The Student Room Group

AQA M2

What did everyone think? Any unofficial mark schemes floating around?


This was posted from The Student Room's iPhone/iPad App
Reply 1
pretty good except from the last question, if you can remember questions we can try
Reply 2
Easy till the end....and i messed up centres of mass because of a dumb mistake you?
Reply 3
I did this today, can't remember exactly what I got for the distance, think it was 13.3 or something? But i remember for the angle I think I got low 50's, maybe 53? You?
Reply 4
Original post by SamHLFC
I did this today, can't remember exactly what I got for the distance, think it was 13.3 or something? But i remember for the angle I think I got low 50's, maybe 53? You?


I got 47.6
doing tan(x)=(30-13.6)/15 I think
I thought the paper was pretty nice, until question 9 :biggrin: did anyone show l as 4acos(2x) / cos(x) ? (x being theta) i managed to get acos(2x) / cos(x) as the length of the rod inside the hemisphere, unsure where the other 3a came from...
Reply 6
What I got (mostly wrong probably):

Q1 (Energy):
a) KE = 9.68J (2 marks)
b) GPE = 7.84J (2)
c) New KE = 17.52J (2)
d) v = 14.8ms-1 (2)

Q2 (Kinematics):
a) Find acceleration (can't remember)
bi) Find F (can't remember) (2)
bii) Magnitude of F = 73.8N (2)
c) r = ((36/pi)sin(pi/3 t)+4)i + (79-3t^3)j (5)

Q3 (Power):
P = 188.9kW (5)

Q4 (Center of mass):
a) Lamina symmetrical (1)
b) 14.3cm (3)
c) Angle = 43.6 deg (4)

Q5 (Differential equation):
a) Show that v = (4-(2/9)t)^(3/2) (6)
b) t = 18s (1)

Q6 (Circular motion):
a) Angle = 54.8 deg
b) v = 3.33ms-1
c) t = 0.436s

Q7 (Circular motion):
a) v = 3.37ms-1
b) T = 57.8N

Q8 (Hooke's Law):
a) Show EPE formula using intgration - limits e and 0 (3)
bi) x = 0.2m
bii) EPE = 44.1J
biii) 1.59m

Q9 (Circular motion):
a) Reaction perpendicular to mg; therefore acts through O (1)
b) Forces diagram (2)
c) Show that l = 4acos2x/cosx (5)
(edited 11 years ago)
Reply 7
Original post by jarasta
I got 47.6
doing tan(x)=(30-13.6)/15 I think


Oh yeah, i've got the angles mixed up and done 15/16.4 :frown: Should have used z angles
Reply 8
Original post by stevey396
What I got (mostly wrong probably):

Q1 (Energy):
a) KE = 9.68J (2 marks)
b) GPE = 7.84J (2)
c) New KE =1.84 (2)
d) v = 4.79ms-1 (2)

Q2 (Kinematics):
v=12cos(tpi/3)i-9t^2j
a) a=-4pisin(tpi/3)i-18tj
bi) F=-16sin(tpi/3)i-72(t)j
bii) Magnitude of F = 216N
c) r = ((36/pi)sin(pi/3 t)+4)i + (79-3t^3)j (5)

Q3 (Power):
P = 188.9kW (5)

Q4 (Center of mass):
a) Lamina symmetrical (1)
b) 13.6cm (3)
c) Angle = 47.6(4)


Q5 (Differential equation):
a) Show that v = (4-(2/9)t)^(3/2) (6)
b) t = 18s (1)

Q6 (Circular motion):
a) Angle = 54.8 deg
b) v = 3.33ms-1
c) t = 1.5s

Q7 (Circular motion):
a) v = 3.71ms^-1
b) T = 61.1


Q8 (Hooke's Law):
a) Show EPE formula using intgration - limits e and 0 (3)
bi) x = 0.2m
bii) EPE = 44.1J
biii) 0.0167m i think

Q9 (Circular motion):
a)reaction acts perpendicular to the hemisphere at A thus it is normal (radial) to the hemisphere(circle) and it will go through the centre (1)
b) Forces diagram (2)
c) Show that l = 4acos2x/cosx (5)

got q1 wrong im pretty sure-edited to what it should be- looking at the rest of the answers im pretty sure they are wrong :/
ill edit it to what i think it should be. that should be correct
(edited 11 years ago)
Original post by stevey396
What I got (mostly wrong probably):

Q1 (Energy):
a) KE = 9.68J (2 marks)
b) GPE = 7.84J (2)
c) New KE = 17.52J (2)
d) v = 14.8ms-1 (2)



I got the same part a and b, but for part c i deducted the GPE from initial K.E to find the new K.E whereas you've added them, if the initial K.E is the only energy on the ball at the moment of impact with the throw then surely 9.68J is the TOTAL energy, therefore at the point where the ball hits the tree the total energy would not exceed 9.68J, resulting in 7.84 GPE + 1.84K.E ?
Reply 10
Original post by matt242o1552
I got the same part a and b, but for part c i deducted the GPE from initial K.E to find the new K.E whereas you've added them, if the initial K.E is the only energy on the ball at the moment of impact with the throw then surely 9.68J is the TOTAL energy, therefore at the point where the ball hits the tree the total energy would not exceed 9.68J, resulting in 7.84 GPE + 1.84K.E ?


you are right, i edited his answers look above your post
Reply 11
Original post by jarasta
Q8 (Hooke's Law):
a) Show EPE formula using intgration - limits e and 0 (3)
bi) x = 0.2m
bii) EPE = 44.1J
biii) 0.0167m i think


So glad someone else got this, only asked one person so far that got the same answer. No doubt I will lose a mark though, i only put 0.017m
Reply 12
Original post by SamHLFC
So glad someone else got this, only asked one person so far that got the same answer. No doubt I will lose a mark though, i only put 0.017m


Why would u lose a mark for 0.017 m...its correct
Reply 13
The first question definitely is KE - gpe. The ball is rising upwards to the height of the tree.

I think this went okay apart from the last bloody question. Hoping for a couple of method marks to bring me up to the 85% I so desperately want.
Reply 14
Original post by kitkat19
Why would u lose a mark for 0.017 m...its correct


Not done to 3.s.f
Reply 15
The last question was VERY hard, i managed to get it...JUST but my working was so messy I don't know if the examiner will be able to make head or tail of it.
My friend who has an offer from Oxford to study maths didn't manage to do it, if that makes anyone feel any better :tongue:
Reply 16
Original post by Matt III
The last question was VERY hard, i managed to get it...JUST but my working was so messy I don't know if the examiner will be able to make head or tail of it.
My friend who has an offer from Oxford to study maths didn't manage to do it, if that makes anyone feel any better :tongue:


id love to see the solution- i started it, then i was like. hhhmm this will be tough i may not do it (20mins left), ill check my paper maybe guarantee 70/75. Good idea as it turned out i had made too very costly errors which i corrected XD
Reply 17
Ok so, obviously you do moments around A.

To find the length from A to C (the two points touching the bowl). You make an isocelese triangle AOB (AO and BO are radius') and then split that in half to find length AC, comes out as AC = 2acos(theta).

Now you need to find the reaction force at C. So because you know that the reaction force at A works through the centre, you can resolve it into horizontal and vertical parts.
So R(vertical) = RSin(2Theta)
R(horizontal)=RCos(2theta)
Then do the same with reaction force at C:
S(vertical) = Scos(theta)S
(horizontal) = SSin(theta).
You know that the vertical parts add up to the weight:
RSin(2theta)+SCos(theta) = mg
and you know that the two horizontal forces must be equal:
SSin(theta) = RCos(2theta)
Rearrange for R: R=(SSin(theta)/cos(2theta)).
Sub that into the vertical equation and you get:
S*Sin(theta)*Sin(2theta)/cos(2theta) + Scos(theta) = mg.
Therefore S=(mg)/(Sin(theta)Tan(2theta)+cos(theta))
Times top and bottom by cos(2theta):
S=(mgcos(2theta))/(sin(theta)*sin(2theta)+cos(theta)*cos(2theta))

Then do moments:
Weight/half-length = reaction force at C*AC
0.5L*cos(theta)*mg = 2aCos(theta)*(mgcos(2theta))/Cos(theta)
(note that Sin(theta)*Sin(2theta)+cos(theta)*cos(2theta) = cos(theta))
mg cancels out. Times by 2. Cos(thetas) cancel out.
L=(4acos(2theta))/cos(theta)).

Done! I think this is probably unreadable, but hopefully you can make some sense of it.
Reply 18
The above is exactly how my Further Maths teacher solved it. Take moments about A, equate horizontal forces, trig and rearranging
Reply 19
Original post by Matt III
Ok so, obviously you do moments around A.

To find the length from A to C (the two points touching the bowl). You make an isocelese triangle AOB (AO and BO are radius') and then split that in half to find length AC, comes out as AC = 2acos(theta).

Now you need to find the reaction force at C. So because you know that the reaction force at A works through the centre, you can resolve it into horizontal and vertical parts.
So R(vertical) = RSin(2Theta)
R(horizontal)=RCos(2theta)
Then do the same with reaction force at C:
S(vertical) = Scos(theta)S
(horizontal) = SSin(theta).
You know that the vertical parts add up to the weight:
RSin(2theta)+SCos(theta) = mg
and you know that the two horizontal forces must be equal:
SSin(theta) = RCos(2theta)
Rearrange for R: R=(SSin(theta)/cos(2theta)).
Sub that into the vertical equation and you get:
S*Sin(theta)*Sin(2theta)/cos(2theta) + Scos(theta) = mg.
Therefore S=(mg)/(Sin(theta)Tan(2theta)+cos(theta))
Times top and bottom by cos(2theta):
S=(mgcos(2theta))/(sin(theta)*sin(2theta)+cos(theta)*cos(2theta))

Then do moments:
Weight/half-length = reaction force at C*AC
0.5L*cos(theta)*mg = 2aCos(theta)*(mgcos(2theta))/Cos(theta)
(note that Sin(theta)*Sin(2theta)+cos(theta)*cos(2theta) = cos(theta))
mg cancels out. Times by 2. Cos(thetas) cancel out.
L=(4acos(2theta))/cos(theta)).

Done! I think this is probably unreadable, but hopefully you can make some sense of it.


Aaah I see its not that bad a question it seems

Posted from TSR Mobile

Quick Reply