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

Original post by CD223

Can some people help me with Q8, Q12 and Q24 on June 2013?

I got answers of C, B, B which is wrong.

Posted from TSR Mobile

Question 8 just uses one ratio of 1:2 (for m and f) so I would use the third type of informal approach in my last answer which is proportions. The other approaches would work but are unnecessary in this 1:something format; I'll show both anyway though. It's hard to distinguish between when it is best to use them but it comes down to how intuitive the question is.
. SHM Question.png

LHS:
First two lines are the given ratios rearranged.
Second two lines show the velocity ratio is equal to the frequency ratio (constant cancels since ratios used) based on the SHM maximum velocity equation.
Last lines substitute last rearranged equations into the kinetic energy equation.

The last stage could have just substituted the direct m2/m1 and v2/v1 ratios E=mv^2 (not constant since ratios) to give 8 but wanted to show it in full.

RHS is intuitive.

(edited 8 years ago)

Reply 1662

Original post by CD223

Can some people help me with Q8, Q12 and Q24 on June 2013?

I got answers of C, B, B which is wrong.

Posted from TSR Mobile

This uses the gravitational field equation with two ratios.

Gravitation Fields Question 2.png

First two equations are the given ratios.
Second two equations rearrange the gravitational field equation for r.
Last two equations substitute the given ratios into the rearranged equation, leaving out the constant since ratios are substituted in (cancel if you like).

(edited 8 years ago)

Reply 1663

OP

Original post by JayCS

Question 8 just uses one ratio of 1:2 (for m and f) so I would use the third type of informal approach in my last answer which is proportions. The other approaches would work but are unnecessary in this 1:something format; I'll show both anyway though. It's hard to distinguish between when it is best to use them but it comes down to how intuitive the question is.
. SHM Question.png

LHS:
First two lines are the given ratios rearranged.
Second two lines show the velocity ratio is equal to the frequency ratio (constant cancels since ratios used) based on the SHM maximum velocity equation.
Last lines substitute last rearranged equations into the kinetic energy equation.

The last stage could have just substituted the direct m2/m1 and v2/v1 ratios E=mv^2 (not constant since ratios) to give 8 but wanted to show it in full.

RHS is intuitive.

Thank you

Reply 1664

OP

Original post by JayCS

This uses the gravitational field equation with two ratios.

Gravitation Fields Question 2.png

First two equations are the given ratios.
Second two equations rearrange the gravitational field equation for r.
Last two equations substitute the given ratios into the rearranged equation, leaving out the constant since ratios are substituted in (cancel if you like).

Again - thanks - do you know how to do Q24?

Posted from TSR Mobile

Reply 1665

Original post by CD223

Again - thanks - do you know how to do Q24?

Posted from TSR Mobile

No problem :smile:

.

Do you mean a different question to Q24? You said you got B for it and that's correct.

Reply 1666

OP

Original post by JayCS

No problem :smile:

.

Do you mean a different question to Q24? You said you got B for it and that's correct.

Lol I'm so stupid... I meant I got C and the answer was B :frown:

ImageUploadedByStudent Room1432050230.509931.jpg

Reply 1667

OP

Does the equation:

[br](n+1) \times {T_{short}} = (n) \times {T_{long}}[br]

Always apply for two pendulums where n= the number of oscillations before they are next in phase?

If so, could I have an explanation of why it works please?

Brain not functioning today!

Posted from TSR Mobile

Reply 1668

Original post by CD223

Lol I'm so stupid... I meant I got C and the answer was B :frown:

I got B because I knew the formulae was Blv and remembered the derivation of it used the same direction (where the rod is moving in the length's direction).

Reply 1669

Lau14

11

Original post by CD223

Lol I'm so stupid... I meant I got C and the answer was B :frown:

For a start it must be one of the rows with horizontal and perpendicular in, because otherwise it won't cut any field lines and no emf would be induced.

Reply 1670

OP

Original post by Lau14

For a start it must be one of the rows with horizontal and perpendicular in, because otherwise it won't cut any field lines and no emf would be induced.

Original post by JayCS

I got B because I knew the formulae was Blv and remembered the derivation of it used the same direction (where the rod is moving in the length's direction).

That's true. Thanks guys!

I think I put vertical because I tried to use the right hand rule or something. Stupid mistake.

Posted from TSR Mobile

Reply 1671

Lau14

11

Original post by CD223

That's true. Thanks guys!

I think I put vertical because I tried to use the right hand rule or something. Stupid mistake.

Posted from TSR Mobile

Some form of Fleming's rule would probably have been my first response too to be honest!

Reply 1672

OP

Original post by Lau14

Some form of Fleming's rule would probably have been my first response too to be honest!

I used the right hand rule which seemed to not work out?

Unless I'm mistaken.

ImageUploadedByStudent Room1432054350.141178.jpg

Reply 1673

sam_97

6

Original post by CD223

I used the right hand rule which seemed to not work out?

Unless I'm mistaken.

ImageUploadedByStudent Room1432054350.141178.jpg

Hi CD223,

The problem with this question is that it can be quite difficult to visualise at first.

You don't need to use any of Fleming's hand rules actually, if you imagine the wire moving upwards (bare in mind the diagram is a side view) it doesn't cut through any lines of flux therefore no emf is induced. The wire can only cut through the flux lines if it moves perpendicular to them.

Hope that helps :smile:

Reply 1674

OP