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OCR Physics A G481 Mechanics [Thursday 17th May 2012]

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Reply 40
anyone fancy doing a crash course in rearranging? for all the suvat/projectile motion questions etc? i know its pretty basic stuff but it always catches me out, and if you cant rearrange you're pretty much stuck...would be really appreciated!
rearranging basically what you do too one side the opposite applies..

v=u+at
to rearrange for a
v=u+at (to get u to the other side, since its a PLUS, MINUS it)
v-u=at (since these two are MULTIPLY, DIVIDE one)
(v-u)/t=a :smile:

now for a more complicated one, eg

s=ut+1/2at^2
lets say we rearrange for a again BUT
s=10
u=0
A=9.81
T=?
it immediately gets rid of ut, as 0*anything = 0 which is very common in both projectiles and kinematics so:
s=1/2at^2 (do the opposite of halve, times by 2)
2s=at^2 (divide by a)
2s/a=t^2
then sub in
2*10 = 20
20/9.81 = 2.04 so t^2 = root 2.04 = 1.43

ermm dunno if that would of helped or not but thats the best i can do to help, i used to struggle with rearrangments, just remember, to get one thing to the other side, do the opposite!
Reply 42
Original post by genetic paradox
rearranging basically what you do too one side the opposite applies..

v=u+at
to rearrange for a
v=u+at (to get u to the other side, since its a PLUS, MINUS it)
v-u=at (since these two are MULTIPLY, DIVIDE one)
(v-u)/t=a :smile:

now for a more complicated one, eg

s=ut+1/2at^2
lets say we rearrange for a again BUT
s=10
u=0
A=9.81
T=?
it immediately gets rid of ut, as 0*anything = 0 which is very common in both projectiles and kinematics so:
s=1/2at^2 (do the opposite of halve, times by 2)
2s=at^2 (divide by a)
2s/a=t^2
then sub in
2*10 = 20
20/9.81 = 2.04 so t^2 = root 2.04 = 1.43

ermm dunno if that would of helped or not but thats the best i can do to help, i used to struggle with rearrangments, just remember, to get one thing to the other side, do the opposite!


thank you so much! you explained it in a way more straightforward way than any text book/my useless teachers :smile:
Reply 43
would anyone mind a quick run through on projectiles,

I know horizontal velocity is constant, what else would we need?
Reply 44
any help on this question? cheers to the two that helped me out earlier.

1/2mv squared is equal to work done. Rearrange for force.
Reply 46
Original post by Sarabande
1/2mv squared is equal to work done. Rearrange for force.


is work done always = to KE then?
Kinetic energy is a value of work done. Work done is energy transfer, so in essence, yes it is always equal. This question makes the assumption that the KE is only converted into energy to provide a reaction force.

Another method for that question would be to use suvat to find the deceleration then use force is equal to mass multiplied by acceleration.
Reply 48
Original post by Sarabande
Kinetic energy is a value of work done. Work done is energy transfer, so in essence, yes it is always equal. This question makes the assumption that the KE is only converted into energy to provide a reaction force.

Another method for that question would be to use suvat to find the deceleration then use force is equal to mass multiplied by acceleration.


thanks! super helpful, again.
Reply 49
Hey guys!
I was wondering, what type of crumple zone questions could they ask?
for seat belts and air bags, the mark scheme is similar and so i memorized the answer pretty well on that (the basic, time for stopping is more...) is this the same answer for crumple zones too? or do i need to learn something else on top?
Thanks xx
Original post by sinkersub
would anyone mind a quick run through on projectiles,

I know horizontal velocity is constant, what else would we need?


Like you said, horizontal velocity is constant, and the vertical velocity decreases because of gravity: the distance travelled is governed by the time taken to rise and fall.

You can work out the time if you know the initial vertical component of velocity using suvat:
you know u, v = 0 and a = -9.81
you want to know t, so use v = u + at
v = 0 so u + at = 0
so at = -u
so t = -u/-9.81 = u/9.81

This gives the time the projectile takes to reach maximum height; multiply this by 2 to give the total flight time.

Multiply flight time by horizontal velocity to get the total distance travelled.

The only other complication you may encounter is if they give the velocity as a single vector V acting at an angle theta, which case you can work out horizontal and vertical components using trigonometry:

Horizontal velocity = Vsin(theta)
Vertical velocity = Vcos(theta)

That's pretty much all there is to know about projectiles, and I'm pretty sure that's really the hardest maths they'll ever make you do in this module :smile:
Original post by sarah-xx
Hey guys!
I was wondering, what type of crumple zone questions could they ask?
for seat belts and air bags, the mark scheme is similar and so i memorized the answer pretty well on that (the basic, time for stopping is more...) is this the same answer for crumple zones too? or do i need to learn something else on top?
Thanks xx


Crumple zones is exactly the same: increased time to stop = reduced deceleration; as F = ma this means reduced force. Also it plastically deforms, absorbing some of the energy of the collison. Occasionally I think you may be able to get a mark for saying that it prevents foreign objects from entering the car, but read the question carefully; sometimes they ask about reducing injury in which case normally anything goes, but a lot they ask for reducing forces upon the driver :smile:
Reply 52
Original post by DavidH20
Crumple zones is exactly the same: increased time to stop = reduced deceleration; as F = ma this means reduced force. Also it plastically deforms, absorbing some of the energy of the collison. Occasionally I think you may be able to get a mark for saying that it prevents foreign objects from entering the car, but read the question carefully; sometimes they ask about reducing injury in which case normally anything goes, but a lot they ask for reducing forces upon the driver :smile:


Oh thanks! Helped a lot. xx
Dear physics exam currently sat in schools waiting to slay us all, please contain lots and lots of projectile marks, lots of young modulus marks, and then 3 marks on work is plenty! Fab, thanks, bye.
Good luck everyone! :biggrin:
Good luck all.
Reply 56
how did you guys find the exam , and what are your predictions for the grade boundaries
How was it then. I loved it.
Original post by ss2012
how did you guys find the exam , and what are your predictions for the grade boundaries


No hard maths questions. Some of the wordy questions were a little bit picky and I guess I will loose a couple there. Apart from that it was really nice
Reply 59
i hope the calculation is alright though my ans is different to quite a lot of my fds -o-

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