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AQA MM1B - Mechanics 1 -Tuesday 21st June 2016

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1. (Original post by Pinocchiolewis)
Attachment 554071

These are the answer that me and my friends got, let me know if you can't read it/if any of them are wrong 😊
i agree with all of them
2. (Original post by GabbytheGreek_48)
full ums usually determined by i think the highest score achieved or sumn so hopefully low
Alright thanks man
3. Did anyone else get 12.8ms^1 for the 7 marker on question 7?
4. Also did anyone actually get that question where it asked you to find the angle between the ramp and the horizontal?
5. (Original post by Ano123)
Velocities were not equal but parallel, so you cannot just equate them as I think you may have done. One was a multiple of the other, but not necessarily a multiple of 1.
How many marks lost for this?
6. From the answers people have piste on here, I think I've done quite well
7. (Original post by Hep123)
did people get 8 for the bearing?
I certainly did, 008 degrees
8. For question 6, where you had to say 49 + Tsin50... how ****ed am i that i thought it was Tcos60...? if i did that for both parts, out of 8 how many would i lose???
9. (Original post by dgkjhl)
Also did anyone actually get that question where it asked you to find the angle between the ramp and the horizontal?
i think so! since there was no resistance forces acting on the toy car i assumed the only force acting was mgsinB, B being the angle to be found.
plug that into f = ma and its mgsinB = m(1.5)
cancel out the m and it's gsinB = 1.5
so sinB = 1.5/9.8
angle comes out to be 8.80... not sure if you had to round it up or if anyone else got similar answers.
10. A-54
B-47
C-41
D-34
E-28
11. (Original post by Arima)
ahhh i made the same mistake, but there should be marks for at least forming the v=u+at equation/s, otherwise AQA is as sadistic as i've always made them out to be
Phew, it is nice to see someone else has the same thought processes as me. I got the 2 v=u+at questions and hopefully yeah like you say maybe 3 marks will be gained for that.
12. Just wondering, what year did you take this in (AS or A2) As I believe question 8 expects you to know what parallel vectors are, which is covered in C3 or C4. Thanks.
13. Unofficial ms http://www.thestudentroom.co.uk/show...php?p=65990395
14. (Original post by wholockonmars)
Just wondering, what year did you take this in (AS or A2) As I believe question 8 expects you to know what parallel vectors are, which is covered in C3 or C4. Thanks.
This is expected knowledge in M1, vectors are parallel if they are scalar multiples of each other.
15. To everyone that got t=5 or t=25, if you put these time values into the velocity vectors for A and B, it doesn't match, whether you got t=5 or t=25 depends on which components you equated, the i component or the j components.
16. (Original post by Ano123)
This is expected knowledge in M1, vectors are parallel if they are scalar multiples of each other.
Yes I am mistaken, it is very briefly mentioned on page 29 of the textbook.
17. (Original post by OturuDansay)
On question 8 there was a way of getting 106 if you had the times 25 and 5. You simple had to find the distances between A and B for both of times and then calculate the difference between those 2 answers, in which you get 106
Could you please show the working with this method. (I didn't equate the j components to get t=25 😬
18. (Original post by Dynamic_Vicz)
Could you please show the working with this method. (I didn't equate the j components to get t=25 😬
What do you have for the 2 j components? This is not the correct way of doing it btw.
19. (Original post by Dynamic_Vicz)
Could you please show the working with this method. (I didn't equate the j components to get t=25 😬
Okay so you get t= 5 and t=25. 5 from equating i components and 25 from j components (as I remember).
You then plug t=5 into S=ut +0.5at^2 for both A and B. Resulting in two distances. You then find the difference between the two distances.
Then you do the same for t=25.
And finally then you find the difference between them to get the final answer.

I definitely to the long route to do it because i didn't know how to interpet parallel velocities. Don't listen to my method, it will just confuse lol
20. (Original post by OturuDansay)
Okay so you get t= 5 and t=25. 5 from equating i components and 25 from j components (as I remember).
You then plug t=5 into S=ut +0.5at^2 for both A and B. Resulting in two distances. You then find the difference between the two distances.
Then you do the same for t=25.
And finally then you find the difference between them to get the final answer.

I definitely to the long route to do it because i didn't know how to interpet parallel velocities. Don't listen to my method, it will just confuse lol
Thanks I really appreciate the response. This was the method i was using without the equating of the j component... Quick question if you subbed t= 5 and t=25 for R(a) and R(b) wouldn't you have 4 position vector equations to find the difference of 😕

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