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The Physics PHYA2 thread! 5th June 2013

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well they could ask to explain an experiment where projectiles are involved, and ask what you could measure and how to work things out, maybe include some graphs. Im not very creative with questions, but im sure aqa could come up with a decent question about projectiles. But to be fair im just completely guessing.
can someone help me with q6bii on jun 12? i dont understand why it is 1/4 of the cycle isnt it half?
Calculate the time taken for the string at point Z to move from maximum displacementback to zero displacement.
wave.png35.0KB
Original post by rainerised
can someone help me with q6bii on jun 12? i dont understand why it is 1/4 of the cycle isnt it half?
Calculate the time taken for the string at point Z to move from maximum displacementback to zero displacement.


Well it is currently at maximum displacement, so in the next quarter cycle it moves down to equilibrium position.

From equilibrium position, the next quarter cycle take Z to minimum displacement. Notice that means in a half cycle it has moved from maximum displacement to minimum displacement.

Then it does the reverse (sort of) minimum displacement to equilibrium in a quarter cycles and then equilibrium back to its original maximum displacement in the next quarter cycles.... therefore it goes from maximum equilibrium all the way back to maximum equilibrium in 1 cycle.

Hoop this helps.
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Hullo
8
This question is pretty simple right? Just trace up/up the line match to the original.. however when I did this the mark scheme said otherwise. It said 1,5,6.. however calcium and strontium had some matching lines (2 and 3)?? :s-smilie:Screen shot 2013-05-28 at 10.04.37 AM.png
Any help please? :smile:
Could anyone quickly just go over the energy changes we need to know and how to apply the, to questions? Thanks :smile:


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image.jpgimage.jpg

bi stuck on exam question 1 can anyone help? thanks
Original post by pearson1995
Well it is currently at maximum displacement, so in the next quarter cycle it moves down to equilibrium position.

From equilibrium position, the next quarter cycle take Z to minimum displacement. Notice that means in a half cycle it has moved from maximum displacement to minimum displacement.

Then it does the reverse (sort of) minimum displacement to equilibrium in a quarter cycles and then equilibrium back to its original maximum displacement in the next quarter cycles.... therefore it goes from maximum equilibrium all the way back to maximum equilibrium in 1 cycle.

Hoop this helps.

thanks :smile:
Original post by BenChard
image.jpgimage.jpg

bi stuck on exam question 1 can anyone help? thanks


Well forces up equal forces down

You need to convert the mass into weight

You can work out the vertical component of the tension


Just remember you have two lots of tensions there.


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Qari
20
Original post by BenChard
image.jpgimage.jpg

bi stuck on exam question 1 can anyone help? thanks


Think It'll be 8*gravity=Tension*cos50
tension= (8*gravity)/cos50
Original post by Qari
Think It'll be 8*gravity=Tension*cos50
tension= (8*gravity)/cos50


One thing, you should have two tensions there, or have 4*gravity


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Could anyone quickly go over the conservation of energy? Please


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Would you ever get a question like a pulley with an inclined plane?


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http://filestore.aqa.org.uk/subjects/AQA-PHYA2-W-QP-JUN10.PDF

How the hell are you meant to find the kinetic energy on 4bi?


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Original post by Jimmy20002012
http://filestore.aqa.org.uk/subjects/AQA-PHYA2-W-QP-JUN10.PDF

How the hell are you meant to find the kinetic energy on 4bi?


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Wouldn't it just be the loss in gravitational potential energy? To he best of my knowledge, when you jump on a trampoline, you don't actually leave the trampoline until the surface unstretches - in this case, once the actual trampoline material thingy reaches that dotted line. My guess would be that when she's below that dotted line she has elastic energy, and the gravitational/kinetic energy only applies between A and B.

EDIT: Actually, having looked at the question, I think you just take the distance to be 3.2m and use the same mgh equation as you did in the previous part. The question seems to disregard elastic energy, strange.
(edited 10 years ago)
Original post by lebron_23
Wouldn't it just be the loss in gravitational potential energy? To he best of my knowledge, when you jump on a trampoline, you don't actually leave the trampoline until the surface unstretches - in this case, once the actual trampoline material thingy reaches that dotted line. My guess would be that when she's below that dotted line she has elastic energy, and the gravitational/kinetic energy only applies between A and B.

EDIT: Actually, having looked at the question, I think you just take the distance to be 3.2m and use the same mgh equation as you did in the previous part. The question seems to disregard elastic energy, strange.


I dort of get it, my teacher didn't teach us about conservation if energy, could you just quickly go through it so I fully understand?


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Original post by Jimmy20002012
I dort of get it, my teacher didn't teach us about conservation if energy, could you just quickly go through it so I fully understand?


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Hmm, I'm probably not the best person when it comes to explaining but I'll have a stab at it. Although I'm not sure whether you mean conservation of energy as a whole topic or just the concept of gravitational/kinetic/potential energy.. Which I think is the same think?

Anyway, kinetic energy, as you know, is given by 1/2(m)v^2. This, I believe, is derive from Newton's second law where you have (m*v)/t = f. So I guess you could say kinetic energy is the energy of an object as a result if its motion.

For gravitational potential, the energy is given by the force (which is mass*gravity) multiplied by the distance moved. This energy is also referred to ad the work done to move an object x amount of meters. So E_p = mgh, bearing in mind this only applies to objects being moved vertically (obviously, gravity is involved).

Lastly, you need to know that the gain in kinetic energy is equal to the loss in potential energy, and visa versa. This is assuming that air resistance is negligible, of course.

Therefore, we can say that 1/2(m)v^2 = mgh and that really is all there is to it.

That's was a terrible attempt at an explanation so I'm really sorry, but as I said, I absolutely suck at explaining things.
Original post by lebron_23
Hmm, I'm probably not the best person when it comes to explaining but I'll have a stab at it. Although I'm not sure whether you mean conservation of energy as a whole topic or just the concept of gravitational/kinetic/potential energy.. Which I think is the same think?

Anyway, kinetic energy, as you know, is given by 1/2(m)v^2. This, I believe, is derive from Newton's second law where you have (m*v)/t = f. So I guess you could say kinetic energy is the energy of an object as a result if its motion.

For gravitational potential, the energy is given by the force (which is mass*gravity) multiplied by the distance moved. This energy is also referred to ad the work done to move an object x amount of meters. So E_p = mgh, bearing in mind this only applies to objects being moved vertically (obviously, gravity is involved).

Lastly, you need to know that the gain in kinetic energy is equal to the loss in potential energy, and visa versa. This is assuming that air resistance is negligible, of course.

Therefore, we can say that 1/2(m)v^2 = mgh and that really is all there is to it.

That's was a terrible attempt at an explanation so I'm really sorry, but as I said, I absolutely suck at explaining things.


So with the other question posted above, how would you work it out?

Also what is elastic and chemical energy?


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Original post by Jimmy20002012
So with the other question posted above, how would you work it out?

Also what is elastic and chemical energy?


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Which other question? Sorry, I'm on my phone and it's a little hard to see some stuff.

And I don't think the syllabus has anything on elastic and chemical energy so I don't think you have to know about them. My bad for bringing it up, I don't even know the spec :colondollar:
Can someone please help on question 2 b) on page 98 of the AQA textbook? The answer is 6.2 but I got 0.8...

It really doesn't make sense to me.
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Jack93o
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Original post by PickwickianGeek
Can someone please help on question 2 b) on page 98 of the AQA textbook? The answer is 6.2 but I got 0.8...

It really doesn't make sense to me.


the pivot point is at the centre of mass of the ruler, which is the middle of the ruler - the 50cm mark

if you draw the diagram of the ruler with all the masses, you should have the 3N and 2N weights on one side of the pivot (the 0-50cm half of the metre rule), and 80N would be on its own in the other side of the pivot (the 50-100cm half of the metre rule). The 3N and 2N have the same direction of rotation (for example clock-wise), and the W weight would have the opposite direction of rotation to oppose them (i.e anti-clockwise)

you assume the whole thing is in equilibrium and the ruler is in balance, you take moments about the pivot point (the middle of the ruler):

3*0.45 + 2*0.25 = W*0.3

1.85 = W*0.3

1.85/0.3 = W

6.1666.... = W

rounded to 2sf would give W as 6.2N

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