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# AQA MM1B - Mechanics 1 -Tuesday 21st June 2016 Watch

1. (Original post by johnapplebottom)
i done the same and worked out time to be like 1.96 or 1.7 something eded up getting 12.3 for the speed, which makes sense cuz it moved by 3m so ofc the speeds wont be so different nor will the times
I got 13.2 for the speed I think. Might be remembering wrong/made a mistake, though.
2. oh ffs i knew something was wrong (335kg too big) but whenever i change my answer i end up getting my crossed out answer right, so i thought w/e and left it
gg(my grade) = x where x = -75 marks
3. Grade Boundary predictions for the exam anyone? The last exam was a killer, I managed to get t=12s but feel that my C4 knowledge helped otherwise I would have been stuck.
4. (Original post by johnapplebottom)
wait wtf
1) the first question i got like 6.8 for the velocity and 335kg for the mass lol what did i do wrong
i did momentum a + momentum b = momentum of particle -> 3(6) + 2(8) = 5(v) v = 6.8m/s
then the second part i done momentum particle = 5+m(0.1) -> 5(6.8) = 0.5 + 0.1m -> m = 335kg

5) how did u wiork out max height of connected particle? is that when v=0?

7) i got 12.3m/s for the speed. i used horizontal displacement formula -> v=d/t and re-arragned to get in terms of t(since t is same for horizontal and vertical_, i ended up with like 16.6m = ucos50 * t -> u(speed) = 16.6/tcos50 and subbed that into s=ut+1/2at^2 to work out what the time was
then used that time in 16.6m = ucos50 * t to get u = 12.3

8) i got t=5 and t=25 but i thought time isnt a vector so i minused them and used that in the equation i ended up with like 138m or 132m, something like that

how many marks would i have lost on the 10 marker, 7 marker if my answer is wrong?
q1) you should take one direction as negative that might help!
5. (Original post by johnapplebottom)
oh ffs i knew something was wrong (335kg too big) but whenever i change my answer i end up getting my crossed out answer right, so i thought w/e and left it
gg(my grade) = x where x = -75 marks
lel dont worry im sure you did well
6. (Original post by PiTheta97)
If the velocities were the same they wouldn't be at the same point because they have different origins and initial speeds. They'd only be at the same point if the displacements from the origin were the same?

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hmm..how did you do it?
7. (Original post by Chickenslayer69)
For Q8 I only did it for the i components and got 80.5m with t=5, but I think people who did it for j components got t=12 and 106m, which is why there are 2 answers many are getting. Not sure which is right.
didnt the question mention that its horizontal?
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8. (Original post by LiesandAlibis)
I got 106m for Q8
The components must give the same angle of the resultant velocity in order to be parallel, no?
Therefore (ia + ja) = x(ib +jb)?
You can then equate the i and j components to form a pair of simultaneous equations.
Solve for t (from v = u +at)
x = 2/3 or 3/2 (can't remember)
t=12 and find displacement and distance from there?
Thank god someone that got same answer as me
9. (Original post by PiTheta97)
Not sure if its remotely correct but the method I used:

•No resisitive forces so acceleration comes soley from component of toys car weight.
•The acceleration down the slope from the previous SUVAT was 1.5ms^-2
•If the slope was vertical (at 90° to horizontal) the acceleration in the SUVAT would be 9.8
• 9.8sinTHETA=1.5
•Angle = sin^-1(1.5/9.8)

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Second that!
Thank god someone that got same answer as me
Quite a few people did, so we're all in the same boat here. I have a feeling it's right though!
11. (Original post by johnapplebottom)
oh ffs i knew something was wrong (335kg too big) but whenever i change my answer i end up getting my crossed out answer right, so i thought w/e and left it
gg(my grade) = x where x = -75 marks
i did the same mistake as u.. so frustrated easy marks lost :/
12. (Original post by AQA-Disgrace)
Second that!
I did this too!^
13. how long had the airplane been in the air for to travel 240 metres? think i read it wrong ffs
Thank god someone that got same answer as me
I got t equalled twelve too but everyone was telling me i was wrong and the answer was 5. Can you remember all the numbers in the question?
15. (Original post by Jgirl2000)
q1) you should take one direction as negative that might help!
Do you think might get error carried forward if mass is correct based on the previous velocity?
16. website just crashed lool
17. (Original post by LiesandAlibis)
Do you think might get error carried forward if mass is correct based on the previous velocity?
youd think you would but sometimes they can be harsh - especially if its the simplest question (sorry!) i think maybe a method mark
18. (Original post by GabbytheGreek_48)
i didnt know but someone explained to me that you know the resultant acceleration =1.5 from one of the earlier answers and the vertical acceleration is 9.8 as its due to gravity or sumthing. then you use Pythagoras to find the horizontal component and then tan with the horizontal and vertical to find the angle hope this is clear
yh i tried finding horizontal and got error sqrt(1.5^2 - 9.8^2)
got error using x= sin-1(9.8/1.5) too
19. (Original post by koolgurl14)
did anyone get 80.6 for the ten marker?
Yes ! I think i got t= 5 s
20. (Original post by koolgurl14)
did anyone get 80.6 for the ten marker?

Yes ! I think i got t= 5 s

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