post-decay speeds

[Jeremy Hurwitz]

According to the amendment:

When a ball decays, it splits into two recently-discovered particles called a "posiball" and an "antiball", each with half the mass (but the same size) of the original ball. This event conserves momentum, but releases an energy equal to m*L2/s2, where m is the mass of the ball. After a ball decay, both the posiball and the antiball have the same speed as the original ball in the direction of the original ball's travel vector. In the direction perpendicular to the original ball's velocity, posiballs always go right, and antiballs always go left (with respect to the original ball's velocity).

Does this mean that the energy is "lost" or is it imparted to the new balls?

[Advay Mengle]

I think it must be imparted to the newly created balls.  The spec says that the speed of the new balls [edit]in the original direction[/edit] is that of the original ball.  It also says the momentum is conserved.  Therefore it would seem that this energy would create the speed in the component perpendicular to the original ball.

[akishore]

Advay,

The speed of the new balls is NOT equal to the speed of the original ball. Read the specs carefully, it says that the speed is equal only in the direction of the original ball's velocity.

So the new balls are actually going faster than the original ball. But I'm having a little trouble understanding what is the magnitude of the new balls' velocities in the perpendicular direction... any help?

Aseem

[Advay Mengle]

Sorry - you're right, I miswrote (I corrected it to avoid confusing anyone else).

I believe the perpendicular speed must be calculated by knowing that momentum is conserved and energy is conserved except for the amount that they say is released (added).

[Vincent Yeung (TA)]

I believe the perpendicular speed must be calculated by knowing that momentum is conserved and energy is conserved except for the amount that they say is released (added).

That's correct.

[akishore]

Great, thanks.

Aseem