(I'm repeating myself a lot here, but it's because I want to make my confusion clear.)
If 2 billiard balls are the same exact mass, and one hits another stationary one head on, I have heard that the hitting ball will often stop entirely while the one which got hit will go off at the original velocity of the first one (ignoring friction and heat and other potential loss of energy).
I understand that this agrees with conservation of momentum. However, now that I am thinking about it, I am a little confused as to how it is possible.
Consider:
Let's say we have 2 balls, Ball 1 and Ball 2. The balls have equal mass. $$M_{1} = M_2$$
Ball 1 is the hitting ball, with an original velocity of $V_1$, and Ball 2 is getting hit, originally stationary.
Once Ball 1 hits Ball 2, it immediately starts accelerating it at the same rate that it decelerates. In other words, Ball 1 exerts the same force on Ball 2 that Ball 2 exerts on Ball 1. In this way, momentum is conserved, so that for any amount of momentum that Ball 2 gains, Ball 1 loses.
$$F_{1 \to 2} = F_{2 \to 1}$$
That's just Newton's third law.
Since they have the same mass, Ball 1 will decelerate at the same rate Ball 2 accelerates.
However, after a certain amount of time, both balls will have the same amount of momentum in the same direction. That is, Ball 1 will have been decelerated to ${V_{i}/2}$ and ball 2 will have been accelerated to ${V_{i}/2}$.
Ball 1 and Ball 2 at this point of equal momentum must have the same velocity, since they have the same mass.
Now, the only way for Ball 1 to exert a force on Ball 2 is for them to be in contact. However, the instant after they gain the same velocity, the 2 balls would no longer be in contact, as Ball 2 would now be moving away from Ball 1, or at least at the same rate as Ball 1.
That being said, it seems impossible that Ball 2 would ever become faster than Ball 1, since Ball 2 would only be able to be accelerated up to the point where it is going at the same velocity as Ball 1.
And it seems even more impossible for Ball 1 to stop completely and Ball 2 to go off at the original velocity of Ball 1.
What am I missing?
EDIT:
After reading some scrupulous comments, (not in a bad way though, I love it when there are new variables introduced and things pointed out, and all the "what if" questions allways make for awesome discussion, thanks for the comments!) I've realized its better to imagine the billiard balls as floating through space than on a pool table, to get rid of the possibility that spin will affect the collision.
Or we could just imagine a frictionless pool table.
But better to imagine them floating through space, because everyone loves space
Answer
Your analysis is correct up to the point when the two speeds are equal. This is also the point when the relative velocity is zero. But at this point the two balls are deformed (elastically). The forces acting on the two balls during approach is an elastic force. However small the deformation is, it is there. So what happens now is that indeed the two bodies start to fall apart (the distance between them increases) and they start to expand back to the original shape. The interaction force starts to decrease (does not drop to zero instantly). The work done during this second part of the process is the same as during the first part so the kinetic energies and velocities change by another on half. This is the difference between perfect elastic and non-elastic collisions. In the later case the two bodies do not expand back to the same shape (or not at all) so the exchange of velocities is not complete.
It may help if imagine a spring between the two bodies. When the velocities are equal, the spring is still compressed. Actually it has maximum compression. So obviously the force exerted by the spring is not zero as it starts to expand.
Note:A nice example is considering a rubber ball bouncing off a ground(suppose perfectly elastically), in that situation there is also a point where the relative velocity with the ground is zero(when the ball is momentarily at rest with the ground) but still the ball rises up due to elastic forces as it gains kinetic energy due to the energy stored through the deformation it had initially endured.
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