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becca-x
did you use the fact that it was projected 3.9m high? I used equations for the vertical s and horizontal s, really mucked that one up i think!

Its looking like I got overall 54-56 :/


Basically, you know that the vertical displacement is -3.6 (it ends up 3.6 m below where it started) so using s=ut+0.5at^2, you get 4.9t^2-8t-3.6=0

Using the quadratic formula, the positive root is 2, so t=2

Then to find the horizontal displacement, you know a horizontally is 0, so s=12t

and 12x2=24 :smile:
Reply 41
Mathematical
Basically, you know that the vertical displacement is -3.6 (it ends up 3.6 m below where it started) so using s=ut+0.5at^2, you get 4.9t^2-8t-3.6=0

Using the quadratic formula, the positive root is 2, so t=2

Then to find the horizontal displacement, you know a horizontally is 0, so s=12t

and 12x2=24 :smile:

Hmm I did a similar thing but substituted x/21 for t in the equation for s, I did this at the last minute with about 4 minutes to spare so I probably made an error somewhere.
Reply 42
after b was moved the tension would have been divided by 2 because F=ma, leading to T+T-250N = 0, so t equalled 125 N i guess. I kinda remember question 6 now, which was easy enough when treating vertical and horizontal components separetely. I think i got t to be 0.89 because u had a quadratic there involving 4.9t^2 something. You had to use the quadratic formula (or a graphic calc. but i didnt have one) to work out t and then like someone else said put that into the s=12t you worked out earlier on.
Reply 43
for the position time graph, what did everyone get for...

Greatest displacement?

I put 0m

Greatest distance?

I put 0m

Time when v = 4m/s ?

I put 0.5 seconds and 1.5 seconds

Also did anyone else get 1.5m/s for the velocity for the last part of the last question?

I put 125N for tension when B had been moved.
Reply 44
For one of the last questions what did you guys get for the equation relating y and t

I got

y=16t^2-112t+192

The curve intercepts the t axis at t=3 and t=4 the minimum to the curve is (3.5, -4)
That paper didn't go badly for me, but some of the answers I've read on here are odd :s-smilie: The displacement was 12, the distance was 16 Tamara.

Don't know, could get an A could get an E. Have to wait :smile:
Reply 46
Vergote
i thought the last 3 parts of the last question were horrible (you had to calculate the acute angle, without any clear information whether the system was in equilibrum or not, what the force of the friction was etc) Just really complicated. Would be nice if someone could explain those last 3 parts (magnitude F, angle theta, showing that F = 661 N when a was 0.5 and then the last bit about the decrease in velocity when F was increased by 200 N).

Was unsure about question 2 (the wall, the rope and the pulley) as well but I think I did the right thing (resolved horizontally and vertically). I assume the tension on the left rope was equal to mg.

And last, the question that involved k. I was confused in the exam because I kept trying to work out a constant but now I realised k was meant to be in vector form as well (I think) because solving the i and j equations involving k separately would give 2 different values...

other than that, I can't complain about the questions


i think with the k question it didnt have to be necessarily in vector form. Because it had to be in the DIRECTION of 7i-9j it could be any multiple eg (14i-18j). If you used these two numbers instead you get k=6 which works...took me a very long time to realise that though!!

And i agree, finding the angle theta question was horrible. i put all the other forces back into vetor form added them so that they equalled (i + 0j) i.e in the "i" direction...:/
ctaylor91
i think with the k question it didnt have to be necessarily in vector form. Because it had to be in the DIRECTION of 7i-9j it could be any multiple eg (14i-18j). If you used these two numbers instead you get k=6 which works...took me a very long time to realise that though!!


If you multiplied 7i-9j by lambda and used simultaneous equations then you got 6 :smile:
ctaylor91
i think with the k question it didnt have to be necessarily in vector form. Because it had to be in the DIRECTION of 7i-9j it could be any multiple eg (14i-18j). If you used these two numbers instead you get k=6 which works...took me a very long time to realise that though!!

And i agree, finding the angle theta question was horrible. i put all the other forces back into vetor form added them so that they equalled (i + 0j) i.e in the "i" direction...:/


The question stated that the resultant force was in the i direction, meaning that vertically, the cpts of the angles had to cancel out.
As for k, I used that the horizontal parts = 7x, vertical = 9x
rearranged to get x and put it back in the question.
Reply 49
ah well the answer i got was correct for the second part of the question where you took 125N away to get 161N (i think? cant remember the exact number). maybe my way of working it out was just a bit skew
Reply 50
so was anyone else really silly and forgot to convert to metres in Q1? or was it just me lol
smiley:)
so was anyone else really silly and forgot to convert to metres in Q1? or was it just me lol


Aw no! I'm sure it wasn't just you, lots of people would have done it :smile:
Reply 52
i wrote a very angry note about the last question involving theta! It was riduclous to lose SIXTEEN marks just because one couldn't find out theta!!
Reply 53
I did the force diagrams for the crates incorrectly (tension in wrong direction). Do you think I wil get follow through marks?
Only other ones i dont think i got was the displacements right at the start of question 7 and the last part of that question seemed too easy for 5 marks so sure got it wrong.
Did anyone get 3.83 for the first question?
Tamara100
for the position time graph, what did everyone get for...

Greatest displacement?

I put 0m

Greatest distance?

I put 0m

Time when v = 4m/s ?

I put 0.5 seconds and 1.5 seconds

Also did anyone else get 1.5m/s for the velocity for the last part of the last question?

I put 125N for tension when B had been moved.


I think your first two are wrong. It was very badly worded tbh and I also put nought first. I *think* what it was asking for the time when the greatest displacement from the place where t was equal to 0, so 12 metres.
I.e my answers were t=1.5 displacement and t=3.5 for distance - basically the furthest point up and down on the y axis, starting from 12 if you see what I mean.

All the rest I got exactly the same as you.

Also for the person who forgot to convert- couldn't you have put 38cm/s as an answer without converting surely it would still work because mass doesn't come into it and gravity will always be the same? I don't know
Reply 55
for the very last question and last part i got the accelerations to be 0.9m/s and 0.41m/s anyone else get those
Reply 56
judd789
for the very last question and last part i got the accelerations to be 0.9m/s and 0.41m/s anyone else get those


the first acceleration i think 0.5

the second acceleartion was negative -0.39...
LisaWilliams
i wrote a very angry note about the last question involving theta! It was riduclous to lose SIXTEEN marks just because one couldn't find out theta!!


:redface: *please be low boundries because of this*! :woo:
Reply 58
smiley:)
so was anyone else really silly and forgot to convert to metres in Q1? or was it just me lol

really oh crap i dont remember the question but i definately didn't covert any units on q1
Reply 59
That wasn't too bad. I didn't get K but got the right working (i hate calculators :P) so I shouldnt lose too many. And also I am not sure about the tension... did you put 250N or 125N