Mathematics: Post your doubts here!

[smzimran]

guyz anyone can help with this its pretty easy just not to my brain r8 now .. here is it just part b ! =p
Two particles A and B are moving on a smooth horizontal plane. The mass of A is km,
where 2 <
k < 3 , and the mass of B is m. The particles are moving along the same straight
line, but in opposite directions, and they collide directly. Immediately before they collide
the speed of
A is 2u and the speed of B is 4u. As a result of the collision the speed of A is
halved and its direction of motion is reversed.
(a) Find, in terms of
k and u, the speed of B immediately after the collision.
(b) State whether the direction of motion of
B changes as a result of the collision,explaining your answer.

(a)
m(A)u(A) + m(B)u(B) = m(A)v(A) + m(B)v(B)
(km)(2u) + (m)(-4u) = (km)(-u) + (m)v(B)
Eliminating m from both sides
2ku - 4u = -ku + v(B)
3ku - 4u = v(B)
v(B) = u(3k - 4)

(b)
We know that k lies in between 2 and 3, so lets assume k = 2.5
v(B) = u(3(2.5) - 4)
v(B) = u(7.5 - 4)
v(B) = 3.5u

v(B) is positive
and it was initially negative so
the direction changes

 

[smzimran]

Winter 2009 P31

http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9709 - Mathematics/&file=9709_w09_qp_31.pdf

Question 7 - Part 4 . I know the first inequality. Not sure about the Arg one.

Help, when you get some spare time please.

For the second inequality, sketch a half line with argument (1/4)π but in the second quadrant because the argument should start with u as the starting point
The shaded region is between this line and the first inequality

 

[DANGERBP]

(a)
m(A)u(A) + m(B)u(B) = m(A)v(A) + m(B)v(B)
(km)(2u) + (m)(-4u) = (km)(-u) + (m)v(B)
Eliminating m from both sides
2ku - 4u = -ku + v(B)
3ku - 4u = v(B)
v(B) = u(3k - 4)

(b)
We know that k lies in between 2 and 3, so lets assume k = 2.5
v(B) = u(3(2.5) - 4)
v(B) = u(7.5 - 4)
v(B) = 3.5u

v(B) is positive so,
direction does not change

apreaciate it .. but yo how did you know k lies bet 2 & 3 ???

 

[smzimran]

apreaciate it .. but yo how did you know k lies bet 2 & 3 ???

The question said so
Read the question again, i think you missed the point

 

[DANGERBP]

apreaciate it .. but yo how did you know k lies bet 2 & 3 ???

pahahah i just saw it r8 now lol !! forget it
but ya you ddnt explain they asked for it so what should we r8 ?

 

[DANGERBP]

The question said so
Read the question again, i think you missed the point

lol ^ ^^^
ya man just saw it .. but where the explain part ?

 

[smzimran]

lol ^ ^^^
ya man just saw it .. but where the explain part ?

Wait wait wait the direction changes as B was initially negative and now its positive i have edited my answer check it out!

And the explanation required is given!

 

[Zephyro]

For the second inequality, sketch a half line with argument (1/4)π but in the second quadrant because the argument should start with u as the starting point
The shaded region is between this line and the first inequality

The first inequality actually makes a vertical line at x = 1 right? I think in your diagram you considered 2 to be the y coordinate by mistake. Also, in your diagram you started the arg angle from the origin. I thought I should start the angle line from ( -2, 1 ) and focus it towards the right till it intersects my vertical x line? That triangle would represent my inequality right?

I've attached a sort of rough sketch. Its quite terrible, I know. Is that how its suppose to look and I assume that I should join the coordinate from u till it touches my vertical line to make my base aswell so it can be a complete triangle which represents the shaded area?

 

[s1294]

Use geometric progression

it gives different answer to when i calculate it the long way- can u please do it both ways n tell me which is the right answer?

 

[rideronthestorm]

S1 may/june 2010 varient 2 ... question 7 ... part iv and v ? how do u do it ? and please explain the basic concept behind it :/...,thanks

 

[umarashraf]

permutation n combination mein waqa koi logic hoti hai hai kia....????

 

[sea_princess]

can someone please explain how to solve this question

 

[salman haider]

why is the angle twice the angle of arg of z......whats the reason.. cant get it .............in nov 2003 q 7 part 3

 

[salman haider]

please someone help me with p3 nov 2004 q 9 part 2.........cant get it......thanks

 

[Silent Hunter]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_11.pdf

Q2 ??

 

[ffaadyy]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_11.pdf

Q2 ??

(1+ax)^6

1 + (ax)(6) + [(ax)^2(6)(5)]/2 + [(ax)^3(6)(5)(4)]/6

The coefficient of 'x' is -30.

6ax = -30x
6a = -30
a = -5

To find the coefficient of 'x^3', substitute the value of 'a = -5' in the equation ' [(ax)^3(6)(5)(4)]/6''.

[(ax)^3(6)(5)(4)]/6
[(5x)^3(6)(5)(4)]/6
-2500x^3

Therefore, the coefficient of 'x^3' is '-2500''.

 

[Silent Hunter]

(1+ax)^6

1 + (ax)(6) + [(ax)^2(6)(5)]/2 + [(ax)^3(6)(5)(4)]/6

The coefficient of 'x' is -30.

6ax = -30x
6a = -30
a = -5

To find the coefficient of 'x^3', substitute the value of 'a = -5' in the equation ' [(ax)^3(6)(5)(4)]/6''.

[(ax)^3(6)(5)(4)]/6
[(5x)^3(6)(5)(4)]/6
-2500x^3

Therefore, the coefficient of 'x^3' is '-2500''.

Thank you but how did you do the highlighted part?

 

[ffaadyy]

please someone help me with p3 nov 2004 q 9 part 2.........cant get it......thanks

r = 2i - j + 4k + s ( i + j - k)

As PQ is perpendicular to 'l', the dot product of PQ and 'l' will be equal to 0.

PQ . l = 0

Now we've to find the direction vector of PQ.

PQ = OQ - OP
PQ = (2i +0j - k) - OP

OP lies on 'l' therefore OP = 2i - j + 4k + s ( i + j - k).

OP = 2i - j + 4k + s ( i + j - k)
OP = (2 + s)i + (-1+s)j + (4-s)k

PQ = (2i - k) - OP
PQ = (2i + 0j - k) - (2 + s)i + (-1+s)j + (4-s)k
PQ = (-t)i + (1-t)j + (-5+t)k

Find out the value of 't' by equating the dot product of 'PQ' and 'l' to '0'

PQ . l = 0
(-t)i + (1-t)j + (-5+t)k . (1)i + (1)j + (-1)k = 0
- s + 1 - s + 5 - s = 0
6 = 3s
2 = s

Put the value of 's=2' in the equation 'OP = (2 + s)i + (-1+s)j + (4-s)k' to find the position vector of P.

OP = (2 + s)i + (-1+s)j + (4-s)k
OP = (2 + 2)i + (-1+2)j + (4-2)k
OP = 4i + 1j + 2k

Therefore, the position vector of 'P' is '4i + 1j + 2k'.

 

[ffaadyy]

Thank you but how did you do the highlighted part?

Using the formula of binomial expansion.

(1 + x)^n = 1 + [(x)( n )]/1 + [(x)^2]( n )(n-1)]/[(1)(2)] + [(x)^3]( n )(n-1)(n-2)]/[(1)(2)(3)] +[(x)^4]( n )(n-1)(n-2)(n-3)]/[(1)(2)(3)(4)] ........

 

[ffaadyy]

why is the angle twice the angle of arg of z......whats the reason.. cant get it .............in nov 2003 q 7 part 3

I don't think that they've asked us to explain why is the angle twice the angle of arg z; In Q7(iii) N03 P3, they've asked us to find the greatest value of arg s for points on this locus. Are you referring to the same question?