Fox 336

Let's mess around little bit of what we got, with a little twist of Physics. Don't blame us college kids! It is the truth that's calling. What an overwhelming call that is!

Do we need to know the object's mass? Probably not.

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Fox 272 - Solution

Yu claims the following:
Without the parabolic track a particle (initially at rest) at the vertex will be in free fall and drops at maximum acceleration of g. With the parabolic track the motion is restricted by the track and the particle is no longer in free fall; it slides down the track with increasing speed. Will it leave the track at some point?

All parabolic tracks are similar (See Fox 270). Consider the track y=-x^2 which ends at the point (x,-x^2).

When a particle slides from rest at the vertex (0,0) to the point (x,-x^2), it acquires speed v=sqrt(2gx) in the direction given by tanθ=dy/dx=-2x, where θ is the angle with the x-axis.
.: cosθ = 1/sqrt(1+4x^2), sinθ=-2/sqrt(1+4x^2).

At point (x,-x^2) the particle is in free fall,
v_H=sqrt(2gx)/sqrt(1+4x^2),
v_y=-2sqrt(2gx^2)/sqrt(1+4x^2).

After a further time Δt, Δx=sqrt(2gx)Δt/sqrt(1+4x^2),
Δy=-2sqrt(2gx^2)Δt/sqrt(1+4x^2) - (1/2)g(Δt)^2
= -2xΔx-(1+(1/4x^2))(Δx)^2,
and the particle is at y=-x^2 - 2xΔx -(1+(1/4x^2))(Δx)^2.

If the track did not end at (x,-x^2) but continued to infinity,
y=-(x+Δx)^2
= -x^2-2xΔx-(Δx)^2 > -x^2 - 2xΔx -(1+(1/4x^2))(Δx)^2.
.: The particle stays on the parabolic track.

Fox 272

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The claim assumes that there is a positive gravity in the given direction.

So, it should hold in Pluto as well.

Himalayan fox: Here is a claim ready to be blown away with a single counterexample. How strong is your theory Polar?
Polar fox: Well, I can smell it.
Himalayan fox: It demands more than that. Are you living in a paper castle ready to be drifted away by the next wind? Or are you a real believer of something that even the strongest quake can not shake you a bit - you stand firmly on the ground?
Polar fox: No problem on my stand man! I am not shaken at all, but... the thing is... the ice is melting!

Fox 253 Solution

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Bleaug goes on:

Energy acquired in free fall from rest position is 1/2.mV2=mg(h0-1/2)

Vertical velocity after bounce is V reduced by a factor equal to cos(2a), where a is the angle of tangent to 1/x at B: Vy = V.cos(2a) and tan(a)=-1/4

Highest point reached after bounce is such that: 1/2.mVy2=mg(h-1/2), so h = 1/2 + cos(2a)2.(h0-1/2)

From simple trigonometry we get cos(2a) = 1-tan(a)2/(1+tan(a)2) = 15/17

So h = 1/2 + (15/17)2 . (h0-1/2) = 1/2 + (225/289). 289/2 = 113.

Fox 253

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The answer is now confirmed! Will post later...

Fox 241

YES, the solution has NOW been confirmed!!

We will post these excellent solutions here later!

There are at least one more slick solution, not identified yet!

Thank you guys for the great work!

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