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Mathematics: Post your doubts here!

Obsidian Fl1ght

Obsidian Fl1ght

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ahmed abdulla said:
any
explaination please ?its easy looking at mark scheme!!
Click to expand...
Equations formed are basically resolving of the forces. First ie 12*cos(80)= Q - P*cos(60) is resolving in the x axis.
Consider all forces and their individual components first.
In x axis:
For Q: It's Q.
For 12N (aka the Resultant) = 12cos(80) N
For P: Pcos(60) N
(60 = 180 - (80 + 40) )
In y axis:
For Q: ZERO - Nada - Zilch.
For 12N (aka the Resultant) = 12sin(80) N
For P: Psin(60) N
Now equate all of these.
Resultant = vector sum of force components
In x axis:
12cos(80) = Q - Pcos(60) (i)
Similarly, y axis:
12sin(80) = Pcos(60) (ii)
First find P from (ii).
Substitute it into (i)
Voila u get Q.
 
ahmed abdulla

ahmed abdulla

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Obsidian Fl1ght said:
Equations formed are basically resolving of the forces. First ie 12*cos(80)= Q - P*cos(60) is resolving in the x axis.
Consider all forces and their individual components first.
In x axis:
For Q: It's Q.
For 12N (aka the Resultant) = 12cos(80) N
For P: Pcos(60) N
(60 = 180 - (80 + 40) )
In y axis:
For Q: ZERO - Nada - Zilch.
For 12N (aka the Resultant) = 12sin(80) N
For P: Psin(60) N
Now equate all of these.
Resultant = vector sum of force components
In x axis:
12cos(80) = Q - Pcos(60) (i)
Similarly, y axis:
12sin(80) = Pcos(60) (ii)
First find P from (ii).
Substitute it into (i)
Voila u get Q.
Click to expand...
tanks bro
 
Pie-man

Pie-man

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Obsidian Fl1ght

Obsidian Fl1ght

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Pie-man said:
well can you please explain to me number 6 (ii) i know the answer and got it right but by complete luck and can't understand how he got it in the marking scheme.

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Click to expand...
Kay so:
We have Vg/Vb = 2.55 or Vg = 2.55 Vb

First we consider the simplest thing (to consider) aka energy change from X to ground.
Gain in KE = Loss in PE
So 1/2 (m) (Vg square - Vx square) = (m) (g) (H)
Since object starts from rest, Vx is zero.
Thus: 1/2 (m) (2.55Vg square) = (m) (10) (H) (i)
Note that we substitute Vg as 2.55 Vb

Now look at energy change from X to B: (We couldn't care less about energy change from B to ground. I don't in any case. Coz to find H, we need two equations of H and another variable ie Vb)
So we have:
1/2 (m) (Vb square - Vx square) = (m) (g) (H-h)
1/2 (m) (Vb square) = (m) (10) (H - 2.2) (ii) (h = 2.2 we get from prev. ans.)

Substitute... and observe the magic errr get the answer.
 
Pie-man

Pie-man

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Obsidian Fl1ght said:
Kay so:
We have Vg/Vb = 2.55 or Vg = 2.55 Vb

First we consider the simplest thing (to consider) aka energy change from X to ground.
Gain in KE = Loss in PE
So 1/2 (m) (Vg square - Vx square) = (m) (g) (H)
Since object starts from rest, Vx is zero.
Thus: 1/2 (m) (2.55Vg square) = (m) (10) (H) (i)
Note that we substitute Vg as 2.55 Vb

Now look at energy change from X to B: (We couldn't care less about energy change from B to ground. I don't in any case. Coz to find H, we need two equations of H and another variable ie Vb)
So we have:
1/2 (m) (Vb square - Vx square) = (m) (g) (H-h)
1/2 (m) (Vb square) = (m) (10) (H - 2.2) (ii) (h = 2.2 we get from prev. ans.)

Substitute... and observe the magic errr get the answer.
Click to expand...
well this will sound really stupid...probably, why did we use (H-h) and not h immediately aren't i supposed to use the height from the ground to point B?
 
Obsidian Fl1ght

Obsidian Fl1ght

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Pie-man said:
well this will sound really stupid...probably, why did we use (H-h) and not h immediately aren't i supposed to use the height from the ground to point B?
Click to expand...
Sometimes people confuse stupidity with curiosity...
Actually, you can...
Depends on you.

If you follow that route, you'll get:
(m)(g)(h) = 1/2 (m)( (2.55Vb)square - (Vb)square) where 2.55 Vb is Vg
You simplify that to get Vb.
Then you find Vg by putting value of Vb into Vg = 2.55 Vb
Then you simply take energy stuff from H to ground. Put Vg as what you've found.

So yeah... you can do that. Definitely. Didn't think of that. Thanks man ahem Pie-man.
 
littlecloud11

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Alice123 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_63.pdf Q6 iii littlecloud11 if u arent busy...... evn my maths teacher cudnt solve this.......
Click to expand...

Okay, this is a tricky one.
Screenshot - 5_10_2013 , 9_22_39 PM.jpg

See the picture. The red seats highlight the ones that Mrs. Lin can occupy. She has to sit directly behind a student, so if she occupies those seats then it's possible for a student to be placed in front of her. The reason I kept one seat in the second row uncolored is because one of the seats in the front row is occupied my Mrs. Brown. A student can only take two seats in the front row.
So for Mrs. Lin the possible combinations are 10P1
She can sit behind any of the 5 students, so it's 5P1, the reason you don't count the possible seats for the student is because it doesn't matter where they are placed as long as one of them is in front of Mrs. Lin
Mrs. Brown has to sit in one of the front seats so 3P1
And finally after Mrs. Lin, a student and Mrs. brown has been placed there are 9 remaining passengers and 11 remaining seats. There is no restrictions. So it's 11P9

Total no. of ways = 10P1 * 3P1* 5P1 * 11P9

If all 12 passangers are seated randomly = 14P12
so probability = 10P1 * 3P1* 5P1 * 11P9/ 14P12 =.0687

Hope you get this!
 
Pie-man

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Obsidian Fl1ght said:
Sometimes people confuse stupidity with curiosity...
Actually, you can...
Depends on you.

If you follow that route, you'll get:
(m)(g)(h) = 1/2 (m)( (2.55Vb)square - (Vb)square) where 2.55 Vb is Vg
You simplify that to get Vb.
Then you find Vg by putting value of Vb into Vg = 2.55 Vb
Then you simply take energy stuff from H to ground. Put Vg as what you've found.

So yeah... you can do that. Definitely. Didn't think of that. Thanks man ahem Pie-man.
Click to expand...
oh good then my only mistake was not noticing that Vx is 0 ( hope i don't do these silly mistakes on monday ) :D thanks for the help
 
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