AQA Mechanics 1B 24 May 2012 Unofficial Markscheme

Original post by louisjevans

M1 one mark for 360 - 22
A1 one mark for 338

Just hope that the fact that I put it as a bearing - 022 that I might get a second M1 mark. :s-smilie:

Reply 62

Original post by tom472

Yeah I did that. I got for that 2.02s, it was 3.01~ - 0.99~ = 2.02~ s dunno if its right but hopefully it is.

Reply 63

Original post by Miyata

Yeah I did that. I got for that 2.02s, it was 3.01~ - 0.99~ = 2.02~ s dunno if its right but hopefully it is.

I got 3.45 seconds.

Reply 64

louisjevans

Original post by tom472

On the projectiles question where it asked for the time when its above 5m, did anyone else use the quadratic formula? I got two values and then subtracted the small value from the large one (there was two values because it reaches 5m on its accent and 5m on its decent?).

btw on the sledge question does anyone remember what the coefficient of friction and the acceleration were? I pretty sure it;s correct, just wanted to check the working :redface:

THE FORMULA :P yeah 3.73-0.27 =3.46 ithink

sledge co eff was 0.3 and cant remember the acceleration :ashamed2:

Reply 65

karchun

Original post by tom472

1st Q's: yes I got 3.45 seconds
2nd Q's: 0.3 I think

This was posted from The Student Room's Android App on my GT-N7000

Reply 66

A bearing is measured from north going clockwise. If you go the other way, I presume you'd have to state it. However, on most past papers, in the mark scheme they usually accept the angle, whether it is giving as a bearing or not.

Reply 67

Anyone know what modelling assumptions were correct?

Reply 68

Sledge question had an acceleration of 0.05, with mass 8.

If I remember correctly, then NR in terms of T was:
NR=78.4-Tsin30 sin30 = 0.5, so:
NR=78.4-0.5T.
T-Fr=ma and as Fr = 0.3 x NR, T-0.3(78.4-05T)=0.05x8
T-23.52+0.15T=0.4
1.15T=23.92
T=20.8N

This is how I done it. Unsure if it is right or worng, though.

Reply 69

Original post by 1platinum

Anyone know what modelling assumptions were correct?

Well no Air resistance is definitely 1 mark, but don't know about others. I mean they gave out way too many. They said its particle, string is light, string is inextensible, and I think something else they have mentioned.. So I didn't know what to put besides air resistance. Plus I didn't want to guess, because in mark schemes they have penalized before for writing a wrong assumption so I left it.

Reply 70

tom472

Original post by karchun

1st Q's: yes I got 3.45 seconds
2nd Q's: 0.3 I think

This was posted from The Student Room's Android App on my GT-N7000

I think I got 3.45 s as well (definitely 3. something).

For the modelling assumptions I put the string it taut and the pulley is light (no weight). They already said the string is light and inextensible and the pulley was smooth (and something else which I can't remember).

For the bearing question I got 338 (360 - angle). I can't remember the values.

Reply 71

Original post by Miyata

They would only penalise if you put more than two assumptions. Anyway my second one was that the friction won't remain constant.

Reply 72

karchun

Original post by Miyata

My friend said you could say it was a particle since the block was not mentioned as a particle :-\

This was posted from The Student Room's Android App on my GT-N7000

Reply 73

Original post by jph12

Thanks!. I stil don't remember if I got that or not.. either I got that or 154~ but I can't see how it would be possible to get 154~. I know I ha 0.4, and I seem to remember 23.52, so hopefully i got it. Thank you!

Reply 74

Original post by 1platinum

They would only penalise if you put more than two assumptions. Anyway my second one was that the friction won't remain constant.

Depends on how mean they are, I did some papers and they said penalize for WRONG assumption, not for more than 2.

Original post by karchun

My friend said you could say it was a particle since the block was not mentioned as a particle :-\

This was posted from The Student Room's Android App on my GT-N7000

Hmm, well if it's true then good for some people probably lol. I just noticed they mentioned particle in the question so I didn't write about it.

Reply 75

tom472

Original post by jph12

Are you sure the coefficient of friction was 0.3? I'm not sure but with those values I get:

I get:

R=8g - Tsin30

F=ma
Tcos30 - 0.3 [8g - Tsin30] = 8 * 0.05
Tcos30 - 23.52 + 0.3Tsin30 = 0.4
Tcos30 + 0.15T = 23.92
T (cos30 + 0.15) = 23.92
T = 23.92 / (cos30 + 0.15)
T = 23.5427...
T = 23.5 (3sf)

Reply 76

Original post by Miyata

What do you think the boundaries will be like? I hope I will manage to get an A... I lost 1 mark for not saying about second assumption, then I didn't get that car constant breaking force, I had no idea what they meant by that... I just did mass times gravity and got 13720... and for second part did this -200, to get 13520 which is apparently bull**** :sss for my sake I hope not, or at least they will Follow through my mistake in second part, well that's -4 marks

I think the grade boundaries will be slightly lower than the past average, because there were a few questions that people didn't understand. However, I can't imagine it being like 50 for an A.

For the braking force use F=ma, and take right as positive, with an acceleration of -2.
Constant braking force, R: -R=ma: -R=1400x-2: =2800N.
Cant braking force with air resistance, AR: -R-AR=ma: -R-200=1400x-2:
-R=-2800+200: R=2600.

Reply 77

karchun

Original post by Miyata

Hmm, well if it's true then good for some people probably lol. I just noticed they mentioned particle in the question so I didn't write about it.

Exactly what I thought, I know I definitely got that one wrong for 1 mark :facepalm:

This was posted from The Student Room's Android App on my GT-N7000

Reply 78