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Mathematics: Post your doubts here!

Esme

Esme

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

5 ii ..?
stuck in it
Click to expand...

For particle P:
u=20, a=-10,
s=ut+0.5at^2
s=20t+(0.5 x -10 x t^2)
s=20t - 0.5t^2

For Particle Q:
u=25, a=-10, time=t-4
s=25(t-4) - 0.5(t-4)^2
After simplifying you get
s=29t + 10.8 - 5t^2

Now equate both the distances :
20t-0.5t^2 = 29t + 10.8 -5t^2
After solving this you'll get t=1.2

For Particle P:
v=u+at
v=20+(-10 x 1.2) = 8m/s

For Particle Q:
v=25-10(t-0.4)
v=25-10(1.2-0.4) = 17m/s

I hope you got it :)
 
Anika Raisa

Anika Raisa

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Khalid Mohammed Okiely said:
The one in the first part, really sorry I forgot to specify.
Click to expand...
Theres nthin to be sorry!
Well 1 lame thing but it works out: take out ur O level bks n go through some geometry n trigonometry basic stuffs or simply surf the net! n practice d simple sums!It worked out 4 me wen i realised i gt stuck in some kind of sums!

1 mr thng read d question thoroughly ... many hints are there already!uderline em!so that u cn mentally work out solutions like puzzle!Use all info in Q.
I gs deres no othr way to clear such general doubts!

As 4 dis sum: D solution is:
u c AMN forms a segment n we knw arc length,s=radius*theta
so here MN=r*x
nw perimeter of the segment=AN+AM+MN=2r+rx
Nw considering, the trianfle ADN
ang AND=x
[Cz ang AND n ang NAM r alternate angs]
hence, using trig for triangle AND
we cn write: sinx=a/r
=> a=rsinx
Nw using the info that perimeter of segment is half of perimeter of rectangle:
Perimeter of rectangle= 2*(3a+a)=8a
Nw perimeter of segment hence= 4a
We had calculated b4 that perimeter of segment=2r+rx
So we gt 2 prsn 4 d perimeter so lets equate both equns n so we get:
r(2+x)=4(rsinx)
so sinx=(2+x)/4 ....
Hope i cud help!
so I thnk practising mr sums lyk dis or goin' through mensuration,trigonometry n geometry sums frm O levels may clear ur doubt! :)
 
Khalid Mohammed Okiely

Khalid Mohammed Okiely

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Anika Raisa said:
Theres nthin to be sorry!
Well 1 lame thing but it works out: take out ur O level bks n go through some geometry n trigonometry basic stuffs or simply surf the net! n practice d simple sums!It worked out 4 me wen i realised i gt stuck in some kind of sums!

1 mr thng read d question thoroughly ... many hints are there already!uderline em!so that u cn mentally work out solutions like puzzle!Use all info in Q.
I gs deres no othr way to clear such general doubts!

As 4 dis sum: D solution is:
u c AMN forms a segment n we knw arc length,s=radius*theta
so here MN=r*x
nw perimeter of the segment=AN+AM+MN=2r+rx
Nw considering, the trianfle ADN
ang AND=x
[Cz ang AND n ang NAM r alternate angs]
hence, using trig for triangle AND
we cn write: sinx=a/r
=> a=rsinx
Nw using the info that perimeter of segment is half of perimeter of rectangle:
Perimeter of rectangle= 2*(3a+a)=8a
Nw perimeter of segment hence= 4a
We had calculated b4 that perimeter of segment=2r+rx
So we gt 2 prsn 4 d perimeter so lets equate both equns n so we get:
r(2+x)=4(rsinx)
so sinx=(2+x)/4 ....
Hope i cud help!
so I thnk practising mr sums lyk dis or goin' through mensuration,trigonometry n geometry sums frm O levels may clear ur doubt! :)
Click to expand...
Many thanks for the answer! <3

And I will go through the basics real quick whenever I stumble across such types of questions. I have the habit of freaking out and leaving this question to the end and sometimes I get it wrong :/. Ty again ^^
 
AlishaK

AlishaK

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mominzahid said:
Hi... can someone please help me with these questions? got my mechanics paper tomorrow. :/
q5 (ii) and q6(i)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_42.pdf
Click to expand...
Using s=ut+0.5at^2, Hp=12t+0.5(-10)t^2 => Hp=12t-5t^2
Hq=7t+0.5(-10)t^2 => Hq=7t-5t^2
Given that 3Hp=8Hq, therefore, 3*(12t-5t^2)=8*(7t-5t^2), I hope you can take it further from here. t=0 or t=0.8s
Now using v=u+gt,
V of P at t=0.8s: v=12+(-10)(0.8), V of Q at t=0.8s: v=7+9-10)(0.8)....Hopefully it hits your brain.
 
AlishaK

AlishaK

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mominzahid said:
Hi... can someone please help me with these questions? got my mechanics paper tomorrow. :/
q5 (ii) and q6(i)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_42.pdf
Click to expand...
Q6,i) See triangle ABR, use cosine rule,
Cos B = 80^2 + 50^2 - 50^2/ (2*80*50).....B=36.9 deg.
both the bas angles A and B are the same, therefore, resolving forces vertically at R: Tsin36.9+Tsin36.9=6 (as T in both the strings is the same)
2Tsin36.9=6....T=4.99~5.0 N

Cheers! Remember me in your prayers. Jazak Allah khair! :)
 
AlishaK

AlishaK

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Mayedah said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_41.pdf
Question no 6 part (iii) why do take speed for one as negitive and other as positive while both will travel downwards?
Click to expand...
Because it says, 'at the instant when the string breaks', that means as P has a greater weight, it will fall towards the ground (downwards) and Q will travel upwards. therefore the velocities will have opposite directions. as P travelling down with for eg +V nd Q upwards with -V...or vice versa.
Jazak Allah khair. Remember me in ur prayers. And did u get it?!
 
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