IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: mousa on April 07, 2010, 06:08:26 pm

Title: M1 HELP NEEDED, RIGHT NOW!!
Post by: mousa on April 07, 2010, 06:08:26 pm: Hello everyone, i need help with those couple of questions.
Please explain your steps in detail and clearly.

Q 7) ii .....Nov 2008

Q6) Nov 2004

Thanks.
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 07, 2010, 06:24:37 pm: Nov 08
Q7

integrate acc, ull get velocity, integrate velocity, ull get distance...

double integration of (10 ? 0.3t)

v=dat=10t-0.3t²/2 +5
s=dvt= 10t²/2 - 0.3 t³/2X3 +5t
= 5t² - 0.3t³/6 + 5t
substitute t=2.5 (since 0.5sec taken to travel uptil A)
s=42.97
total s= 42.97+1.25
=44.2m
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 07, 2010, 06:40:21 pm: nov 2004

one of my fav qs

i) F<= uR
<=0.75 8000
F<= 6000

ii) F max on top box= uR=.4 x4000=1600=400a a<=4

P-6000=3200
pmax=9200
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: mousa on April 07, 2010, 07:19:06 pm: Quote from: nid404 on April 07, 2010, 06:40:21 pm
nov 2004

one of my fav qs

i) F<= uR
<=0.75 8000
F<= 6000

ii) F max on top box= uR=.4 x4000=1600=400a a<=4

P-6000=3200
pmax=9200

i am confused about the 2nd part, can you explain more?
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 07, 2010, 07:25:39 pm: Fmax on top box<=1600
F=ma
hence a<=4

P-6000=ma
Pmax-6000=800X4
P=3200+6000
=9200

If there has to be no slipping in the system, both should move together....and that will happen only when they both have the same accn
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: mousa on April 07, 2010, 07:36:14 pm: Quote from: nid404 on April 07, 2010, 07:25:39 pm
Fmax on top box<=1600
F=ma
hence a<=4

P-6000=ma
Pmax-6000=800X4
P=3200+6000
=9200

If there has to be no slipping in the system, both should move together....and that will happen only when they both have the same accn

Thank you niddy. :)
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 07, 2010, 07:36:40 pm: no prob
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on April 08, 2010, 11:13:21 am: could someone answer this plzzz rili urgent :S"
a stone is dropped from the top of a tower. one second later another stone is throw vertically downwards
from the same point witha velocity of 14m/s .
if theyb hit the ground together find the height of the tower.

thx a ;Dloot
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 08, 2010, 04:45:40 pm: assuming the stone is of the same mass

s= ut+1/2 gt²
s=14(t+1)+ 1/2 g(t+1)²
equate the two

ut+1/2 gt²=14(t+1)+ 1/2 g(t+1)² u=0
So
1/2 gt²=14(t+1)+ 1/2 g(t+1)²
find the value of t
substitute to get s
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on April 08, 2010, 05:00:45 pm: Quote from: nid404 on April 08, 2010, 04:45:40 pm
assuming the stone is of the same mass

s= ut+1/2 gt²
s=14(t+1)+ 1/2 g(t+1)²
equate the two

ut+1/2 gt²=14(t+1)+ 1/2 g(t+1)² u=0
So
1/2 gt²=14(t+1)+ 1/2 g(t+1)²
find the value of t
substitute to get s

thx a bunch man! life saver u r!
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 08, 2010, 05:29:44 pm: Ur welcome
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on April 14, 2010, 01:16:55 pm: hello
could someone help me with this question please!!!!!!!!!
m1 edexcel AS
jan 2009
Q3,b,c
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: astarmathsandphysics on April 14, 2010, 04:55:56 pm: 2 hours
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 14, 2010, 05:00:00 pm: Quote from: halosh92 on April 14, 2010, 01:16:55 pm
hello
could someone help me with this question please!!!!!!!!!
m1 edexcel AS
jan 2009
Q3,b,c

put the paper up plz
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on April 14, 2010, 05:07:17 pm: there u go
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 14, 2010, 05:37:53 pm: also the ms if possible...i'd like to cross check...cuz i do cie
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on April 14, 2010, 05:46:26 pm: By the way i want to ask u..which one is tougher cie or edexcel?
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 14, 2010, 06:12:50 pm: Quote from: halosh92 on April 14, 2010, 05:46:26 pm
By the way i want to ask u..which one is tougher cie or edexcel?

I've just solved old M1 edexcel papers...considerably easier I'd say :-\
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 14, 2010, 06:22:53 pm: srry for the delay...
glad i ws right ;D

ok...so basically momentum is a vector quantity so u have to take into the direction of motion

so in collision where momentum is conserved

m1u1+m2u2=m1v1+m2v2

km2u - 4mu = -kmu + mv
v=u(3k - 4)
v>0 that's because k >2 if k>2 then v>+2u implies v>0
change in sign....earlier mometum=? 4mu now=+u(3k ? 4) implying change in direction

ok if k=7/3
impulse is the change in momentum
m[u(3k - 4)] - (- 4mu)
3muk-4mu+4mu
3mu X7/3 = 7mu
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on April 14, 2010, 06:30:04 pm: Quote from: nid404 on April 14, 2010, 06:22:53 pm
srry for the delay...
glad i ws right ;D

ok...so basically momentum is a vector quantity so u have to take into the direction of motion

so in collision where momentum is conserved

m1u1+m2u2=m1v1+m2v2

km2u ? 4mu = ?kmu + mv
v=u(3k ? 4)
v>0 that's because k >2
change in sign....earlier mometum=? 4mu now=+u(3k ? 4) implying change in direction

ok if k=7/3
impulse is the change in momentum
m[u(3k ? 4)] - (? 4mu)
3muk-4mu+4mu
3mu X7/3 = 7mu

no problem could you just tell me whats the "?" in ure explanation.....and plz explain this part for me "v>0 that's because k >2"
thx ;D ;D
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 14, 2010, 06:36:01 pm: edited my post
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on April 14, 2010, 06:43:46 pm: Quote from: nid404 on April 14, 2010, 06:36:01 pm
edited my post

thx a bunch dude ;D ;D
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: nid404 on April 14, 2010, 06:45:48 pm: Quote from: halosh92 on April 14, 2010, 06:43:46 pm
thx a bunch dude ;D ;D

dudet :P
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: sameee on April 17, 2010, 05:26:07 pm: Hello everyone, i need help with those couple of questions: In the cie M1 book pg: 142, Q7 and 8,
and the question on page 143.

Thanks in advance
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: astarmathsandphysics on April 17, 2010, 06:48:53 pm: Dont have the book post the Questions
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on April 17, 2010, 07:32:31 pm: M1 edexcel AS maths
2004 june
q6 d
q6 e

could someone explain this step by step i dont get anything form the ms.
THXXX
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: astarmathsandphysics on April 17, 2010, 07:40:40 pm: That question doesnt have a d or e
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: Saladin on April 17, 2010, 09:12:23 pm: (a) $2ukm+-4um=-ukm+mv$

$3ukm-4u=v$

$v=u(3k-4)$

(b) We have assumed that $v$ is towards the right. And therefore, as we know that $k>2$ , let us use this to determine whether the direction changes or remains the same.

If $k>2$

Then $v>0$ , as we can see from calculation, in (a).

(c) $R(\rightarrow)$ using B

Therefore, $I=m(v-u)$

$m(u(3k-4)--4u)$

Thus $7mu$
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: Saladin on April 17, 2010, 09:33:16 pm: oops, i guess i was a bit late huh?? :-\
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on April 19, 2010, 03:53:03 pm: Quote from: astarmathsandphysics on April 17, 2010, 07:40:40 pm
That question doesnt have a d or e

sorry :S
its quetsion 7
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: astarmathsandphysics on April 19, 2010, 11:29:19 pm: d)After 7 s speed =8,4m/s
F=ma for P
uR=ma so -2/7*4g=4a so a=-2.8m/s^2
v=u+at
0=8.4-2.8a so a=3
e)F=ma
40-2/7*6g=40-2/7*6*9.8=40-16.8=23.2=6a
a=23.2/5=3.973m/s^2
v=u+at=8.4+3*3.87=20m/s
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: Phosu on May 21, 2010, 05:24:26 pm: jun 09 5bi and ii)
i rote the reaction force as R = 1.1gCos a - PSin a
can someone plz explain how to do it?
cuz the force of P that im getting turns out to be R in the ms
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: halosh92 on May 21, 2010, 05:50:31 pm: Quote from: Phosu on May 21, 2010, 05:24:26 pm
jun 09 5bi and ii)
i rote the reaction force as R = 1.1gCos a - PSin a
can someone plz explain how to do it?
cuz the force of P that im getting turns out to be R in the ms

R+ P sin 50 = 15g
R= 15g -P sin 50
P cos 50 - friction
P cos 50= 0.2 ( 15g- P sin 50)
P cos 50= 29.4- P 0.2 sin 50
P cos 50 + P 0.2sin 50 = 29.4
P( cos 50 + 0.2 sin 50) = 29.4
P= 37 N
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: Phosu on May 21, 2010, 05:54:39 pm: im soooooooooo srry i rote the rong month, its jan 09
Title: Re: M1 HELP NEEDED, RIGHT NOW!!
Post by: cooldude on May 21, 2010, 06:49:17 pm: Quote from: halosh92 on May 21, 2010, 05:50:31 pm
R+ P sin 50 = 15g
R= 15g -P sin 50
P cos 50 - friction
P cos 50= 0.2 ( 15g- P sin 50)
P cos 50= 29.4- P 0.2 sin 50
P cos 50 + P 0.2sin 50 = 29.4
P( cos 50 + 0.2 sin 50) = 29.4
P= 37 N

let coefficient of friction=c and alpha=x
Psinx+1.1gcosx=R
resolving horizontally (parallel to the plane)-->
Pcosx+F=1.1gsinx, F=cR, F=0.5R
Pcosx+0.5(Psinx+1.1gcosx)=1.1gsinx
Pcosx+0.5Psinx+0.55gcosx=1.1gsinx
P(cosx+0.5sinx)=1.1gsinx-0.55gcosx
P(cosx+0.5sinx)=11sinx-5.5cosx
P=2