***2015 LC Physics - June 15th***

smartz

Broseph wrote: »

I didnt do the extra question in the end I just did the extra part in Q12. Hopefully it payed off because I got full marks in the ski question and did b and c too I dont know how I did in c but I did awfully bad in b lol

How did you do the last bit of the ski Q?

Broseph

smartz wrote: »

How did you do the last bit of the ski Q?

the force? v=u+at , let v=o and you have the time value, so acceleration = whatever value that comes out as, and fill it into f=ma, you have the mass and you can find the force

smartz

there was 2 different forces though. one the girl on the snow drift and the other the snowdrift on the girl.

Broseph

smartz wrote: »

there was 2 different forces though. one the girl on the snow drift and the other the snowdrift on the girl.

I think it's the same force (equal but opposite forces, newtons third law) thats what I said anyway could be wrong

smartz

Yeah I thought that at first too, but then I thought they wouldn't be that sly would they?
Anyway, I got the acceleration of the girl downhill from the first bit, used that to find her force on the snowdrift.
Then I added that to the force the snowdrift exerts in making her stop(your answer), to find the force of the snowdrift on here.
But then again, I may have massively over-complicated that.

Broseph

smartz wrote: »

Yeah I thought that at first too, but then I thought they wouldn't be that sly would they?
Anyway, I got the acceleration of the girl downhill from the first bit, used that to find her force on the snowdrift.
Then I added that to the force the snowdrift exerts in making her stop(your answer), to find the force of the snowdrift on here.
But then again, I may have massively over-complicated that.

ah yeah I get you, if after the 400 metres shes still going downhill you're right and if it flattens out my way is right so I'd say I assumed wrong. But you'd minus the forces not add them for your way

Chickennuggets

What was the answer for the last part of Q 12(d)

DarraghF197

It was a weird phrasing as well. One asked what force the girl exerted and then the next part asked what is the force IIRC. I think they might have been asking for the name of the force for one of them with their awkward phrasing.

smartz

Chickennuggets wrote: »

What was the answer for the last part of Q 12(d)

I got 4.39x10^12

Broseph

smartz wrote: »

I got 4.39x10^12

Is there a marking scheme that anyone knows of?

Peregrine

The force she exerts on the snowdrift can be calculated by F = (her mass)(the component of her acc. due to gravity parallel to the slope calculated from the previous part)

The force the wall exerts on her is F = (her mass)(her decceleration)

Edit: Ignore this. The second part is the net force anyway and I'm still not sure about the whole method overall.

The_N4sir

smartz wrote: »

I got 4.39x10^12

I assume that is a typo and should be minus 12?:P

smartz

Nim wrote: »

The force she exerts on the snowdrift can be calculated by F = (her mass)(the component of her acc. due to gravity parallel to the slope calculated from the previous part)

The force the wall exerts on her is F = (her mass)(her decceleration)

why is the force the wall exerts on her not:
(her mass)(her decceleration) + the force she exerts on the wall.

shoopdeboop

The_N4sir wrote: »

I assume that is a typo and should be minus 12?:P

Was it not the amount of atoms? it would still be pretty high, no?

smartz

The_N4sir wrote: »

I assume that is a typo and should be minus 12?:P

no, if it was a minus it would only be a fraction of an atom.

The_N4sir

shoopdeboop wrote: »

Was it not the amount of atoms? it would still be pretty high, no?

Sorry I misread the post completely. Thought that was the answer to the last part of particle physics

Broseph

The_N4sir wrote: »

Sorry I misread the post completely. Thought that was the answer to the last part of particle physics

yeah its minus for the last part of particle physics your answers right

Cr4pSnip3r

smartz wrote: »

why is the force the wall exerts on her not:
(her mass)(her decceleration) + the force she exerts on the wall.

Should the two forces not be the same because of Newton's third law of motion no? lol, I just ****ted on for like 10 pages during that exam

Broseph

Cr4pSnip3r wrote: »

Should the two forces not be the same because of Newton's third law of motion no? lol, I just ****ted on for like 10 pages during that exam

I said the same I thought it was right at the time but now I'm not sure anyone sure of it?

smartz

Actually the answer for the radius in the particle physics is 9 x 10^3.
You get x10^-12 if you put in Mega Tesla instead of milli Tesla. It's a small 'm'

Broseph

smartz wrote: »

Actually the answer for the radius in the particle physics is 9 x 10^3.
You get x10^-12 if you put in Mega Tesla instead of milli Tesla. It's a small 'm'

that wouldnt make any sense though would it? as it's a cloud chamber and that radius is 9000m?
but yeah maybe it goes in a line thats almost straight with a tiny curve which would form a circle of that radius IF there was the space

DarraghF197

smartz wrote: »

why is the force the wall exerts on her not:
(her mass)(her decceleration) + the force she exerts on the wall.

Isn't that sort of saying, why isn't the force acting on a football not the force of the kick and the mass times acceleration. The force equals mass times acceleration, I think you're double counting here.

It's simply an equal but opposite force (something I didn't understand for a long time!). It was testing your understanding of Newton's laws in this question.

smartz

Broseph wrote: »

that wouldnt make any sense though would it? as it's a cloud chamber and that radius is 9000m?
but yeah maybe it goes in a line thats almost straight with a tiny curve which would form a circle of that radius IF there was the space

Sorry i meant x10 ^-3

Broseph

smartz wrote: »

Sorry i meant x10 ^-3

yeah ok thats right how many marks would you lose for saying mega instead of milli?

smartz

Broseph wrote: »

yeah ok thats right how many marks would you lose for saying mega instead of milli?

3 max

DarraghF197

Broseph wrote: »

yeah ok thats right how many marks would you lose for saying mega instead of milli?

I'd say you'd only lose one mark. It's similar to a mathematical slip or an ommission of units, which both only lose you one mark.

smartz

DarraghF197 wrote: »

Isn't that sort of saying, why isn't the force acting on a football not the force of the kick and the mass times acceleration. The force equals mass times acceleration, I think you're double counting here.

It's simply an equal but opposite force (something I didn't understand for a long time!). It was testing your understanding of Newton's laws in this question.

Yeah, that's true. But I still don't think the two forces are just the same. I think Nim has it right up above.

Peregrine

If the force she exerts on the snowdrift was equal to the force the snowdrift exerts on her, she wouldn't stop. She would stop accelerating and keep moving at a constant velocity, as per Newton's 1st Law of Motion. Since there is decceleration in this case, there would have to be a net force working against the motion of the skiier i.e. the force exerted by the snowdrift must be greater than the force she exerts on it.

smartz wrote: »

why is the force the wall exerts on her not:
(her mass)(her decceleration) + the force she exerts on the wall.

You have made me question everything I know about mechanics :P I can't think straight, I'll try and give you a proper answer tomorrow.

qweerty

Nim wrote: »

If the force she exerts on the snowdrift was equal equal to the force the snowdrift exerts on her, she wouldn't stop. She would stop accelerating and keep moving at a constant velocity, as per Newton's 2nd Law of Motion. Since there is decceleration in this case, there would have to be a net force working against the motion of the skiier i.e. the force exerted by the snowdrift must be greater than the force she exerts on it.

You have made me question everything I know about mechanics :P I can't think straight, I'll try and give you a proper answer tomorrow.

Nim, I don't think you're right. Only had a quick look at the question, and I'm on my phone and my mind is somewhat impaired by alcohol, but the force she exerts is as a result of being decelerated. If it were a car which creates its own driving force you'd be correct. But because the drift causes her to decelerate, they each exert the same force on each other. At least, I think.

Regarding second para, one is asked for the (strictly speaking, average) force exerted over the 0.8 interval in order to bring the skier to rest, and not the force required to subsequently keep it in place (in which case you'd in some way account for W after t=0.8). The skier's weight is effectively built into the equations.

rainbowtrout

Broseph wrote: »

Is there a marking scheme that anyone knows of?

Marking schemes for all exams are not released until the end of August, the weekend of the viewing of the papers.

MmmPancakes

Jammysticks wrote: »

Tin-whistle length was .579 divided by 2, wasn't it?

Fundamental frequency for an open pipe: wavelength = 1/2 length
c = f(lambda)
lambda = 340/587

Divide that by 2 to get the pipe length no? (I multiplied by 2 in the exam )

Broseph

rainbowtrout wrote: »

Marking schemes for all exams are not released until the end of August, the weekend of the viewing of the papers.

sorry I meant a general marking scheme with predicted marks just to get a general idea of what you got

Peregrine

qweerty wrote: »

Nim, I don't think you're right. Only had a quick look at the question, and I'm on my phone and my mind is somewhat impaired by alcohol, but the force she exerts is as a result of being decelerated. If it were a car which creates its own driving force you'd be correct. But because the drift causes her to decelerate, they each exert the same force on each other. At least, I think.

You could be right. I'll have another look at it tomorrow and try and make sense of it :P

rainbowtrout

otoolej2 wrote: »

What did people say for the advantages of kelvin as a S.I unit? odd question.

I would say that because the temperature scale starts at absolute zero, then the temperature rise on the Kelvin scale is in proportional with the energy of whatever substance you are dealing with. Zero degrees celsius does not mean zero energy, it's the freezing point of water which is an arbitrary way to measure a temperature scale (so too is the boiling point for that matter). When you go into the negative end of the celsius scale objects still possess heat/energy.

Or standard temperature as in STP, which is stated in many physics (and chemistry questions) is given as 0 degrees celsius. You have to convert it to Kelvin to be able to perform calculations as you can't divide by zero and multiplication by zero would yield a zero result.

Broseph wrote: »

sorry I meant a general marking scheme with predicted marks just to get a general idea of what you got

There's no way of predicting a marking scheme. you might come up with a perfectly valid marking scheme and someone else on this thread might come up with an equally valid marking scheme and weight the marks within a question totally differently. Some answers which you might think are valid may not be without extra information, some answers which you didn't think of might be correct.

Don't do a post mortem on it trying to predict down to the mark what you will get, it's impossible.

uzpuz

rainbowtrout wrote: »

There's no way of predicting a marking scheme. you might come up with a perfectly valid marking scheme and someone else on this thread might come up with an equally valid marking scheme and weight the marks within a question totally differently. Some answers which you might think are valid may not be without extra information, some answers which you didn't think of might be correct.

Don't do a post mortem on it trying to predict down to the mark what you will get, it's impossible.

I do that all the time including for exams and the mark I predict is always within 5% error. I predict I got 80 % but I will get within 75-80%. It isn't accurate but nevertheless gives a good indication

DarraghF197

Nim wrote: »

If the force she exerts on the snowdrift was equal to the force the snowdrift exerts on her, she wouldn't stop. She would stop accelerating and keep moving at a constant velocity, as per Newton's 2nd Law of Motion. Since there is decceleration in this case, there would have to be a net force working against the motion of the skiier i.e. the force exerted by the snowdrift must be greater than the force she exerts on it.

Newton's third law states every action has an equal but opposite reaction. If you take a gun: there is a force that causes the bullet to be shot, and then there is an opposite reaction on the gun. The opposite reaction is equal to the mass of the bullet multiplied by the acceleration. The force in the gun is equal. Another way of looking at it is Newton's second law. The rate of change of momentum for the gun equals the rate of change of momentum for the bullet.

My opinion is unavoidably biased as I hope I wrote down the right answer, and part of me feels like I'm battling a lost case!

Do You Even Squat

Does the same rule apply in physics as it does in maths where if you write down a bunch of calculations, you can still pick up the marks if you wrote down the correct answer muddled in with a few meaningless/wrong calculations?

2345

What was the nuclear equation representing beta decay.

smartz

n = p + e + v

with the mass numbers and charge numbers as well.

Peregrine

qweerty wrote: »

Nim, I don't think you're right. Only had a quick look at the question, and I'm on my phone and my mind is somewhat impaired by alcohol, but the force she exerts is as a result of being decelerated. If it were a car which creates its own driving force you'd be correct. But because the drift causes her to decelerate, they each exert the same force on each other. At least, I think.

Regarding second para, one is asked for the (strictly speaking, average) force exerted over the 0.8 interval in order to bring the skier to rest, and not the force required to subsequently keep it in place (in which case you'd in some way account for W after t=0.8). The skier's weight is effectively built into the equations.

Just had another look and I still can't make sense of it. Any chance you could do the question out?

qweerty

Nim wrote: »

Just had another look and I still can't make sense of it. Any chance you could do the question out?

Yeah, give me a sec. I think you're over-complicating it, though.

Q12 (a)

The answer to the first part of the question is that energy is converted from potential to kinetic, which suggests that they're encouraging the first method I used.

I guess only the last three parts are contentious and that part of the confusion arises from wanting to split the force the skier exerts into its components. That is over-complicating it. Put simply, there is an object with motion. In order to bring it to rest, a set of forces whose magnitudes vary over the time-interval act on the object. We're being asked to find the average resultant force (i.e. the combination of all the forces, taking the deceleration as constant): simply, the product of the object's mass and deceleration.

Hope that helps

Disclaimer: even when I'm sure, I'm never sure!

Broseph

smartz wrote: »

n = p + e + v

with the mass numbers and charge numbers as well.

And Kinetic energy on the right for the neutrino no?

PotmBottom

for 12(a) I just said the force of the skiier was equal to change in momentum/time taken and got an answer of 3727.5N (very close to the solution above). I ignored the minus on the velocity as that would only indicate direction.

qweerty

PotmBottom wrote: »

for 12(a) I just said the force of the skiier was equal to change in momentum/time taken and got an answer of 3727.5N (very close to the solution above). I ignored the minus on the velocity as that would only indicate direction.

F = dp/dt is just an earlier stage of F = ma. In fact, your way is the most efficient because you don't need to find a, but I only had enough room for one solution, so went with the way I guess most would do it.

My answer was the same as yours, I just rounded

Peregrine

qweerty wrote: »

Yeah, give me a sec. I think you're over-complicating it, though.

Q12 (a)

Yeah, I got that as the net force and went on to do this..

I think I'm interpretting 2nd Law wrong.

qweerty

Nim wrote: »

Yeah, I got that as the net force and went on to do this..

I think I'm interpretting 2nd Law wrong.

Yeah, I think you are. Momentum before = momentum after. F = dp/dt. Skier's momentum goes from p to 0, therefore drift's momentum goes from 0 to p. As the change is the same, the force is the same.

Having to justify (if only to myself) my method is making my head spin. Tbh, this is what I wish physics did. App Maths offers an interesting test at times, but only a tricky physics question requires you to apply all your physics knowledge. It rarely does, tho, and when it does (like with the underwater goggles question) you're not expecting it to require lateral thinking. (Edit: just to clarify, I'm not saying I think this is a tricky question!)

I hope enough time has passed to say that, bluntly, that was a decidedly easy exam. Literally, everything bar a handful had come up before. For my money, the only question-parts that had never been on before were the capacitance short question and particle physics question about radius; neither of them was difficult. (That particle physics question could have been made into one about a circular particle accelerator and been made more challenging. Shame they didn't.)

Peregrine

qweerty wrote: »

Yeah, I think you are. Momentum before = momentum after. F = dp/dt. Skier's momentum goes from p to 0, therefore drift's momentum goes from 0 to p. As the change is the same, the force is the same

That bit there makes sense Cheers

Conaaaa

How much percent is it if you learn all experiments laws etc besides the equations ?