Qualification > Math
m1 edexcel!
zic2009:
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?
Igcseboy:
--- Quote from: zic2009 on May 26, 2010, 09:06:58 am ---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?
--- End quote ---
maybe because its unofficial ..... lets see ...concentrate on the next exam .....
Ijantis:
--- Quote from: Igcseboy on May 26, 2010, 08:45:23 am ---i also got 0.42 with a similar method :S:S
and i was very confident abt it ::S:S:S
--- End quote ---
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
frfoosh:
can some 1 answer my question please ??? :(
Namenth:
--- Quote from: Igcseboy on May 26, 2010, 08:45:23 am ---i also got 0.42 with a similar method :S:S
and i was very confident abt it ::S:S:S
--- End quote ---
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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