Hockey Puck Physics

[Maybeso]

The second puck will win because it spends time travelling faster than the first puck.

Originally Posted by Wackerbarth

The horizontal speed of the second puck is NEVER less than that of the first puck.

Not true, the transfer of vertical velocity to horizontal velocity using the downslope works the other way on the upslope.

Consider the case where the second puck travels over a parabolic hill above the height of the first puck -- horizontal velocity is changed to vertical, which becomes potential energy, which becomes vertical velocity going down the other side, which becomes horizontal again as it levels out ...

Now apply this to a parabolic valley. Steepen the upslope side to *near* vertical. The second pucks horizontal velocity will be less than the first pucks. I will restate your rule with a qualifier:

If the horizontal speed of the second puck is EVER less than that of the first puck, it will become airborne when it exits the valley.

So, to make the second puck lose; start both pucks with a fairly high initial velocity, drop the second puck down a bit, then give it a ski-jump that kicks it way up in the air to land right on the finish line. From the end of the jump to the finish line, its horizontal speed is less than the first puck, so it will lose.

[epatka]

I think the first one must be faster. The second has longer to travel and will be slowed down by the up and down movement.

I think the slopes are not smooth enough to let it glide freely. It must lose time and momentum when it hits the bottom and then the other side of the slope.

[epatka]

If the horizontal speed of the second puck is EVER less than that of the first puck, it will become airborne when it exits the valley.

Yes!!!!!!! :)

[cheesehead]

Okay, I think it's been established that ee meant the pucks to be in a completely frictionless environment. The pucks are completely rigid, so that there is no energy lost to deformations, and perfectly elastic, so that no energy is lost when a puck bounces.

However, in the absence of other idealizations, the shape of the second puck's downslope and upslope can affect the answer. Some upslope profiles could stop the second puck's rightward horizontal motion (e.g., when it ran into the bottom of a vertical or more-than-vertical (more than 90 degrees) upslope) by collision. Also, certain less-than-90-degree upslopes could convert so much of the puck's horizontal speed to vertical speed (upon bounce) that it would spend enough time in the air for the first puck to pass it anc cross the finish line first.

So apparently we need a further idealization: both pucks are constrained to travel along the (frictionless) surfaces, never above/below them, in a manner which does not cost either puck any energy or allow either puck to bounce. Also, we do need a constraint on the slope profiles in the second puck's case - the upslope can't be vertical or lean backwards (to the left), and neither can the downslope be pathological, for instance.

[eepiccolo]

Originally Posted by cheesehead

Okay, I think it's been established that ee meant the pucks to be in a completely frictionless environment. The pucks are completely rigid, so that there is no energy lost to deformations, and perfectly elastic, so that no energy is lost when a puck bounces.

However, in the absence of other idealizations, the shape of the second puck's downslope and upslope can affect the answer. Some upslope profiles could stop the second puck's rightward horizontal motion (e.g., when it ran into the bottom of a vertical or more-than-vertical (more than 90 degrees) upslope) by collision. Also, certain less-than-90-degree upslopes could convert so much of the puck's horizontal speed to vertical speed (upon bounce) that it would spend enough time in the air for the first puck to pass it anc cross the finish line first.

So apparently we need a further idealization: both pucks are constrained to travel along the (frictionless) surfaces, never above/below them, in a manner which does not cost either puck any energy or allow either puck to bounce. Also, we do need a constraint on the slope profiles in the second puck's case - the upslope can't be vertical or lean backwards (to the left), and neither can the downslope be pathological, for instance.

Thanks, cheesehead. Though I am an electrical engineer with a pretty decent background in physics, you've made a better technically sound description than I could of what's going on.

Anyway, even as I give the answer now, I'm sure some may still contend otherwise, but indeed the second puck wins.

Due to everything chessehead and I have stated, the total energy of the pucks remains constant. But the second puck gains kinetic energy due to the loss of potential energy, due the going in the dip. So the second puck pulls ahead while it is in the dip. Once it leaves the dip, it slows back down to the original speed, but that speed is identical to the first puck's and the second puck is already ahead, so the second wins.

I realize now that we could have avoided a lot of confusion if I had defined that the puck never leaves the track (in other words, it never goes airborne), but that's just the way it goes.

[epatka]

Hi eepiccolo,

I am not trying to argue with anyone, because I am a layperson, but has this been tried and proven in real life?

Hockey pucks are quite heavy. I can't imagine that a hockey puck would follow this slope. It looks more that once it reaches the slope it will fly in the air and land almost in the middle of the slope... Of course it depends on the speed of the puck. And for some reason the puck on your picture is too wide to follow the down curve of the slope.....

I am not questioning what you and cheesehead said, but if this is true we should tell the speed skaters about it:)

Thanks for the mental stimulation anyway :)

[eepiccolo]

Originally Posted by epatka

Hi eepiccolo,

I am not trying to argue with anyone, because I am a layperson, but has this been tried and proven in real life?

Hockey pucks are quite heavy. I can't imagine that a hockey puck would follow this slope. It looks more that once it reaches the slope it will fly in the air and land almost in the middle of the slope... Of course it depends on the speed of the puck. And for some reason the puck on your picture is too wide to follow the down curve of the slope.....

I am not questioning what you and cheesehead said, but if this is true we should tell the speed skaters about it:)

Thanks for the mental stimulation anyway :)

I'm not a great artist :( , the picture is really just to give a rough idea of what's going on.

The most important things are the assumptions given, which include that the puck doesn't leave the surface.

The easiest way to try this in real life is with rolling marbles, since that would have the least amount of friction. We actually tried this in my first physics class at college, and trust me, it works.

The reason I don't use rolling marbles in the example is the unavoidable problem of angular momentum, but I won't explain that unless someone asks.

[epatka]

Thanks for the explanation. I think I should learn to keep my mouth shut :)

[eepiccolo]

Originally Posted by epatka

Thanks for the explanation. I think I should learn to keep my mouth shut :)

Not at all! There's no such thing as a stupid question.


(Well, that's not entirely true, but none of the questions posted so far here have been anywhere near stupid :) )