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AQA AS M1 Mechanics 1 21st June 2016 [Exam Discussion Thread]

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howd u get 12 seconds
Reply 21
Original post by hsadiku
Distance I got 106, got t=12 and subbed into the position vectors, found the difference between them and took the modulus of that value


What if u got a different value of T sub it in and do the rest, do u lose rest of the marks
Reply 22
Original post by johnapplebottom
howd u get 12 seconds

Basically all you had to do is make an equation for the final velocity for each helicopter using v = u + at leaving t the same in both equations.

Then since the velocity is parallel at t the resultant of the velocities will be acting at the same angle to the horizontal so you could say that tan(angle) = J/I for both the velocities. Then since they both equal tan(angle) you can set J/I for both velocities equal to eachother and solve to get t.

Then from this you can find displacement of the two helicopters at time t. Then finding the distance between the two to give 106 m 3.s.f
(edited 7 years ago)
Alright, here's my working out for Q8. I missed out on the last bit where you work out SB and left it as -105.625i -125j in the actual exam, so I won't get full marks for it :/

https://imgur.com/a/qLIc8
Reply 24
Original post by thestickystickma
Alright, here's my working out for Q8. I missed out on the last bit where you work out SB and left it as -105.625i -125j in the actual exam, so I won't get full marks for it :/

https://imgur.com/a/qLIc8


Thats wrong. First off you cant square a vector so v^2 = u^2 + 2as will produce a false answer
Original post by Crizmikk
Thats wrong. First off you cant square a vector so v^2 = u^2 + 2as will produce a false answer


Oh ffs :frown:
(edited 7 years ago)
Original post by johnapplebottom
how did u get 13.2 i got 12.3 lol
You sure you never mixed the numbers around (I got 13.2)? the numbers are similar
From the look of things on here, I think I've done quite well :biggrin:
Reply 28
It was going so well up until question 7!! And then I just froze and couldn't do 7c at all (doesn't help that I'm not great at projectiles anyway). And then question 8 happened and I just could not even with that...surprise 10 marker and my worst part of mechanics yay...WHY COULD THE 10 MARKER HAVE NOT BEEN ON MOMENTUM?!
Reply 29
Original post by KV848
It was going so well up until question 7!! And then I just froze and couldn't do 7c at all (doesn't help that I'm not great at projectiles anyway). And then question 8 happened and I just could not even with that...surprise 10 marker and my worst part of mechanics yay...WHY COULD THE 10 MARKER HAVE NOT BEEN ON MOMENTUM?!


Haha I found the question with the angle between the ramp and horizontal hard but q7 and 8 were lightweight, should of expected a 10 or 9 marker on position vectors since they came up last year
Reply 30
how should 7)c) be done?
I initially subbed in the time to part a or b put, and worked out 'u' from that. This gave me like 14.1m/s but this didn't seem to make the vertical height right.
So I made the wonderful mistake of a last minute change of my working. The horizontal distance was just +3 to the distance from the previous part, but after the vertical height seemed wrong, I created two simultaneous equations which cancelled out 'u' (what I called the initial velocity), and gave me a time. I then subbed this time into the equation which gave me 6.4 or something, which seems unlikely, as it should surely increase if the ball is to travel 3m more horizontally?? Is either remotely the right method??
Reply 31
Original post by yelash
how should 7)c) be done?
I initially subbed in the time to part a or b put, and worked out 'u' from that. This gave me like 14.1m/s but this didn't seem to make the vertical height right.
So I made the wonderful mistake of a last minute change of my working. The horizontal distance was just +3 to the distance from the previous part, but after the vertical height seemed wrong, I created two simultaneous equations which cancelled out 'u' (what I called the initial velocity), and gave me a time. I then subbed this time into the equation which gave me 6.4 or something, which seems unlikely, as it should surely increase if the ball is to travel 3m more horizontally?? Is either remotely the right method??


t is would be different in part 7c because its traveled a different distance at a different speed. All you had to do is make an equation for horizontal displacement(16.something + 3) in terms of t and v (the initial velocity) and an equation for the vertical displacement in terms of t and v (the initial velocity). Then from the first equation express v in terms of t and sub into the second equation to get 13.2 m/s
Reply 32
Original post by Crizmikk
t is would be different in part 7c because its traveled a different distance at a different speed. All you had to do is make an equation for horizontal displacement(16.something + 3) in terms of t and v (the initial velocity) and an equation for the vertical displacement in terms of t and v (the initial velocity). Then from the first equation express v in terms of t and sub into the second equation to get 13.2 m/s


Hmm seems my changed method was right even if my answer wasn't. Might pick up a few method marks hopefully

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