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The Proof is 'not-so' Trivial - Physics Edition

This thread is a Physics based companion to the maths thread here.

Similar rules apply. This is not a thread to ask homework questions, use the main Physics sub section for that.

If you post a question, please ensure it's possible to answer, and you already know the solution for it.



Post questions in this syntax:


Heisenberg
Problem 0*


A car starts from rest and accelerates uniformly over a time of 7.98 seconds for a distance of 100m. Determine the acceleration of the car.


Dirac

Solution 0


s=ut+12at2⇒a=2st2 s = ut + \dfrac{1}{2}at^2 \Rightarrow a = \dfrac{2s}{t^2}

Therefore:

a=2×1007.982≈π ms−2 a = \dfrac{2\times 100}{7.98^2} \approx \pi \, ms^{-2}




Like in maths, the asterisks denote the knowledge required of the set problem (not difficulty), please use them accordingly. The legend is as follows:

* - A-Level knowledge required.
** - Knowledge beyond A-Level spec required, but shy of undergraduate level.
*** - Undergraduate knowledge required.



Every problem and solution must use LaTex.

To reiterate, please post a question with a definite solution, refrain from posting something such as:


Hawking


How big is the universe?





All hints must be placed correctly into spoilers, and any long solutions too.



Reserved for problems and solutions. If this becomes too long I'll make an online spread sheet.
(edited 9 years ago)

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Reply 1
Avatar for Phichi
Phichi
OP
11
Problem 1***

When a proton, p, collides at sufficiently high energy with another proton, a proton anti-proton pair can be created in addition to the original protons.

p+p→p+p+(p+pˉ) p + p \rightarrow p + p + (p + \bar{p})

How much addition energy is required to create these particles?
Original post by Phichi
Problem 1***

When a proton, p, collides at sufficiently high energy with another proton, a proton anti-proton pair can be created in addition to the original protons.

p+p→p+p+(p+pˉ) p + p \rightarrow p + p + (p + \bar{p})

How much addition energy is required to create these particles?


Ok I tried again and this probably isn't right either but I got 2097 MeV?
Original post by Phichi
Problem 1***

When a proton, p, collides at sufficiently high energy with another proton, a proton anti-proton pair can be created in addition to the original protons.

p+p→p+p+(p+pˉ) p + p \rightarrow p + p + (p + \bar{p})

How much addition energy is required to create these particles?


Solution 1

No calculations really needed.

Consider the centre of mass frame of the system. After the collision, assuming the initially colliding protons had the minimum possible energy to achieve pair creation, there is no left-over kinetic energy and all the energy is contained in the rest mass of the four (anti)protons. The kinetic energy prior to the collision must therefore be equal to the rest energy of two new protons, divided equally between each initial proton.

Now, I suspect you might have wanted the frame where one initial proton is stationary, although you didn't explicitly ask for it. So if in the COM frame the initial protons are moving together at speed v, the frame at which one of the initial protons is stationary is at speed v relative to the COM frame. By considering the physics in the COM frame it is obvious that the products are are moving with the same velocity in the new frame, also v. This v is the speed required to give each proton a kinetic energy equal to its rest mass, which is obviously equivalent to a gamma of 2. This means all 4 final (anti)protons have a gamma of 2.

Now, consider energy in this new frame. The energy of the products is 4x2 mc^2. The energy of the initial stationary proton is mc^2. By conservation of energy the total energy of the other proton is 7mc^2.
(edited 9 years ago)
Problem 2*

A sphere of mass 3 kg is rolling along a surface (without slipping). It moves with a constant velocity of 2 ms-1 until it reaches a ledge and lands in a fluid where all of its energy is converted into rotational energy and it stays spinning in the fluid. Given the radius of the sphere is 7 cm, calculate the angular velocity of the sphere after it lands in the fluid and which direction the angular velocity vector points (into screen/out of screen)

Spoiler

(edited 9 years ago)
Problem 3***

Consider a quantum mechanical particle with energy greater than , traveling in the +ive x direction, passing over a potential well of length L such that
V(x)=VoV(x)= V_o when x<L x < L, 0 0 when 0≤x≤L 0 \leq x \leq L and Vo V_o when x>L x > L
What are the forms of the eigenfunctions in each of the regions defined by the potential? Determine an expression for the reflection coefficient.
(edited 9 years ago)
Original post by Rinsed

By considering the physics in the COM frame it is obvious that the products are are moving with the same velocity in the new frame, also v. This v is the speed required to give each proton a kinetic energy equal to its rest mass, which is obviously equivalent to a gamma of 2. This means all 4 final (anti)protons have a gamma of 2.

I don't quite see how if the initial velocity of the moving proton before the collision is v, then this implies that the products after the collision are also moving at velocity v?
Reply 7
Avatar for Phichi
Phichi
OP
11
Original post by Rinsed
(..)


Would be quicker, IMO, and more rigorous to consider the conservation of four momentum.


Posted from TSR Mobile
Original post by langlitz
I don't quite see how if the initial velocity of the moving proton before the collision is v, then this implies that the products after the collision are also moving at velocity v?


Different frames. The products are all at rest in the COM frame. The COM frame is at speed v relative to the frame where one initial proton is stationary.

Original post by Phichi
Would be quicker, IMO, and more rigorous to consider the conservation of four momentum.


Posted from TSR Mobile


Erm, that's perfectly rigorous thank you. Energy is one component of 4-momentum, the conservation of which is considered. The momentum components are taken care of by noting the behaviour in the COM frame. To set it up a 4-vector equation, you need to consider the COM frame anyway to note that the products necessarily coalesce for minimum allowable energy. At that point you've solved it in that frame, and it's trivial to transform to other frames. There's no reason to use invariants or whatever in such a simple example.

I will concede it takes a little more explaining than writing out 4-momenta, but calculation wise it couldn't really be simpler.
Reply 9
Avatar for lerjj
lerjj
13
Problem 4:

Probably too easy but still: show that for an object to float, it must have a lower density than the surrounding fluid without assuming Archimedes' principle (i.e. show that the up-thrust equals the weight of the displaced fluid).

Difficulty: * (could just about be done with GCSE knowledge...)
(edited 9 years ago)
Original post by lerjj
Problem 4:

Probably too easy but still: show that for an object to float, it must have a lower density than the surrounding fluid without assuming Archimedes' principle (i.e. show that the up-thrust equals the weight of the displaced fluid).

Difficulty: * (could just about be done with GCSE knowledge...)

Solution 4

Spoiler

(edited 9 years ago)
Reply 11
Avatar for lerjj
lerjj
13
Original post by langlitz
Solution 4

Spoiler



Yep.
Problem 5*/**

After a long day at work Walt returns home with a pizza to share with his wife Skyler. However, on discovering that Walt is the cook, Skyler rejects his kind offer of pizza. In his fury Walt steps outside and launches the pizza onto the roof of the garage.


Walt throws the pizza from a horizontal distance of 2 m from the garage door, which has a roof starting at a height of 3m above the ground. Given that he projects it from a vertical height of 1m, and can throw a pizza at 10 ms-1, calculate the minimum angle of projection required in order for the pizza to reach the roof.


When launched, the pizza rotates at a constant rate of 90 rpm. Calculate the angular distance traveled, in radians, during the time of flight.


The garage roof is sloped at an angle of 40o to the horizontal. Given that the coefficient of static friction between roof and pizza is 1.0 determine whether or not the pizza will slide off the roof.

Assume gravity to be 10 ms-2 and ignore air resistance for all questions.
Reply 13
Avatar for lerjj
lerjj
13
Original post by langlitz
Problem 5*/**

After a long day at work Walt returns home with a pizza to share with his wife Skyler. However, on discovering that Walt is the cook, Skyler rejects his kind offer of pizza. In his fury Walt steps outside and launches the pizza onto the roof of the garage.


Walt throws the pizza from a horizontal distance of 2 m from the garage door, which has a roof starting at a height of 3m above the ground. Given that he projects it from a vertical height of 1m, and can throw a pizza at 10 ms-1, calculate the minimum angle of projection required in order for the pizza to reach the roof.


When launched, the pizza rotates at a constant rate of 90 rpm. Calculate the angular distance traveled, in radians, during the time of flight.


The garage roof is sloped at an angle of 40o to the horizontal. Given that the coefficient of static friction between roof and pizza is 1.0 determine whether or not the pizza will slide off the roof.

Assume gravity to be 10 ms-2 and ignore air resistance for all questions.


Solution 5
Is this an actual episode? Spoilers, seriously!

Spoiler



Was there a nicer way to do this? The other two parts should be pretty easy, I'll write them up in a sec.

Spoiler



Spoiler

(edited 9 years ago)
Reply 14
Avatar for lerjj
lerjj
13
Problem 6 **

You've just bored a large tunnel through the Earth from pole to pole. The tunnel runs straight through the centre because we've modelled the Earth as a uniform spherical mass of radius R and mass M (and no lava/radioactive iron in sight). When you drop a mass down the tunnel, it eventually comes back up to you can starts heading back down again.

1) Why didn't you bore the tunnel at the equator? Why does this not work there?

2) Show that the motion of the mass is simple harmonic and find the period. What size orbit does this correspond to?


For 1) The model of the Earth is still rotating in space. What effect does this cause at the equator?
(edited 9 years ago)
Reply 15
Avatar for Phichi
Phichi
OP
11
Original post by lerjj
Is this an actual episode? Spoilers, seriously!

Spoiler



Was there a nicer way to do this? The other two parts should be pretty easy, I'll write them up in a sec.

Spoiler



Spoiler



You said find the minimum angle of which the pizza will slide. The pizza will slide when the component of weight down the slope is greater than friction, cos≥sinθcos \geq sin \theta would yield that it won't slide between 0≤θ<450\leq \theta<45

Also, add the solution title like in the OP, so I can keep track.
(edited 9 years ago)
Reply 16
Avatar for Phichi
Phichi
OP
11
Original post by lerjj

Spoiler



Solution 6


1). Will hit the walls during its descent, due to the rotation of the planet. If you bore through the axis of rotation, as suggested, this won't occur.

2).

Simple:

Spoiler



Will update with a more complex solution revolving more around polar co-ords soon.
(edited 9 years ago)
Original post by lerjj
Is this an actual episode? Spoilers, seriously!


Yep sorry haha spoilers. I didn't actually make the question and also not my solution however I do agree with their solution...

Is that 0.685 radians or degrees? The ^c is a little confusing...

Your approach makes sense, and is the way I did it before looking at his solution, however it yields a different answer

This is their solution for the first bit:

Spoiler

Reply 18
Avatar for lerjj
lerjj
13
Original post by Phichi
All seems correct apart from the last parts wording. You said find the minimum angle of which the pizza will slide. The pizza will slide when the component of weight down the slope is greater than friction, cos≥sinθcos \geq sin \theta would yield that it won't slide between 0≤θ<450\leq \theta<45

Also, add the solution title like in the OP, so I can keep track.


I think what I wrote makes sense. I wrote the conditions for it to slide, and that turns out to be that you need an angle greater than 45, which was not the case. Hence 40 is less than the minimum angle needed and so it won't slide. But if you have a re-wording suggestion, I'll change it.
Reply 19
Avatar for Phichi
Phichi
OP
11
Original post by langlitz
Yep sorry haha spoilers. I didn't actually make the question and also not my solution however I do agree with their solution...

Is that 0.685 radians or degrees? The ^c is a little confusing...

Your approach makes sense, and is the way I did it before looking at his solution, however it yields a different answer

This is their solution for the first bit:

Spoiler



Reading through that, I agree, looking at lerjj's solution, I probably fell under the same trap and ignored the horizontal displacement.

Original post by lerjj
I think what I wrote makes sense. I wrote the conditions for it to slide, and that turns out to be that you need an angle greater than 45, which was not the case. Hence 40 is less than the minimum angle needed and so it won't slide. But if you have a re-wording suggestion, I'll change it.


It makes sense, you just did it backwards :P You said minimum for it to slide, but found the angle for it to stick.
(edited 9 years ago)

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