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yitriana
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Why is the angle between products in non head on, perfectly elastic collisions always 90 degrees?
I think this is only true for two equal masses. You can get there like this:yitriana said:Why is the angle between products in non head on, perfectly elastic collisions always 90 degrees?
The recoils of the two products are always 90 degrees apart if both incident masses are the same. It is proved somewhere in Goldstein's book on Classical Mechanics. Both masses have to be equal, and one has to be stationary before the collision.yitriana said:Why is the angle between products in non head on, perfectly elastic collisions always 90 degrees?
You are correct. Just imagine what the velocities would be relative to an observer who is flying past at a speed, in the original frame, that is 100 times greater than either of the particles' speeds.Nanyang said:This moving off perpendicularly thing seems to only be possible from the frame of reference where originally one of the particles are at rest. I'm not sure about angles being preserved under a frame transformation... because if I consider relative to K' the velocities are perpendicular, but relative to K I have to add the relative velocity between K' and K and this would imply that relative to K the velocities are not perpendicular. I don't know if I'm correct on this point.
...not true.A.T. said:If this holds true in the frame of reference where B is initially at rest, it has to be true in every frame, because angles are preserved under frame transformations.
Yeah you're right, angles between velocities are of course frame depended. What was I thinking. So it's 90° only when one mass is at rest. Still useful in billiard.DrGreg said:Just imagine what the velocities would be relative to an observer who is flying past at a speed, in the original frame, that is 100 times greater than either of the particles' speeds.
I used to believe that billiad ball collisions were ideal two-body collisions. But 2/7 of the total kinetic energy of a billiard ball is rotational, not linear (translational) kinetic energy, because their moment of inertia is 2/5 mR2.A.T. said:Yeah you're right, angles between velocities are of course frame depended. What was I thinking. So it's 90° only when one mass is at rest. Still useful in billiard.
An elastic collision is a type of collision between two objects where there is no loss of kinetic energy. This means that the total energy of the system before the collision is equal to the total energy after the collision.
The angle of 90 degrees in an elastic collision signifies that the two objects involved in the collision are perpendicular to each other. This means that the collision is happening directly from one side to the other, rather than at an angle.
The angle in an elastic collision affects the direction of motion of the objects after the collision. If the angle is 90 degrees, the objects will continue to move in the same direction as before the collision, but with a change in velocity.
The angle in an elastic collision can be affected by the mass and velocity of the objects involved. A higher mass or velocity can result in a larger angle, while a lower mass or velocity can result in a smaller angle.
The angle in an elastic collision can be calculated using the law of conservation of momentum and the law of conservation of energy. By solving for the angle using these equations, the angle can be determined for any given elastic collision.