/sci/ - Science & Math

B.

Because it's velocity would increase (provided the orientation of your picture is vertical) as it reaches the downward slope. Although they're identical, just flipped, the increased velocity initially would allow it to overcome its rise at greater speed.

top one

>>5282022
Your teacher is trying to trick you, OP.
protip: they'll get there at the same time.

This is due to <span class="math">SU(N_5)[/spoiler] disparity over the field <span class="math">\mathbb{K}_\zeta[/spoiler]. Kid's stuff.

>>5282026
But the tracks are equal, and the distance traveled is equal, and the height difference is equal just opposite

>>5282029
If the top one isn't moving fast enough to get over the bump, they won't get there atr the same time.

Depends on if the balls are in a gravitational field and what direction the field is pulling. If there is no field then both marbles would reach the end of the track at the same time.

>>5282030
OP here, the top one does have enough speed to get over the bump

>>5282031
G is in a 90 degree angle clockwise from the direction the balls are traveling

>mfw OP is in first year physics or even high school
>mfw OP can't use vector analysis
>mfw gravity only affects up/down motion, not forward or backward

>>5282028
still on babby stuff here, if the marble doesn't have enough momentum to get it over the hill wouldn't it just not finish, while b would finish?

>>5282034
seems legit, so as long as it can get over the hill they'll finish at the same time?
>n0

>>5282035
Yes. Also, there is no conditional 'as long as they both get over the hill'. If one can get over the bump, so can they other with a matching speed.

>>5282035
of course, why shouldn't they?

>>5282032

Seems you can't into <span class="math">\mathbb{K}_\zeta[/spoiler] fields or even <span class="math">[\cdot, \cdot] \circ (\mathrm{id} + \tau_{A,A}) = 0[/spoiler].

>>5282037
well, if you tip marble b over gently into the halfpipe it'll make it to the top on the other side
if you tip marble a into the hill, it won't roll up the hill
wouldn't there have to be the condition?

>>5282039
>feelsbadman

>>5282038
sorry i wasn't using seems legit sarcastically

It's not the same. Provided they really do follow the drawn path, the upper one would come to the goal later. Because the average velocity of it on the bumo is less the the start velocity and for the othe one it's bigger than the start velocity, due to centripetal forces acting (or whatever you'd like to call them).

Assuming the A goes over the bumb, A will overtake B at some point. (wat point exactly depends on the initial speed)

The only difference is the air resistance, which is exponential. Because B will reach just a little higher speed then A (because it accelerates earlier)
it will lose more energy over the whole trajectory.

>>5282048
pardon my fucking dyslexia

>>5282045
>dem american commercials
>"are you a fucking retard? then you absolutely need product <span class="math">X[/spoiler]!"

>>5282052

>implying your written English isn't great
>apologizing for a nonexistent issue

>>5282055

>mfw /sci/ cant do a simple physics 1 question

protip: E=E0

>>5282058

>>5282059
It's not. there are forces acting on the particle in the direction of initial velocity. If you want the body to follow a path like that, there needs to be a centripetal force etc. figure it out

>>5282062
but there isnt any resistance on the ball while it goes along the track, only G and the initial velocity is working on the ball

>>5282063
Energy is preserved exclusively in closed system, ie if there aren't any forces coming from out of system. If there is a force which is not a result of particle acting on itself it's not a closed system

>>5282048
WINRAR

>>5282065
Lets hear your forces that magically appear from outside on these tracks

Both lagrangians have the form <div class="math">L = \frac{1}{2}m \left( \frac{\mathrm{d} s}{\mathrm{dt}}\right)^{2} - mg(z+C)</div> and <span class="math">ds=Rd\theta[/spoiler] for both cases so they give the same equations of motion.

Question is, whats is the speed of each at the instant they enter the circular track

>>5282034
>>5282035
maybe i should have been using this sarcastically, wtf just thought about this, of course gravity affects "forward and backward" motion
like a ball rolling down several different angle inclines, they'll end up at the same speed, but the steepest incline will reach the bottom first

they'll end up at the same velocity, but B will get there quicker is my final answer, since it's average speed will be greater

>>5282028
>>5282028
i like you man, but i don't even know anymore

>>5282074
ain't nobody can read that

>>5282069
wat
There's only gravity and the elastic force wich is the reaction of the surface. The combination of the two gives the centriptel force in the bumps.

>>5282076
>they'll end up at the same velocity, but B will get there quicker is my final answer, since it's average speed will be greater
this

B

if A bounces off

>>5282039
>He can't even into sigma mu manifolds


<div class="math">\sum_{\mu}^{\alpha }\; \Gamma\cdot \frac {\varepsilon_{\mu}\cdot \widehat{\alpha}\bigotimes \ss }{\lambda\Cup \L }=\zeta(\AA )</div>

Since no mention of gravity, and the inside of a circle is smaller than the outside, I'd say B, only because the center of each ball is either on the inside of the curve or the outside. You know what I mean?

>>5282039
>He's still in highschool
This is how you SHOULD solve it.

<div class="math">\sum_{\mu}^{\alpha }\; \Gamma\cdot \frac {\varepsilon_{\mu}\cdot \widehat{\alpha}\bigotimes \amalg }{\lambda\Cup \L }=\zeta(\AA )</div>

I believe it's B, because think about it: When A hits the bump, it's velocity will decrease beneath its starting velocity and then on its way down go back up to it.

When B hits the dip, it's velocity will increase LARGER than its starting velocity, and then go back down to it.

Not a Physics major, simply my (un)educated guess.

<span class="quote deadlink">>>5282123[/spoiler]

Son, do you even into <span class="math">\displaystyle \int\!\!\!\!\int_{A\,\subset\mathbb R^2} \left (\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, d\mathbf{A}=\oint_{\partial A} \left ( L\, dx + M\, dy \right )[/spoiler]?

>>5282034
They travel at an angle at some points, and the normal vectors resulting from contact with the ground affect right and left movement.

>>5282129
>2012
>still stuck on babby-tier equations

I did the proof of <div class="math">\sum_{\mu}^{\alpha }\; \Gamma\cdot \frac {\varepsilon_{\mu}\cdot \widehat{\alpha}\bigotimes \amalg }{\lambda \cup \omega }=\zeta(\Phi )</div> in middle school.
I'm onto bigger stuff now like <div class="math">\int \left (\prec \mu \succ,\models \mathbf{Z}\asymp \circ \right )d\alpha = \Psi (\beta )</div>

>>5282048
THIS
/thread

>>5282168
(Fuck, I would <span class="math">love[/spoiler] to understand notation such as that.)

>>5282168

>showing me notation I've never seen before
><span class="math">\asymp[/spoiler]

Well played, brother, well played.

>>5282271
you mean made up notation? its not hard, it means whatever you want it to mean

if you hit both marbles with just a little more impulse so that A reaches critical velocity, then:

A will almost be zero in the middle, and B will be almost twice as fast. Both marbles velocity will return to the start velocity at the end of the hill/valley.

That should let you easilly answer.

>>5282022
This is stupid.

The kinetic energy lost by A in the up slope is equal to the to the kinetic energy gained by B in the down slope, and vice versa. They end up with the same kinetic energy after the slopes, hence the same velocity.

<span class="quote deadlink">>>5282278[/spoiler]
how does that in any way answer OPs question? are you a moron? did he ask which one is the fastest affter the curve? no. did you even read his post? Dont just retardedly assume some shit based on an image and then act like youi are smart by spewing irreverent garbage that has nothing to do with the question. why are you on sci? you couldn't possibly have knowledge in any field related to science or math if you just go around half reading shit then making up your own rules and pretending to be better then other people. fuck off and die.

>mfw people argue with words
>mfw nobody argues with mathematics

>>5282276
I actually make up notation all the time. It makes me feel clever. I am a retard though, probably wouldn't understand half the symbols I look up on Wikipedia and mash into my "equations".

>>5282296
around what problem?

>>5282296
>implying words and mathematics are mutually exclusive

Check out this babby.

>>5282313

Oh, I don't know.
Could it be I refer to the problem the whole thread is centered around arguing with?

>racistball cannot into reading comprehension

>>5282317

>implying any mathematics beyond fake stuff is used in this thread

Assuming the velocity is high enough to overcome A's climb:
They reach the end at the same time.
A and B both lose the same amount of velocity climbing vertical height (A the beginning of the hump, and B the second half of its dip), and they both gain the same amount of velocity, (A as it falls out of it and B as it falls into the hump).

The only problem I see is if the B ball goes airborne out of the dip, but let's assume it doesn't.

>>5282380
Think about it like this. If they both travel both the rise and the fall, it won't matter which one happens first, they will traverse them in the same time. (Not just average time, also)

>>5282380
>Assuming the velocity is high enough to overcome A's climb:
The OP already said that it is.

>They reach the end at the same time.
No.
B is the faster track.
See: >>5282026

Look, bitches, I had this question, first semester undergrad physics. The correct answer is B and people above have spoken the turth, some of them at least.

THE END

>>5282393
I demand video proof of this strange phenomenon.

gravity plays no role so both at the same time, the end

>>5282403
I am looking.
There is one on Youtube; I have seen it before.

>>5282168
>dat External Direct Product
>dat Union
10/10 would laugh again

>>5282136
This guy is write. Let's assume that the marble rolls smoothly along the track with no friction. This means that the marble's energy is conserved the whole way through, and is simply converted from kinetic to potential and vice versa. The marbles will have the same kinetic energy (speed) at the end as at the beginning. The top marble loses kinetic energy as it goes up the hump, and then its speed goes back to where it is at the beginning as it comes down. So at every point on the track, its speed was either less than or equal to its starting speed. The reverse is true for the bottom marble, going down the dip lets its potential energy be converted to kinetic energy to gain speed. When it climbs back out, its speed goes down to marble. So the bottom marbles speed is always either greater than or equal to its starting speed.
The bottom marble is faster. Think of it this way, if the starting speed is low and the hump is big, the top marble might not even make it over the hump.

>>5282422
>This guy is write.
*Right.

>>5282422
>Think of it this way, if the starting speed is low and the hump is big, the top marble might not even make it over the hump.
Again, the OP already stated that it does:
>>5282033

>>5282392
Are you fucking retarded? Did you even bother reading my post that you are replying to?

For some reason you take into account the increased velocity from B's initial fall but not from A's back fall.
B's increased velocity is slowed down by the climb, whereas A's increased velocity from the back fall is all put into horizontal movement.

>>5282022
Do not get confused OP.
No force would do any kind of work on the marbles during their motion, except for gravity , which is conservative.
Therefore we can easily conserve energy , and say that the sum of gravitational potential energy and kinetic energy would be constant throughout the motion of the blocks.
Change in Gravitational Potential energy would be zero , since after going through depression/climb the marbles would return to their original position ( in vertical sense )
Therefore , the change in kinetic energy will also be zero, so velocity would be same , so both will reach at the same time.

This however assumes that the velocity of the upper block is sufficient enough to climb over the hill or whatever.

>>5282453
For A: the hill lowers the velocity during the climb, and then increases it again to what it was originally, on the fall at the other end.
The part during the hill climb and fall has a lower velocity than the starting velocity.

For B: the dip increases the velocity during the fall, and then decreases it again to what it was originally, on the rise at the other end.
The part during the dips fall and rise has a higher velocity than the starting velocity.

And you do not need to be rude.

>>5282461

You cannot infer that the travel time is the same, from the fact that the speeds at the beginning and endpoints are the same.

>>5282544
My apologies , I did not read the question properly.

In that case , the lower ball would arrive first since it had its velocity increased and then decreased whereas in the case of upper the velocity was first decreased then increased.

>>5282575
Exactly. :)

There is no friction, so you can ignore velocity, starting velocity is the same as end velocity. It's just a steady rate. Gravity wasn't mentioned so inserting it into the question is silly, neither were any other forces. Assuming the track has a greater than zero width, in the picture it certainly does, then the paths followed by balls A and B are not the same length.
The tracks are the same length, but the path followed by the balls are not. You're counting at a steady rate towards two different sums. So basically which path is shorter? Path B. If the the track has zero width, the balls get there at the same time, because the path's followed mirror each other.

>>5282586
You seemed to have missed the point.

The OP has already stated that the tracks are of equal length, and it is assumed that gravity does apply, and all of this occurs on Earth.

>>5282617
Gravity or not, the answer is the same.

>>5282702
No it is not. Read:
>>5282575

interesting question. I am ashamed to admit I was stumped between if it was B or the same time....but it turns out the right answer is B. I like the way this guy put it the best:
>>5282125

but here's proof B is right:
http://ken.duisenberg.com/potw/archive/arch97/970530sol.html

http://apcentral.collegeboard.com/apc/members/courses/teachers_corner/42197.html

see question 10

>>5282432
What he is telling is, the top marble loses kinetic energy and then recovers it
The bottom marble gains kinetic energy and then it loses it.
The top one has less or equal X kinetic energy over time.
And the bottom one has more or equal X kinetic energy over time.
Is easier this way to tell B is the right answer.

>>5282758
That has already been stated multiple times, and also by myself earlier in the thread.

Bringing up an unhelpful 'what-if?' that has already been pre-emptively answered by the OP is just a waste of time, and shows that one has posted without reading the thread.

>Doesn't give mass of the marbles.
>Doesn't give initial velocity.
>Doesn't give any function for defining the slopes.
>Doesn't say if the marble sticks to the ground or jumps.
>Doesn't say if we should consider the "bump" of the upper marble against the slope
OP you are a retard-

B, because it's always moving at the same or greater speed than A.

>>5282839

None of that shit matters in the face of the main effect. Saying "I can't answer that" isn't a way to seem smart.

>>5282889
that's not true though, when a is at the bottom of the slope it could be moving faster than b at the top of the dip

>>5282992
No. The dip and the slope are the same size.
The speed at when a is at the bottom of the slope and b is at the top of the dip is the same as both their initial velocities, which are equal.

The answer would be B. Plotting a velocity curve vs time and getting the area under the curve would show B reaching the finish line first. Ball A loses KE and bounces back to it's original KE. Ball B gains KE and bounces back to it's original KE.

>>5282992

Incorrect. After both the slope and the hump, both balls return to their starting height, which means their potential energy is the same as when they were started. This means their velocities at that point are the same as when they started, ie equal.

Obviously B will arrive first. It has been explained very well in this thread, but let me write a simple example for the retards:

If two cars are driving at 100 km/h (or mph if you're amerifag or britfag) and car A hits the brake to slow down to 50 km/h and then goes back to 100, while car B starts at the same speed but accelerates to 150 and then goes down to 100, what do you think who will cross the finish line first?
Sure they will have the same speed at the end, but car B will be way ahead.

/thread

>>5283299
While correct, your explanation is not exactly any clearer than the others in this thread.

Any 'retards' still left in this thread will be just as confused as before.

This is clear and simple: >>5282125

>>5283305
>While correct, your explanation is not exactly any clearer than the others in this thread.

I haven't looked at the other ones but his explanation sure is clear, what the fuck more do you need? Especially when the explanation you linked says exactly the same thing less the intuitive example.

No lie, this exact problem was given to me in my Cambridge interview. Couldn't do it, but they still let me in.

As far as I remember, the answer is the bottom one, can't remember why exactly though.

>>5283335

After reading through this thread I feel like an idiot for not being able to figure out why its B

>>5282125
perfect

>>5282839

This exact question was in the British School Physics Olympiad at one point, a paper written by people much smarter than you, so no, it is very answerable.

Here's a different question. A just barely has enough speed to make it to the top of the bump, whereas B crashes through the bottom easily. Now is it clear why B makes it first?

>>5283335

>Couldn't do it, but they still let me in

It strikes me as odd that they didn't at least steer you towards a right answer. Shitty interviewers, or just short on time?

>>5282022
It's obvious the answer is B. Maybe /sci/ thinks this is some sort of trick question or something but it isn't. The distance for each is the same, and the velocity can be written as a function of height. So it's obvious that B has a higher average velocity than A, so B gets there in less time. I should know, I'm a physics.

>>5283381

Yeah they did do that, can't remember exactly how it went, was a few years ago now.

Solved it.

just wondering, how would you explain this to a 10 year old kid, or younger?

>>5285197

With words.

The answer is the bottom one.

Plot a distance speed diagram.

Bottom one will have a speed greater than starting.

Top one will never have a speed greater than starting.

They both finish at original velocity.