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AQA Mechanics 1B 24 May 2012 Unofficial Markscheme

I saved all my answers on my graphical calculator. I do not remember the questions, and will edit this thread according to the general feedback of the answers. Some of these may be wrong, again, these are simply my answers.
Updated version, 09:11 25th May 2012. I have included some very general "questions", so people can jog their memories to the question. A small summary follows the x), and the answer is the bit after the colon [:]. Again, any errors, please let me know so I can edit them. Question 6c) changed according to the general consensus. I have added what I can remember to be the marks allocated per question. I do not know if they're right, I have done them purely from memory. I have laid it out in a format people might be able to understand a bit easier. Prior to this it was a bit of a mess, rofl.

1 - Ship on water:
a) Question: Find the resultant velocity (2 marks):
Answer: 5.39
b) Question: Find the bearing (3 marks):
Answer: 338

2 - Momentum:
2) Questions: Find the mass of B (3 marks):
Mass: 0.5

3 - Car braking to a halt:
ai) Question: Find the acceleration of the braking car (3 marks):
Answer: -2
3aii) Question: Can't remember (2 mark)
Answer: 10
3aiii) Question: Braking force when no other forces present (2 marks):
Answer: 2800N
3b) Question: Braking force when air resistance of 200N is present (2 marks):
Answer: 2600N

4 - Particle in equilibrium. Force of 20N going at an angle of alpha, and in the opposite direction, a force of 10N.
a) Question: Find alpha (2 marks):
Answer: 60
b)Find W, (Weight) (3 marks):
Answer: 17.3
c)Find the mass of the particle (2 marks):
Answer: 1.77

5 - Block and particle system.
a) Question: When smooth, find the acceleration (3 marks):
Answer: 5.88
bi) Question: When rough, find the tension (3 mark):
Answer: 122.4
bii)Question: Find the Reaction Force(?) (2 mark):
Answer: 117.6
biii) Question: Find the coefficient of friction (5 mark):
Answer: 0.735
c) Make two assumptions not included in question (2 mark):
Answers: No air resistance, block is uniform in mass, tension is constant throughout the string, block is modeled as particle, string does not snap, string does not hit the ground before the block hits the peg.

6 - Some stupid girl with a sledge, who has most probably cost a lot of people a few marks, for forgetting the Tcos30.
a) Question: Draw a diagram of the forces acting on the sledge when modelled as a particle (2 marks).
b) Question: Find NR in terms of T (3 marks):
Answer: NR = 78.4 - 0.5T
c) Question: Find tension when acceleration is 0.05 (6 marks):
Answer: 23.5

7 - Vector question. Unsure of the following questions. i and j vectors.
a) Question: Unsure. Maybe express t in terms of s? (2 marks):
Answer: s = (-i+3j)+0.5(0.1i-0.2j)t^2
b) Question: Again, unsure (3 marks):
Answer: 30
c) Question: Find the speed when going SE. Needed to form an equation of v=u+at, not s = (-i+3j)+0.5(0.1i-0.2j)t^2 (6 marks):
Answer: 1.41

8 - Projectile. She fires at an angle of alpha with speed 22.4.
a) Question: Find alpha (2 marks):
Answer: 61 degrees
b) Question: Find maximum height (3 marks):
Answer: 19.6
c) Question: Find horizontal distance(?) (3 marks):
Answer: 43.4
d) Question: Find the length of time at which the particle is above 5m. Most people seem have gotten 3.46 seconds, so I have changed it to that (6 marks):
Answer: 3.46
e) Find the speed at which the particle hits the ground (2 marks):
Answer: 10.8
(edited 11 years ago)

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Sick, looks like I did pretty well if these answers are right :biggrin:
Original post by jph12
I saved all my answers on my graphical calculator. I do not remember the questions, and will edit this thread according to the general feedback of the answers. Some of these may be wrong, again, these are simply my answers.

1a)5.39
b)338

2)0.5

3ai)? (Seemed to have not entered anything in for this question, though I did do it.
3aii)10
3aiii)2800N
3b)2600

4a)60
b)17.3
c)1.77

5a)5.88
bi)122.4
bii)117.6
biii)0.735
c)No air resistance, block is uniform in mass.

6a)Diagram, I think
b)NR = 78.4 - 0.5T
c)20.8

7a) s = (-i+3j)+0.5(0.1i-0.2j)t^2
b)30
c)1.41

8a) Angle was 61 degrees
b)19.6
c)43.4
d)3.69
e)10.8


You beat me too it, i put my answers in my calc too YAY!!! :biggrin:

For 6 c i got 23.5 and for 8d i got 3.46 otherwise i put the same as you but i didnt put a second assumption for 5c though :/


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(edited 11 years ago)
For 5C, I said only resistant force is friction/no air resistance, model block as a particle and the string does not break. Just in case one of the three was wrong.
Original post by jph12
I saved all my answers on my graphical calculator. I do not remember the questions, and will edit this thread according to the general feedback of the answers. Some of these may be wrong, again, these are simply my answers.

1a)5.39
b)338

2)0.5

3ai)? (Seemed to have not entered anything in for this question, though I did do it.
3aii)10
3aiii)2800N
3b)2600
4a)60
b)17.3
c)1.77

5a)5.88
bi)122.4
bii)117.6
biii)0.735
c)No air resistance, block is uniform in mass.

6a)Diagram, I think
b)NR = 78.4 - 0.5T
c)20.8

7a) s = (-i+3j)+0.5(0.1i-0.2j)t^2
b)30
c)1.41

8a) Angle was 61 degrees
b)19.6
c)43.4
d)3.69
e)10.8


Werent those two minuses.
Plus 7c i got 5.38
(edited 11 years ago)
Original post by darkshadow1111
Werent those two minuses.


Magnitude would make it positive, no? I dunno! I was unsure on those two.
Reply 6
Original post by darkshadow1111
Werent those two minuses.
Plus 7c i got 5.38


I got minuses because weren't we working out the braking force?
Original post by 1platinum
I got minuses because weren't we working out the braking force?


Yeah, same here.
Original post by 1platinum
I got minuses because weren't we working out the braking force?


Depends which way you put as positive direction, i put positive backwards therefore the negative forward acceleration became positive to form positive breaking force.


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Reply 9
I got 5.88 for 7(c) and for 8(d) 3.45.
Reply 10
Sweet, seems like I've got most things correct. I can't remember my answer for question 7, but can you explain briefly how you did the south east thing? I'm not sure if I did it correctly!

Edit: Just remembered my 7c, I put 2 point something... could you enlighten me as to how you got your answer? :smile:
(edited 11 years ago)
Original post by louisjevans
Depends which way you put as positive direction, i put positive backwards therefore the negative forward acceleration became positive to form positive breaking force.


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I think I used -2 as the acceleration.
Original post by 1platinum
I think I used -2 as the acceleration.


Youll get all the credit :smile:


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Original post by stupefy!
Sweet, seems like I've got most things correct. I can't remember my answer for question 7, but can you explain briefly how you did the south east thing? I'm not sure if I did it correctly!

Edit: Just remembered my 7c, I put 2 point something... could you enlighten me as to how you got your answer? :smile:


Use v=u +at, sub in u and a

i component equals negative j component because its SE.

So t = 20

Therefore v= i - j

Speed = root (1^2 + 1^2)
= 1.41





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Reply 14
Original post by 1platinum
I got 5.88 for 7(c) and for 8(d) 3.45.


+1

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Reply 15
bleh it seems that for a few of these i put the right answer but then changed it to something else, however i didnt cross the right answers out just in case, will i still get the marks for those???
Reply 16
The magnitue of a number is the size of it. In other words, turn it into positive. You will learn more about it in A2.
Original post by poyyo
bleh it seems that for a few of these i put the right answer but then changed it to something else, however i didnt cross the right answers out just in case, will i still get the marks for those???


No, this is not trial and error. You dont just put as many answers as you want and then get the marks.
Reply 18
Original post by louisjevans
Use v=u +at, sub in u and a

i component equals negative j component because its SE.

So t = 20

Therefore v= i - j

Speed = root (1^2 + 1^2)
= 1.41



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yeah, I was trying to do this right now as well, but I forgot what's u and a. I got same answer as you.
What did you get for 8d)? time length? and how did you do it? Thanks.

Also what was the value for u and R in question 6? I want to just if I got 20.8 or as I can't remember the question. I think I remember part b).
(edited 11 years ago)
Original post by louisjevans
Use v=u +at, sub in u and a

i component equals negative j component because its SE.

So t = 20

Therefore v= i - j

Speed = root (1^2 + 1^2)
= 1.41





This was posted from The Student Room's iPhone/iPad App


Can't you do v=u+at?

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