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AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread]

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Amplitude must be small

Posted from TSR Mobile

I said the angle between the equilibrium position of the string at rest and the the position of the line at the amplitude must be 10 degrees when I measured out when to drop the ball. Sounds acceptable?

Reply 1521

-Gifted-

Original post by ombtom

Instead of using v and V I'll use v = 2, V = -1 for the velocities.

Everyone's saying mv - VN = 0
So V = mv/N
-1 = 2m/N

Since m/N > 0, this implies that -1 = 2 x positive constant.

What's going on? :redface:

If V=-1 the formula would be mv+VN=0. However, considering right to be positive mv-VN=0 is the correct momentum conservation.

Reply 1522

DesignPredator

To be honest it will probably end up with them accepting both positive and negative due to all the confusion.

(edited 7 years ago)

Reply 1523

tanyapotter

How many marks was the radius calculation? Was it worth 2 or 3?

Reply 1524

Cadherin

Original post by marioman

In your dreams.

If only... I think more like 57 for an A* if we're lucky. 62 or 63 for 120 UMS depending on standard deviation.

And I think that's me being conservative.

Reply 1525

js.int

Original post by 1017bsquad

Ok i got scared for a moment xD, I didn't do well on this paper :/ couldnt even answer the time constant question... and the kinetic energy fml

Yeah the E_alpha derivation got me too. I only managed to get to something like:

E_alpha = E(1 - ((1/2N)m^2v^2)/E)

:s-smilie:

Reply 1526

smisocm

Does anyone have an unofficial markscheme?

Reply 1527

Jay1421

Original post by English-help

Thanks you! :smile:

I need 85+ in Unit 5 now which will be hard but i'll have to get it for my offer! :/

Best of luck to you! What is your offer for, which grades and where? :smile:

Reply 1528

-Gifted-

All these multiple choice answers with many D's, I don't really think its right. I barely had D's...

Reply 1529

username1967507

Original post by -Gifted-

Defo 2F and X to Y.

if they are both positive, then the force on X has direction Y to X (repulsive).

Reply 1530

Jay1421

Original post by js.int

Yeah the E_alpha derivation got me too. I only managed to get to something like:

E_alpha = E(1 - ((1/2N)m^2v^2)/E)

:s-smilie:

Yeah, that was the one question where I just couldn't see how to get to it.

Everything else was either a case of "Oh I know this" or "Ahhh I wish I revised this more" but.. this one.. I just couldn't recall anything about it at all and I couldn't derive the right equation :frown:

Reply 1531

Tokhmehsag

What was the momentum of the alpha particle ? 8.6x10^-19???

Reply 1532

-Gifted-

Original post by Sid1234

if they are both positive, then the force on X has direction Y to X (repulsive).

Initially it was negative and was y to x. after, the potential was positive, so must have changed direction .

Reply 1533

-Gifted-

Original post by Tokhmehsag

What was the momentum of the alpha particle ? 8.6x10^-19???

I got like 2*10^-20

Reply 1534

Jay1421

Original post by -Gifted-

All these multiple choice answers with many D's, I don't really think its right. I barely had D's...

Just so you know, for the past 10 past papers, there has always been an even distribution of answers A, B, C and D. In every paper so far, there's been 6 answers for 3 letters and then 7 answers for 1 letter.