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Mathematics: Post your doubts here!

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I think i may have found a mistake in the mark scheme.

Question 3 on Oct/Nov 2010 Qp 12

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf

The mark scheme doesn't multiply the dx/dt by the power of the bracket.

Link to mark scheme:

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_12.pdf

I need help on this ASAP.

EDIT: Guys i need help with this ASAP, exam's tommorow.

iKhaled
hussamh10
minato112
nightrider1993

I solved the questions. I did it step-by-step as far as possible. Hope this clears your doubt. If you still can't understand, notify me.
 

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Please help with part (ii)
Why is the MS skipping all the way to finding the angle???

Thanks in advance
 

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Please help with part (ii)
Why is the MS skipping all the way to finding the angle???

Thanks in advance

Of course, the ms will never go through the trouble of solving the questions completely. They'll only write the essential part that'll lead to the answer.

Here are the answers :
 

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Of course, the ms will never go through the trouble of solving the questions completely. They'll only write the essential part that'll lead to the answer.

Here are the answers :
wow, just when i thought i was lost with trig, hahaha, u really r a life saver.
Thanks
 
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Can I know how to find the PE gain?
Why the height is (AB/20)??? I don't get this step..

Question:
Screen shot 2012-10-17 at 9.12.48 PM.png

Answer:
Screen shot 2012-10-17 at 9.13.40 PM.png
 
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Can I know how to find the PE gain?
Why the height is (AB/20)??? I don't get this step..

Question:
View attachment 16611

Answer:
View attachment 16612
Firstly you should consider the work rule ,i.e. WD = gain in PE - loss in KE + WD against resistance
After that substitute given information: loss in KE = 0.5 * 16000 * (15^2-12^2)
Gain in PE = mgh = 160000h (h is the VERTICAL height of B above A)
WD against resistance = 1240s (s is the distance AB)
therefore, 1200000 = 160000h - 648000 + 1240s
Furthermore, consider the right angled triangle with sides 1 and hyptenous of 20 where 1 is opposite to theta.
Now if you imagined that triangle on the slope, concluding the h=1 and s=20
remember that this is a ratio, so h/s = 1/20
rearrange ratio to subject h (h=s/20) (this is the subject of your question)
Moreover, Substitute this later equation into your work rule and you will get distance (s) = 200 m

Hope I have helped you, ;)
 
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can someone explain me on work-energy principle??
i tend to get confuse..
for eg.. one question needs to use K.E - P.E = W.d - W.d against resistance
theres also other question using P.E - K.E = W.d - W.d against resistance
thats different :(
pls help
 
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can someone explain me on work-energy principle??
i tend to get confuse..
for eg.. one question needs to use K.E - P.E = W.d - W.d against resistance
theres also other question using P.E - K.E = W.d - W.d against resistance
thats different :(
pls help
I know this difference but i'm confused with W.D against friction part. Plz anyone out there clear this formula for work Energy.
 
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I know this difference but i'm confused with W.D against friction part. Plz anyone out there clear this formula for work Energy.
W.D - WDr = KE + PE

thats the formula..thats it! KE gets a negative value if its kinetic energy loss and it gets a positive value if its kinetic energy gain and some goes to the PE..
 
Messages
870
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can someone explain me on work-energy principle??
i tend to get confuse..
for eg.. one question needs to use K.E - P.E = W.d - W.d against resistance
theres also other question using P.E - K.E = W.d - W.d against resistance
thats different :(
pls help
W.D - WDr = KE + PE

thats the formula..thats it! KE gets a negative value if its kinetic energy loss and it gets a positive value if its kinetic energy gain and some goes to the PE..
 
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