How do I use this confusing elastic collision equation?

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The discussion centers on the application of the elastic collision equation for relative velocities, specifically Va1x-Vb1x = -(Va2x-Vb2x), which is valid only in one dimension. Participants clarify that this equation does not extend to two or three dimensions, as it only addresses x-direction collisions. When considering initial conditions where objects have different velocities in 2D, the equation cannot be applied without violating conservation of momentum principles. The conversation concludes that for systems with no initial net momentum in the y or z directions, the equation is not applicable. Overall, the consensus is that the equation is limited to one-dimensional scenarios.
SecretSnow
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Hi guys, this is the equation of relative velocities in elastic collisions:
Va1x-Vb1x = -(Va2x-Vb2x)
I know that this means that the relative velocity of A to B before an elastic collision is the same in magnitude but opposite in direction after the collision. However, this applies only in the x direction (for this particular equation). My questions are:

1. Can it be applied to y and z directions?

2. If so, then if initially A and B are of different velocity in 2D (x and y directions only) and I compare their relative velocities of x and y before and after, does the equation applies?

3. If initially A is only moving in the x direction and B is at rest, and after the elastic collision A and B are scattered in an x-y plane, does the equation for relative velocity of y still apply? Note that initially A does not have any y velocity so if I apply the equation, they should have equal y velocities magnitude. Yet if they don't have the same mass, then the conservation of momentum in the y direction is broken since there's a net y momentum from 0 y momentum initially (note that the question doesn't have external forces at all) does this mean that it is not possible to use this equation in y and z direction? So that a system without initial net y velocity or momentum cannot use the equation given above even if its an elastic collision??

Thanks a lot guys!
 
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That equation works for one-dimensinal collisions only.
In two (or more) dimensions, you can get similar equations using momentum and energy conservation.
 
mfb said:
That equation works for one-dimensinal collisions only.
In two (or more) dimensions, you can get similar equations using momentum and energy conservation.

Oh you i just realized that. That equation don't work for anything more than 1D. Thanks!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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