IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: frfoosh on May 25, 2010, 03:17:41 pm
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can we discuss bout the exam now? i think 24hours are over ?
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The exam was easier that i expected,
Anyone remember what value they got for T on the question with the speed-time graph and the two vehicles?
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i messed up that questions big time i got it rong for sure but i think it was something like 55? not sure...
any1 remembers the mass of the particle in the question were 100N was acting on a a partice on a
rough plane? inclined at 30degrees ??
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the mass of the particle was 12.6
i sort of messd up dat question too
constant velocity is 0 accelration----it dint hit to me then -_-
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i got the mass as 10.44 something like that i thought that since velocity was constant so ma was just m ? so i did 100cos30-friction=m and i got 10.44 did any1 get that?
wat bout the moment 1 ? i got 1.2
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M1 June 2010 unofficial MS (https://studentforums.biz/index.php?action=dlattach;topic=7281.0;attach=7500)
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M1 June 2010 unofficial MS (https://studentforums.biz/index.php?action=dlattach;topic=7281.0;attach=7500)
hw did u get the paper??? weird!!
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Damn....Question 5 with the speed-time graph i pretty much messed up :/ At the time it seemed more complicated but now i realise its quite simple. I managed to get to 64 when finding T, but i didn't take away the 55 cos my graph was wrong...FFS!
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guys wt did u get for q 8 the last part which asks about time?
appreciate ur reply ;)
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Damn i got everything correct except i forgot to ad 49 to 11.025 .. bleh its jst like 1 mark .. thank god
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Damn....Question 5 with the speed-time graph i pretty much messed up :/ At the time it seemed more complicated but now i realise its quite simple. I managed to get to 64 when finding T, but i didn't take away the 55 cos my graph was wrong...FFS!
same here i too messed up tht question .. wat do u think ho many marks r we gonna lose in it
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Damn i got everything correct except i forgot to ad 49 to 11.025 .. bleh its jst like 1 mark .. thank god
exactly the same here!! soo annoying!
getting 99% instead of 100% is like the difference between getting a U and an A
:(
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wow, iam getting lke 73 % :( , for question 3 , i did 100cos30-Friction=m and i my reaction= mg + 100sin30
i did everything correct other than i equated it to m, how many marks do u guys think will i gain?
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Mechanics was the worst 4 me :'(
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Mechanics was the worst 4 me :'(
Same here mate. I didn't like it. at all.
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lol
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isnt the last ans of the paper 0.42sec??????
is there anything wrong with the marking scheme?
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i did everything right, just that i got 0.42 instead of 0.566s in the last ans of the paper. my working and everything is right.
how much will i lose for only the wrong value?
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i still have a feeling i am right because, i'll show you the working:
s=1.175 u=0.7 a=9.8 t=? v=?
if we calculate v first
then
v^2=u^2+ 2as
=0.49 + [2*9.8*1.175]
= 23.52
therefore v=4.849742261 ~ 4.85
now using v=u+at
t= (v-u)/a
=(4.85-0.7)/ 9.8
=0.423223087 ~0.42s
am i right????
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i still have a feeling i am right because, i'll show you the working:
s=1.175 u=0.7 a=9.8 t=? v=?
if we calculate v first
then
v^2=u^2+ 2as
=0.49 + [2*9.8*1.175]
= 23.52
therefore v=4.849742261 ~ 4.85
now using v=u+at
t= (v-u)/a
=(4.85-0.7)/ 9.8
=0.423223087 ~0.42s
am i right????
i also got 0.42 with a similar method :S:S
and i was very confident abt it ::S:S:S
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and not just me, all the toppers in my class, the ones who always get full marks also got 0.42.
even i was very confident, but after seeing the marking scheme, i'm kinda depressed.
do you think there must be something wrong with the marking scheme?
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and not just me, all the toppers in my class, the ones who always get full marks also got 0.42.
even i was very confident, but after seeing the marking scheme, i'm kinda depressed.
do you think there must be something wrong with the marking scheme?
maybe because its unofficial ..... lets see ...concentrate on the next exam .....
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i also got 0.42 with a similar method :S:S
and i was very confident abt it ::S:S:S
I also got the 0.566 answer and i was pretty confident about it.
You had to have calculated the travel of the block in 3 steps.
1. While the string is still intact.
2. After the string breaks going upwards.
3. After the string breaks going downwards.
you then add up t for 2 and 3 using the exact values of it and you get roughly 0.566
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can some 1 answer my question please ??? :(
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i also got 0.42 with a similar method :S:S
and i was very confident abt it ::S:S:S
You problem is using a positive value for your initial velocity.
You have to use either a negative initial velocity combined with positive acceleration and displacement
OR
You use positive initial velocity combined with negative acceleration and displacement
I did similar workings to you... working out Final Velocity first.
u=-0.7
a=9.8
s=1.175
v=?
t=?
Working out V
V^2=U^2 + 2as
v^2=(-0.7)^2 + 2x9.8x1.175
v= 4.849
Then working out t
v= 4.849 ( positive because its now traveling in the same direction as i choose as positive earlier)
u= -0.7
a=9.8
4.849= (-0.7)+9.8t
t= 0.566s
Make any sense?
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the last question answer was 0.423 i'm sure of that 100% cuz they asked for the further time only
and how does one take a negative initial velocity, u really cant cuz its only acceleration and displacement that can be actually taken as negative values and only in certain cases but here the distance was 1.175 and the acceleration is positive value of g cuz its moving down so positive 9.8 anywayz lets wait n see
i guess the mass question was 12.75
nd the graphs question was 9 seconds cuz u have to use the 55 seconds in the same way to find that 9 sec for P
nd the first question 17 i guess
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You problem is using a positive value for your initial velocity.
You have to use either a negative initial velocity combined with positive acceleration and displacement
OR
You use positive initial velocity combined with negative acceleration and displacement
I did similar workings to you... working out Final Velocity first.
u=-0.7
a=9.8
s=1.175
v=?
t=?
Working out V
V^2=U^2 + 2as
v^2=(-0.7)^2 + 2x9.8x1.175
v= 4.849
Then working out t
v= 4.849 ( positive because its now traveling in the same direction as i choose as positive earlier)
u= -0.7
a=9.8
4.849= (-0.7)+9.8t
t= 0.566s
Make any sense?
how much marks will i lose for that? i mean i did everything the same way, just that i took initial vel positive and got 0.42s.
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how much marks will i lose for that? i mean i did everything the same way, just that i took initial vel positive and got 0.42s.
I am pretty confident that it IS 0.42...even the top scorers in my class got 0.42!
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same here i too messed up tht question .. wat do u think ho many marks r we gonna lose in it
Not sure, hopefully we'll lose only 1-2 marks for the slight error :/