Ramos' penalty miss features in a third year university physics exam paper

No offense, but Brasilians are bigger assholes to us than we are to them

is this really for 3rd year physics major problem :O?

braylann wrote:sorry, that was out of line. Were all joking in here, peace and love. No hate.

One love

braylann wrote:No offense, but Brasilians are bigger assholes to us than we are to them

Eh, in my experience, my Brazilian friends have been pretty cool considering the situation. There's banter back and forth but nothing that made me think worse of them.

Yeah well the ones on this website aren't

braylann wrote:Yeah well the ones on this website aren't

Catrollcho disagrees.If Corinthians win he will troll you....forever

Catrollcho

Trollrinthians

If he doesn't I'm going to troll him, all the crap he talked will come back to bite him in the ass...he better hopes they win

everyone is cheering for Boca here in Brazil.Every Corinthians fan is like Catracho

riquelme is south america's only hope against anti football

If Boca was out Id like Corinthians. Catrollcho is a hater

why does everything always come back to talking about Brasilians

THIS IS A PHYSICS THREADDDDDDD

ABOUT RAMOS PENALTY MISS

braylann wrote:

beatrixasdfghjk. wrote:

This is what I think...
Bear in mind I haven't actually done this topic :L. I forgot to mention where braylann says gravity becomes negligible and it breaks free, but it doesn't get that high anyway. And I don't know how to adjust it for the varying gravitational field strength...
Maybe dos can help and do it relativistically .
And yes, I completely ignored significant figures :L.

Hey guy. Your number is wrong. I also used constant acceleration equation, so I agree with your approach, it was exactly the same as mine. I see what I did wrong. I accidentally doubled the distance and forgot to take into consideration the drop. In fact, it is half of what I first said.

3.19 x 10^6 or 3,190,000 M

Not enough to hit the space station.

What did I do wrong?
.

Im not sure what you did wrong considering you didn't show your work, but your number for displacement is off. And you are using the same number for velocity as me. You used Sin(pi/4) where it should havebeen Cos(pi/4) for the vertical component.

my diagram might throw you off, I copied it in a hurry and shows me applying Cos to the wrong side, lol. But if you read the question you see that Cos is the vertical component

is shot forming an angle Pi / 4 with the vertical

it's the cosine component of the velocity vector. Usually, people make theta with respect to the horizontal ground.

braylann wrote:

your angle theta is wrong, should be with respect to the vertical axis. But the formula for Vx, Vy are correct.
After that, one should use the kinematic equation: 0.5at^ + Vup*t = x(t) (Vup is negative in this case)
take the derivative and set it = 0, that should give you the time where the ball is at the maximum height with the given velocity of Vup.
To find the height, plug in the t (usually 2 values due to quadratic nature, but you can easily rule out one for it being too obvious), and solve for the maximum height x. That will give you the height of the ball.

If the ball ever was to orbit the Earth, we can safely assume that it does (or else chances are, it won't ever hit the space station). We had obtained the maximum height of the ball, draw a diagram with the earth as a circle. We shall see where the ball is with respect to the Earth and the space station.

If one wishes to find the centripetal acceleration, we can also use
F = ma = mv^2/r.
Gravitational potential is G m(ball)m(earth)/r^2


If so, G m(ball)m(earth)/r^2 = mv^2/r, solve for v
and you shall have it. I somewhat don't understand what they meant by constant movements of the ball....

oh, and a in the kinematic equation is the acceleration of the Earth, which is g = 9.8 in this case (should be the opposite sign of Vup, because the direction are opposite).

beatrixasdfghjk. wrote:

braylann wrote:

beatrixasdfghjk. wrote:

This is what I think...
Bear in mind I haven't actually done this topic :L. I forgot to mention where braylann says gravity becomes negligible and it breaks free, but it doesn't get that high anyway. And I don't know how to adjust it for the varying gravitational field strength...
Maybe dos can help and do it relativistically .
And yes, I completely ignored significant figures :L.

Hey guy. Your number is wrong. I also used constant acceleration equation, so I agree with your approach, it was exactly the same as mine. I see what I did wrong. I accidentally doubled the distance and forgot to take into consideration the drop. In fact, it is half of what I first said.

3.19 x 10^6 or 3,190,000 M

Not enough to hit the space station.

What did I do wrong?
.

to find the function of gravity as a distance away from Earth, simply use the Newton gravitational formula.

F = GM1M2/ r^2

However, to a good approximation, we can assume that the ball experience a constant g for the kinematic equation part. However, if you have a massive object as big as the Earth, that would totally change the center of Mass, and the gravitational "field" gets weakened as F is proportional to 1/r^2 matters. Hence, binary stars, and blackholes, ect....anything with an elliptical orbit as opposed to a circular one like our solar system's planets (aside from pluto and maybe neptune). If that's the case, i'm afraid the centripetal acceleration won't even hold. This problem is supposed to be classical mechanics, not astromonical statastic mechanics.

Potential wrote:Although I admire your work, you're wrong.

11180 is not the speed that the ball leaves the earth atmosphere.

You cannot use that law, using the conservation of energy laws:



The rest should be correct though if you used the right velocity rather than the one you miscalculated.

Either way, good job if you're not a physics major

I don't think this is right either.
First, P stands for momentum, which doesn't equal to that, it's p = mv.
I think you meant evergy, which is E = K + U. K is the kinetic and U is the potential energy.

Generally, we use this with conservation of energy, which is fine. However, we need a momentum set of equations here also. Pinitial = P final.
Conservation of momentum only works when objects aren't being affected by the outside force, hence : no acceleration. Think about it, when you throw something in the air, V1 =/= V2. Sure, you can account for the horizontal component of velocity, which isn't affected by the gravity force (g). However, if you draw the diagram out with the Earth being a sphere, not a flat surface, you will see that there is definitely another component that is being affected as the ball travels far enough.

Hence, you would need to calculate the r component with respect to the radius of the Earth (r hat vector in polar coordinate). After that, you would need to find the angle with respect to that r^ vector, which is very complicated, and can be impossible even.

braylann wrote:Im not sure what you did wrong considering you didn't show your work, but your number for displacement is off. And you are using the same number for velocity as me. You used Sin(pi/4) where it should havebeen Cos(pi/4) for the vertical component.

my diagram might throw you off, I copied it in a hurry and shows me applying Cos to the wrong side, lol. But if you read the question you see that Cos is the vertical component

..
It hardly makes a difference, sin pi/4 = cos pi/4 .
Like I said, I haven't done gravitation yet...

technically, it's wrong. Professors wouldn't want to see you get mixed up on that
But yea, gravitational force is extremely simple. It's just a force F = GMm/r^2, and you usually apply it with centripetal acceleration or other things like sidereal year of a planet (generalized Kepler's law can be derived from it).

I wasn't sure what they were asking, but i guess I worked in the right direction

vizkosity wrote:technically, it's wrong. Professors wouldn't want to see you get mixed up on that
But yea, gravitational force is extremely simple. It's just a force F = GMm/r^2, and you usually apply it with centripetal acceleration or other things like sidereal year of a planet (generalized Kepler's law can be derived from it).

I wasn't sure what they were asking, but i guess I worked in the right direction

Don't you need the mass of the ball then?
How would you calculate the average gravitational force then, seeing as the radius is always changing?

beatrixasdfghjk. wrote:

vizkosity wrote:technically, it's wrong. Professors wouldn't want to see you get mixed up on that
But yea, gravitational force is extremely simple. It's just a force F = GMm/r^2, and you usually apply it with centripetal acceleration or other things like sidereal year of a planet (generalized Kepler's law can be derived from it).

I wasn't sure what they were asking, but i guess I worked in the right direction

Don't you need the mass of the ball then?
How would you calculate the average gravitational force then, seeing as the radius is always changing?

Sergio Ramos, due to the pressure he felt, miscalculates the parameters of a penalty, and the ball, with mass m

Yes, they said mass m.

Gravitational force is always changing, yes. However, you can easily approximate it by using the kinematic equation 1/2 gt^2+.......

it is generally acceptable at this level. However, if i must point out, the whole classical mechanics of newton is completely WRONG (aside from the gravitational force part). Einstein did it relativistically, and there is a gamma factor or beta fraction that u would need to consider (usually 2nd year U.S. physics major). To a good approximation, a small object orbiting the earth at that level can be done that way. Imagine throwing a snowball upward, with g being a function of x (height). It doesn't make our lives easier does it?

There is no calculating the average gravitational force. When it orbits the Earth, we assumed* that it's a perfect circle (or close to be 1), which we can then use r as 1 value (the highest pt the ball can reach upward+ the radius of the Earth). Once it's in orbit, we know it must follow F = mv^2/r, which is the force that makes thing go in the circle. If you don't do this, it's impossible to estimate when the Earth isn't a "perfect" circle (it's more like an onion with tides being pulled). As well as the fact that both objects orbit the center of mass, not the Earth's center, it's very hard to find the precise number.

Hope that helps

Also, the fact that they only have every number as variables (Vescape, m mass, Rt radius of the Earth), our answer will most likely be a bunch of variables combined. I mean...who would memorize the escape velocity of the Earth, the mass of the Eath, or the radius of the Eath b4 the exam....


my english sucks, it's too obvious, but i can explain

Don't they give you data sheets?
.

beatrixasdfghjk. wrote:Don't they give you data sheets?
.

the data sheet for a typical physics test only contain information like constants, g =9.8 m/s^2, planck's constant (quantum mechanics), speed of light c, charge of electron e, ect...

also, a physics test sometimes must be done without a calculator (not everyone can afford a good one, and professors try to be fair), so they will use a lot of variables like Vo (initial velocity), Xo, c, e, and put them all together at the end like an alphabet soup. The last part is to plug them in a calculator, which might/might not be asked.

The fact that they used m and Rt as special letter gives away the fact that they want an alphabet answer. However, if they gave u Rt numerically, why can't they give you m numberically? Now, that doesn't make any sense does it? m is indeed, extremely important when you are dealing with Force, just like how Rt is important to give you the orbit of the ball numerically:P (or else gl with 3Rt space station and not knowing what 1 Rt is while having an arbitrary value of the radius of orbit for the ball


We physicists are extremely lazy :I Everything must be written in variables as often as possible (unless using numbers can simplify the algebra a lot more).

And yes, I hope you will have fun studying physics later. Personally, Classical mechanics is my favorite subject (ever) in high school and college. None of that bullcrap electromagnetic, modern physics interest me as much as CM. It's very awesome though, but it's not just for me

.
Well here, you don't have a calculator, then you just borrow one...
Our data sheet has mass of the Earth, and a whole bunch of formulae :L.

beatrixasdfghjk. wrote:.
Well here, you don't have a calculator, then you just borrow one...
Our data sheet has mass of the Earth, and a whole bunch of formulae :L.

i guess it's very different how the U.S. operates. We use MKS units (kg, m, second), yet we measure and machine with inches, pounds.....
/facepalm

Calculators are indeed, borrowable. However, most of U.S. universities/colleges have like hundreds of people rolling in 1 physics class (lower div), as it is a requirement for many engineers, biologists, chemists, and many more. It's not practical buying like 10-20 calculators just in case. You can ask them to buy one, or have a cheat sheet with all the formula they can put in. That's fine too, but the best way is to deal with it alphabetically. It will also be easier for graders (imagine sig. fig throwing your answers slightly off, or you copied 1 number wrong onto your paper, which doesn't affect that many points anyways.). They tend to care more about the process, not the "answer". Most students wouldn't get all the answers right anyways.