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

Is this the best way to do these kind of questions? Jan 12, question 15 multiple choice? Thanks

That's how I do them. You can also use the fact that the ratio of the charges/masses is equal to the ratio of the distances squared, then square root (obviously negative root isn't applicable in this case), then you have a linear equation.


Posted from TSR Mobile

thanks!

Why do I not include the radius of the earth in this? I thought newtons law was the graviational force of attraction the centre of 2 masses??

Thanks


Posted from TSR Mobile

The orbit distance takes into account the radius.


Posted from TSR Mobile

I thought this too! But surely that would be to easy for a 6 marker ??


Posted from TSR Mobile

Hi I don't suppose anyone knows the answer to 2 a i)

The answers aren't in the back of the book. And this one confused me. Is it to do with electric field strength being proportional to potential difference?

Thanks

Posted from TSR Mobile

So if I ever see the word orbit, that means from the centre of the earth?



Posted from TSR Mobile

Yup! You assume it's a uniform field between the two plates.

$[br]E = \dfrac{V}{d}[br]$

So increasing the potential difference means a greater E.

As the force on the charge (treating the shuttle ball as a point charge here) is proportional to E:

$[br]F = EQ[br]$

Then that means that a greater force is experienced, causing a larger change in momentum with each impact between the two walls, giving the effect that the shuttle oscillates faster


Posted from TSR Mobile

thank you!

No worries


Posted from TSR Mobile

i am had the same dream hahaha

I would interpret that, yeah. But I've emailed my teacher to make sure - I'll let you know what he says!


Posted from TSR Mobile

Just out of curiosity, what grade are you aiming for in physics (if you don't mind me asking)?

Just out of curiosity, what grade are you aiming for in physics (if you don't mind me asking)?

Thanks!! You've helped me so much!! If I see a similar question I'll see what it says


Posted from TSR Mobile

I never knew you could do it using this method.... I always used the rms equations