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M1 ON 13 MAY ,WHO HAS THE BEST PREPATION,Post your doubts here?

AlishaK

AlishaK

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Sheikh Nahiyan said:
Someone please explain to me Question no. 2 from May-June 2012-Paper 42. I have a serious problem in my concept and need a thorough explanation. And also question 4 and 5! Please guys.

If possible upload a podcast here (like explain it in and record your voice and send here)!!!

Please guys. In total chaotic situation right now :(
Click to expand...
Yay! I did that paper. :D
 
AlishaK

AlishaK

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Sheikh Nahiyan said:
Great! Explain me those questions now 8|
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Ok for Q2, I made a right-angle triangle. Notice a triangle can b formed, with 15 N force as the hypotenuse, 12 N as the short leg and F N as the longer leg (base), You gotta elaborate it. then using trig, I found out alpha: sin (alpha)= 12/15, (alpha)=53.1 deg.
Now find the other angle by 90-53.1=36.9 ( i hope u got this). Then for ii) I used sine rule: F/sin36.9=12/sin53.1....F=9N *Ta-da* ....I hope u got it.
 
Sheikh Nahiyan

Sheikh Nahiyan

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AlishaK said:
Ok for Q2, I made a right-angle triangle. Notice a triangle can b formed, with 15 N force as the hypotenuse, 12 N as the short leg and F N as the longer leg (base), You gotta elaborate it. then using trig, I found out alpha: sin (alpha)= 12/15, (alpha)=53.1 deg.
Now find the other angle by 90-53.1=36.9 ( i hope u got this). Then for ii) I used sine rule: F/sin36.9=12/sin53.1....F=9N *Ta-da* ....I hope u got it.
Click to expand...

I did the same thing but can find where to put alpha.. like can you draw the fig and give here? :\
 
AlishaK

AlishaK

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Sheikh Nahiyan said:
Great! Explain me those questions now 8|
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Q4) Resolve the force T. Therefore, Normal contact force (N)= 40+Tsin25 (where 40 is the weight of the ring). Friction force (F)=Tcos25.
ii) Coeff of fric=F/N:
0.4 = Tcos25/40+Tsin25...Solving this gives u... 0.906T-0.169T=16, 0.737T=16, T=21.7 N.
I hope I was helpful. Anyhow, remember me in your prayers! :)
 
Sheikh Nahiyan

Sheikh Nahiyan

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AlishaK said:
Q4) Resolve the force T. Therefore, Normal contact force (N)= 40+Tsin25 (where 40 is the weight of the ring). Friction force (F)=Tcos25.
ii) Coeff of fric=F/N:
0.4 = Tcos25/40+Tsin25...Solving this gives u... 0.906T-0.169T=16, 0.737T=16, T=21.7 N.
I hope I was helpful. Anyhow, remember me in your prayers! :)
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THANKS :D
 
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sagar65265

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AlishaK

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Silent Hunter said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_42.pdf

its 5 (ii) ..... the problem is why should we take (t-0.4) for Q if we take 't' for P ? I mean at any point the time taken by P to reach there is 't' then the time taken for Q to reach must t+0.4 cuz its coming later !! ?? am confused :\ :confused:
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Hala! I just did that. 2 mins ago! Well, see initial time for p is 0. but initial time for q is 0.4 as it starts 0.4 s later than p.
That is why for p time will be t-0=t, and for q t-0.4.
Got it?!
Jazak Allah khair!
 
Silent Hunter

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AlishaK said:
Hala! I just did that. 2 mins ago! Well, see initial time for p is 0. but initial time for q is 0.4 as it starts 0.4 s later than p.
That is why for p time will be t-0=t, and for q t-0.4.
Got it?!
Jazak Allah khair!
Click to expand...

means whenever its like somthing is released later.... its gonna be t (for later released) = t - (the value) ?
 
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rawraxel1111

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Guys, In your opinion what are the hardest exams in the past papers? I'll really appreciate a reply. Thank you.
 
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sagar65265

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AlishaK said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_42.pdf
Is the graph for Q7 ii, an inverted mountain? Please reply asap!
Thanks!
Click to expand...

According to what the marking scheme says, the graph is more like an upward slope, then an inverted mountain:

If t = 0, the initial velocity = 5, so the graph must start at t = 0 and v = 5.

After that, from t till T1, the question states that the velocity is increasing. This can be shown on the graph as a curve going upwards to a positive maximum velocity.

From T1 till T2, the velocity decreases, so the graph should then show a decreasing velocity (however, the velocity must NOT be negative because the velocity only decreases, it doesn't change in direction) till the positive minimum, which they've asked you to calculate - it's the velocity at T2, 5 ms-1.

Lastly, since it says that the velocity increases after T2, the graph should show an increasing velocity curve.

This is what i've been able to gather from the mark scheme, which is not so strict on the values at which it goes to a maximum and minimum, etc.

Hope this Helped!
Good Luck for all your exams!
 
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