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

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Reply 80
https://5505b400f275fd9f29b9b4679ca512236789cf2e.googledrive.com/host/0B1ZiqBksUHNYZEhDNnU3TUxGaG8/June%202012%20QP%20-%20Unit%202%20AQA%20Physics.pdf
For question 3b(i) you need to use v^2 = u^2 + 2as. But I'm a bit confused as to why you cant use s=vt, then use the t value in a=v-u/t. Can anybody clarify this for me, thanks :smile:
Original post by boyyo
https://5505b400f275fd9f29b9b4679ca512236789cf2e.googledrive.com/host/0B1ZiqBksUHNYZEhDNnU3TUxGaG8/June%202012%20QP%20-%20Unit%202%20AQA%20Physics.pdf
For question 3b(i) you need to use v^2 = u^2 + 2as. But I'm a bit confused as to why you cant use s=vt, then use the t value in a=v-u/t. Can anybody clarify this for me, thanks :smile:


Velocity × time only works if velocity is contant....v in this case is the final velocity....so it doesnt work

10ms for 10 seconds you go 100m....but the sprinter only arrives at 9.3ms at the end....you would need the average velocty times time to get s
Reply 82
Original post by philo-jitsu
Velocity × time only works if velocity is contant....v in this case is the final velocity....so it doesnt work

10ms for 10 seconds you go 100m....but the sprinter only arrives at 9.3ms at the end....you would need the average velocty times time to get s


oh right ok thanks. I asked this earlier but no one replied; you see the suvat equations, they are only used when acceleration is constant right? Here it isnt, so how come we can use a suvat equation?
I think you can't really revise much except for uncertainties and practicals if you know the physics, not sure what to do honestly :/ Hopefully I can get about 20+ on the multiple choices
Original post by philo-jitsu
It was tricky...the diffraction spectra was one ive never seen before...but after looking at the answer it actually makes perfect sense...the trampoline section was easy to mess up on...but the grade boundaries did match the difficulty.


Could you do the trampoline question? If you could, would you explain it to me please? I don't understand the mark scheme and i got 4/14 :/
Original post by metrize
I think you can't really revise much except for uncertainties and practicals if you know the physics, not sure what to do honestly :/ Hopefully I can get about 20+ on the multiple choices


Theres no multiple choice on this paper, do you want the new spec or unit 4??
Original post by McDerdactyl
Theres no multiple choice on this paper, do you want the new spec or unit 4??


Oh whoops I'm doing new spec
Original post by metrize
Oh whoops I'm doing new spec


thought so haha, at least you didn't have to do an EMPA :') very jealous
Original post by boyyo
oh right ok thanks. I asked this earlier but no one replied; you see the suvat equations, they are only used when acceleration is constant right? Here it isnt, so how come we can use a suvat equation?


It says average acceleration, which is essentially saying if the acceleration was constant it would be the average acceleration...the key word is average, otherwise your right you couldnt use the eqution
Original post by McDerdactyl
Could you do the trampoline question? If you could, would you explain it to me please? I don't understand the mark scheme and i got 4/14 :/


There are alot of parts, which parts did you drop marks on...

The acceleration part, you just take the velocity from the previous question and input into (V-U)/T =A...you use that velocity as in oder to gain the required height you must leave with the same velocity as the gymnist hit on the way down....because if you throw a ball upwards at 2ms, when it returns on the way down it will be going 2ms again.....so because we worked out the velocity when she hit the trampoline on the way down when she first reached the height required, we can use that velocity for this question.... which other parts did you struggle with?
Original post by metrize
I think you can't really revise much except for uncertainties and practicals if you know the physics, not sure what to do honestly :/ Hopefully I can get about 20+ on the multiple choices


Damn, I hate MC....always run out of time in unit 4....I always second guess myself
Reply 91
Original post by philo-jitsu
It says average acceleration, which is essentially saying if the acceleration was constant it would be the average acceleration...the key word is average, otherwise your right you couldnt use the eqution


Thank you for clearing that up for me:smile:
Original post by philo-jitsu
There are alot of parts, which parts did you drop marks on...

The acceleration part, you just take the velocity from the previous question and input into (V-U)/T =A...you use that velocity as in oder to gain the required height you must leave with the same velocity as the gymnist hit on the way down....because if you throw a ball upwards at 2ms, when it returns on the way down it will be going 2ms again.....so because we worked out the velocity when she hit the trampoline on the way down when she first reached the height required, we can use that velocity for this question.... which other parts did you struggle with?


I wasnt sure how to get the kinetic energy because I didn't have velocity and even with Ep=Ek i end up needing v. I think if I'd have gotten that I might've been more confident with the rest of the questions but because i stressed over that, i lost confidence and made stuff up :/ I do M1 for maths so I'm trying to use this knowledge to help but physics mechanics and maths mechanics are so different from each other, its really stressful...
[QUOTE=McDerdactyl;65513473]I wasnt sure how to get the kinetic energy because I didn't have velocity and even with Ep=Ek i end up needing v. I think if I'd have gotten that I might've been more confident with the rest of the questions but because i stressed over that, i lost confidence and made stuff up :/ I do M1 for maths so I'm trying to use this knowledge to help but physics mechanics and maths mechanics are so different from each other, its really stressful...

Hi yeah I'm doing m1 as well.....hate it lol

I haven't checked but can't you use mgh=grav potential...but use 1 meter as the height as that's height increase.
Can anyone please help, why does the graph look like what's shown in the mark scheme in this question june 12 q2b?
(edited 7 years ago)
Please help!
Reply 97
Original post by ryandaniels2015
Can anyone please help, why does the graph look like what's shown in the mark scheme in this question june 12 q2b?


Because from point B to A the ball is accelerating, since its losing EP and gaining KE. So the first part of the graph would be an increase in velocity, so a steepish line.
I think the questions says air resistance is negligable, so at B and onwards there is no resultant force therefore it travels at constant spped which would be a horizontal line.
Since the speed decreases steadily, it would be a curve.

If I made a mistake in my explanation, someone please correct me.
Reply 98
I have this question my teacher made up, but I didnt get to see the answers. Could someone double check these for me:

1.A car of mass 70,000kg travelling horizontally, experiences a total horizontal decelerating force of 1.9 x 10^5 N.
What is the deceleration of the car? ANS: F=ma, 2.71 m/s^2 ?

2.The car travels down a straight road at a speed of 68m/s. Assuming that the decelrating force acting on car is constant, find the distance the car travels along the road until it comes to rest. ANS: a=v-u/t, t=25, s=0.5(u+v)t, s=853m
Reply 99
Original post by boyyo
I have this question my teacher made up, but I didnt get to see the answers. Could someone double check these for me:

1.A car of mass 70,000kg travelling horizontally, experiences a total horizontal decelerating force of 1.9 x 10^5 N.
What is the deceleration of the car? ANS: F=ma, 2.71 m/s^2 ?

2.The car travels down a straight road at a speed of 68m/s. Assuming that the decelrating force acting on car is constant, find the distance the car travels along the road until it comes to rest. ANS: a=v-u/t, t=25, s=0.5(u+v)t, s=853m


looks good to me!

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