# g485 questions

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#1
in a g485 paper a question about a charge in a magnetic field that went into circular motion appeared,, annd the question asked why does the magnetic field make no difference to the velocity of circular motion, i understand that its because the force is perpendicular to direction of motion but the second mark required the statement no work is done, i know why because w=fxcostheta but whats the point of making that statement, i dont get it.

also a galaxy has a parallac of 0.275 or arc second so they find the distance in pc by doing 1/0.275, i know the long way to get the answer but i dont understand how they used just this straightforward method, even the examiner report say its straightforward as this. ive not come across any quick formula or quick tip to get to the answer quickly like this.
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#2
also does anyone know where i can get the g484 and g485 june 2013 mark schemes ive got the papers but no answers.
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7 years ago
#3
This would be much easier if I could actually see the questions either by a link or an image. However I can answer the first question for you because all of the relevant information is here.
So you've already said you know the formula W = F s cos (theta), so I wont go into how to use that equation. By the definition of Work done, a Force when applied over a distance, we can tell if there is a change in velocity. We know that no work is done, and as such a Force has not been applied over a distance, due to Newtons first law of motion which states an object will remain at a constant velocity when in a inertial frame unless acted upon by a external Force. We have proved that there is no Force acting on the object by showing that no work has been done, thus have proved no change in velocity.
As for why you have to say that no work has been done, it's to expand on your answer and show you understand why there is no change.

Hope that helps.
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7 years ago
#4
(Original post by nmjasdk)
i understand that its because the force is perpendicular to direction of motion
Another example of why I hate A-level physics. You're exactly right. At undergrad level this answer would most certainly be fine. The extra bit about no work being done... It's correct but is so finicky.

also a galaxy has a parallac of 0.275 or arc second so they find the distance in pc by doing 1/0.275, i know the long way to get the answer but i dont understand how they used just this straightforward method, even the examiner report say its straightforward as this. ive not come across any quick formula or quick tip to get to the answer quickly like this.
I presume the reasoning behind the calculation is trigonometrical.

A parsec means literally 'a parallax of one arcsecond' (wiki). This may explain why you can do such a seemingly basic calculation. It's just a manipulation of SOHCAHTOA.
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#5
(Original post by Zaros)
This would be much easier if I could actually see the questions either by a link or an image. However I can answer the first question for you because all of the relevant information is here.
So you've already said you know the formula W = F s cos (theta), so I wont go into how to use that equation. By the definition of Work done, a Force when applied over a distance, we can tell if there is a change in velocity. We know that no work is done, and as such a Force has not been applied over a distance, due to Newtons first law of motion which states an object will remain at a constant velocity when in a inertial frame unless acted upon by a external Force. We have proved that there is no Force acting on the object by showing that no work has been done, thus have proved no change in velocity.
As for why you have to say that no work has been done, it's to expand on your answer and show you understand why there is no change.

Hope that helps.
the second question im not really sure which paper its from, but that was literally all the information they gave.

OK so from what im reading the work done formula needs to be mentioned to prove that f=0 but if w= fxcostheta and if you rearrange it you would never be able to find F and say its zero for sure its like 0=x*0 x could be a number but theirs no way to know.
i think i better way to word my question is how does algaebraically no work done mean no change in velocity wheres that equation linking them.

btw thanks a lot for the help,
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7 years ago
#6
(Original post by nmjasdk)
the second question im not really sure which paper its from, but that was literally all the information they gave.

OK so from what im reading the work done formula needs to be mentioned to prove that f=0 but if w= fxcostheta and if you rearrange it you would never be able to find F and say its zero for sure its like 0=x*0 x could be a number but theirs no way to know.
i think i better way to word my question is how does algaebraically no work done mean no change in velocity wheres that equation linking them.

btw thanks a lot for the help,
Well, relating Work done to velocity would have to be W = (mv^2)/2 in other words the formula for Kinetic Energy. You've got a constant mass (constant enough for it not to be relevant anyway) so we can say that m will be constant. As such, there just isn't any other number than v can be in order for W to also be 0. That would be the algebraic proof.

Hope that helps.
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#7
(Original post by Pessimisterious)
Another example of why I hate A-level physics. You're exactly right. At undergrad level this answer would most certainly be fine. The extra bit about no work being done... It's correct but is so finicky.

I presume the reasoning behind the calculation is trigonometrical.

A parsec means literally 'a parallax of one arcsecond' (wiki). This may explain why you can do such a seemingly basic calculation. It's just a manipulation of SOHCAHTOA.
(Original post by Zaros)
Well, relating Work done to velocity would have to be W = (mv^2)/2 in other words the formula for Kinetic Energy. You've got a constant mass (constant enough for it not to be relevant anyway) so we can say that m will be constant. As such, there just isn't any other number than v can be in order for W to also be 0. That would be the algebraic proof.

Hope that helps.
OK i get it now forgot about that fromula relating the energy to speed, I think its because many formulas indicate change in speed (f-ma) its only when you relate it to energy you can show the speed doesnt effect it.

and also for the parallax thing i get that sin cos tan manipulation but the MS and examiner report show it specifically being as simply as 1/0.275 with all that sin cos tan thing being unnecassary? any reasons how they got this so simply?
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7 years ago
#8
(Original post by nmjasdk)
in a g485 paper a question about a charge in a magnetic field that went into circular motion appeared,, annd the question asked why does the magnetic field make no difference to the velocity of circular motion,
As another poster has said, it's better if you post the actual question if you want a reasoned and worthwhile explanation.
For example, this cannot have been the original question because it's not true that there is no change in velocity.
There is no change is speed.
The velocity is constantly changing.
In physics the distinction is very important. It's not just nit-picking.
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#9
oh yeah its speed sorry i was telling myself to write speed specifically when i type out the question mustve forgot when typing it out.
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7 years ago
#10
(Original post by nmjasdk)
also does anyone know where i can get the g484 and g485 june 2013 mark schemes ive got the papers but no answers.
Do you have a link to the g485 paper?
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#11
pm me
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7 years ago
#12
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7 years ago
#13
(Original post by nmjasdk)
pm me

Posted from TSR Mobile

Done
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#14
(Original post by Jamesnorniron)
Posted from TSR Mobile

Done
I'll get it to u tomorrow for sure
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