# OCR M3 (non MEI) Wednesday 8th JuneWatch

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Thread starter 3 years ago
#1
Couldn't find a thread for this exam and as its approaching I thought I would start a thread for us all to share resources, ask questions etc.
1
3 years ago
#2
Hey, not sure how to feel about m3, I feel like I understand all of the theory and a large amount of the applications, some questions do seem to crop up and catch me off guard however, how do you feel? In order of decreasing difficulty I'd rate the topics:

1) SHM
2) Circular motion
3) Rigid bodies
4) Impulse and momentum
5) Hookes law
6) Variable forces
1
3 years ago
#3
A few little things definitely worth remembering across the topics:

SHM) if you set up your SHM scenario but get an acceleration proportional to displacement or theta rather than -displacement or -theta, then you are simply taking your direction in the opposite of the actual, this can be easily fixed.

Circular motion) conditions for there to be a circular motion,
String: T>0N (otherwise we are dealing with a projectile and M2 projectile motion can be used)
Cylinder: R>0N (same as above)
Rigid rod: v>0 at top (the rod can't become slack)
Outside of sphere: R>0N (projectile)

Variable forces) remember the integral of f'(x)/f(x) = ln(f(x)) and you can save some time, recall all of your C4 integration and separation of variables.
Velocity = dx/dt
Acceleration = dv/dt = vdv/dx (a result from product rule)

DON'T forget your constants

Hookes law) recall that T= (modulus of elasticity)(extension)/(natural length), this can also be written as F=kx where k is stiffness however this is rare (I imagine) and just thing of k as (lambda)/(L)

Don't forget EPE= (lambda)(extension)^2 / 2(natural length)
Or EPE= 1/2 kx^2

Rigid bodies) remember if the two rods are identical, and the system is symmetrical in the horizontal plane (e.g two identical rods jointed at top and with same angle to floor/ horizontal) then you only have to worry about horizontal forces on pin joint. However in most cases (unless if they felt particularly nice) it is not a symmetrical system.
Remember it is best to find an expression for cos(theta) and sin(theta) early on to help with calculations.

Impulse and momentum) I=mv-mu

If you have a ball hitting a horizontal smooth surface then the parallel component of velocity remains constant whereas vertical component is dictated by coefficient of restitution (remember to get the directions right). And for two balls colliding then velocity perpendicular to line of centres is the same, but parallel components found using coefficient of restitution and momentum.

Don't forget impulse triangles, they can be fiddly to set up and you must therefor eve careful with the angles used.

Good luck
1
Thread starter 3 years ago
#4
(Original post by tobibo)
A few little things definitely worth remembering across the topics:

SHM) if you set up your SHM scenario but get an acceleration proportional to displacement or theta rather than -displacement or -theta, then you are simply taking your direction in the opposite of the actual, this can be easily fixed.

Circular motion) conditions for there to be a circular motion,
String: T>0N (otherwise we are dealing with a projectile and M2 projectile motion can be used)
Cylinder: R>0N (same as above)
Rigid rod: v>0 at top (the rod can't become slack)
Outside of sphere: R>0N (projectile)

Variable forces) remember the integral of f'(x)/f(x) = ln(f(x)) and you can save some time, recall all of your C4 integration and separation of variables.
Velocity = dx/dt
Acceleration = dv/dt = vdv/dx (a result from product rule)

DON'T forget your constants

Hookes law) recall that T= (modulus of elasticity)(extension)/(natural length), this can also be written as F=kx where k is stiffness however this is rare (I imagine) and just thing of k as (lambda)/(L)

Don't forget EPE= (lambda)(extension)^2 / 2(natural length)
Or EPE= 1/2 kx^2

Rigid bodies) remember if the two rods are identical, and the system is symmetrical in the horizontal plane (e.g two identical rods jointed at top and with same angle to floor/ horizontal) then you only have to worry about horizontal forces on pin joint. However in most cases (unless if they felt particularly nice) it is not a symmetrical system.
Remember it is best to find an expression for cos(theta) and sin(theta) early on to help with calculations.

Impulse and momentum) I=mv-mu

If you have a ball hitting a horizontal smooth surface then the parallel component of velocity remains constant whereas vertical component is dictated by coefficient of restitution (remember to get the directions right). And for two balls colliding then velocity perpendicular to line of centres is the same, but parallel components found using coefficient of restitution and momentum.

Don't forget impulse triangles, they can be fiddly to set up and you must therefor eve careful with the angles used.

Good luck
That's great, thanks!
0
3 years ago
#5
Another thing:

Don't forget the Core basics, I recently did a circular motion question where you get R in terms of theta and sin(theta)

e.g R=a(theta) + bsin(theta)

It then asked for the angle that gives the biggest value of R......
I was stumped for quite a few minutes but remembered that if i do dR/d(theta) and set it =0 I will get the max point on the R vs theta graph and therefore value of theta for biggest R!
0
Thread starter 3 years ago
#6
(Original post by tobibo)
Another thing:

Don't forget the Core basics, I recently did a circular motion question where you get R in terms of theta and sin(theta)

e.g R=a(theta) + bsin(theta)

It then asked for the angle that gives the biggest value of R......
I was stumped for quite a few minutes but remembered that if i do dR/d(theta) and set it =0 I will get the max point on the R vs theta graph and therefore value of theta for biggest R!
I remember that paper, I think I did it with an iteration.
0
3 years ago
#7
hyped af, how many people are actually taking this?
0
3 years ago
#8
(Original post by drandy76)
hyped af, how many people are actually taking this?
not too many for sure.
1
3 years ago
#9
there's about 15 people in my class taking this exam.. I'm hoping it isn't going to be as bad as the M2 paper was :/
1
3 years ago
#10
I'm not neccesarily hyped but I feel that I know all of the theory, I'm just worried about curveball questions popping up, there are 4 including me in my school taking it
0
3 years ago
#11
Does anyone know which was the hardest paper so far?
0
3 years ago
#12
Hoping it's a hard paper this year, the last 2 years have been ridiculously easy and the grade boundaries outrageously high because of it, 69/72 for 90% last year! M2 was better this year so hopefully they'll do the same with M3/M4
1
3 years ago
#13
Anyone fancy sharing last minute tips?
At Vmax a=0 and such
0
3 years ago
#14
Anyone have a mark scheme for the 2015 paper?

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0
Thread starter 3 years ago
#15
I think that the most recent papers have all been fair so hopefully they'll also be kind this year, I feel they need to after the M2 paper that I barely passed
0
3 years ago
#16
How did everyone find that?
0
3 years ago
#17
(Original post by Miracle-)
How did everyone find that?
More of a pure mathematician anyway

Posted from TSR Mobile
0
3 years ago
#18
(Original post by drandy76)
More of a pure mathematician anyway

Posted from TSR Mobile
Lol,
what sort of answers did you get?
0
3 years ago
#19
(Original post by Miracle-)
Lol,
what sort of answers did you get?
Dunno the exam was all a blur to me

Posted from TSR Mobile
0
3 years ago
#20
Good solid paper I thought, I think many people would've struggled with the R=73/50W and the last question. Didn't get the last one fully, only got the time of Q falling and the amplitude of their motion, but that was it before running out of time.
0
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