Question

Question below
please correct my other answers
because i'm really not sure

Answer 1

oh i changed my answers
a.) the Y path
b.) because it will have more speed due to....

Answer 2

b.) The Y track has a steeper slope at the beginning which will give the ball a much higher speed than the ball on X-track.

Answer 3

For d.) the speed will also be the same? ?

Answer 4

I agree with your new (a) and (b)! And the speed will be the same. Can you tell me why? (Hint: energy)

Answer 5

uhm.. maybe because energy cannot be destroyed nor created only transformed into other types of energy?. Like when a ball is released from a height, it has gravitational potential energy and then it transforms into kinetic energy until it falls down. But the mechanical energ of the ball will still the same.

not sure, if this explains it

Answer 6

It pretty much does! Energy is always conserved. Mechanical energy is often conserved (kinetic friction is one of the exceptions).

So, the gravitational potential energy is the same for both balls before and after. They're balls, rolling, so there's no kinetic friction stealing the energy. So we assume all the potential energy lost is turned into kinetic energy.

If the balls start with the same kinetic energy (not moving), and are lowered the same amount, they will gain the same kinetic energy and still be the same.

Answer 7

Well, they might not gain the same amount of kinetic energy, because it depends on the mass. But they will have the same speed.

Answer 8

so the mass doesn't matter?
so if the other ball is bigger, they will still have the same speed?

Answer 9

Well! Let's look. (I'm only 99.9% sure, so I'll have to prove it to myself anyway).
Let's use \(M\) for a big mass, and \(m\) for a little mass.
For the big mass,
\(PE_{initial}=Mgh\)
If we set it up to have the "reference" at the bottom of the ramps, we are saying that the potential energy is \(0\) there. And, when we do that, it's convenient because then there is no potential energy there. Meaning it's all kinetic because of the energy change! :)
So, \(PE_{before}=KE_{after}\)
\(KE_{after}=\frac12Mv_f^2\)

Now we use the "\(PE_{before}=KE_{after}\)"
\(PE_{before}=KE_{after}\)
\(\Downarrow\)
\(Mgh=\frac12Mv_f^2\)
\(\Downarrow\)
\(\cancel Mgh=\frac12\cancel Mv_f^2\)
\(\Downarrow\)
\(gh=\frac12v_f^2\)
Well, looks like we don't need to go any further! Forget \(m\), and forget \(M\), because they don't affect the final velocity!

The thing is, each ball has potential energy proportionate to its mass. And the kinetic energy is also proportional to that mass. So, when the balls loses potential energy and gains kinetic energy, the exchange is sort of.. balanced.

Answer 10

I have to go! I'll see any questions later, though. Good luck!

Answer 11

ooh, now i understand that part :)

i understand your explanation why they will have the same same at the end of the tracks, but I cannot absorb it, how are they going to have the same speed?


oh ok thanks btw!

Answer 12

The quick version is that
\(gh=\frac12v^2\implies v=\sqrt{2gh\ }\)
so mass is not a factor.

Otherwise...

It's tough to absorb because my explanation was more math than concepts. That's sort of appropriate, though, because the whole idea of energy is a mathematical concept. It only takes abstract meanings when we look at how it "behaves."

Pretty much, energy is made up and always right.

Buuut we can try to make sense of it.
When we look at the balls at the top, we say they have the potential to gain energy. This is like from the work-energy theorem. Gravitational force has the \(potential\) to do work. When that work is actually being done as the ball rolls down, we see that the potential energy is less, but just as much "work" was done, giving the ball kinetic energy.

Recapping:
So, gravity gives the marble initial potential energy, we say.
Then, when gravity does some actual work, the marble goes down, and has less potential energy. The work results in kinetic energy.

Now, the potential energy is given to the ball depending on its mass. It loses this energy, and is granted the same amount of energy as kinetic energy based on its mass.

Since the potential energy and kinetic energy of an object is based on how much mass (stuff) the object has, it doesn't make a difference between the energies. So \(gh\) translates to \(\frac12v^2\) directly. The energy it has, goes right to the velocity. Velocity squared, really...

Answer 13

oh okay, i tried to do the math part for letter d (i followed what you explained) and i understand it better :))

sorry for being a slow learner :/

For part B, are there more any explanations for it?
can you check my answer please.

P.S. I really appreciate your time and especially your effort and patience explaining this to me

Answer 14

My pleasure! I'm a slow learner, which is why I like to help. I might just be a more particular learner or something - I don't know for sure.

Anyway! I like you're math.. But I'm trying to find a mistake, and here's why I think there is one:
\(F_N=\dfrac{mg}{sin\theta}\qquad\)I follow it... but this allows \(F_N\rightarrow\infty\) as \(\theta\rightarrow0\) or \(\theta\rightarrow\pi\). That doesn't seem right.

Another result of this is something you pointed out:
\(\vec a\rightarrow\infty\)
The most that \(\vec a\) can be is \(-9.8\ \rm m/s^2\).

But I do like how you pursued this mathematically! I usually work with the coordinate system likewhich is probably why I see something weird here.

And then the last thing I want to point out is the last paragraph. Since the ball is at rest and then starts rolling, its velocity changes. That means it experiences acceleration. So just look at that paragraph and see if there's something you want to change :)

But your general reasoning is right! The ball gains kinetic energy and so velocity much more quickly, so it has more time with a greater velocity to reach the bottom faster.

Answer 15

Oh! I think I found the issue.
\(0\neq F_N\sin\theta-mg\)
I'm not at my A-game at the moment, but that's it.
I understand the force diagram, and how it looks. For the y-components,
But you can't assume that they cancel! And that's where your mistake was. There is actually a net force in the y-direction also, which is why the downward velocity of a rolling ball will change.

It's hard to work with coordinates where both have non-zero forces. That's why I like to use the coordinate system I drew above! It's okay to use any coordinate system you like as long as you make sure everything is in terms of that coordinate system.

I'm still impressed with the work you did. You used \(\cot\theta\), even! But I hope this helps you see what is going on!

So
\(F_N\sin\theta-mg=\Sigma F_y\neq0\)

I hope your teacher wouldn't require all of the math to explain what you said about the velocity, but every teacher is different.

Take care! Feel free to ask questions as necessary!

Answer 16

thanks!
i changed my answer and i tried to follow your explanation

sorry for my messy hand writing

Answer 17

Hi! I don't have too much time, but I have two things to say!

First, I wouldn't say \(\sum F=Fg\cos\theta\). You're good saying \(\sum F=mg\cos\theta\) :)

Second, the ball on the flat incline still has acceleration. It will get faster and faster.

I guess it's probably enough to say that, even though the length of the curved track is a little greater, the ball on that path gets fast very quickly where it is steep.

And, since the balls are the same speed at the end, you can see how the ball on the flat incline never gets faster than the other ball.

Answer 18

How's it going? I'll be able to look at my computer intermittently.

Answer 19

okay thanks for your help!
I'm gonna change it again xD
I have to submit this today, so i'll just hope that i'll get full marks