Talking of puzzles, here's one that is actually something to do with classical physics. It involves kinetic energy and frames of reference so it's rather more relevant to this thread that the Monty Hall problem! (I'll also own up now to not really solving it correctly when I first heard it, although I think I was "heading in the right direction"!
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First, "
the rules" (it's an honour system)!
1. You may not post a solution on this thread after reading any other posted solution, and you agree to "hide" your solution between [spoiler]...[/spoiler] tags.
2. No public discussion for 24 hours after this is posted to give others a chance to post their solution.
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Imagine that we have a cart (an ordinary wheeled trolley, no propeller or anything complicated like that!) with mass m near the top of a sloping hill. As we're viewing it the hill slopes down from upper left to lower right where the ground gently levels out at a height of h metres (vertically) lower than the initial position of the cart. The cart is initially at rest on the hill.
We want to look at this in terms of conservation of energy.
Initially (say at time 0) the cart has no kinetic energy (w.r.t. ground) and potential energy of mgh (if we take the level area at the bottom of the hill as "zero" to keep things simple). When the cart is released, it begins to roll down the hill and ends up rolling at constant velocity v along the level area at some fixed time (t) after being released. At that stage it has zero potential energy and kinetic energy of mv^2/2. Our cart is very streamlined and has very low friction bearings, etc., so we'll ignore the tiny losses from that stuff during the roll down the hill.
At time 0, total energy = mgh
At time t, total energy = mv^2/2
Therefore, conservation of energy tells us that:
mgh = mv^2/2
Pretty much a standard high school type problem so far.
Now we'll look at the same thing again but this time use a frame of reference that moves to the right with the same velocity (v) that the cart ends up with in the earlier description. This means that in our new frame the cart's initial velocity (at the top of the hill) becomes -v, which also means it has non-zero kinetic energy to start with. At time t it will have zero velocity (it's at rest in our new frame) and so its kinetic energy is also zero at that point. Our new energy budget looks like this:
At time 0, total energy = mgh + mv^2/2
At time t, total energy = 0
Therefore, if energy was conserved we'd have:
mgh + mv^2/2 = 0
Clearly that can't be right. The cart had non-zero energy at the top of the hill but zero at the end. What has gone wrong?
You're beginning to sound a bit like humb there Tunny.
