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Official January 2010 M1 Discussion !! watch

1. The Moments Question. W was equal to 280 correct ?
2. (Original post by malteser12345)
for the tension of A, the second part of the moments question, i got something like
(1/6 W -40/30)N.
it was (1/6W -40/3) that was right because using the fact that moment up = moment down C and A had to equal W+20, so adding C and A you got 5/6W + 1/6W + 100/3 -40/3 = W -60/3 = W-20
3. (Original post by laez90)
guys here's what....for the distance question i.e the value of T...it said 'it travels at a constant velocity for T seconds' and not 'upto T seconds'.....so you gotta do like 75-(T+5) for the last triangle....at least that's what i think....this way you get 50 secs and not 55....rest all was ok and i think i rememberr my answers to majority of the questions....lemme know if anyone is interested
yeah i got 50 too
4. (Original post by man u)
wt did ppl get for T in speed time graph
I got 55

Also wt did ppl get for the tension of BC

Wt did ppl get the distance travelled by B in pulleys qs
i don't think it was 55 seconds. i think the question said it accelerate for 5 seconds, then stays at a constant speed for T. this means the constant was for T seconds, ending at T+5, so T was 50? We need to see a paper and a mark scheme.
5. Here is what I got:

Momentum Q:
I= 18Ns
m=1.5 kg

Speed time:
graph: trapezium
T=50s

Moments:
(a) was a proof
(b) (W/6 -40/3)N
(c) W=280N

Inclined surfaces force question:
coefficient= 0.5066..... = 0.507
X=12N

Vectors!!
speed = 5km/h
4got the rest...
last part of vectors, T=1hr and T=5hrs

Pulleys:
k=3
total distance = 3.969 m

there was another question.. but i 4got wat it was about
6. (Original post by onawara)
Here is what I got:

Momentum Q:
I= 18Ns
m=1.5 kg

Speed time:
graph: trapezium
T=50s

Moments:
(a) was a proof
(b) (W/6 -40/3)N
(c) W=280N

Inclined surfaces force question:
coefficient= 0.5066..... = 0.507
X=12N

Vectors!!
speed = 5km/h
4got the rest...
last part of vectors, T=1hr and T=5hrs

Pulleys:
k=3
total distance = 2.205 m

there was another question.. but i 4got wat it was about
i dont think the total distance on pulleys is that. you got how far it was off the plane by seeing how far A dropped (because they were equal). you also got how far it moved when both sides were still moving (ie how far it dropped again), and then work out the extra bit. it was like 1.67+1.67+0.2 or something like tht, i got around 3.6m?
7. (Original post by bibekpd)
Yeh...T was 55s

For tension i got 12 N after rounding off

and the distance in the pulley question i got till where you get 5.3 and then i didn't know what else to do afterwards so I guess its wrong... : /

anyway what do people think the grade boundaries will be for this paper as it was fairly easy ?
I think T was 50 as the runner had been running for 5s at the start

I also got the tension as 12N

I reckon grade boundaries will be around 63/75 for A, 56 for B
8. (Original post by PHOLIO)
I think T was 50 as the runner had been running for 5s at the start

I also got the tension as 12N

I reckon grade boundaries will be around 63/75 for A, 56 for B
I hope not
9. absol-u-tely!!!
10. absol-u-tely!!!
11. Dose anyone know the answer for the last question of the paper (last part of the last question (vectors))??
12. (Original post by titsmcgee)
i dont think the total distance on pulleys is that. you got how far it was off the plane by seeing how far A dropped (because they were equal). you also got how far it moved when both sides were still moving (ie how far it dropped again), and then work out the extra bit. it was like 1.67+1.67+0.2 or something like tht, i got around 3.6m?
i think that the value of W was 40N
value A ws 1/6 W and C was 8 x A
so C was 4/3
so 4/3 plus 1/6 = value B and W = 20N and W
3/2 = W + 20n
3/2W -W = 20 n
1/2 W = 20 n
W = 40 N
13. (Original post by Maximus112)
i think that the value of W was 40N
value A ws 1/6 W and C was 8 x A
so C was 4/3
so 4/3 plus 1/6 = value B and W = 20N and W
3/2 = W + 20n
3/2W -W = 20 n
1/2 W = 20 n
W = 40 N
I beg to differ Its 280 N , i revised it with alot of ppl And all said the same. u didnt Put (g)
14. so what was the maximum height gained by the particle.
ps. did u have to account for the particle acting under gravity aswel.
i got around 3. something
15. (Original post by Bassiouni)
I beg to differ Its 280 N , i revised it with alot of ppl And all said the same. u didnt Put (g)
Yup it surely is 280 N
16. (Original post by Robbyv)
To clear up the athlete question i'll try and do it by memory but it may be wrong:

s=1/2bh + bh + 1/2bh
500=(1/2 x 8 x 5) + (8T) + (0.5 x 8 x (75 - T - 5)
500=20 + 8T + 300 - 4T -20
200=4T
T=50s

Thats what i got anyway so ya know
That's correct according to Arsey.
17. (Original post by Bassiouni)
I beg to differ Its 280 N , i revised it with alot of ppl And all said the same. u didnt Put (g)
18. (Original post by onawara)
Here is what I got:

Momentum Q:
I= 18Ns
m=1.5 kg

Speed time:
graph: trapezium
T=50s

Moments:
(a) was a proof
(b) (W/6 -40/3)N
(c) W=280N

Inclined surfaces force question:
coefficient= 0.5066..... = 0.507
X=12N

Vectors!!
speed = 5km/h
4got the rest...
last part of vectors, T=1hr and T=5hrs

Pulleys:
k=3
total distance = 2.205 m

there was another question.. but i 4got wat it was about
Pulley distance is wrong, I got 2.205 but the weight was alread initially above the ground so the greatest height reached was something different..
19. so wat is the maximum height of the particle???
did any1 actuaaly do it right?
20. if I get 69/75 in raw mark how much will I get in ums/

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