AQA Maths Mechanics 3 M3 June 18th 2013

Original post by Mydwinter

Hmm right, I shall hope to update this as I remember more questions, but heres what I got
Q1 : T= 2sec
Impulse = 26Ns
Q2: think this was basketball, if it is then the first two parts were show thats and the last direction i got as cos theta = 6.403/3.4619 so direction of velocity =61.something degrees below horizontal (same as -61. whatever with horizontal or bearing of 151.x degrees)

Q3 Dimensional analysis (may have been Q5) - just did DA with each separate component of equation, all of which gave unit of power => ML^2T^-3 therefore dimensionally consistent

Q4
I think this was the only part i couldnt do which was prove vb = 9/2 or something.
the say 0<e
next part find e. e=5/8

Q5

Q6

Q7 this one was annoying.
a
200 sin theta = -240 sin 40 so theta = 50.47 and bearing = 120 - theta. Therefore bearing = 069.5 deg to 1d.p?
Time taken = 3.86min

b
the last part you had to use geometrical something or other, i needed extra paper for this and got theta = 51.1 deg , but the bearing was theta - 20 because h was 40 deg from the distance line. so Bearing = 031.3

I agree with all of the answers. Ive updated some bits

(edited 10 years ago)

Reply 25

crap I made an error on the first question, I miscalculated the impulse to be 12 instead of 8.
How many marks was the first question and how many would I have lost? Of course everything I did that followed was incorrect.

Reply 26

Right I updated some stuff on my earlier post, does anyone remember Q5 ?

Reply 27

Original post by Mydwinter

Right I updated some stuff on my earlier post, does anyone remember Q5 ?

Do you remember how many marks the 1st question was worth, i messed the impulse up so wrong time :rolleyes:

. I wanna know how much I've lost.
I'm confused on which question is which. But to do the proof to 9/2i, you can add it appropriately
You have to consider the new direction of the ball relative to the horizontal which was 90-x from the original one. The original direction have arctan(3/2) if i remember correctly. So the new direction had arctan(2/3) as the tan of the angle.
Use the fact that j component stayed the same as 3.
the new i componet a lets call it.
3/a=2/3
re arranging gives a=9/2

The projecticle first part was find U in terms of T, theta and g etc.
U=Tgcos(a)/2sin(theta)
something along those lines

then prove 2u^2sin(a)cos(a-x)/gcos^2(a)
phew you nearly gave me a heart attack, when you said sub the earlier part into the second part, i thought you were saying they wanted OP in terms of T and U
I think you have mixed up your answers, the answers to q4) are correct except the part a about 9/2, thats for q5) I believe which was the oblique sphere, q4) was 1d spheres in a line and it had the answers you have down.
end of q5 and q4) i think was finding the impulse on B

(edited 10 years ago)

Reply 28

never used/ idk what arc tan is.
Q1 was worth 6 marks

Reply 29

Original post by Mydwinter

never used/ idk what arc tan is.
Q1 was worth 6 marks

arctan is just another name for inverse tan or tan^-1

I did it the same way as jarasta, i think. I remember having tan(theta)=3/2 where theta is the bearing of the final velocity minus 90 degrees or in other words the angle below the horizontal

Reply 30

maybe stupid but why tan of the velocity? what does tan have to do with it?

Reply 31

Original post by Mydwinter

maybe stupid but why tan of the velocity? what does tan have to do with it?

This was my method:
ImageUploadedByStudent Room1371662137.217298.jpg

Can't really think of any other straightforward at the moment

Posted from TSR Mobile

Reply 32

Original post by Mydwinter

maybe stupid but why tan of the velocity? what does tan have to do with it?

You use tan to work out directions of 2x2 vector generally as an angle when you draw the vector on a quadrant or something. Its in m1 I think if that jogs your memory.

Reply 33

Original post by Mydwinter

never used/ idk what arc tan is.
Q1 was worth 6 marks

Do you reckon I've gotten about 4 marks for my error? :/
1 mark lost for the incorrect Impulse
Method maths for working the integral and quadratic 4.
1 lost on the answer mark

Reply 34

Original post by Allahu Akbar

This was my method:
ImageUploadedByStudent Room1371662137.217298.jpg

Can't really think of any other straightforward at the moment

Posted from TSR Mobile

yeah, i havent a clue what you just did, ahh well.

Reply 35

Original post by jarasta

Do you reckon I've gotten about 4 marks for my error? :/
1 mark lost for the incorrect Impulse
Method maths for working the integral and quadratic 4.
1 lost on the answer mark

well the marks would probably be for I =dmv= 2(5-1) = 8
then if I=8 then 8 = integral of (3t+1) .dt between 0 and T
integrate to get 8 = [3/2 T^2 + T ]
therefore 3T^2 + 2T - 8 = 0
quad from that stuff to get T=2 or T=-3/8 or something
write that obv T >0 so T=2 seconds.

Reply 36

Bric

Original post by Mydwinter

Hmm right, I shall hope to update this as I remember more questions, but heres what I got
Q1 :
T= 2sec as Impulse was 8 Ns
AGREE

Q2: Q2 is the dimension question
Think this was basketball, if it is then the first two parts were show thats and the last direction i got as cos theta = 6.403/3.4619 so the angle of direction as 61 deg below horizontal. Boom.
AGREE

Q3
threw me at first coz it said power, i was like wtf thats m2.

Anyway power = force meters per second which = L^2 M T^-3 so after 5 min of jibber jabber you prove theat [RHS] = [LHS] therefore dimentionally constant. Boom.
AGREE

Q4
I think this was the only part I couldn't do which was prove Vb = 9/2 or something. ceebs spending 20 min on a 4 mark question
-then say 0<e , hence Vb is max 2.5 and obv 2.5<3
Va = 1/2u (5e-3) ACTUALLY bit in brackets is (5-3e)
Vb = 1/2u (e+3) <------- something like these brackets is (5+e)
-next part find e.
e=5/8
No. This question gave e as 2/3, leading to impulse as 15mu

Q5 was projectiles
I think Q5 was spheres on a straight line? with 2kg and 4kg? or was this a pp q i did?
this one needed to find e = 5/8
leading to impulse on B of 13Ns

Q6 was spheres
I think this was projectile on a plane?
If it was then a = (gsinƟ)i + (-gcosƟ)j
I think basically all parts of this were show thats
and the last bit was where x=OP and it wanted it as (2(u^2)cos (alpha-Ɵ))/(gcos alpha) ? somethign liek that, you had to rearrange the answer before to get T , then do T^2 and sub into the formula for OP.

projectiles was all show that

Q7 this one was annoying.
a
200 sin theta = -240 sin 40 so theta = 50.47 and bearing = 120 - theta. Therefore bearing = 069.5 deg to 1d.p? agree

the time when they met was like t = 0.00469 or something so as t is in hours i got T = 3 mins 51.4 sec (Was this meant to be all in min ? e.g 3.86 min? idk )
3 mins 51 secs

b
the last part you had to use geometrical something or other, i needed extra paper for this and got theta = 51.1 deg , but the bearing was theta - 20 because h was 40 deg from the distance line. so Bearing = 031.3

agree

You've a lot right there

Reply 37