🧵 Untitled Thread
Anonymous at Fri, 21 Jun 2024 11:07:13 UTC No. 16245409
Any physicist worth his salt should be able to solve this in 5 seconds.
Anonymous at Fri, 21 Jun 2024 11:16:02 UTC No. 16245419
>>16245409
homework thread.
fuck off
Anonymous at Fri, 21 Jun 2024 11:20:23 UTC No. 16245426
>>16245409
given curvature isn't provided i will assume that the skateboard always remains in contact with the ground. it is left as an exercise to the reader to determine why it doesn't matter where the thruster is fired. it's obvious.
Anonymous at Fri, 21 Jun 2024 12:02:09 UTC No. 16245490
>>16245426
>it is left as an exercise to the reader to determine why it doesn't matter where the thruster is fired.
wrong
Anonymous at Fri, 21 Jun 2024 12:06:07 UTC No. 16245497
>>16245409
by analogy with gravitational slingshot my guess would be right at the bottom, but I don't actually know if that analogy applies here
Anonymous at Fri, 21 Jun 2024 12:24:52 UTC No. 16245526
>>16245497
That's more about the path than increasing speed(the rocket has a set amount of potential energy to convert to kinetic energy)
Anonymous at Fri, 21 Jun 2024 12:31:49 UTC No. 16245533
The best place to fire the rocket depends on whether you are trying to maximize your speed or maximize the distance you travel.
To maximize your speed, you should fire the rocket just before you reach the bottom of the hill. This is known as the Oberth effect. The Oberth effect refers to the fact that the final velocity of a rocket is greatest when the propellant is ejected at the highest exhaust velocity relative to the initial velocity of the vehicle.
Here's why it works:
Firing the rocket downhill adds the exhaust velocity to the skateboard's downhill velocity.
Firing the rocket near the bottom means the skateboard already has a higher downhill velocity to add the exhaust velocity to.
Therefore, the greatest final speed is achieved by maximizing the sum of the exhaust velocity and the downhill velocity. This is achieved by firing the rocket just before reaching the bottom of the hill.
However, if you want to maximize the distance you travel, you should fire the rocket sooner, at point A in the image. This way, you can use the rocket exhaust to propel yourself for a longer distance down the hill.
Anonymous at Fri, 21 Jun 2024 12:32:25 UTC No. 16245534
>>16245409
With no friction, fire at point B, because you want to spend the most time naturally accelerating, and then fire the thrusters before you start decelerating that way you spend the least time decelerating.
Anonymous at Fri, 21 Jun 2024 13:03:11 UTC No. 16245570
>>16245533
AI post? Kinda impressive though I'm not sure it's coherent everywhere.
Anyway it does make me think of this simple explanation: if you fire it at the bottom, you get to profit from the kinetic energy of the unspent fuel going downhill.
Anonymous at Fri, 21 Jun 2024 13:26:17 UTC No. 16245606
>>16245409
>using rockets on land
Physician's first mistake
Anonymous at Fri, 21 Jun 2024 13:49:11 UTC No. 16245631
>>16245409
Isn't this a typical situation where you need both the energy balance equation and the F = m * a equation to see how you can maximize upwards F relative to downward F?
Anonymous at Fri, 21 Jun 2024 13:58:52 UTC No. 16245639
>>16245570
can AI already read images
Anonymous at Fri, 21 Jun 2024 14:55:01 UTC No. 16245720
>>16245533
>Violates conservation of energy (of system)
>Acknowledged as a paradox
>Only qualitative explanation used to explain paradox
>No equations in the textbooks proving energy is still conserved
What a joke.
Anonymous at Fri, 21 Jun 2024 15:12:33 UTC No. 16245743
>>16245409
the problem can't be solved because you don't have the amount of chemical energy released by the rocket during the pulse.
Anonymous at Fri, 21 Jun 2024 15:26:56 UTC No. 16245767
>>16245743
wherever the rocket is fired it increases the momentum of the vehicle by the same amount.
Anonymous at Fri, 21 Jun 2024 15:35:28 UTC No. 16245777
>>16245720
Conservation of energy only applies to the whole system skateboard + exhaust. To maximise the energy of the skateboard you have to minimize the energy of the exhaust which is of course zero if the exhaust velocity is equal to speed of the negative of the velocity of the skateboard. Generally you can expect the exhaust velocity to be much higher then velocity at any point so you release the exhaust when the skateboard is as fast as possible which is point B.
Anonymous at Fri, 21 Jun 2024 15:43:25 UTC No. 16245786
>>16245720
I can also read Wikipedia
Anonymous at Fri, 21 Jun 2024 15:48:34 UTC No. 16245794
>>16245767
then it's on B
on point B the rocket has maximum momentum.
Anonymous at Fri, 21 Jun 2024 15:49:33 UTC No. 16245795
>>16245720
this is the caliber of poster that "thinks" the ideal gas law debunks the greenhouse effect
Anonymous at Fri, 21 Jun 2024 15:49:35 UTC No. 16245796
>>16245767
>>16245794
or kinetic energy, I don't like to call it momentum.
Anonymous at Fri, 21 Jun 2024 16:16:13 UTC No. 16245817
Point A in pristine textbook conditions
In real life it depends on many factors like the soil, how is the rocket strapped to that skateboard, etc
Anonymous at Fri, 21 Jun 2024 17:58:02 UTC No. 16245993
>>16245777
No equations. No quantitative proof energy of the skateboard+exhaust system is conserved. That's my point.
Anonymous at Fri, 21 Jun 2024 18:37:52 UTC No. 16246045
>>16245409
including you
Anonymous at Fri, 21 Jun 2024 18:56:27 UTC No. 16246064
>>16245795
Finding out that Venus is hot because of the relationship between atmospheric pressure and temperature really seems to have triggered you badly, you're carrying your anger over to completely unrelated topics.
You should be happy that you learned something new, not upset. The anger that you're feeling is cognitive dissonance, you learned some new information (thats the cognition part) which was out of tune (theres the dissonance) with your preconceived notions.
Instead of angrily rejecting the new information and holding on to your less informed previous beliefs, you should accept the new information and incorporate it into a new and more well informed worldview. Thats what learning is all about.
Anonymous at Fri, 21 Jun 2024 19:01:37 UTC No. 16246071
i don't really care about textbook physics as they are oversimplified mathgarbage and does not be able to simulate a grass field.
in reality if fired on A, it will get airborne and crash to bank.
If fired on B, it will kinda crash againt to that bank.
If fired on C, it actually is aimed where it should realistically get a boost.
Anonymous at Fri, 21 Jun 2024 19:34:27 UTC No. 16246120
>>16245794
it does have maximum momentum at B but your explanation doesn't make sense.
Anonymous at Fri, 21 Jun 2024 19:35:33 UTC No. 16246123
>>16245993
rigorous proofs don't require equations
Anonymous at Fri, 21 Jun 2024 19:38:13 UTC No. 16246127
>>16245409
I think B is the most right: you want to minimize the amount of rocket thrust that's spent throwing the rocket in the wrong direction to be corrected by the slope, losing energy to friction & heat in the wheels and slope, while maximizing the amount of speed you get from gravity at the beginning. Once you start rolling up the hill, you're losing speed without the rocket. Depending on the strength of the rocket vs the strength of gravity, I think the best ignition point may vary slightly, but that's just my intuition telling feelings to me.
Anonymous at Fri, 21 Jun 2024 19:45:57 UTC No. 16246139
>>16246064
>Venus is hot because of the relationship between atmospheric pressure and temperature
nta but this is wrong. Venus is essentially hot because it's closer to the sun.
🗑️ Anonymous at Fri, 21 Jun 2024 20:08:39 UTC No. 16246166
>>16245533
What if you had this situation and you wanted to reach the maximum height on the hill before the skateboard would fall back to the bottom? It's not the same problem but I'm actually curious
Anonymous at Fri, 21 Jun 2024 20:16:19 UTC No. 16246172
If you fire the rocket downhill, then you have potential energy + rocket energy added to the total kinetic energy simultaneously. But if you fire the rocket on flat ground at the bottom, then you have first, the potential energy added and only then the rocket energy (so not simultaneously but both are still added to the kinetic energy).
So, why would t matter if the two types of energy are added simultaneously or one after the other? Would you, in any case, be left with the same amount of kinetic energy right at the moment when the uphill begins?
Anonymous at Fri, 21 Jun 2024 20:48:35 UTC No. 16246206
>>16246123
Good b8
Anonymous at Fri, 21 Jun 2024 20:50:30 UTC No. 16246208
Not enough information
Anonymous at Fri, 21 Jun 2024 20:53:52 UTC No. 16246211
>>16246071
>in reality if fired on A, it will get airborne and crash to bank.
no, that depends on other factors like the weight and thrust of the rocket and how is it strapped to the skateboard. In ideal conditions you get maximum energy by firing at A, unless gotcha trollface factors
Anonymous at Fri, 21 Jun 2024 20:56:00 UTC No. 16246215
>>16246206
nice argument.
>>16246208
there is enough information. Obviously we assume no friction etc
Anonymous at Fri, 21 Jun 2024 21:46:04 UTC No. 16246286
Work(done by thrust f to the board)=f*d, thrust f stays constant
distance d=v*t, duration t is constant, thus v has to be maximized, which is at lowest point B
Anonymous at Fri, 21 Jun 2024 21:48:02 UTC No. 16246289
>>16246286
finally a correct answer
Anonymous at Sat, 22 Jun 2024 04:22:12 UTC No. 16246807
>>16246286
>distance d=v*t, duration t is constant, thus v has to be maximized, which is at lowest point B
D=vt if a = 0 which clearly isn't true
>>16246289
Lol imagine not knowing distance eqn was wrong ::skull::
Anonymous at Sat, 22 Jun 2024 04:27:26 UTC No. 16246816
it doesn't need a rocket, any average skater would be able to rock the halfpipe back and forth a few times and eventually reach the top.
Anonymous at Sat, 22 Jun 2024 04:54:47 UTC No. 16246857
On the way up
t. has good intuition so does not need math
Anonymous at Sat, 22 Jun 2024 04:58:35 UTC No. 16246863
>>16245409
Neither. Just when it's over the edge. Why? BRACHISTOCHRONE. That's why.
🗑️ Anonymous at Sat, 22 Jun 2024 07:02:17 UTC No. 16247024
>>16246863
>BRACHISTOCHRONE
so you're assuming that the rocket would fly in a straight line without following the curve at all?
Anonymous at Sat, 22 Jun 2024 07:15:58 UTC No. 16247037
you fire at C
Anonymous at Sat, 22 Jun 2024 07:25:34 UTC No. 16247048
>>16246286
>duration t is constant
This is extremely wrong. How tf do you think time of transit is constant? How does a person understand the concept of speeds but thinks it always takes the same time to cross some distance?
Anonymous at Sat, 22 Jun 2024 07:39:05 UTC No. 16247061
C because it reduces the propulsion lost to friction and because denser objects slow down slower
Anonymous at Sat, 22 Jun 2024 13:26:32 UTC No. 16247379
I'm going to imagine that the rocket skateboard has the mass of Jupiter, such that once this thought experiment begins, the ground wraps around the rocket skateboard, trapping it as the center of mass of the new system.
The rocket skateboard goes nowhere, because there is nowhere to go.
Anonymous at Sat, 22 Jun 2024 13:50:41 UTC No. 16247408
>>16247048
Time is relative to a reference point and without reference point time is subjective. For a solipsist variable distances divided by variable speeds can have a constant result.
Anonymous at Sat, 22 Jun 2024 14:05:51 UTC No. 16247417
>>16245409
babby's first energy conservation problem
Anonymous at Sat, 22 Jun 2024 15:56:44 UTC No. 16247571
>>16245409
Either in A or in C because a given amount of energy will turn into more speed when initial speed is low. Probably A because you lose mass before being decelerated and get more speed from the overall energy (height+combustion)
Anonymous at Sat, 22 Jun 2024 16:07:28 UTC No. 16247592
>>16246807
It is not true being a rule of thumb calculation, but how false is it to not use integration?
The acceleration inline with the track direction is the component of gravity and rocket acceleration.
The realer equation is Dist=int[t1,t1+trkt](v(t))dt=int[t1
You can assume acceleration is constant, a(t)=a, but again it does change a bit from the changing gravity component becuase the track changes direction. Luckily most track is designed to be as smooth a possible so the assumption is safe.a=arkt+agrv
On the otherhand rocket loses mass and change thrust through out the burn. So the real real acceleration won't be like that.
Dist=int[t1,t1+trkt](v(t1)+a*trkt)d
The different is in the 0.5a*trkt^2,
If we plug this back to f*d:
Work=f*d=m*arkt*d=m*artk*(v(t1)*trk
Is the conclusion changed? Trkt is assumed small(brief impluse) to reduced the influence of squared acceleration terms due to squared trkt attached, having high arkt deminishes gravity component agrv's significance, and having higher v(t1) still simply gives you more work done.
Anonymous at Sat, 22 Jun 2024 16:15:52 UTC No. 16247620
>>16245409
Just after /b/
Anonymous at Sat, 22 Jun 2024 16:16:12 UTC No. 16247622
>>16247592
i don't have a horse in this race, but nobody is going to read that. /sci/ has latex support for a reason.
Anonymous at Sat, 22 Jun 2024 16:19:11 UTC No. 16247627
>>16247622
Yeah
Anonymous at Sat, 22 Jun 2024 17:11:43 UTC No. 16247715
AI bots have been a thing for years, anon. Even before the AI boom.
Anonymous at Sat, 22 Jun 2024 19:06:18 UTC No. 16247957
>>16247379
I didn't know your mom was a skater.
Anonymous at Sat, 22 Jun 2024 19:14:01 UTC No. 16247965
>>16245409
I think I got it:
The rocket impulse will add a fixed amount of momentum independent of when the impulse is fired; hence, a fixed amount of speed (mass is assumed constant, not worrying about eg fuel)
You want the greatest increase in (kinetic) energy; since KE is proportional to speed^2, the greatest increase in kinetic energy will be when the speed is greatest;
which is at the point of lowest height, B
Anonymous at Sat, 22 Jun 2024 20:20:10 UTC No. 16248046
>>16247965
Wrong. Gravity adds constant acceleration which increases speed. To maximize the amount of energy gravity adds to the rocket, it is best to trigger the impulse when the rocket is moving in a smooth downhill region. Assuming ideal friction on ground and wind conditions.
Anonymous at Sat, 22 Jun 2024 21:13:38 UTC No. 16248159
C is literally the only point where you aren't wasting thrust firing into/away from the ground.
Anonymous at Sat, 22 Jun 2024 21:59:41 UTC No. 16248255
Assuming the rocket’s acceleration is independent of the position where you fire it (and ignoring friction) wouldn’t the average acceleration along the path be the same no matter where the rocket fires?
Anonymous at Sat, 22 Jun 2024 22:25:37 UTC No. 16248292
>>16245409
My guess is C.
Assuming no or little friction, the cart will roll to C on its own. Any force added, only adds to this.
However, only at C is the resultant force vector of the rocket along the path of the cart. At any other point, the rocket has a resultant that is either perpendicular to the direction of the cart, or directly opposed to it, wasting the force by driving the cart into the ground, and not adding to its velocity.
Anonymous at Sat, 22 Jun 2024 23:14:53 UTC No. 16248397
>>16246863
At any point it adds up to inertia but on the propulsion force vector has the desired direction,
Anonymous at Sat, 22 Jun 2024 23:16:03 UTC No. 16248399
>>16248397
on C
Anonymous at Sat, 22 Jun 2024 23:20:39 UTC No. 16248414
wtf is this thread.
the rocket clearly is assumed to point in a direction co-linear with the velocity of the cart
the rocket therefore does maximal work on the cart anywhere, causing a maximal increase in kinect energy.
conservation of energy then implies anywhere is good (neglecting things like friction and air resistance).
Anonymous at Sun, 23 Jun 2024 00:04:15 UTC No. 16248470
If you ever wonder why /sci/ is so fucking shit, just look at this thread and count the number of correct answers (hint: "it doesn't matter" is not the correct answer).
Gym Shaman !5zilbEj30g at Sun, 23 Jun 2024 00:14:44 UTC No. 16248480
>>16245409
>>74559396
>>74559429
It's worded quite poorly, the diagram sux, an aspie grad student definitely spent too much time making this thinking they're clever, and the prof definitely either sighed at the impending shitstorm, or snickered gleefully at the suffering inflicted b/c all the students are going to panic b4 realizing they don't need to calculate anything, but you can solve this without any additional constraints, beyond assuming the rocket ship is constrained to move along the track perfectly, and begins at rest.
What you do is compute the potential energy required to surmount the hill with height d_{b}, PE = mgd_{b}, and realize that wherever you choose to fire, you need to obtain enough speed at that point so that your instantaneous kinetic energy at that location becomes equal to this value.
Since you're starting a height (d_{b} - d_{a}) above ground, the max kinetic energy you will obtain without application of impulse is 0.5 * m * v_{b}^{2} = mg(d_{b} - d_{a}).
This occurs at point B, when all the initial gravitational potential energy has been converted into kinetic energy, giving a velocity v_{b} = sqrt(2g (d_{b} - d_{a}). The speed you'd have at the top of the hill if all the grav. pot. energy you'd have there was converted to kinetic energy would be v_{H} = sqrt(2gd_{b}), so you can fire at point B and crest the hill if you obtain v_{I} = v_{H} - v_{b} worth of speed.
You can figure out your instantaneous kinetic energy at any reachable point along the track using conservation of energy: however far you've travelled below your initial height, which is CANCEROUSLY written as d_{b} - d_{a}, that change in gravitational potential energy turns into kinetic energy.
In general, you can fire at any point that is reachable without firing, so all the points b/w A and C, and make it so long as you obtain the required speed when you fire, which you can calculate using an energy principle (i.e., cons. of energy), as stated previously.
Gym Shaman !5zilbEj30g at Sun, 23 Jun 2024 00:15:57 UTC No. 16248483
>>16248480
Any /plg/ bros in here, I got a three day ban because the tripfags couldn't handle the sauce, and got buttmad
Gym Shaman !5zilbEj30g at Sun, 23 Jun 2024 00:23:01 UTC No. 16248494
>>16248483
Also riceballman is a confirmed larper who can't even solve this, the max speed in the absence of firing occurs at B as that's when all the initial potential energy has turned into kinetic.
Without firing, there will be no speed at C because you'll be at the same height you were at initially so all the energy will be back into potential energy again.
Only a brainlet larper would not know this
>like riceballman
Anonymous at Sun, 23 Jun 2024 01:31:29 UTC No. 16248577
>>16248159
Thrust is increasing the speed of the rocket. Acceleration from gravity at higher speeds is worth more energy than acceleration at lower speeds because energy increases at the square of velocity.
Anonymous at Sun, 23 Jun 2024 04:13:49 UTC No. 16248730
>>16245426
If it remains in contact with the ground you would want to fire at B.
The faster velocity means that less energy goes into the propellant and more goes into the rocket.
Anonymous at Sun, 23 Jun 2024 04:15:01 UTC No. 16248732
>Firing at A means less time accelerating due to gravity
>Firing at C means more time decelerating due to gravity
>Therefore firing at B is optimal
Anonymous at Sun, 23 Jun 2024 05:36:26 UTC No. 16248808
>>16245639
Yup, and far better than >>16245533 and most other answers posted here.
This was first try and the only hint I gave it was that it was a physics problem.
yeah, I'm thinking gookmoot should delete /sci/ and replace it with a link to Claude 3.5 or GPT4-o. unless your IQ is >130 you are dead weight.
Anonymous at Sun, 23 Jun 2024 16:58:58 UTC No. 16249431
>>16248480
it's actually worded quite appropriately
Anonymous at Sun, 23 Jun 2024 17:01:02 UTC No. 16249435
>>16248808
if AI can already do this we're finished
Anonymous at Sun, 23 Jun 2024 17:32:40 UTC No. 16249496
>>16249435
AI can 'do' the Monty Hall problem if you just filter the majority of the time it gets it wrong.
Ideally AI would be better at more niche problems since the training data is less likely to include reddit posts
Anonymous at Mon, 24 Jun 2024 02:45:03 UTC No. 16250251
I was trying to figure out if it was B or does-not-matter
If you think the rocket adds the same energy no matter where it is, then it's does-not-matter
If a rocket fixed delta-V is the most efficient at the lowest V, then it's B
Another thing to consider:
Let's say that with no rocket, the car spends 10 seconds being pulled by gravity as it rolls back up the hill
With the rocket it goes super fast and only spends 1 seconds going up the hill and fighting gravity. A lot less total gravity is applied in this scenario
So B
Anonymous at Mon, 24 Jun 2024 03:00:45 UTC No. 16250268
>>16250251
>adds the same energy
It adds the same --Impulse--
The energy added is ~impulse*speed
Anonymous at Mon, 24 Jun 2024 03:22:24 UTC No. 16250291
>>16248808
what version is this?
Anonymous at Mon, 24 Jun 2024 06:44:25 UTC No. 16250442
>>16250291
GPT-4o, it's not much smarter than 4, but it can understand visuals and audio.
Anonymous at Mon, 24 Jun 2024 06:48:01 UTC No. 16250445
>>16249435
yeah it's basically fucking over no matter what your IQ is. I was just coping when I said IQ >130
Anonymous at Mon, 24 Jun 2024 14:21:03 UTC No. 16250846
>>16250442
But it must be significantly smarter than 3.5 at least? I am on a free 3.5, and 4 gets enabled occasionally as a promotion or something, but mostly 3.5, and it is just terrible at math and physics. It hallucinates and makes stuff up.
Anonymous at Mon, 24 Jun 2024 14:24:05 UTC No. 16250851
>>16247957
Heh!
Anonymous at Mon, 24 Jun 2024 17:27:36 UTC No. 16251096
>>16245409
When you reach the highest freefall speed, in a start of that speed momentum gain when the wheels angle is maxed
Anonymous at Mon, 24 Jun 2024 17:32:07 UTC No. 16251101
>>16251096
When this angle is greatest
Anonymous at Tue, 25 Jun 2024 02:07:36 UTC No. 16252036
>>16245409
This can be worked out through trial and error. Don't need any physics degree for it.
Anonymous at Tue, 25 Jun 2024 06:54:38 UTC No. 16252275
>>16252036
yeah doc, just build and pay 3+ rockets instead of one, duh!
Anonymous at Tue, 25 Jun 2024 07:34:06 UTC No. 16252321
>>16249435
oh noooo it can do the leddit problem
Anonymous at Tue, 25 Jun 2024 07:36:45 UTC No. 16252325
>>16247048
retarded? :(
Anonymous at Tue, 25 Jun 2024 11:23:54 UTC No. 16252547
1. starts falling from A with some potential energy U being converted to kinetic energy K
2. reaches B which is where all potential U from point A has been converted to kinetic energy K.
Assuming there was enough kinetic energy to reach point C and the thrust is enough to overcome the gravitational force at that point, then it should be point C.
Note: This is a faggot question as no numerical values of height, distance, mass, air resistance, friction etc is given meaning that it is essentially guess work.
Anonymous at Tue, 25 Jun 2024 22:50:49 UTC No. 16253799
>>16247592
refine your autism using latex
Anonymous at Wed, 26 Jun 2024 00:40:54 UTC No. 16254006
>>16245409
At B its velocity is perfectly equal to the tangent at the slope and so the quick impulse thruster acceleration is maximized in the direction of movement
...RIGHT? THATS THE RIGHT ANSWER RIGHT?
Anonymous at Wed, 26 Jun 2024 01:17:28 UTC No. 16254052
Let's simplify the system. An object is dropped from a height of 1 at point A/C. Gravity is 1. It has a perfectly elastic collision with the ground at B. At A, B, or C, its speed can be increased by 1. Which increase results in the highest upward velocity at C?
At A or C, the result is a velocity of 1. At B, the velocity goes from sqrt(2) to sqrt(2)+1. So the final velocity at C is sqrt(2sqrt(2)+1). That's more than 1.
B wins.
Anonymous at Wed, 26 Jun 2024 01:25:41 UTC No. 16254064
>>16254006
>is perfectly equal to the tangent at the slope
true, but i that applies anywhere for this problem