AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

OP

Original post by muyiwaaiyenuro

http://filestore.aqa.org.uk/over/stat_pdf/AQA-A-LEVEL-GDE-BDY-JUNE-2014.PDF

Any idea what those second grey grade boundaries suggest?

They're the individual sections' breakdown I guess. Even though your grade is actually worked out by the sum total of your marks.

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Reply 2263

Can someone help me with question 13 please?http://filestore.aqa.org.uk/subjects/AQA-PHYA41-QP-JUN14.PDF

Reply 2264

OP

Original post by Adangu

Can someone help me with question 13 please?http://filestore.aqa.org.uk/subjects/AQA-PHYA41-QP-JUN14.PDF

[br]\displaystyle F = \dfrac{Qq}{4 \pi {\epsilon_0} r^2}[br]

This means

\displaystyle F \propto \dfrac{1}{r^2}

.

The original distance was

d

when the force was

F

.

The force is halved to become

0.5 F

, so the previous distance must be

20mm \div \sqrt{2} = 14mm

.

Does that help?

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(edited 8 years ago)

Reply 2265

Original post by CD223

[br]\displaystyle F = \dfrac{Qq}{4 \pi {\epsilon_0} r^2}[br]

This means

\displaystyle F \propto \dfrac{1}{r^2}

.

The original distance was

d

when the force was

F

.

The force is halved to become

0.5 F

, so the previous distance must be

20mm \div \sqrt{2} = 14mm

.

Does that help?

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Sorry to bother you more, but where does the root 2 come from?
My brain doesn't seem to be working with me haha!

Reply 2266

OP

Original post by Adangu

Sorry to bother you more, but where does the root 2 come from?
My brain doesn't seem to be working with me haha!

As you have:

[br]\dfrac{1}{2}F [br]

On one side of the equation, you must have

[br]\dfrac{1}{\left(\sqrt{2} d \right)^2}[br]

On the other side - does that make sense?

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Reply 2267

OP

How would you go about mathematically proving this?

I got it right by guessing, because:

[br]{m_{before}} = 200g[br]{m_{after}} = 200g + 300g = 500g[br]

[br]KE = \dfrac{1}{2}mv^2[br]

[br]\therefore \dfrac{{m_{after}}}{{m_{before}}} = \dfrac{500}{200}[br]

[br]\therefore \dfrac{{KE_{after}}}{{KE_{before}}} = \dfrac{5}{2}[br]

But this gives the inverse of the correct answer?

I know KE depends on m as well as v, but I just didn't think enough info was given in the question about the velocity? It just says the initial velocity is U, then it coalesces with a stationary object, without saying the velocity after?

Confused.

(edited 8 years ago)

Reply 2268

OP

EDIT: Never mind! All sorted

The frequency has to reduce by two thirds so

f \rightarrow \frac{2}{3}f

So the period would have to increase by 1.5, so

T \rightarrow \frac{3}{2}T

As

T = 2 \pi \sqrt{\dfrac{m}{k}}

This means m would have to increase by 1.5 squared so

m \rightarrow 2.25m

\therefore 0.40kg \rightarrow 0.90kg

meaning 0.50kg needs to be added.

(edited 8 years ago)

Reply 2269

OP

Original post by Mehrdad jafari

Use this equation as this is quicker and less stressful in the exam :smile:

ImageUploadedByStudent Room1433167827.772755.jpg

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Edit: oops, in the first equation its K/m and not m/k

Aha yeah I was gonna ask about the k/m!

Thanks :smile:

Do you get the KE question?
I just didn't think there was enough info on the velocity :s-smilie:

Reply 2270

lebanon95

1

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JAN10.PDF Question 11 Lads, I can't seem to get my head around where to start off

EDIT: No problem just got it by forming an equation of the earths GFS and another for the planets GFS, rearranging both so M = blah blah and dividing the pair to get X^2Y. Just wondering if there would be any quicker method or realizing some sort of ratio without having to do all of that?

(edited 8 years ago)

Reply 2271

OP

Original post by Mehrdad jafari

In case if you haven't got it

ImageUploadedByStudent Room1433168379.847663.jpg

Great! Thank you :smile:

really should have got that one >.<

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Reply 2272

OP

Original post by lebanon95

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JAN10.PDF Question 11 Lads, I can't seem to get my head around where to start off

EDIT: No problem just got it by forming an equation of the earths GFS and another for the planets GFS, rearranging both so M = blah blah and dividing the pair to get X^2Y. Just wondering if there would be any quicker method or realizing some sort of ratio without having to do all of that?

Unfortunately that method is the quickest way I know of. Form an equation for M:

[br]M = \dfrac{gr^2}{G}[br]

Then sub in the values x and y, and divide.

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Reply 2273

Original post by CD223

As you have:

[br]\dfrac{1}{2}F [br]

On one side of the equation, you must have

[br]\dfrac{1}{\left(\sqrt{2} d \right)^2}[br]

On the other side - does that make sense?

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Sorry it still doesn't, I don't see where the root 2 comes from.
I don't know how I don't get it!

Reply 2274

OP

Original post by Adangu

Sorry it still doesn't, I don't see where the root 2 comes from.
I don't know how I don't get it!

F is is inversely proportional to the distance, d, squared. This means that, if you look at it another way:

ImageUploadedByStudent Room1433174430.075968.jpg

ImageUploadedByStudent Room1433174430.075968.jpg

Reply 2275

gcsestuff

ImageUploadedByStudent Room1433175122.890440.jpg

Not sure on why the angular velocity is the same. Using w=v/r means that if the radius is smaller the speed should be greater??

Or am I supposed to be using the g=f/m. Which shows acceleration only depends on force and mass of the object.

Thanks again 😀😀

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Reply 2276

Original post by CD223

F is is inversely proportional to the distance, d, squared. This means that, if you look at it another way:

ImageUploadedByStudent Room1433174430.075968.jpg

Thank you so much! Seriously thank you!

Looking at it like that I understand it!

Reply 2277

OP