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# 16May 2018 Edexcel M3 Unofficial Mark Scheme watch

1. 2a. 11.3
2b. 2.5
3. 2R
4. k=1.86
5b. Centre of Mass = 9a(16-k)/4(k-8)
5c. k=16
6b. u>=sqrt(3ag)
6c. u=sqrt(5ag)
7b. a=-49x therefore SHM
7c. v=2.42
7d. Time taken = 0.547

Please tell if there are any mistakes
2. 2. 11.3
B. 2.5
3. R/5
4. 1.7
5. 9a(k-16)/4(8+k)
b. 35.2
6. Sqrt 3ag
b sqrt 5ag
7. -49x
b V=2.4
c t=0.55
3. Same as amber but 3R/4 for Q3
4. I got 1.86 for 4, 3sf
5. How did you get Q5b and what was Q4?
6. Q4 was when the particle went down a slope with friction
7. 2a. λ=(98 root 3) divided by 15
5b k=16
I got 1.86 for 4, 3sf
same coz it was to 3sf

1. Show that...

2.(a) 2g/sqrt(3) = 11.3 (3 s.f.)
2.(b) omega = 2.51 rad s^-1 (3 s.f.)

3. Distance = R/5

4. k = 1.72 (3 s.f.)
EDIT: I made a silly mistake - the answer was actually k = 1.86.

5.(a) Prove centre of mass = 3r/8 from O (I couldn't remember how to do it, FFS!)
5.(b) Distance from O = Modulus of {3a(16-k)}/{16(8+k)}
5.(c) k = 16

6.(a) Show that...
6.(b) Minimum u = sqrt(3ag)
6.(c) u = sqrt(5ag)

7.(a) Show that l = 1.2
7.(b) Prove SHM (acceleration = -49x)
7.(c) v = 2.42 ms^-1 (3 s.f.)
7.(d) Time taken = 0.547 seconds (3 s.f.)
10. (Original post by Pitsillides)
2a. λ=(98 root 3) divided by 15
5b k=16
I got k=16 too!!
11. I swear question 4 k was 1.86
12. (Original post by mupsman2312)

1. Show that...

2.(a) 2g/sqrt(3) = 11.3 (3 s.f.)
2.(b) omega = 2.51 rad s^-1 (3 s.f.)

3. Distance = R/5

4. k = 1.72 (3 s.f.)

5.(a) Prove centre of mass = 3r/8 from O (I couldn't remember how to do it, FFS!)
5.(b) Distance from O = Modulus of {3a(16-k)}/{16(8+k)}
5.(c) k = 16

6.(a) Show that...
6.(b) Minimum u = sqrt(3ag)
6.(c) u = sqrt(5ag)

7.(a) Show that l = 1.2
7.(b) Prove SHM (acceleration = -49x)
7.(c) v = 2.42 ms^-1 (3 s.f.)
7.(d) Time taken = 0.547 seconds (3 s.f.)
I got pretty much the same as you! I couldn't remember 5a too, I tried to do a circle of radius r around centre (r,0) but i'm not sure it worked lol. I did however get that the distance was 0 because my x = R and it was from the surface, seems i may have gotten that one a tad wrong lol.
13. (Original post by mupsman2312)

1. Show that...

2.(a) 2g/sqrt(3) = 11.3 (3 s.f.)
2.(b) omega = 2.51 rad s^-1 (3 s.f.)

3. Distance = R/5

4. k = 1.72 (3 s.f.)

5.(a) Prove centre of mass = 3r/8 from O (I couldn't remember how to do it, FFS!)
5.(b) Distance from O = Modulus of {3a(16-k)}/{16(8+k)}
5.(c) k = 16

6.(a) Show that...
6.(b) Minimum u = sqrt(3ag)
6.(c) u = sqrt(5ag)

7.(a) Show that l = 1.2
7.(b) Prove SHM (acceleration = -49x)
7.(c) v = 2.42 ms^-1 (3 s.f.)
7.(d) Time taken = 0.547 seconds (3 s.f.)
Q 4 is defo wrong! the ans is 1.86 🙃
14. (Original post by boodledoodle123)
I got k=16 too!!
got k=16 as well but seems many ppl get other things?
15. How did you get 7)d)?
16. (Original post by gnehcnosna)
got k=16 as well but seems many ppl get other things?
Me and my friend both got k=16
17. (Original post by gnehcnosna)
got k=16 as well but seems many ppl get other things?
Well, to get to 16 I didn't do much? Like it was 2 marks? So lol I hope it's 16 xD if not oh dearie me
18. 2. Lamda = 11.3
Omega = 2.51
3. Distance = R/5 (was from surface)
4. k = 1.86
5. (C) k=16
6. (B) Minimum u = sqrt(3ag)
(C) u = sqrt(5ag)
7. (B) a = -49x therefore SHM
(C) v = 2.42 m/s
(D) t = 0.55 secs
19. (Original post by gnehcnosna)
got k=16 as well but seems many ppl get other things?
K=16 is right
20. (Original post by Legend123456)
How did you get 7)d)?
I got 0.443 for 7d so haha idk

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