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# AQA A Level Maths MM2B Mechanics 2 Unofficial Mark Scheme 25 June 2018 watch

1. I'll start it below
2. Q1

(a) Find the x distance. [3 marks]

, 52 cm

(b) Find the y distance. [2 marks]

, 40.5 cm

(c) Find the mass of the zebra. [4 marks]

, 2.67 kg

Q2

(a) Calculate the KE at the top. [2 marks]

(b) (i) Calculate the KE at the bottom. [3 marks]

42 + = 1688.4 J = 1690 J (3 sf)

(ii) Calculate the speed at the bottom. [2 marks]

(c) Find . [3 marks]

By Newton's second law,

(3 sf)

Q3

(a) Find . [2 marks]

(b) (i) Find F newtons. [2 marks]

(ii) Find the magnitude of F newtons at t = 0. [2 marks]
F = (24i + 24j) N
|F| = 33.9 N

(c) Find t when F newtons is acting due north. [2 marks]

(d) Find r, if r = ??? when t = 0. [5 marks]

Q4

(a) Show that the frictional force is 1620 N. [2 marks]

By Newton's second law, since ,

(b) Find the least value of . [2 marks]

By Newton's second law,

Q5

The van experience a resistive force of newtons, and has a maximum wattage of 32 kilowatts.

(a) The maximum speed is . Show that . [2 marks]

By Newton's first law,

So, using the wattage equation,

(b) The wattage is reduced to zero.

(i) Show that . [3 marks]

The wattage is zero, so the engine provides no force

Using Newton's second law, the component of weight parallel to the slope is , and the resistive force is

By Newton's second law,

(ii) Find in terms of . [5 marks]

When the wattage is reduced to zero, the speed is still 18ms^-1, so

Using v(0) = 18,

(iii) Find the time, to 3 sf, for the speed to increase to 22 ms^-1. [3 marks]

= 12.96... = 13.0 s

Q6: I'll post workings later

(a) Find in terms of and . [6 marks]

(b) Find the ratio . [2 marks]

(0.894 to 3 sf)

Q7

The string AP has modulus of elasticity 160 newtons, and natural length of 2 m; BP has modulus of elasticity 240 newtons and natural length of 3 m.

(a) Show that the total elastic potential energy is 160 J. [2 marks]

The EPE can be derived using Hooke's law .

Extension in AP is 0 m, so EPE is 0 J.

Extension in BP is 5 - 3 = 2 m, so EPE is .

Total EPE = 0 + 160 = 160 J

(b) Find . [5 marks]

Conservation of energy:

(c) Find when the speed is greatest. [2 marks]

Q8 Find in terms of . [9 marks]

This question was the same as January 2013 Q8 (paper: http://filestore.aqa.org.uk/sample-p...-QP-JAN13.PDF; mark scheme: http://filestore.aqa.org.uk/sample-p...W-MS-JAN13.PDF), but here the radius was (not ), and the angle was specified as .

Hence, using the derived formula, .
3. The last question was find r in terms of l, wasn’t t
4. Agh I missed one of the forces on 7B so it was all wrong. Honestly dropped like 18 marks on those last 3 questions so I need the rest of the paper to be completely perfect for me to stand a chance at an A
5. if my method was 100% right for the last question but i made a mistake when subbing one equation into another, how many marks would i drop out of 9?
6. Thanks so much! Also I think HARvey1233 is right I think it was find r in terms of l so just rearrange for r.
7. Whoops on 7b forgot to divide the 8ug when simplifying so I had the v expression and the max speed distance with a 8ug instead of 2ug. But 'twas alright apart from that. I vividly remembered doing the last question on the 2013 paper, got a bit excited at that.
8. Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
16ag:20ag
4:5

Or is that wrong
9. r = 4i - 2j, for 3d) if you still need it
10. (Original post by Integer123)
Q2

(a) Calculate the KE at the top. [2 marks]

(b) (i) Calculate the KE at the bottom. [3 marks]

42 + = 1688.4 J = 1690 J (3 sf)

(ii) Calculate the speed at the bottom. [2 marks]

(c) Find . [3 marks]

By Newton's second law,

(3 sf)

Q3

(a) Find . [2 marks]

(b) (i) Find F newtons. [2 marks]

(ii) Find the magnitude of F newtons at t = 0. [2 marks]
F = (24i + 24j) N
|F| = 33.9 N

(c) Find t when F newtons is acting due north. [2 marks]

(d) Find r, if r = ??? when t = 0. [5 marks]

Q4

(a) Show that the frictional force is 1620 N. [2 marks]

By Newton's second law, since ,

(b) Find the least value of . [2 marks]

By Newton's second law,

Q5

The van experience a resistive force of newtons, and has a maximum wattage of 32 kilowatts.

(a) The maximum speed is . Show that . [2 marks]

By Newton's first law,

So, using the wattage equation,

(b) The wattage is reduced to zero.

(i) Show that . [3 marks]

The wattage is zero, so the engine provides no force

Using Newton's second law, the component of weight parallel to the slope is , and the resistive force is

By Newton's second law,

(ii) Find in terms of . [5 marks]

When the wattage is reduced to zero, the speed is still 18ms^-1, so

Using v(0) = 18,

(iii) Find the time, to 3 sf, for the speed to increase to 22 ms^-1. [3 marks]

= 12.96... = 13.0 s

Q6: I'll post workings later

(a) Find in terms of and . [6 marks]

(b) Find the ratio . [2 marks]

(0.894 to 3 sf)

Q7

The string AP has modulus of elasticity 160 newtons, and natural length of 2 m; BP has modulus of elasticity 240 newtons and natural length of 3 m.

(a) Show that the total elastic potential energy is 160 J. [2 marks]

The EPE can be derived using Hooke's law .

Extension in AP is 0 m, so EPE is 0 J.

Extension in BP is 5 - 3 = 2 m, so EPE is .

Total EPE = 0 + 160 = 160 J

(b) Find . [5 marks]

Conservation of energy:

(c) Find when the speed is greatest. [2 marks]

Q8 Find in terms of . [9 marks]

This question was the same as January 2013 Q8 (paper: http://filestore.aqa.org.uk/sample-p...-QP-JAN13.PDF; mark scheme: http://filestore.aqa.org.uk/sample-p...W-MS-JAN13.PDF), but here the radius was (not ), and the angle was specified as .

Hence, using the derived formula, .
I think it's when t=0, r =4i-2j for question 1
11. (Original post by HarryGCSE)
Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
16ag:20ag
4:5

Or is that wrong
I mean you can't just go from root16:root20 to 4:5 because root20 is not 5
12. (Original post by HarryGCSE)
Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
16ag:20ag
4:5

Or is that wrong
Unfortunately you can't go from √(16ag):√(20ag) to 16ag:20ag because √(16ag) is not equal to √(20ag). If it was √(16ag) = √(20ag), squaring both sides would be valid.
13. (Original post by OllyBishop)
Unfortunately you can't go from √(16ag):√(20ag) to 16ag:20ag because √(16ag) is not equal to √(20ag). If it was √(16ag) = √(20ag), squaring both sides would be valid.

Oh damn, silly me. Oh well it's only 2 marks and I think that they may be the only marks I lost
14. (Original post by HarryGCSE)
Oh damn, silly me. Oh well it's only 2 marks and I think that they may be the only marks I lost
Yeah, and to be honest you'll probably get a method mark for getting that far. I think I only got 7c wrong so hopefully only two marks dropped too!
15. for 6b would it be ok to put the ratio as 4 : 2root5 without cancelling the 2? and i think i forgot to square root the v^2 expression in 7b lmao does that lose 1 or 2 marks?
16. (Original post by levi ackerman)
for 6b would it be ok to put the ratio as 4 : 2root5 without cancelling the 2? and i think i forgot to square root the v^2 expression in 7b lmao does that lose 1 or 2 marks?
Not simplifying the ratio further might lose you 1 mark, not square rooting the v^2 expression will lose you 1 mark (as it will just be an answer mark). So 1-2 marks lost.
17. (Original post by ppdtg)
if my method was 100% right for the last question but i made a mistake when subbing one equation into another, how many marks would i drop out of 9?
hmm not sure tbh, because the question from the 2013 paper was 5 marks and this one was 9 marks i think? but based on past exam questions on moments, the majority of the marks come from a. taking moments about a point and b. resolving in 2 directions, so either vertically + horizontally or parallel + perpendicular to the rod. i'm guessing there's also marks for knowing which way the reaction forces act. so maybe you only drop 1-2/3 marks? basically not many i shouldn't think!
18. (Original post by HarryGCSE)
Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
16ag:20ag
4:5

Or is that wrong
That's wrong, the ratio between the squares was 16:20 so between u:v it will be root both of them
19. Any ideas what the 100ums and A* grade boundary will be?
20. (Original post by wa17)
Any ideas what the 100ums and A* grade boundary will be?
tbh probably 75 for 100 again :/

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