AQA June 2011 Mechanics 1B 26th May

I got t=40 for the heli.
m=9 and v=7 for the balls.
250 for the first part of projectiles.
253 or something for the second part (sqrt(x+y))
angle around 8 degrees

Tension was T-mg=ma, so working the downwards for 1 ball has to equal the upwards of the other ball... my answers came out the same so managed to get the tension.

Last question was a joke though. The whole paper was definitely harder than most (changing around of questions like projectiles etc, using tan and s=ut+1/2at^2 about 3 times)... hopefully should push the grade boundaries down but IMO the last question was the only hard one.

can anyone remember what questions 1/2/3 were about? o_O

I am kicking myself so bad right now, but i didn't put my answers to 3 sf all of the time :/ does anyone know how many marks they will knock off for this? Will it be one for every question?

Hey guys! Did this exam today and I am gonna try and put as much solutions as I can possibly think of. If you think you didn't do too well, it might not be a good idea to read on, and that is to avoid disappointment

For the first question, I remember finding the tension which was 8,000 N.

The second question, where you had to sketch the graph of speed-time, it was 64 seconds for it to get from its contstant speed, to decelerate to 18 m/s. The graph should have looked like this: A straight line going for 12.5 seconds at 32 m/s, then the line would slope down linearly to 76.5 seconds at 18 m/s The average speed was the whole distance of 2000 meters, divided by the time taken which was (64 + 12.5), answer was something like 26.1 m/s

Not sure on the question number for this one, but it is the one with the momentum. V=7. M=9.

Connected particles over a pulley was nice and simple. 2.45 m/s^2 acceleration (given in the question) and the rest of it was Suvat, where you had to calculate the time it took to drop to the surface, and then you calculate the velocity as it hits the surface by doing ROUTE( 2 x acceleration x displacement)

The helicopter question (I screwed this one up, but i know the right answer) you were supposed to find only the horizontal velocity, which was 12.5 You would then use that to find the time at the end, the time was 40 seconds (I did my calculation another way and got 39.9, absolutely gutted)

The projectiles question was nice and easy. The velocity was 250. The angle at the end was 8.91, which makes sense really.

Finally, the last question, which when i saw i thought there was no way i'd be getting this one right, It was actually quite do-able. You had to find an equation for Q vertically and horizontally. Then you would simply be left with simultaneous equations, where you would have two equations which equal Q. The Q would disappear and you will be left with P and g/a bla bla. you would work your way until you get 2P= 250g/sin80 + 250a/cos80 (or something like that) and then you simply divide everything by 2, then you pull the 125 out to put it in the form they wanted.

Finally, the last part was finding the acceleration when Q was zero. For this you had to take the equation for P whereby it was the one which had the "250g" in and not "250a" because you wouldnt know the a (obviously). You would find the value for P, and then you simply do F=ma. Where it was Pcos80 = 250a. Answer was 1.73 m/s/s

I am pretty ticked off about screwing up the helicopters question. I have honestly never done a question like that before, and had simply no idea how to get the right answer

Bear in mind, i did this exam after TWO HOURS AND A HALF of Religious studies non-stop writing.

Not all my solutions are 100% correct. I could be wrong. So far, I don't see why they would be.

I hope that helped!

The projectiles question was nice and easy. The velocity was 250. The angle at the end was 8.91, which makes sense really.

Got exactly this, but one thing I remember and you missed is the height of the cliff the gun/arrow was shot at, the height h was 78.4m if I remember correctly?

Also, Where you have V=7 and M=9
I think I got 17 for the velocity, don't remember too well, may just have been a typo or I'm wrong.

The velocity was 253ish, not 250. 250 was just the horizontal component. Unlucky on the 39.9...

For the bearing did anyone else get 037?

I got 053
A friend got 057 hmm

For the bearing did anyone else get 037?

I did, but somebody else told me they got a different answer :/ hmmm

I'm pretty sure we're right, because it's to the vertical, right? If it were to the horizontal it would remain 53? But obviously bearing start at North.

But wasn't it 37 to the horizontal? or did I mix up my i's and j's

Nope it was 53 to the horizontal I know this because it was the number i got when I used tan-1 on the triangle I equated. Now I look back you could found the bearing straight away, but I am certain 037 is the correct bearing.