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# AQA AS Physics Unit 2 (PHYA2) June 9th 2016 Resit paper

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1. I am so screwed for this... I know the stuff but I'm hopeless at applying it...
2. (Original post by Stallzy)
looks good to me!
Thanks

3. June 2012 Question 6aii:

Supposedly X and Y are 180 degrees out of phase and X and Z are in phase, but X isn't at the peak amplitude (90 degrees) like Z is so I can't see why they'd be exactly in phase.

Can anyone clear this up?
4. what is everyone predicting for the 6 marker.

last year it was on stationary waves

i think that this year it will be on the youngs modulus experiment
anyone else?
5. (Original post by micycle)

June 2012 Question 6aii:

Supposedly X and Y are 180 degrees out of phase and X and Z are in phase, but X isn't at the peak amplitude (90 degrees) like Z is so I can't see why they'd be exactly in phase.

Can anyone clear this up?
If an even number of nodes separates two points, then they are in phase with one another. If an odd number of nodes separates them, they're 180o out of phase (or pi radians out of phase )
6. (Original post by Exams987)
what is everyone predicting for the 6 marker.

last year it was on stationary waves

i think that this year it will be on the youngs modulus experiment
anyone else?
I think it'll be a diffraction grating experiment comparing two different colours ...
7. (Original post by micycle)

June 2012 Question 6aii:

Supposedly X and Y are 180 degrees out of phase and X and Z are in phase, but X isn't at the peak amplitude (90 degrees) like Z is so I can't see why they'd be exactly in phase.

Can anyone clear this up?
Its because its a standing wave.....amplitudes vary at each point so while x is lower than peak it is at its peak amplitude ....while y is at its lowest peak amp.....
8. (Original post by Gapyearstudent54)
I think it'll be a diffraction grating experiment comparing two different colours ...
do you have a possible model answer as i tend to really struggle with the refraction grating
9. In Jan 12
Qs 2bii says state two causes of wasted energy for a person cycling !
Can someone tell me which part of the bike would cause energy loss by friction ?
Thanks
10. (Original post by kother2015)
In Jan 12
Qs 2bii says state two causes of wasted energy for a person cycling !
Can someone tell me which part of the bike would cause energy loss by friction ?
Thanks
The tires against the ground- the coefficient of friction between tires and ground in the opposite direction to the motion and also the drag from air resistance again in the exact opposite direction to the resultant motion.
11. Just did Mechanics paper, hyped for this yeeeee
12. (Original post by micycle)

June 2012 Question 6aii:

Supposedly X and Y are 180 degrees out of phase and X and Z are in phase, but X isn't at the peak amplitude (90 degrees) like Z is so I can't see why they'd be exactly in phase.

Can anyone clear this up?
Think the diagram is poorly printed,it should have been at max amplitude
13. (Original post by TSRPAV)
The tires against the ground- the coefficient of friction between tires and ground in the opposite direction to the motion and also the drag from air resistance again in the exact opposite direction to the resultant motion.
I don't think you get mark for saying friction between tires and ground , as it is keeping bike on the ground, it is useful .

Thanks
14. (Original post by ryandaniels2015)
Here's the question
The graph is a velocity time graph man so the gradient of the graph is acceleration .At top of ramp once the ball is released it accelerated due to gravity,however as it is in contact with a ramp that curves (and the component of gravity is always directly down to ground) its component changes as it is on each angle of the ramp so basically the acceleration component constantly changes( from its maxium g=9.81 m/s to zero when the ball is horizontal) and as there is a rate of change of velocity(this is acceleration) there is a rate of change(steepness of graph) that curves off to a straight line as at the end the ramp is horizontal so the component of gravity is directly vertical(this means it does not effect horizontal motion) and as there are no other forces (friction-less ramp and no "pushing"/thrust on the ball) it is in equilibrium so there is no resultant force so no acceleration so constant velocity be it zero m/s or xm/s

15. (Original post by kother2015)
I don't think you get mark for saying friction between tires and ground , as it is keeping bike on the ground, it is useful .

Thanks
Nah you will, the cyclist is doing work against friction to maintain velocity/accelerate (depends what Q is asking) so has to waste energy to do work to maintain velocity/acceleration
16. (Original post by TSRPAV)
Nah you will, the cyclist is doing work against friction to maintain velocity/accelerate (depends what Q is asking) so has to waste energy to do work to maintain velocity/acceleration
it even says it in mark scheme

"mention of friction and appropriate location given " --> that 1 mark

the other was air resistance
17. When do you know when to use s=vt and s=0.5vt because I was always under the assumption s=vt is for when acceleration is constant such as when a=g and so is 9.81 but one of the past papers said I had to use s=vt for a horizontal calculation
18. Does anyone know if there will be any sort of model answers for this or anything??
19. (Original post by GdotMdot)
When do you know when to use s=vt and s=0.5vt because I was always under the assumption s=vt is for when acceleration is constant such as when a=g and so is 9.81 but one of the past papers said I had to use s=vt for a horizontal calculation
isn't the formula S=0.5(v+u)t
20. (Original post by TSRPAV)
The graph is a velocity time graph man so the gradient of the graph is acceleration .At top of ramp once the ball is released it accelerated due to gravity,however as it is in contact with a ramp that curves (and the component of gravity is always directly down to ground) its component changes as it is on each angle of the ramp so basically the acceleration component constantly changes( from its maxium g=9.81 m/s to zero when the ball is horizontal) and as there is a rate of change of velocity(this is acceleration) there is a rate of change(steepness of graph) that curves off to a straight line as at the end the ramp is horizontal so the component of gravity is directly vertical(this means it does not effect horizontal motion) and as there are no other forces (friction-less ramp and no "pushing"/thrust on the ball) it is in equilibrium so there is no resultant force so no acceleration so constant velocity be it zero m/s or xm/s

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