometry?
 
  • #4
Tim667 said:
Is there a way to do this with just the velocity vectors? Without the trigonometry?
Possibly a counter-example will suffice. Reduce the problem to two dimensions. A pool table with conveniently frictionless felt and perfectly elastic balls.

We have the eight ball at rest in the middle of the table. We have the cue ball in motion on a course which will result in a head-on impact. For an elastic impact, your prediction is upheld. The two balls exchange velocities. The cue ball comes to a stop and the eight ball moves on with the same velocity that the cue ball had.

But if the impact is glancing so that the cue ball merely grazes the eight ball, what happens? Do the two balls still exchange velocities?

However, you ask whether we can somehow eliminate the trigonometry. Perhaps so. Let us adopt the center-of momentum frame of reference so that our coordinate system is anchored with its origin at a point midway between the two balls. Now rotate the coordinate system so that at the moment of impact, one ball is on the negative z axis, the other is on the positive z axis and the two balls momentarily touch at the origin. The impact imparts a purely vertical impulse to each ball.

Now you have each ball retaining their original x and y velocities and inverting their z velocities in the selected coordinate system. Still some trigonometry, but perhaps only a tolerable amount.
 
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Tim667 said:
Is there a way to do this with just the velocity vectors? Without the trigonometry?
What's wrong with trigonometry?
 
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Tim667 said:
Is there a way to do this with just the velocity vectors? Without the trigonometry?
When the distance between the centres of two elastic spheres is equal to (or very slightly less than) the sum of their radii, the line between their centres is normal to the plane of contact at the point of reflection. You can solve that collision and the outcome with vectors in 2D or 3D.
https://en.wikipedia.org/wiki/Elastic_collision#Two-dimensional

PeroK said:
What's wrong with trigonomery?
Spelling. Trigonometry was invented by the Devil, so fascist teachers could fail otherwise sane students. Trigonometry also gives, otherwise competent engineers, several places to get a sign wrong. God created vectors to save us.
 
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Well, but vectors in Euclidean space also use trigonometric functions like defining the angle through the scalar product ##\vec{x} \cdot \vec{y}=|\vec{x}| |\vec{y}| \cos \theta## or you need polar coordinates in the plane, cylindrical, and spherical coordinates in 3D space, etc.
 

1. What is an elastic collision?

An elastic collision is a type of collision between two objects where there is no loss of kinetic energy. This means that the total kinetic energy of the objects before and after the collision remains the same.

2. How does an elastic collision between two spheres differ from an inelastic collision?

In an elastic collision between two spheres, the total kinetic energy of the spheres remains the same before and after the collision. In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound.

3. What factors affect the outcome of an elastic collision between two spheres?

The outcome of an elastic collision between two spheres can be affected by factors such as the masses of the spheres, their velocities, and the angle at which they collide. The elasticity of the spheres and any external forces acting on them can also impact the outcome.

4. How is momentum conserved in an elastic collision between two spheres?

In an elastic collision between two spheres, the total momentum before the collision is equal to the total momentum after the collision. This is known as the law of conservation of momentum, which states that the total momentum of a closed system remains constant.

5. How is the coefficient of restitution related to an elastic collision between two spheres?

The coefficient of restitution is a measure of the elasticity of a collision. In an elastic collision between two spheres, the coefficient of restitution is equal to 1, indicating a perfectly elastic collision where there is no loss of kinetic energy. A lower coefficient of restitution indicates a less elastic collision where some kinetic energy is lost.

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