AQA MM1B - Mechanics 1 -Tuesday 21st June 2016

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

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?

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

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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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.

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?

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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Thank god someone that got same answer as me

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

Second that!

Thank god someone that got same answer as me

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?

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

did anyone get 80.6 for the ten marker?

did anyone get 80.6 for the ten marker?