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

Original post by JJBinn

Oh man I was setting M to zero in that equation because I was so fixated on the fact it was weightless.

On a side note, for this equation, is the mass representative of both objects? Ie gravitational attraction occurs between 2 masses, do you add the masses together for example if it was a satellite orbiting the earth, in using that equation do you add the mass of the satellite to the earth's mass?

I see. The mass M is representative of the mass of the object creating the field. In this case m is around

5.9\times10^{24}

as that is the mass of the earth only. The satellite is the small test mass orbiting the mass creating the field. In equations, capital M is the mass creating the field, lower case m is the small test mass (eg: satellite) in that field.

Original post by Mehrdad jafari

me too. To be honest i cannot remember studying unit 5 lool.
In theory definitely :smile:

Hahaha. Need to start revising it properly for sure. Thanks for that - really hoping i didn't do badly :frown:

Original post by Kiki05

Ay yo
Any1 out there plz help me with this
Q 14 June 10

I dont know y the answer is A
can some1 plz explain y its A, it would b much appreciated

Thanks

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Original post by Paul Dirac

Thanks

For June 2014 Section B question 3bii would this be correct? (see attached)

june 2014.png

It would be correct. They're usually quite vague when it comes to diagram mark schemes but as long as the general shape is there (positive or negative) they usually give the mark.

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Reply 1783

Jimmy20002012

Cn anyone help me in these momentum questions. For the hose pipe one I have found the mass by using the density equation but haven't idea on how to get velocity from the given information. For the truck momentum one, seem really easy but can't seem to get it I know the initial momentum is 20,000 now what? Thanks

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Reply 1784

Jimmy20002012

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Reply 1785

Jimmy20002012

Original post by Mehrdad jafari

For the first question i would suggest you to work out with units to figure out the velocity because that can be something of a challenge if you don't.
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Oh how did I not get that momentum question, thinking about it they were pretty easy. Thanks, the units really helped on the hosepipe question quite a lot. :smile:

Reply 1786

Absent Agent

Original post by Jimmy20002012

Oh how did I not get that momentum question, thinking about it they were pretty easy. Thanks, the units really helped on the hosepipe question quite a lot. :smile:

It's cool

. Yeah, the wording can catch you out as well

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Reply 1787

looks simple but i cant get to the answer

Reply 1788

Original post by Jimmy20002012

Oh how did I not get that momentum question, thinking about it they were pretty easy. Thanks, the units really helped on the hosepipe question quite a lot. :smile:

I remember doing those questions thinking it was odd - that hose pipe question has appeared twice now on papers. The exact same question and answer!

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and this one

Original post by CD223

I remember doing those questions thinking it was odd - that hose pipe question has appeared twice now on papers. The exact same question and answer!

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Yeah I know think they rebank at least 3 questions each year. I have a feeling that they could do put that in the written section as well. As in June 2013 they put the phase difference question of two pendulums in section B and it appeared in section A the previous year.

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Reply 1791

Original post by AR_95

looks simple but i cant get to the answer

[br]v=\omega r[br]

[br]v = \dfrac{108 \times 10^{3}}{3600}[br]

[br]\omega = \dfrac{v}{r}[br]

[br]\therefore \omega = \dfrac{30}{0.2}[br]

[br]\therefore \omega = 150\ rads^{-1}[br]

Reply 1792

Original post by CD223

[br]v=\omega r[br]

[br]v = \dfrac{108 \times 10^{3}}{3600}[br]

[br]\omega = \dfrac{v}{r}[br]

[br]\therefore \omega = \dfrac{30}{0.2}[br]

[br]\therefore \omega = 150\ rads^{-1}[br]

What I don't get is why is the 108000 divided by 3600 and not multiplied. Or is it different in this case because the units stated ^-1?

Reply 1793

Original post by AR_95

What I don't get is why is the 108000 divided by 3600 and not multiplied. Or is it different in this case because the units stated ^-1?

It's because 108 kilometres per hour is equal to 108,000m in 3600 seconds. Dividing one by the other gives a speed in metres per second.

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Reply 1794

Original post by CD223

It's because 108 kilometres per hour is equal to 108,000m in 3600 seconds. Dividing one by the other gives a speed in metres per second.

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lmao that's just silly from me. Forgot the meaning that it was km PER HOÚR not km x hoúr in seconds haha

Reply 1795

Original post by AR_95

and this one

The answer is D. On the right I've explained what the formula breaks down into.

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Reply 1796

Original post by AR_95

lmao that's just silly from me. Forgot the meaning that it was km PER HOÚR not km x hoúr in seconds haha

No problem! The best of us make that mistake a thousand times over. Comes back to the classic "read the question" mistake we all make :biggrin:

Reply 1797

frankiejayx

would anyone mind helping me with Q2biii in June 2012? I dont understand why the proton's acceleration increases whereas the electron's decreases?

Reply 1798

Done for the day. Hoping someone can help me with these two

Reply 1799