In burst Beyblades make each other skip teeth and then, burst. 

But a left spin vs right spin match would make them skip teeth backwards due to the fact the bursting gimmick requires both beys pushing in the same direction to actually make them skip teeth. Despite this, I have seen left vs right spin battles where one of them bursts and some where they act according to my thinking.

Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by y.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey

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Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by x.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey

Bravo, great post.

Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by x.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey

Bravo, great post.

Thanks ;D

Beys are full of physics

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Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by y.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey

This thing applies when somebody is using Drain Fafnir ( rubber appilies recoil to other beys when DF is launched lightly )

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 THE   Beyblader   from   ODISHA 

Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by y.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey

This thing applies when somebody is using Drain Fafnir ( rubber appilies recoil to other beys when DF is launched lightly )

Increases Friction with Soft Plastic.

Even the dragon Heads do the same

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Recoil.
a bit of physics, let the speed of slower spinning bey be X, and Speed difference be y
the faster spinning beys has speed, x + y , which is greater than by Slower spinning by y.
So if you Subtract Slower speed from both,
i.e., x-x=0; x+y-x= y
that means at some point of contact Slower spinning bey will act as stationary object, and faster bey will skip it's teeth out recoil from "stationary" bey

Well i like ur thinking 
Its great to so physics out here bcos i myself is bachelor in physics first year
So good eqn man
But not to be rude
If the bey is spun at an equal speed without any contact then it can be considered as
Same speed of spin - X
And the depth contact from opponent -Y
Then it would be said to be X =Y
But in case of rubber the momentum would be reduced to 0 by spin equalization
So anything cannot be calculated by theory
The external forces such as stadium,design of bey ,metal, would affect it ..
So anyways sorry if i sound nerd ish
Ur eqn is correct
Sorry if im being rude