Post by: Saladin on May 04, 2010, 12:43:39 pm
I do Edexcel M1.
Post by: halosh92 on May 04, 2010, 01:48:48 pm
the string passes over a smooth fixed pulley . the system is released from rest with both masses a distance of 2m
above a horizontal floor. find how long it takes for particle Q to hit the floor.
assuming particle P does not reach the pulley find its greatest height above the floor
i got the time which is 1.11 but i cant get the distance.
Post by: Saladin on May 04, 2010, 01:55:12 pm
Post by: halosh92 on May 04, 2010, 04:07:07 pm
Post by: halosh92 on May 04, 2010, 04:10:08 pm
??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ???
Post by: astarmathsandphysics on May 04, 2010, 04:18:46 pm
For 1 kg t-g=a
Add there two equations g=3a so a=g/3
When string breaks particle falls under gravity so a=g
Post by: astarmathsandphysics on May 04, 2010, 04:23:11 pm
Post by: halosh92 on May 04, 2010, 04:26:58 pm
Height=1.5+0.5*g/3*0.5^2 then use this height with s=1/2at^2 to find find t
how did u exactly get this?
0.5*g/3*0.5^2 and for s=1/2at^2 do we use 9.8 for the acceleration?
thx
Post by: halosh92 on May 04, 2010, 04:56:37 pm
find:
a) the distance the particles haved moved when the string broke
b) the velocity of the particles when the string breaks
c) the furthur time that elapses before A reaches the floor, given that the table is 0.9m high.
how to do part c ??
Post by: Saladin on May 04, 2010, 04:59:10 pm
Post by: halosh92 on May 04, 2010, 05:02:13 pm
can you please send the exact question paper. I need to see that.
its from my textbook actually :S
Post by: Khey [Rainbow] on May 04, 2010, 05:13:02 pm
By modelling the crate as a particle,
(a) Find the value of [mue], the cofficient of frction between the crate and the ground.
The rope remains at 30 degree to the ground but the tension init is increased to 200N.
(b) Find the acceleration of the crate.
Post by: Saladin on May 04, 2010, 05:14:09 pm
Post by: Saladin on May 04, 2010, 05:17:57 pm
two particles A and B of masses 0.5kg and 0.7kg respectively are connected by a light inextensible string. particle A rests on a smooth horizontal table 1m from a fixed smooth pulley and B hangs freely 0.75m above the floor. the system is released from rest with the string taut. after 0.5s the string breaks.
find:
a) the distance the particles haved moved when the string broke
b) the velocity of the particles when the string breaks
c) the furthur time that elapses before A reaches the floor, given that the table is 0.9m high.
how to do part c ??
u can also send me the book name and page number.
Post by: halosh92 on May 04, 2010, 05:28:45 pm
u can also send me the book name and page number.
heinemann modular mechanics for edexcel AS and A level
pg113
q11
Post by: nid404 on May 04, 2010, 05:31:48 pm
A large crate of mass 40kg lies at rest on rough horizontal ground. One end of the rope is attached to the crate and the rope makes an angle of 30 with ground as shown in the diagram. The tension of the rope is 150N and the crate is on the of moving along the plane.
By modelling the crate as a particle,
(a) Find the value of [mue], the cofficient of frction between the crate and the ground.
The rope remains at 30 degree to the ground but the tension init is increased to 200N.
(b) Find the acceleration of the crate.
a) u= F/R
F=150cos30
R=400-150sin30
u=130/325
=0.4
b) horizontal force=Tcos30=200cos30=173.3N
173-F(uR)=40a
173-130=40a
a=3.4m/s2
Post by: Saladin on May 04, 2010, 05:33:39 pm
Post by: Khey [Rainbow] on May 04, 2010, 05:53:13 pm
but both of u have differnt answers so m confused bout who's answer is right..but its alryt i will ask my teacher
once again Thanks alot ;D
Post by: nid404 on May 04, 2010, 05:54:19 pm
i dunno....I think R= mg-150sin30
Post by: Saladin on May 04, 2010, 05:58:17 pm
Post by: MasterMath on May 04, 2010, 07:22:31 pm
Post by: MasterMath on May 04, 2010, 07:26:12 pm
Post by: halosh92 on May 04, 2010, 08:03:17 pm
two particles P and Q of masses 1kg and 2kg respectively are hanging from the ends of alight inextensible string which passes over a smooth fixed pulley. the system is released from rest with both particles a distance 1.5m above a floor. when the masses have been moving fir 0.5 s the string breaks. find the furthur time that elapses before P hits the floor
??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ???
can someone solve this in more details please??
Post by: Sue T on May 05, 2010, 10:18:25 am
can ny1 solve the last question 7(f)?
Post by: halosh92 on May 05, 2010, 06:07:23 pm
where X=arcsin 0.6
a second particle B of mass "2m" hangs freely attached to a light inextensible string which passes over a smooth pulley fixed at P. the other end of the string is attached at A . the coefficient of friction between A and the plane is 1/4.
B is initially hanging 1m above the ground and A is 2m from the pulley. when the system is released from rest with the string taut A moves up the plane
a) find the initial acceleration of A
b) calculate the total distance moved by A before it first comes to rest.
Post by: cashem'up on May 05, 2010, 07:16:11 pm
Thanks to "nid404" and "the mysterious dude"
but both of u have differnt answers so m confused bout who's answer is right..but its alryt i will ask my teacher
once again Thanks alot ;D
no i am pretty sure nid404 is correct because to find the the Normal Contact u have to find the resulting vertical downward force that box exerts on ground and then the normal will be same as that.
If u also think logically when ur pulling something at such an angle u will be lifting some of the weight off........ dont take this literally but logically so less force will be exerted on the ground...... u get wat i am saying
Post by: cashem'up on May 05, 2010, 07:26:22 pm
two particles P and Q of masses 1kg and 2kg respectively are hanging from the ends of alight inextensible string which passes over a smooth fixed pulley. the system is released from rest with both particles a distance 1.5m above a floor. when the masses have been moving fir 0.5 s the string breaks. find the furthur time that elapses before P hits the floor
??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ???
hey halosh is the answer possibly 1.01 seconds
Post by: cashem'up on May 05, 2010, 07:50:33 pm
a particle A of mass "m" which can move on the rough surface of an inclined plane at an angle X to the horizontal.answer to (a) should be 4m/s2
where X=arcsin 0.6
a second particle B of mass "2m" hangs freely attached to a light inextensible string which passes over a smooth pulley fixed at P. the other end of the string is attached at A . the coefficient of friction between A and the plane is 1/4.
B is initially hanging 1m above the ground and A is 2m from the pulley. when the system is released from rest with the string taut A moves up the plane
a) find the initial acceleration of A
b) calculate the total distance moved by A before it first comes to rest.
and to (b) 1.99 m
if iam correct i will explain , dont want to confuse you with wrong answrs.......:P
Post by: halosh92 on May 05, 2010, 08:10:15 pm
answer to (a) should be 4m/s2
and to (b) 1.99 m
if iam correct i will explain , dont want to confuse you with wrong answrs.......:P
this is the exact answer:
3.92 m/s
1.5 m
thankyou ;D
Post by: immortal on May 05, 2010, 11:13:46 pm
here u go.using this method i got a different ans.
u=0.3997
200cos30-(u*(40g-200sin30)=40a
a=1.33
If i replace da red wit 150sin30, i get ur ans.
Post by: cashem'up on May 06, 2010, 08:16:07 am
this is the exact answer:
3.92 m/s
1.5 m
thankyou ;D
this is the exact answer:hey halosh i got it..... i initially took g=10 lolzz.....
3.92 m/s
1.5 m
thankyou ;D
now we have a diagram something like this
ok now for the first part it will be easier to take the system as a whole
so lets find
the total resisting force
1.Weight is 9.8m
2. downward force due to weight is 0.6x9.8m = 5.88m
3. due to friction is F= 1/4 x 0.8(cos of arcsin0.6) x 9.8m= 1.96m
Total Resisting- 7.84m
Total Forward = 2mg = 19.6m
therefore resultant is
19.6m - 7.84m = 3ma
11.76=3a
a=3.92 m/s2
Post by: halosh92 on May 06, 2010, 10:27:19 am
hey halosh i got it..... i initially took g=10 lolzz.....
now we have a diagram something like this
ok now for the first part it will be easier to take the system as a whole
so lets find
the total resisting force
1.Weight is 9.8m
2. downward force due to weight is 0.6x9.8m = 5.88m
3. due to friction is F= 1/4 x 0.8(cos of arcsin0.6) x 9.8m= 1.96m
Total Resisting- 7.84m
Total Forward = 2mg = 19.6m
therefore resultant is
19.6m - 7.84m = 3ma
11.76=3a
a=3.92 m/s2
thx alot :)
sorry for the trouble
;D ;D
Post by: cashem'up on May 06, 2010, 10:41:35 am
Post by: Sue T on May 06, 2010, 12:42:36 pm
jan 2007:
can ny1 solve the last question 7(f)?
???no1? ???
Post by: halosh92 on May 06, 2010, 01:39:40 pm
At the same time ship B has position vector (-i + 4j)km relative to the lighthouse and velocity (2i + 3j)km/h . find, after "t"h
a) the position vector of A relative to the lighthouse.
b)the position vector of B relative to the lighthouse.
c)the position vector of A relative to B.
d) find the time when A is due north of B.
can someone just solve part (d) how to come about such questions wat do they mean exactly?
Post by: Phosu on May 06, 2010, 02:25:30 pm
(a) the speed of P,
(b) the direction of motion of P, giving your answer as a bearing.
At time t = 0, P is at the point A with position vector (7i – 10j) m relative to a fixed origin O. When t = 3 s, the velocity of P changes and it moves with velocity (ui + vj) m s–1, where u and v are constants. After a further 4 s, it passes through O and continues to move with velocity (ui + vj) m s–1.
(c) Find the values of u and v.
(d) Find the total time taken for P to move from A to a position which is due south of A.
plz tell me how to solve part d
Post by: halosh92 on May 06, 2010, 02:44:08 pm
A particle P is moving with constant velocity (–5i + 8j) m s–1. Findhow did u do part (c) first of all?
(a) the speed of P,
(b) the direction of motion of P, giving your answer as a bearing.
At time t = 0, P is at the point A with position vector (7i – 10j) m relative to a fixed origin O. When t = 3 s, the velocity of P changes and it moves with velocity (ui + vj) m s–1, where u and v are constants. After a further 4 s, it passes through O and continues to move with velocity (ui + vj) m s–1.
(c) Find the values of u and v.
(d) Find the total time taken for P to move from A to a position which is due south of A.
plz tell me how to solve part d
Post by: Phosu on May 06, 2010, 03:02:13 pm
(7i-10j) +(-5+8j)3 = -8i+14j
after the boat reaches this vector position it changes speed, and travels for 4s until it reaches origin O which is O so
0=(-8i+14j)+(ui+vj)4
0=(-8+4u)i+(14+4v)j
now solve seperately
-8+4u=0 u=2
14+4v=0 v=-3.5
new velocity (2i-3.5j)
Post by: halosh92 on May 08, 2010, 09:15:08 pm
at 2.p.m. the position vector relative to a lighthouse of ship A is (3i+j)km and its velocity is (-i + 5j)km/hcould someone plzz solve this ?
At the same time ship B has position vector (-i + 4j)km relative to the lighthouse and velocity (2i + 3j)km/h . find, after "t"h
a) the position vector of A relative to the lighthouse.
b)the position vector of B relative to the lighthouse.
c)the position vector of A relative to B.
d) find the time when A is due north of B.
can someone just solve part (d) how to come about such questions wat do they mean exactly?
:-[ :-\
Post by: T.Q on May 08, 2010, 09:48:26 pm
could someone plzz solve this ?
:-[ :-\
(i) component of A will be equal to the (i) component of B
Post by: halosh92 on May 08, 2010, 09:53:02 pm
(i) component of A will be equal to the (i) component of B
ok but is there some kind of rule for these questions they come very often...i mean i dont rili understand when the say
"north of B" or "east of B"
shouldnt one of the components be = 0 in such cases :-\
Post by: astarmathsandphysics on May 09, 2010, 06:39:21 am
Post by: halosh92 on May 09, 2010, 08:06:26 pm
There is a page on my website astarmathsandphysics.com>a level maths notes> relative positions more on relative positions
thankyou :)
Post by: Uchia on May 10, 2010, 10:11:24 am
and thanks
Post by: astarmathsandphysics on May 10, 2010, 10:41:05 am
Post by: Saladin on May 10, 2010, 11:05:54 am
2007 january paper question 7 last part
and thanks
I am quite sure its Edexcel.
We need to find out the final velocity that the particle reaches after the other particle has fallen.
Now we need to find the time it takes for it to stop, as the resultant force is not towards going down. Thus we have to find the acceleration.
Using this, calculate the time taken for the particle to come to rest.
The time taken to go up, is the time taken to go down, as there is same acceleration.
Thus the total time taken is
Thus the total time taken is 0.51 seconds,
Post by: Uchia on May 10, 2010, 11:46:08 am
Post by: Saladin on May 10, 2010, 12:38:35 pm
lol yeah thnxs i used to solve it in the same way but the last step i am always confused about cz some old questions you didn't need to multiply by 2 but this question here required to do so y??
Because it said that how long it takes for it to be taught again. That means up and down.
Post by: Phosu on May 10, 2010, 01:37:20 pm
i just straight off rote R=6gCos30+PSin30
but marking scheme has some other calculations, can someone plz explain how to get that.
Post by: nid404 on May 10, 2010, 02:17:52 pm
Post by: astarmathsandphysics on May 10, 2010, 02:19:17 pm
Post by: melony on May 14, 2010, 10:49:55 am
???
thank yu! :)
Post by: nid404 on May 14, 2010, 11:40:36 am
check the diagram
A=7+3.2=10.2
B=3.83N
C=5.2N
D= 3N
Resolving these
A-C=5N
B-D=0.83N
Now find the resultant
R2=
R=5.09N
tan
Post by: melony on May 14, 2010, 08:15:21 pm
;)
Post by: Fenomenoe9 on May 15, 2010, 09:04:46 am
Its in June 2008 M1 paper question 5.
Two forces P and Q act on a particle at a point O. The force P has magnitude 15 N and
the force Q has magnitude X newtons. The angle between P and Q is 150°, as shown in
Figure 1. The resultant of P and Q is R.
Given that the angle between R and Q is 50°, find
(a) the magnitude of R, (4)
(b) the value of X. (5)
the diagram is something like this its in the question :
P
\
\
\
\
\._________________ Q
O
angles is 150 ^^
Post by: astarmathsandphysics on May 15, 2010, 09:22:56 am
Post by: Fenomenoe9 on May 15, 2010, 10:00:13 am
Post by: wakemeup on May 15, 2010, 07:08:50 pm
Q7(d)
Post by: astarmathsandphysics on May 15, 2010, 07:14:04 pm
http://astarmathsandphysics.com/a_level_maths_notes/M1/a_level_maths_notes_m1_least_distance_between_two_moving_points.html
Post by: melony on May 15, 2010, 09:53:09 pm
Quote
Can any1 help in this question... please.Hey! i knw astars alredy dunt buh i gav it a shot usin a diff method if it helps.. :)
Its in June 2008 M1 paper question 5.
Two forces P and Q act on a particle at a point O. The force P has magnitude 15 N and
the force Q has magnitude X newtons. The angle between P and Q is 150°, as shown in
Figure 1. The resultant of P and Q is R.
Given that the angle between R and Q is 50°, find
(a) the magnitude of R, (4)
(b) the value of X. (5)
the diagram is something like this its in the question :
P
\
\
\
\
\._________________ Q
O
angles is 150 ^^
Post by: halosh92 on May 15, 2010, 10:30:27 pm
jan 2005 Q5b
Post by: cooldude on May 16, 2010, 06:19:51 am
edexcel
jan 2005 Q5b
cud u please post the link of the paper as i'm not familiar with edexcel so i might end up solving the wrong question.
Post by: halosh92 on May 16, 2010, 08:06:34 am
cud u please post the link of the paper as i'm not familiar with edexcel so i might end up solving the wrong question.
Post by: cooldude on May 16, 2010, 08:38:13 am
a=3.2
for the 0.8 kg mass---->
8-T=0.8a
8-T=0.8*3.2
T=5.44 N
Post by: melony on May 16, 2010, 06:26:09 pm
Quote
Insert Quote
edexcel
jan 2005 Q5b
There you go! ;D
Post by: Fenomenoe9 on May 16, 2010, 07:08:20 pm
It's 2007 January Q7 part (f).
Appreciate the help peace..
Post by: Fenomenoe9 on May 16, 2010, 07:14:21 pm
Post by: Fenomenoe9 on May 16, 2010, 07:15:39 pm
here m and thanks oh yahh the answer is 0.51s
Post by: cooldude on May 16, 2010, 07:29:14 pm
here m and thanks
okay for this basically u find out the time the for P to come to rest and then double that value---.
as the string is now slack for this period there is no tension in the string so the deceleration of P will be-->
F=ma, -30sin30=3a
a=-10sin30=-5
the time taken for Q to reach the ground is-->
0.8=0.5*0.98*t^2
t=1.28 sec
therefore final speed of P=
v=0+(1.28*0.98)
v=1.24 m/s
0=1.24-5t
t=0.248 sec
this is the time P takes to come to rest therefore the time for it to become taut again will be double that=0.248*2=0.496 sec, By the way there might be some inaccuracies in the final answer as i did not use exact values
Post by: cooldude on May 16, 2010, 07:32:05 pm
okay for this basically u find out the time the for P to come to rest and then double that value---.
as the string is now slack for this period there is no tension in the string so the deceleration of P will be-->
F=ma, -30sin30=3a
a=-10sin30=-5
the time taken for Q to reach the ground is-->
0.8=0.5*0.98*t^2
t=1.28 sec
therefore final speed of P=
v=0+(1.28*0.98)
v=1.24 m/s
0=1.24-5t
t=0.248 sec
this is the time P takes to come to rest therefore the time for it to become taut again will be double that=0.248*2=0.496 sec, By the way there might be some inaccuracies in the final answer as i did not use exact values
take t as 4root5/7 sec and ull get the exact answer
Post by: Fenomenoe9 on May 16, 2010, 07:43:56 pm
Post by: astarmathsandphysics on May 16, 2010, 07:48:33 pm
Post by: cooldude on May 16, 2010, 07:49:22 pm
ok but why does the time taken for P to rest double?
cuz c wen P reaches rest, Q is still incapable of movement as the string is still slack, but then when P goes back down the plane and when it reaches the point where it initially was i.e. 0.51 sec previously, the string becomes taut again and Q rises again, we double the time because we consider both the period when P travels up the plane and back down. P moves down the plane because of its weight, there is no other force acting on P as the tension in the string is non-existent for the duration of the time the string is slack, i.e. the interval of 0.51 sec, hope u get it.
Post by: cooldude on May 16, 2010, 07:50:55 pm
cos it has to travel the same distance with the same acceleration - motion in reverse
sorry sir, didnt notice you'd already answered :-[
Post by: Fenomenoe9 on May 16, 2010, 07:54:26 pm
Post by: hesho21 on May 17, 2010, 02:20:08 pm
Post by: nid404 on May 17, 2010, 02:46:08 pm
Since the system is in equilibrium, the sum of horizontal components should be equal to zero.
That is
W2sin60o- W1sin40= 0 Call this eq 1
The sum of vertical components should also be zero
W1cos40+ W2cos60 - 5N= 0 Call this eq 2
Now you rearrange eqn 1 to get either W1 in terms of W2 or the other way round. I'll go with the first
Eq 1 W2sin60o= W1sin40
W1=W2sin60o/ sin40
Substitute W1 in eqn 2
W2sin60o/ sin40 (cos 40) + W2cos60= 5
1.53 W2= 5
W2=3.26N
Now substitute the value of W2 in eq 1
W1sin40= 3.26 sin 60
W1=4.4N
Post by: immortal on May 17, 2010, 06:23:09 pm
By the way is it ok 2 solve this using da data 2 form a triangle.
V wud get 5/sin80=W1 /sin60=W2/sin40.
Post by: nid404 on May 17, 2010, 06:25:47 pm
Post by: hesho21 on May 17, 2010, 07:56:27 pm
Post by: solo_G on May 17, 2010, 11:20:31 pm
Post by: astarmathsandphysics on May 17, 2010, 11:24:38 pm
Post by: solo_G on May 17, 2010, 11:33:15 pm
Something wrong with that entire paper. My copy too. Just picures with no questionssorry about that.....imma upload another copy
Post by: solo_G on May 17, 2010, 11:43:12 pm
Post by: astarmathsandphysics on May 17, 2010, 11:59:39 pm
Post by: hesho21 on May 18, 2010, 12:06:38 am
parallel to the 10 N force u have 8cos(theta)
perpendicular to the 10N force u have 8sin(theta)
Now u want to find the resultant
(a) parallel to the 10 N force ==> 10-8cos(theta)
(b)perpendicular is just 8sin(theta)
Hope it helped
Post by: solo_G on May 18, 2010, 12:34:43 am
So for the 8N forceaha .....that was helpful hesho21.....i got it now....makes sense ;D thanks man!
parallel to the 10 N force u have 8cos(theta)
perpendicular to the 10N force u have 8sin(theta)
Now u want to find the resultant
(a) parallel to the 10 N force ==> 10-8cos(theta)
(b)perpendicular is just 8sin(theta)
Hope it helped
Post by: falafail on May 18, 2010, 12:42:34 am
sorry people but i have another doubt...its from CIE nov 08...the first question part (i) and part (ii).....i know its just a two marks question but its driving me crazy.....thanks in advance.
(i)(a) you have to resolve the forces horizontally, so its 10 - 8cos?
(i)(b) you have to resolve the forces vertically, so it's 8sin?
(ii) the x-component squared + the y-component squared will = the resultant squared.
(10 - 8cos?)2 + (8sin?)2 = 82
then just simplify and you'll get cos? = 5/8
Post by: solo_G on May 18, 2010, 12:57:21 am
(i)(a) you have to resolve the forces horizontally, so its 10 - 8cos?yeah falafail.....i got it and i did the last question just like you said....thanks for replying ...ehm ehm i wana ask smthn else....in nov09 varient 1 question 6 part(iii).......can u explain?!
(i)(b) you have to resolve the forces vertically, so it's 8sin?
(ii) the x-component squared + the y-component squared will = the resultant squared.
(10 - 8cos?)2 + (8sin?)2 = 82
then just simplify and you'll get cos? = 5/8
Post by: falafail on May 18, 2010, 01:28:00 am
yeah falafail.....i got it and i did the last question just like you said....thanks for replying ...ehm ehm i wana ask smthn else....in nov09 varient 1 question 6 part(iii).......can u explain?!right okay,
so in part (ii) you found the height above the ground of P and Q, and their speed, when the string breaks.
that's sP= 3 m, sQ= 7 m, and their speed is 2 ms-1.
you use the equation s = ut + 1/2 at2, except here they're accelerating due to gravity only so a=10
so for P:
3 = 2t + 1/2 (10)t2
solving for t you get t = 0.6 or
for Q:
7 = -2t + 1/2 (10)t2
here it's -2 because it's in the opposite direction to P. remember this is velocity, you have to take into consideration the speed AND the direction.
so yeah, solve for this one you get t=1.4 or
1.4 - 0.6 = 0.8
Post by: hesho21 on May 18, 2010, 02:01:07 am
Post by: solo_G on May 18, 2010, 02:14:47 am
right okay,yeah i know about the solving but the thing that i dont get is that when the string breaks both particles will move downwards and so whats the point of using u=-2.....soory dude bt am lil bit bad at mechanics :-\
so in part (ii) you found the height above the ground of P and Q, and their speed, when the string breaks.
that's sP= 3 m, sQ= 7 m, and their speed is 2 ms-1.
you use the equation s = ut + 1/2 at2, except here they're accelerating due to gravity only so a=10
so for P:
3 = 2t + 1/2 (10)t2
solving for t you get t = 0.6 ort = -1(obviously the second is invalid)
for Q:
7 = -2t + 1/2 (10)t2
here it's -2 because it's in the opposite direction to P. remember this is velocity, you have to take into consideration the speed AND the direction.
so yeah, solve for this one you get t=1.4 ort=-1
1.4 - 0.6 = 0.8
Post by: falafail on May 18, 2010, 11:12:18 am
yeah i know about the solving but the thing that i dont get is that when the string breaks both particles will move downwards and so whats the point of using u=-2.....soory dude bt am lil bit bad at mechanics :-\oh right sh*t my bad. it's not negative because they're in opposite directions, they're in the same direction -_- sorry.
okay so you know that velocity is a vector quantity; you take into consideration the speed AND the direction.
here the direction matters especially because, in the first parts of the question, we took the downwards velocity of P as positive and UPWARDS of Q as positive.
so in this part, since P is going down and so is Q, the velocity of P remains positive as it's still int he downwards direction, but the velocity of Q becomes negative, since it is now going downwards (opposite to what we took as positive)
geddit?
Post by: Saladin on May 18, 2010, 02:15:11 pm
oh rightsh*tmy bad. it's not negative because they're in opposite directions, they're in the same direction -_- sorry.
okay so you know that velocity is a vector quantity; you take into consideration the speed AND the direction.
here the direction matters especially because, in the first parts of the question, we took the downwards velocity of P as positive and UPWARDS of Q as positive.
so in this part, since P is going down and so is Q, the velocity of P remains positive as it's still int he downwards direction, but the velocity of Q becomes negative, since it is now going downwards (opposite to what we took as positive)
geddit?
You did not have to use that word did you?
Post by: falafail on May 18, 2010, 05:15:39 pm
You did not have to use that word did you?what word? >__>
Post by: solo_G on May 18, 2010, 08:11:57 pm
oh right sh*t my bad. it's not negative because they're in opposite directions, they're in the same direction -_- sorry.Aye aye i got that!!! now it makes sense....thanks alot falafail for ur help! ;D
okay so you know that velocity is a vector quantity; you take into consideration the speed AND the direction.
here the direction matters especially because, in the first parts of the question, we took the downwards velocity of P as positive and UPWARDS of Q as positive.
so in this part, since P is going down and so is Q, the velocity of P remains positive as it's still int he downwards direction, but the velocity of Q becomes negative, since it is now going downwards (opposite to what we took as positive)
geddit?
Post by: falafail on May 18, 2010, 08:37:48 pm
Aye aye i got that!!! now it makes sense....thanks alot falafail for ur help! ;Dno problem :)
Post by: AL*Eagle on May 18, 2010, 08:49:53 pm
would someone care to explain?
A 5Kg box lies on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.5 .
A force P is applied to to the box to pull or push it horizontally along the floor.
Find the Magnitude of P which is necessary to achieve this if:
a P is applied at an angle x above the horizontal, where tan x = 3/4
b P is applied at an angle x below the horizontal, where tan x = 3/4
Id really appreciate anyone explainin this??
Post by: falafail on May 18, 2010, 09:06:10 pm
I have the answer of this question.. but i dont get it..
would someone care to explain?
A 5Kg box lies on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.5 .
A force P is applied to to the box to pull or push it horizontally along the floor.
Find the Magnitude of P which is necessary to achieve this if:
a P is applied at an angle x above the horizontal, where tan x = 3/4
b P is applied at an angle x below the horizontal, where tan x = 3/4
Id really appreciate anyone explainin this??
is it P > 19.2 N for part a and P > 35.7 N for part b? i'll explain how i got those if i'm right :P
Post by: astarmathsandphysics on May 18, 2010, 09:12:49 pm
Post by: immortal on May 18, 2010, 09:21:49 pm
Post by: AL*Eagle on May 18, 2010, 09:24:58 pm
is it P > 19.2 N for part a and P > 35.7 N for part b? i'll explain how i got those if i'm right :P
Answers are
a 22N
b 49
(2s.f.)
close to yours
Post by: AL*Eagle on May 18, 2010, 09:25:58 pm
try this
thanks for your effort man
Post by: falafail on May 18, 2010, 09:30:05 pm
Answers are
a 22N
b 49
(2s.f.)
close to yours
right well i found out what i did wrong, but it's still not what you have. maybe you rounded wrong? 'cause i get 22.7 N and 50 N (exact answers are somewhere around 22.727.... and 50.000... so i'm pretty sure i'm right :P)
Post by: immortal on May 18, 2010, 09:33:00 pm
right well i found out what i did wrong, but it's still not what you have. maybe you rounded wrong? 'cause i get 22.7 N and 50 N (exact answers are somewhere around 22.727.... and 50.000... so i'm pretty sure i'm right :P)I did da same, wat u did wrong waz 2 tke g as 10 & not 9.8
Post by: falafail on May 18, 2010, 09:37:46 pm
I did da same, wat u did wrong waz 2 tke g as 10 & not 9.8erm no actually, i did P > F instead of Pcosx > F :-[
anyway, here's the solving.
Post by: AL*Eagle on May 18, 2010, 10:06:09 pm
erm no actually, i did P > F instead of Pcosx > F :-[
anyway, here's the solving.
Thank you very much.. appreciated
Post by: falafail on May 18, 2010, 10:08:08 pm
Thank you very much.. appreciated
you're welcome ;D
Post by: hesho21 on May 18, 2010, 11:08:03 pm
Post by: astarmathsandphysics on May 18, 2010, 11:20:40 pm
Post by: hesho21 on May 18, 2010, 11:27:50 pm
Post by: solo_G on May 19, 2010, 02:08:41 am
Post by: falafail on May 19, 2010, 06:37:44 am
can someone explain Q 7 cz i have trbles wed vector quantities....i mean in this question what shud we take as negative the U or da acceleration? please answer ASAPi used u as negative for Q... my friend said you could use the acceleration as negative too but idk. does it matter? supposedly you get the answer either way :-\
Post by: falafail on May 19, 2010, 06:45:28 am
guys 6(iv) in June 09 on the paper attached here , shouldnt it be 0.72m? cuz in the mark scheme its slightly larger than that ? Plz explain , thank u
no, the markscheme's right ::)
after A hits the floor, B keeps moving upwards with a= -10 and u=1.2
to calculate the maximum height reached, you use v=0
so you get s=0.072 AFTER A HITS THE FLOOR
that means that B is already 0.36 + 0.36 m above the ground.
so 0.36 + 0.36 + 0.072 = 0.792 m.
woops, i didn't notice astar's post. oh well, at least i got practice i guess :P
Post by: Fenomenoe9 on May 19, 2010, 04:47:30 pm
Thanks
Post by: wakemeup on May 20, 2010, 03:42:58 am
Here is a video on the very basics of vectors:
[yt=425,350]pimr9I92GZY[/yt]
Post by: Sweet_03 on May 20, 2010, 01:22:16 pm
M1 , jan2006
can someone please explain how to do :
all of Question 4
and Question 6 part (d) only
:-[
Post by: halosh92 on May 20, 2010, 01:41:50 pm
Q7c- SMALL DOUBT
could someone just explain why is the direction reversed??
THX ;D
Post by: halosh92 on May 20, 2010, 07:21:10 pm
! anyone has the ms for jan 2010 m1 edexcel
VERY URGENT THXXXX
Post by: vanibharutham on May 20, 2010, 07:51:57 pm
Post by: halosh92 on May 20, 2010, 08:21:14 pm
http://www.scribd.com/doc/29694750/Edexcel-GCE-January-2010-Mechanics-M1-Marking-Scheme (http://www.scribd.com/doc/29694750/Edexcel-GCE-January-2010-Mechanics-M1-Marking-Scheme)
thankyouuuuuuuuu
Post by: Phosu on May 21, 2010, 10:53:26 am
7) d)
Post by: vanibharutham on May 21, 2010, 11:18:36 am
Position of S - Position of L
=> [(3t+9)i + (4t-6)j] - [ 18i + 6j ]
LS = (3t-9)i + (4t-12)j
therefore, the time from when DISTANCE, (not displacement), between L and S is 10 km is given by:
LS² = (3t - 9)² + (4t - 12)²
(10)² = 25t² - 150t + 225
0 = 25t² - 150t + 125
t² - 6t + 5 = 0
t = 5 and t = 1
Therefore,
T = 5 or T = 1
Post by: halosh92 on May 21, 2010, 12:45:26 pm
jan 2006 -edexcel
Q7c- SMALL DOUBT
could someone just explain why is the direction reversed??
THX ;D
coudl someone solve this plz!
Post by: halosh92 on May 21, 2010, 12:55:19 pm
Another particle B of mass m kg is moving along the same straight line, in the opposite
direction to A, with speed 8 m s–1. The particles collide. The direction of motion of A
is unchanged by the collision. Immediately after the collision, A is moving with speed
3 m s–1 and B is moving with speed 4 m s–1. Find
(a) the magnitude of the impulse exerted by B on A in the collision
(b) the value of m.
i keep getting the impluse as negative
i took this -------------> as (+) and this <---------- as (-)
Post by: halosh92 on May 21, 2010, 02:22:46 pm
thx
Post by: astarmathsandphysics on May 21, 2010, 02:44:16 pm
Another particle B of mass m kg is moving along the same straight line, in the opposite
direction to A, with speed 8 m s–1. The particles collide. The direction of motion of A
is unchanged by the collision. Immediately after the collision, A is moving with speed
3 m s–1 and B is moving with speed 4 m s–1. Find
(a) the magnitude of the impulse exerted by B on A in the collision
(b) the value of m.
i keep getting the impluse as negative
i took this -------------> as (+) and this <---------- as (-)
conservation of momentum
12*2-8m=3*2+4m
24-8m=6+4m
24-6=8m+4m
18=12m so m=1.5
impulse=mv-mu for A
2*3-2*12=-18
Post by: astarmathsandphysics on May 21, 2010, 02:48:21 pm
r(t)=start point +t*velocity
=16i+5j+3*(-2.5i+6j)=8.5i+23j
Post by: halosh92 on May 21, 2010, 03:09:53 pm
A particle A of mass 2 kg is moving along a straight horizontal line with speed 12 m s–1.
Another particle B of mass m kg is moving along the same straight line, in the opposite
direction to A, with speed 8 m s–1. The particles collide. The direction of motion of A
is unchanged by the collision. Immediately after the collision, A is moving with speed
3 m s–1 and B is moving with speed 4 m s–1. Find
(a) the magnitude of the impulse exerted by B on A in the collision
(b) the value of m.
i keep getting the impluse as negative
i took this -------------> as (+) and this <---------- as (-)
conservation of momentum
12*2-8m=3*2+4m
24-8m=6+4m
24-6=8m+4m
18=12m so m=1.5
impulse=mv-mu for A
2*3-2*12=-18
sir this is my problem...the impulse is 18 not -18 :SS am always getting it negative in all papers. :s
Post by: nid404 on May 21, 2010, 03:22:19 pm
Post by: halosh92 on May 21, 2010, 03:24:27 pm
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 3u. The magnitude of the impulse received by each particle in the
collision is 7mu/2
Find
(a) the speed of A immediately after the collision,
(b) the speed of B immediately after the collision
Post by: astarmathsandphysics on May 21, 2010, 03:28:53 pm
mV-mu for B
1.5*4--1.5*8=18
Post by: 7ooD on May 21, 2010, 05:36:33 pm
Post by: astarmathsandphysics on May 21, 2010, 10:23:06 pm
Post by: Muho on May 21, 2010, 11:30:48 pm
I have the mark scheme i just don't understand how they freakn get there =/
Post by: astarmathsandphysics on May 22, 2010, 12:02:13 am
Post by: falafail on May 22, 2010, 12:03:41 am
Someone help me with january 2009 Q7 (b)..
I have the mark scheme i just don't understand how they freakn get there =/
CIE or edexcel?
Post by: Saladin on May 22, 2010, 07:28:59 am
Post by: Sweet_03 on May 22, 2010, 07:41:51 am
hello
M1 , jan2006
can someone please explain how to do :
all of Question 4
and Question 6 part (d) only
:-[
Post by: Saladin on May 22, 2010, 08:19:59 am
Answering it soon.
Jan 2009 m1
7(b)
This is basically asking you to find the resistance force that R give Q. The thing thatprevents it from having an acceleration of 9.8 ms-2
So
Post by: halosh92 on May 22, 2010, 09:15:28 am
jan 2001
Q6d
URGENT! thx
Post by: Saladin on May 22, 2010, 09:24:28 am
M1 edexcel
jan 2001
Q6d
URGENT! thx
Post all your working upto 6 (c)
or else i will have to do the whole thing
Post by: halosh92 on May 22, 2010, 09:54:50 am
Post all your working upto 6 (c)
or else i will have to do the whole thing
sure
b) v=u +at
v=at
v = 9.8 *2 = 19.6 m/s
c) 1/2 * 2 *19.6 = 19.6m
1/2 * (19.6 + 4) *5 = 59
59 + 19.6= 78.6m
Post by: astarmathsandphysics on May 22, 2010, 11:31:46 am
Post by: halosh92 on May 22, 2010, 12:01:02 pm
sure
b) v=u +at
v=at
v = 9.8 *2 = 19.6 m/s
c) 1/2 * 2 *19.6 = 19.6m
1/2 * (19.6 + 4) *5 = 59
59 + 19.6= 78.6m
could someone solve this plz :(
Post by: Phosu on May 22, 2010, 12:31:53 pm
this is the total distance under the graph, since its the total distance that the woman covers before reaching the ground
out of which 78.6m is the distance between the woman in air, and the helicopter
125=78.6+4T
4T=46.4
T=11.6
total time taken = 2+5+11.6=18.6s
Post by: Ghost Of Highbury on May 22, 2010, 12:38:31 pm
Post by: halosh92 on May 22, 2010, 12:47:09 pm
the distance that the helicopter is above the ground is 125m
this is the total distance under the graph, since its the total distance that the woman covers before reaching the ground
out of which 78.6m is the distance between the woman in air, and the helicopter
125=78.6+4T
4T=46.4
T=11.6
total time taken = 2+5+11.6=18.6s
how did u get this equation exactly??
Post by: Phosu on May 22, 2010, 12:57:41 pm
how did u get this equation exactly??
which one?
the calculation of T or the total time taken?
Post by: halosh92 on May 22, 2010, 01:07:52 pm
which one?
the calculation of T or the total time taken?
no calculation for T
Post by: Phosu on May 22, 2010, 01:36:45 pm
125= 78.6+x
find x
which is 48.6m
so she travels 48.6m for t seconds
now to find t
48.6 = 4 (t+7)-7
48.6 =4t
t=48.6/4 = 11.6s
Post by: cooldude on May 22, 2010, 01:48:53 pm
A load of weight 7kN is being raised from the rest with constant acceleration by a cable. After the load has been raised 20 metres, the cable suddenly becomes slack. The load continues to move upwards for a distance of 4 metres before coming to an instantaneous rest. Assume no air resistance, find the tension in the cable before it became slack.
this is for the first part of the journey, let the tension in the string be T and the velocity of the load after it has travelled 20m--->
T-7000=700a
v^2=u^2+2as
v^2=2*20*((T-7000)/700)
for the second part of the journey (when the string is slack), note in the second line, v^2 is referred to as the velocity found above-->
v^2=u^2+2as
0=v^2+2*-10*4
v^2=80
now substitute the value of v^2 in the above equation-->
80=40((T-7000)/700)
T=1400+7000=8400 N
Post by: halosh92 on May 22, 2010, 01:57:48 pm
see we kno total distance is 125m
125= 78.6+x
find x
which is 48.6m
so she travels 48.6m for t seconds
now to find t
48.6 = 4 (t+7)-7
48.6 =4t
t=48.6/4 = 11.6s
thxxx aloooottt
all clear :D
Post by: Ghost Of Highbury on May 22, 2010, 03:20:06 pm
this is for the first part of the journey, let the tension in the string be T and the velocity of the load after it has travelled 20m--->
T-7000=700a
v^2=u^2+2as
v^2=2*20*((T-7000)/700)
for the second part of the journey (when the string is slack), note in the second line, v^2 is referred to as the velocity found above-->
v^2=u^2+2as
0=v^2+2*-10*4
v^2=80
now substitute the value of v^2 in the above equation-->
80=40((T-7000)/700)
T=1400+7000=8400 N
ok got it , thanks a lot
Post by: wakemeup on May 22, 2010, 03:31:04 pm
http://www.xtremepapers.net/Edexcel/Mathematics/2009%20Jan/2009%20Jan%20QPM1.pdf
Post by: Darkstar3000 on May 22, 2010, 03:51:29 pm
a) The value of u (i can find this)
b) The value of T
part b is my problem because in the marking scheme they used V=u + at and took the initial speed (u) as negative, why do we take it as a negative because in my calculation i wrote it like this: 17.5=10.5 - 9.8t
this is from May 2008 Q2
Post by: cooldude on May 22, 2010, 04:06:44 pm
(m doin the a part as i need the value of u)
a) highest point reached by the object-->
v^2=u^2+2as
0=u^2+(2*-10s)
u^2=20s
b) v^2=u^2+2as
17.5^2=0+(2*10*(10+s)
s=5.3125
u=10.31 m/s
b) for the upward journey time--->
v=u+at
0=10.31-10t
t=1.031 sec
for the downward journey time-->
v=u+at
17.5=0+10t
t=1.75 sec
therefore T=1.75+1.031=2.78 sec
and now ur qs, in these type of qs its best to divide ur answer into two parts (the upward part and the downward part), uve assumed that g is negative for the whole journey but that is only true for the upward journey, thus u will not get the correct answer
Post by: Darkstar3000 on May 22, 2010, 05:21:01 pm
darkstar3000--->
(m doin the a part as i need the value of u)
a) highest point reached by the object-->
v^2=u^2+2as
0=u^2+(2*-10s)
u^2=20s
b) v^2=u^2+2as
17.5^2=0+(2*10*(10+s)
s=5.3125
u=10.31 m/s
b) for the upward journey time--->
v=u+at
0=10.31-10t
t=1.031 sec
for the downward journey time-->
v=u+at
17.5=0+10t
t=1.75 sec
therefore T=1.75+1.031=2.78 sec
and now ur qs, in these type of qs its best to divide ur answer into two parts (the upward part and the downward part), uve assumed that g is negative for the whole journey but that is only true for the upward journey, thus u will not get the correct answer
thank you very much
Post by: wakemeup on May 22, 2010, 08:29:08 pm
Man, edexcel jan 2009 is insane. Q5a. Why is the friction in the upward direction? Q5b. I have no idea what's going on here. Q7b and c. Again, no idea at all. Please help...Please, people. Need answers urgently.
http://www.xtremepapers.net/Edexcel/Mathematics/2009%20Jan/2009%20Jan%20QPM1.pdf
Post by: Igcseboy on May 22, 2010, 08:41:42 pm
Jan 2009 qustion 5 b i and ii
May 2008 6 b
May 2008 5 a and b
Jan 2008 7 c
Jan 2008 6 d
Any1 able to answer these plzzz do so
May God bless us All
Post by: Saladin on May 22, 2010, 08:52:03 pm
Hello every1 plzz i need answers to the following questions
Jan 2009 qustion 5 b i and ii
May 2008 6 b
May 2008 5 a and b
Jan 2008 7 c
Jan 2008 6 d
Any1 able to answer these plzzz do so
May God bless us All
2moro!
Post by: halosh92 on May 22, 2010, 09:03:07 pm
plzzz someone explainnn
thx a bunch URGENTTTT
Post by: astarmathsandphysics on May 22, 2010, 10:33:51 pm
Post by: astarmathsandphysics on May 23, 2010, 07:23:36 am
Resolving vertically-6.75g-6.75gcos36.87=-12.15g
resolving horizontally -6.75gsinn36.87=-4.5g
R=sqrt((-12.15g)^2+(-4.5g)^2)=12.96g
Post by: Saladin on May 23, 2010, 09:14:58 am
This is 7c
Resolving vertically-6.75g-6.75gcos36.87=-12.15g
resolving horizontally -6.75gsinn36.87=-4.5g
R=sqrt((-12.15g)^2+(-4.5g)^2)=12.96g
I owe you one.
Post by: Uchia on May 23, 2010, 09:58:24 am
Post by: Igcseboy on May 23, 2010, 09:59:23 am
Post by: astarmathsandphysics on May 23, 2010, 10:47:38 am
Post by: astarmathsandphysics on May 23, 2010, 10:54:25 am
before A hits It accelerates at g/4 for 1.2s
s=1/2at^2=1/2*g/4*1.2^2=0.18g
When A hits floor, it and B are travelling with speedv=at=g/4*1.2=0.3g
Now B moves freely under gravity
v^2=u^2+2as
0=(0.3g)^2+2*-g*s
s=0.045g
height reached=0.18g+0.045g=0.225g
Post by: georgianna on May 23, 2010, 11:09:58 am
http://www.srepapmaxeeeerf.org/A%20Level/Maths/Edexcel/M1/M1%20Mock.pdf (http://www.srepapmaxeeeerf.org/A%20Level/Maths/Edexcel/M1/M1%20Mock.pdf)
how do you get 8103.2 for 5c??
I only got 7280
Post by: Uchia on May 23, 2010, 11:17:58 am
Post by: Uchia on May 23, 2010, 11:19:03 am
Post by: astarmathsandphysics on May 23, 2010, 11:22:15 am
speed of both togeter is 5.6*78/84=5.2
deceleration=5.2/0.06=87
use F=ma to get R-84g=84*87 so R=84g+86*87
Post by: astarmathsandphysics on May 23, 2010, 11:25:23 am
Post by: Uchia on May 23, 2010, 11:28:45 am
Post by: Uchia on May 23, 2010, 11:36:01 am
Post by: Igcseboy on May 23, 2010, 12:12:48 pm
jan 2010 question 6 d.... plzzz
before A hits It accelerates at g/4 for 1.2s
s=1/2at^2=1/2*g/4*1.2^2=0.18g
When A hits floor, it and B are travelling with speedv=at=g/4*1.2=0.3g
Now B moves freely under gravity
v^2=u^2+2as
0=(0.3g)^2+2*-g*s
s=0.045g
height reached=0.18g+0.045g=0.225g
but its different in the markscheme :S:S:S
Post by: perrychan527 on May 23, 2010, 02:36:28 pm
dont understd the question and marking. thx..
Post by: Saladin on May 23, 2010, 02:50:11 pm
M1 Jan 08 6d plzz
dont understd the question and marking. thx..
answering now.
Post by: Saladin on May 23, 2010, 02:53:28 pm
Post by: Saladin on May 23, 2010, 03:16:06 pm
Thus, we have to see, how long it takes for the
Thus, we got that the speed is
So, the
Thus,
The total time to get to the point
Thus you simply add the new value to
Post by: perrychan527 on May 23, 2010, 05:11:52 pm
TheThx, i got it. ;) ;) ;)co-ordinates have to be the same in order for something to be due south of something.
Thus, we have to see, how long it takes for the
Thus, we got that the speed is
So, the.
Thus,
The total time to get to the pointis 7 seconds,
.
Thus you simply add the new value to
Post by: raimunts on May 23, 2010, 05:35:17 pm
And do we have to write answeres which involve g in terms of g, or in decimals? please help
Post by: astarmathsandphysics on May 23, 2010, 06:01:43 pm
I always use 9.8 but you may be told to use a value in the question
In terms of g is always best
Post by: 7ooD on May 23, 2010, 06:12:51 pm
Post by: zabady on May 23, 2010, 08:55:04 pm
Post by: astarmathsandphysics on May 23, 2010, 08:57:33 pm
Post by: Saladin on May 23, 2010, 09:00:07 pm
plzzz someone question 6 a b c for they said R is parallel to vector (i-2j) i understand by parallel that it means R=K * (i-2j).....,and for b and c i cant get them
which year?
Post by: zabady on May 23, 2010, 09:01:23 pm
question says
6. Two forces, (4i – 5j) N and (pi + qj) N, act on a particle P of mass m kg. The resultant
of the two forces is R. Given that R acts in a direction which is parallel to the
vector (i – 2j),
(a) find the angle between R and the vector j,
(3)
(b) show that 2p + q + 3 = 0.
ms says
tan? = 2
1
?? = 63.4
angle is 153.4
(4 + p)i + (q ? 5)j
(q ? 5) = ?2(4 + p)
2 p + q + 3 = 0 *
Post by: Saladin on May 23, 2010, 09:02:40 pm
looool sry 2009 january....m1 edexcel
sir, i need to go to sleep. I have been helping all day.....
Pls take this one.
Post by: astarmathsandphysics on May 23, 2010, 09:07:28 pm
Post by: zabady on May 23, 2010, 09:10:35 pm
Post by: astarmathsandphysics on May 23, 2010, 09:16:32 pm
of the two forces is R. Given that R acts in a direction which is parallel to the
vector (i – 2j),
(a) find the angle between R and the vector j,
(3)
(b) show that 2p + q + 3 = 0.
a)cos x=j.(i-2j)/|i||i-2j|=1aqrt(5) so x=cos^-1 1/sqrt5
4i-5j+pi+qj=k(i-2j)
4+p=k (3)
-5+q=-2k (4)
(3)*2+(4) gives2p + q + 3 = 0
Post by: raimunts on May 23, 2010, 09:20:35 pm
Post by: zabady on May 23, 2010, 09:22:54 pm
Post by: 7ooD on May 23, 2010, 09:35:44 pm
Post by: astarmathsandphysics on May 23, 2010, 10:34:55 pm
solve tan x=2
Post by: raimunts on May 23, 2010, 10:42:34 pm
Post by: 7ooD on May 23, 2010, 11:14:37 pm
Post by: astarmathsandphysics on May 24, 2010, 07:39:46 am
only at the start and most important at the end
Post by: Sweet_03 on May 24, 2010, 08:40:44 am
M1 june 2009 - edexcel
Q6 (c)
Post by: Saladin on May 24, 2010, 09:21:09 am
can someone please help
M1 june 2009 - edexcel
Q6 (c)
https://studentforums.biz/index.php/topic,5617.0.html
Post by: astarmathsandphysics on May 24, 2010, 09:29:46 am
-200-T=200a
-100-200=200a so a=-1.5
for car -F-400+100=800a
F=-400+100-800*-1.5=-1500N
Post by: Igcseboy on May 24, 2010, 09:31:09 am
Post by: Sweet_03 on May 24, 2010, 09:43:13 am
i hav another Q :(
june2007 ..
Q6 (d)
so v for p was 4.2
therefore u for Q will be 4.2
a= 9.8
and isnt v=0
?
Post by: astarmathsandphysics on May 24, 2010, 09:51:04 am
yes the thrust is in the opposite direction to tension - not always. It depends on the question
One mo
Post by: astarmathsandphysics on May 24, 2010, 09:53:29 am
Post by: Sweet_03 on May 24, 2010, 09:59:15 am
Post by: Igcseboy on May 24, 2010, 10:07:42 am
if they ask for velocity, find the direction too.
yes the thrust is in the opposite direction to tension - not always. It depends on the question
One mo
we will find direction using velocity components rite???
Post by: astarmathsandphysics on May 24, 2010, 10:13:31 am
speed when string goes slack is 4.2
find time to top of fight
v=u+at
0=4.2-9.8t so t=3/7
double this to get the answer
Post by: abuelzouz on May 24, 2010, 10:15:26 am
part b i get it right. part a noo.. :S can somebody help ? i dnt get the answer like the ms
Post by: Sweet_03 on May 24, 2010, 10:17:57 am
Post by: Sweet_03 on May 24, 2010, 10:21:59 am
MJ 09 q 3
part b i get it right. part a noo.. :S can somebody help ? i dnt get the answer like the ms
3a)
I = -7mu/2
m= 2m
v= ?
u= 2u
I = m(v-u)
-7mu/2 = 2m ( v-2u)
-7u/4 = v - 2u
v = u/4
Post by: abuelzouz on May 24, 2010, 10:23:12 am
3a)
I = -7mu/2
m= 2m
v= ?
u= 2u
I = m(v-u)
-7mu/2 = 2m ( v-2u)
-7u/4 = v - 2u
v = u/4
Thanks bt y did u use I in minus?
Post by: Sweet_03 on May 24, 2010, 10:30:26 am
I : is in this direction <--
u and v : are in this direction -->
Post by: AL*Eagle on May 24, 2010, 11:16:32 am
I dont understand it
If anyone has a useful link for evector explanation please share it..
I dont have specific question about it.. i want the whole topic explained
THanks
Post by: Uchia on May 24, 2010, 11:39:44 am
Post by: AL*Eagle on May 24, 2010, 11:48:41 am
young man call me i might help a little
imma call u now..
Post by: abuelzouz on May 24, 2010, 01:26:52 pm
Post by: abuelzouz on May 24, 2010, 01:39:39 pm
Post by: abuelzouz on May 24, 2010, 01:50:55 pm
well thnk u!!
when i need the forum i cnt find it..
im so leavin it.. as if u care abt some1 who leaves or doesnt..
!!
Post by: astarmathsandphysics on May 24, 2010, 03:46:23 pm
Post by: 7ooD on May 24, 2010, 05:01:16 pm
Post by: Saladin on May 24, 2010, 05:07:12 pm
m1 exam shouldn't be for as math such an eazzzy relly the best exam ever
That depends on person to person. Please dont discuss any content of the recent exam.
Post by: abuelzouz on May 24, 2010, 06:10:47 pm
m1 exam shouldn't be for as math such an eazzzy relly the best exam ever
wttttttttttt? it was haaaaaaaaaaaaaaaaaaaaaaard!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Post by: 7ooD on May 24, 2010, 06:15:51 pm
Post by: Uchia on May 24, 2010, 06:18:52 pm
Post by: Saladin on May 24, 2010, 06:19:17 pm
dude it is hard if u didnt practice on papers like 2004-2010
Please be respectful to others. Comments such as these may be hurtful to other people's sentiments and efforts.
Post by: 7ooD on May 24, 2010, 06:24:20 pm
Post by: Saladin on May 24, 2010, 06:42:31 pm
ummm dunno i respect them but if u didnt solve papers then dnt say it was hard but if u solved then found it hard it means u didnt understand them anywayz good luck for alll :D
I solved loads. But many did not. Dont make them feel bad please.
Post by: abuelzouz on May 24, 2010, 06:48:00 pm
Post by: Igcseboy on May 24, 2010, 07:02:40 pm
Post by: Saladin on May 24, 2010, 07:07:52 pm
it was no easy thts for sure it was tough and seeing tht i had C1 only 30 min. b4 tht so it was very tiring day
Yes, just say whether it was easy or hard, but nothing more please.
Post by: Igcseboy on May 24, 2010, 07:12:02 pm
Yes, just say whether it was easy or hard, but nothing more please.
hey edexcel has grade boundaries each year rite??
how r the raw marks converted to uniform marks???
Post by: Saladin on May 24, 2010, 07:13:09 pm
hey edexcel has grade boundaries each year rite??
how r the raw marks converted to uniform marks???
This will help you.
Post by: AL*Eagle on May 24, 2010, 07:25:05 pm
Please be respectful to others. Comments such as these may be hurtful to other people's sentiments and efforts.
......BAHAHAHAHAHAH!!! sorry couldnt hold it..
no offense man but... u sounded like a lil emotional girl xP
altho I liked that UMS converter By the way Thanks
+REP
Post by: Saladin on May 24, 2010, 07:26:16 pm
......BAHAHAHAHAHAH!!! sorry couldnt hold it..
no offense man but... u sounded like a lil emotional girl xP
altho I liked that UMS converter By the way Thanks
+REP
Perhaps giving you wa stern ban will make me seem more masculine.
Post by: AL*Eagle on May 24, 2010, 07:32:49 pm
Perhaps giving you wa stern ban will make me seem more masculine.
No.. that would make you seem like you were REALLY offended by my comment :)
insted of not givin a shi*t [ which is only if what a said was untrue ;) ]
Post by: Saladin on May 24, 2010, 07:40:11 pm
No.. that would make you seem like you were REALLY offended by my comment :)
insted of not givin a shi*t [ which is only if what a said was untrue ;) ]
I pick the third. To warn you that, to keep your derisory comments to yourself.
I do not like speaking in this tone to members, but I do my best to be polite.
I have to give a "sh*t" because I care for what people do and say here.
If you cannot keep your comments to yourself. Then I suggest that you go elsewhere.
Post by: Sweet_03 on May 24, 2010, 10:06:29 pm
Post by: astarmathsandphysics on May 24, 2010, 10:43:58 pm
Post by: 7ooD on May 24, 2010, 11:17:23 pm
Post by: 7ooD on May 25, 2010, 01:15:44 am
Post by: wakemeup on May 25, 2010, 02:39:13 am
Post by: Sweet_03 on May 25, 2010, 07:52:11 am
i found the paper to be pretty easy :")
Post by: Saladin on May 25, 2010, 08:56:36 am
0.423 was gotten by a mistake.
Post by: T.Q on May 25, 2010, 09:14:20 am
Post by: Sweet_03 on May 25, 2010, 10:32:26 am
for the last Q ..
For B:
u=+0.7 (downwards)
and a=+9.8 (downward)
s= 1.175
right?
Post by: Sweet_03 on May 25, 2010, 10:36:33 am
except for Question 3 ..
i forgot about the vertical component [100sin30] :( :( :( :(
i hope i dont lose all of the 7 marks :(
guys do you have any ideas about the UMS of M1 ?
Post by: 7ooD on May 25, 2010, 11:13:05 am
Post by: AL*Eagle on May 25, 2010, 11:14:56 am
M1 june 2010 SOLUTIONS!!!!!!
yo dude u sure bout the first question?
thanks for the paper tho man appreciate it (Y)
Post by: T.Q on May 25, 2010, 11:21:37 am
yo dude u sure bout the first question?
thanks for the paper tho man appreciate it (Y)
these answers are made by an M1 teacher
Post by: 7ooD on May 25, 2010, 11:21:56 am
Post by: 7ooD on May 25, 2010, 11:22:43 am
Post by: AL*Eagle on May 25, 2010, 11:33:51 am
but relly they should add parts in jan 2010 their paper got parts a b c and marks where divided but in this one 9 marks on a stupid question wdf
Messed me UP :(
Post by: Saladin on May 25, 2010, 11:55:38 am
Post by: Ghost Of Highbury on May 27, 2010, 10:09:31 pm
Given the acceleration is 2.4m/s^2, calculate its mass . (I've done this)
Find also the magnitude of the normal contact force exerted on the trolley by the ground (Need help in this)
Thank you.
Post by: astarmathsandphysics on May 27, 2010, 11:09:26 pm
upwards force =R
downwards forces=mg+20sin15
now use the mass
Post by: 7ooD on May 28, 2010, 09:39:10 pm
Post by: Saladin on May 29, 2010, 07:05:20 am
Post by: 7ooD on May 30, 2010, 04:31:48 pm
2005
(http://i109.photobucket.com/albums/n63/7ood/5dfbc803.jpg)
2007
(http://i109.photobucket.com/albums/n63/7ood/601e6e6d.jpg)
Post by: egymoha on May 30, 2010, 10:20:47 pm
please someone reply
thanks my friend
Post by: Meticulous on May 31, 2010, 01:01:35 am
what does this sign | means i always see in the ms when giving a range and how to know where the Y component and x component of the reaction of the wall on the hinge is acting because in 2005 june q 6 x and y were acting x ----- left and y up in 2007 june q 5 it was x ------- right y ---- down im relly confused about this here is the images of the diagrams
2005
(http://i109.photobucket.com/albums/n63/7ood/5dfbc803.jpg)
2007
(http://i109.photobucket.com/albums/n63/7ood/601e6e6d.jpg)
The sign means the modulus; i.e. always positive
and you ALWAYS take Y and X values of the reaction. Even if there's only one, lets see an X component only, then Y will be zero. Therefore, to avoid confusion, always take Y and X components.
Post by: Meticulous on May 31, 2010, 01:03:06 am
In most of the coofficient of restitution questions i see (when they say find the range of value of e ) ,, in the mark scheme and the answers ,, they usually say v1>0 ,, why do they use V1 rather than V2 .
please someone reply
thanks my friend
It depends on the question. For example, if there is further collision or not.
Post by: 7ooD on May 31, 2010, 01:09:21 am
Post by: Meticulous on May 31, 2010, 01:14:18 am
thx a.f but u misunderstood i meant how to kno in which direction does y act and x act cuz as i gave example in the question above one had y acting up and the other had y acting down same thing happened wiz x one was acting left other was acting right thats what i meantDon't worry mate! Put them in the direction you want, up or down ya wadee3 :P
Caues at last, if you put it in the wrong direction, your answer would be in negative, and you would know ;)
Post by: 7ooD on May 31, 2010, 01:38:43 am
Post by: egymoha on May 31, 2010, 09:44:29 am
It depends on the question. For example, if there is further collision or not.
ok so how do i solve the questions of range of values of e ,, cause i dunno how to solve them ??
how should i think ??
thanks friends
Post by: Meticulous on May 31, 2010, 12:28:24 pm
Post by: 7ooD on May 31, 2010, 03:19:00 pm
Post by: egymoha on June 01, 2010, 09:35:50 am
Post by: crazed2009 on June 02, 2010, 07:01:40 am
Don't worry mate! Put them in the direction you want, up or down ya wadee3 :P
Caues at last, if you put it in the wrong direction, your answer would be in negative, and you would know ;)
actually A.F tht doesnt always work.. i was also coming to ask the same question - which way do we put the reactional force?????... c this Q:
a window of mass 15kg and height 120cm is hinged along its top edge. It is kept open by the thrust frm a light strut of lenght 50cm attached to the wall and perpindicular to the lower edge of the window. By modelling the window as a uniform lamina, calculate the thrust in the strut and the magnitude an direction of the force exerted on the window by the hinge? (u hv to draw the diagram)
i attached the answer... i got T but i dnt really understand the rest.. :S...
y is the reactional force acting upwards meaning tht at the hinge the horizontal force is to the left and the vertical force is upwards???? if u put them any other way the correct answer doesnt cm... :S:SS:
\HELP PLEEEEEEEEEEEEZ
Post by: Meticulous on June 02, 2010, 08:15:12 am
Post by: Meticulous on June 02, 2010, 08:59:25 am
what about June 2006 8b ?
First of all:
VB = u(e+1)/5
And e between B and the wall = 4/5
Therefore, the velocity of B after rebounding= 4/5 * u(e+1)/ 5 = 4u(e+1)/25
There is a further collision between A and B, and so VBrebound > VA
PLUS, VA > 0 -----> u(4e-1)/5 >0 and so e>1/4
here is a further collision between A and B, and so VBrebound > VA :
e<9/16
combine the two inequalities to give you
1/4 < e < 9/16
Was I clear ::) ?
Post by: Meticulous on June 02, 2010, 09:04:49 am
actually A.F tht doesnt always work.. i was also coming to ask the same question - which way do we put the reactional force?????... c this Q:
a window of mass 15kg and height 120cm is hinged along its top edge. It is kept open by the thrust frm a light strut of lenght 50cm attached to the wall and perpindicular to the lower edge of the window. By modelling the window as a uniform lamina, calculate the thrust in the strut and the magnitude an direction of the force exerted on the window by the hinge? (u hv to draw the diagram)
i attached the answer... i got T but i dnt really understand the rest.. :S...
y is the reactional force acting upwards meaning tht at the hinge the horizontal force is to the left and the vertical force is upwards???? if u put them any other way the correct answer doesnt cm... :S:SS:
\HELP PLEEEEEEEEEEEEZ
Pls post the original q to see if i have the answer to it, and save time ;)
Post by: crazed2009 on June 02, 2010, 10:15:53 am
Post by: raimunts on June 04, 2010, 08:38:48 pm
Edexcel AS Modular Mathematics M2
Ex.1A No. 16, page: 11
Ex.1B No. 12, page: 19
Any help will be really appreciated. Thanx a lot in advance!
Post by: raimunts on June 05, 2010, 08:47:47 am
Post by: cooldude on June 05, 2010, 01:00:15 pm
Hello. Can someone PLEASE help me solve these two kinematics questions? They are from the Edexcel AS Modular Mathematics M2 book:
Edexcel AS Modular Mathematics M2
Ex.1A No. 16, page: 11
Ex.1B No. 12, page: 19
Any help will be really appreciated. Thanx a lot in advance!
if u posted the question there is a higher probability you would get the answer as the textbook you are referring to, is not in the possession of every person who has the capability of answering your doubt
Post by: 7ooD on June 05, 2010, 06:02:56 pm
Post by: 3oo3a on June 10, 2010, 05:44:32 pm
I couldn't find the answer of question 8 (c) january 2010 paper on the internet, i tried finding it alone, and i got (5i)(c^2 +1)
anyone got the same? :)
Post by: 7ooD on June 10, 2010, 05:51:36 pm
Post by: 7ooD on June 10, 2010, 06:03:24 pm
http://www.scribd.com/doc/28007685/Edexcel-GCE-January-2010-Mechanics-2-M2-Marking-Scheme and
here is how to solve whole question 8
http://i109.photobucket.com/albums/n63/7ood/bf1dcc50.jpg
http://i109.photobucket.com/albums/n63/7ood/ca603707.jpg
http://i109.photobucket.com/albums/n63/7ood/a5e4a210.jpg
http://i109.photobucket.com/albums/n63/7ood/ec107c11.jpg
Post by: shushuIG on June 11, 2010, 02:26:55 am
Post by: astarmathsandphysics on June 11, 2010, 08:09:26 am
Post by: astarmathsandphysics on June 11, 2010, 08:34:24 am
Post by: 7ooD on June 11, 2010, 11:21:18 am
Post by: anachronism on June 11, 2010, 11:42:12 am
done the exam now it was tricky not easy i kno it was easier than jan 2010 but still was tricky who did the last part in last question correct? im not discussing anything about the paper so plz dnt edit the post
Whatever you said above IS discussing the exam ::)
POST MODIFIED
Post by: 7ooD on June 11, 2010, 11:50:26 am
Post by: anachronism on June 11, 2010, 11:58:08 am
i didnt discuss the contents just said i left last part in last question that wont make someone cheat or something and nobody asked ur opinion u r reported for offensive language
Report me for whatever you want you prick.
I made a simple statement that you did deserve 'moron of the month'. I was simply commenting on an already existing title. And by saying ANYTHING about the exam, you basically discussed it [by definition of discuss]. So there :) You're just angry that I embarrassed you.
[this is a post worth reporting for offensive language]
Post by: VR on June 11, 2010, 12:01:12 pm
When do the Edexcel GCE Mathematics, May-June 2010 papers and mark schemes go up?
Post by: VR on June 11, 2010, 12:02:40 pm
Post by: 7ooD on June 11, 2010, 12:13:48 pm
Post by: astarmathsandphysics on June 11, 2010, 12:17:47 pm
Post by: anachronism on June 11, 2010, 12:26:59 pm
@ anachro u have previous history of disrespect for other members https://studentforums.biz/index.php/topic,8186.msg222631.html#msg222631 here and this post shows how u r very careful not to cheat https://studentforums.biz/index.php/topic,8186.msg222525.html#msg222525
I love to be mean to dumb people who look for it, what it to you? ::)
And you quoted by out of context, if you did do your research/stalking of my posts you'd see that I do not support cheating. In that post, I simply pointed out that CIE in fact won't ever know or they can't do anything about it.
Post by: VR on June 11, 2010, 12:56:27 pm
As soon as I get papers they will go up.
Yes, and when is that usually?
Post by: astarmathsandphysics on June 11, 2010, 02:11:42 pm
Post by: 7ooD on June 12, 2010, 12:47:35 am
Post by: anachronism on June 12, 2010, 01:30:40 pm
here iz the paper wiz solutions sry t.q this time :P posted before u https://studentforums.biz/index.php/topic,9101.msg268262.html#msg268262
Yea, but just like TQ you didn't produce that document so I dunno what you are celebrating about now :(
Post by: Meticulous on June 15, 2010, 04:10:19 pm
His post is modified now. :)
Post by: spiderman on August 29, 2010, 07:58:51 am
A boy of weight 294N falls 2.5 m onto a trampoline, which brings him to rest after he has descended a further distance of 0.2m. Calculate the k.e of the boy when he reaches the trampoline, and the energy stored in the trampoline when the boy is at rest?
Post by: S.M.A.T on August 29, 2010, 02:08:57 pm
E.P.E=0.5Fx=0.5×294×0.2=29.4J
Post by: spiderman on August 29, 2010, 03:39:54 pm
Post by: nid404 on August 29, 2010, 03:43:46 pm
hey guys..........................please cud ull help me wit this sum!!
A boy of weight 294N falls 2.5 m onto a trampoline, which brings him to rest after he has descended a further distance of 0.2m. Calculate the k.e of the boy when he reaches the trampoline, and the energy stored in the trampoline when the boy is at rest?
Total energy stored= 294 X (2.5+0.2)= 793.8 J
Post by: S.M.A.T on August 29, 2010, 03:48:35 pm
Total energy stored= 294 X (2.5+0.2)= 793.5 J
Can you please explain the answer,the formula for the elastic potential energy is 0.5Fx,where x is the extension.But how come the answer is 793.5J
Post by: nid404 on August 29, 2010, 03:54:18 pm
Well apparently the potential energy is converted to the elastic potential energy in the trampoline.
Post by: S.M.A.T on August 29, 2010, 04:07:02 pm
Post by: nid404 on August 29, 2010, 04:07:50 pm
Post by: nid404 on August 29, 2010, 04:25:52 pm
A racing car of mass 2000kg accelerates with a driving force of 480(t-10)2 newtons until it reaches its maximum speed after 10seconds. Find its maximum speed,and the distance it travels in reaching this speed.
I don't know even know how to begin -_-
Post by: nid404 on August 29, 2010, 04:32:28 pm
Post by: cooldude on August 29, 2010, 05:13:07 pm
I have a doubt ::)
A racing car of mass 2000kg accelerates with a driving force of 480(t-10)2 newtons until it reaches its maximum speed after 10seconds. Find its maximum speed,and the distance it travels in reaching this speed.
I don't know even know how to begin -_-
F=ma
a=480(t-10)^2/2000
we integrate with the limits 10 and 0 to find the max vel, and then integrate the velocity again with limits 10 and 0 to find the distance travelled.
Post by: S.M.A.T on August 29, 2010, 05:25:16 pm
Post by: nid404 on August 29, 2010, 05:27:10 pm
Post by: haris94 on September 08, 2010, 08:10:40 pm
the question is killing me
as the car passes a point A, its speed is 10m/s. the car moves with constant acceleration 'a'm/s2 along the road for 'T' seconds until it reaches a point B where the speed is 'V'm/s. the car travels at this speed for a further 10 seconds to reach a point C. from C, it travels for a further 'T' seconds with constant acceleration 3'a'm/s2 until it reaches a speed of 20m/s at point D. sketch a tv graph of motion and show that 'V' is 12.5m/s. given that the distance between A and D is 675m, find the value of 'a' and 'T'.
pls answer asap
Thanks
Post by: astarmathsandphysics on September 09, 2010, 03:50:48 pm
Post by: haris94 on September 10, 2010, 11:41:28 am
Post by: thecandydoll on September 14, 2010, 02:59:35 am
find co-eff of friction between B and floor.
Mass of a -250
mass of b- 200
A horizontal force of
magnitude PN is applied to B. The boxes remain at rest if P ? 3150 and start to move
if P > 3150.
The coefficient of friction between the two boxes is 0.2. Given that P > 3150 and that no sliding takes
place between the boxes,
(ii) show that the acceleration of the boxes is not greater than 2ms?2,
(iii) find the maximum possible value of P.
How do you go about this kind of sums :(
Post by: nid404 on September 14, 2010, 07:43:57 am
when object A is placed on object B which is placed on the floor(rough horizontal floor)
find co-eff of friction between B and floor.
Mass of a -250
mass of b- 200
A horizontal force of
magnitude PN is applied to B. The boxes remain at rest if P ? 3150 and start to move
if P > 3150.
The coefficient of friction between the two boxes is 0.2. Given that P > 3150 and that no sliding takes
place between the boxes,
(ii) show that the acceleration of the boxes is not greater than 2ms?2,
(iii) find the maximum possible value of P.
How do you go about this kind of sums :(
i) P max= uR
P<= 0.75 x 8000
P<=6000N
ii)F=0.4X4000=1600N
1600>=400a
a<=4m/s^2
Pmax-1600=3200
Pmax=9200 N
Post by: houmeira on September 25, 2010, 02:50:51 pm
two forces of magnitudes P N and 5 N act on a body. The angle between the two forces is theta and their resultant has magnitude 7N
when the direction of the 5 N force is reversed the magnitude of the new resultant is sqrt 19 N
calculate (i) the value of P
(ii) the value theta
Post by: houmeira on September 25, 2010, 02:57:12 pm
can anyone plz help with this question:
two forces of magnitudes P N and 5 N act on a body. The nagle between the two forces is theta and their resultant has magnitude 7N
when the direction of the 5 N force is reversed the magnitude of the new resultant is sqrt 19 N
calculate (i) the value of P
(ii) the value theta
theta is between P N and 5N
Post by: Deadly_king on September 25, 2010, 04:17:22 pm
can anyone plz help with this question:I'll take theta as A
two forces of magnitudes P N and 5 N act on a body. The angle between the two forces is theta and their resultant has magnitude 7N
when the direction of the 5 N force is reversed the magnitude of the new resultant is sqrt 19 N
calculate (i) the value of P
(ii) the value theta
Using cosine rule :
72 = 52 + P2 - 2(5)(P)cosA
49 = 25 + P2 - 10PcosA
24 = P2 - 10PcosA
That is the first equation.
Again use the cosine rule
19 = 25 +P2 -10Pcos(180-A)
-6 = P2 - 10Pcos(180-A)
This is the second equation. Now you just need to solve simultaneously :)
Post by: houmeira on September 25, 2010, 04:36:28 pm
need some help to solve it><
Post by: houmeira on September 25, 2010, 05:24:28 pm
yeah thanks i've tried itbefore and it's the simultaneous eqtn..i'm getting trouble to solve:S
need some help to solve it><
yay i got the answer!!#
thanks deadly king:P
Post by: Deadly_king on September 25, 2010, 05:41:21 pm
yay i got the answer!!#
thanks deadly king:P
Ok then :)
you welcome :)
Post by: astarmathsandphysics on October 04, 2010, 09:06:40 pm
A light elastic string has natural length 2.5m and modulus elasticity 15 N. A particle P of mass 0.5kg is attached to the string at the point K where K divides the unstretched string in the ratio 2:3. The ends of the string are then attached to the points A and B which are 5m apart on a smooth horizontal surface. The particle is then pulled aside and held at rest in contact with the surface at the point C where AC = 3m and ACB is a straight line. The particle is then released from rest.
a) show that P moves with simple harmonic motion of period pi/5 root2
b) find the amplitude of the motion
Post by: thecandydoll on October 17, 2010, 05:11:01 am
MAy june 2010.
paper 4,mechanics
Variant 43
Question 4.
We have to find the co-eff of friction of b.
We know Tension.
calculate for above subquestion.
it says the B moves up,so does this mean it accelerate up?
There fore T and a must be in the same direction?
Friction in the opposite direction?
I dont get it.
Please clear this for me.
Post by: nid404 on October 17, 2010, 05:54:12 am
So whether the particle is moving up or down, they both have the same acceleration.
I've drawn the free body diagram, hope it helps.
attached.
Post by: thecandydoll on October 17, 2010, 07:27:02 am
mechanics cie
paper 41
Q4!
Post by: nid404 on October 17, 2010, 07:35:16 am
Nov 09 Q4 --> https://studentforums.biz/math-146/mechanics-question/15/
Post by: thecandydoll on October 17, 2010, 09:09:07 am
And one more.
CIE.
MECHANICS.
PAPER 42. MAY/JUNE 2010.
Q6 (ii)
In the markscheme it says
0.2a2 = –0.06g
how do they get the -0.06
Post by: thecandydoll on October 17, 2010, 09:10:11 am
TTHANKK YOU
scratch that
Post by: Ghost Of Highbury on October 19, 2010, 09:46:16 am
Q4..
(AS level)
Could someone plz include diagrams and explanations in the answer...thanks a lot..
Post by: Deadly_king on October 19, 2010, 10:37:41 am
winter 2009, paper 41..
Q4..
(AS level)
Could someone plz include diagrams and explanations in the answer...thanks a lot..
i) First you need to resolve vertically so as to find the reaction.
R = 80 x cos 20 = 75.2 N
Since the block is at rest ----> a = 0
Using Newton's second law of motion : F = ma
T + Fr - mgsin x = ma
where T : tension in the string,
Fr : frictional from between block and the inclined plane
mgsin x : component of weight down the plane and x being the angle of inclination of the plane
Given T =13 and a=0 ----> 13 + Fr - 80sin 20 = 0
Hence Fr is found to be 14.4 N
ii) String is cut off but block is still at rest ----> Fr = mgsin x
Fr = uR where u is the coefficient of friction between block and plane
Hence 80sin 20 = u(75.2) -----> u = 0.364
Sorry I could not add a diagram :-[
Hope it helps :)
Post by: Ghost Of Highbury on October 19, 2010, 11:21:50 am
i) First you need to resolve vertically so as to find the reaction.
R = 80 x cos 20 = 75.2 N
Since the block is at rest ----> a = 0
Using Newton's second law of motion : F = ma
T + Fr - mgsin x = ma
where T : tension in the string,
Fr : frictional from between block and the inclined plane
mgsin x : component of weight down the plane and x being the angle of inclination of the plane
Given T =13 and a=0 ----> 13 + Fr - 80sin 20 = 0
Hence Fr is found to be 14.4 N
ii) String is cut off but block is still at rest ----> Fr = mgsin x
Fr = uR where u is the coefficient of friction between block and plane
Hence 80sin 20 = u(75.2) -----> u = 0.364
Sorry I could not add a diagram :-[
Hope it helps :)
erm, is that question 4 of winter 2009 M1 (paper 41) paper? :-/
Post by: The Golden Girl =D on October 19, 2010, 01:36:35 pm
Because i got this really Low grade in the Quiz and I don't want to show my Dad :S Since I'm dead meat for sure :-X :-X
Thank You In Advance .
Post by: elemis on October 19, 2010, 02:42:48 pm
Post by: SpongeBob on October 19, 2010, 03:19:01 pm
winter 2009, paper 41..
Q4..
(AS level)
Could someone plz include diagrams and explanations in the answer...thanks a lot..
I hope you understood the post above :/
Nov 09 Q4 --> https://studentforums.biz/math-146/mechanics-question/15/
Is this the one?
Post by: Deadly_king on October 19, 2010, 04:14:48 pm
erm, is that question 4 of winter 2009 M1 (paper 41) paper? :-/
That was June 09 p4 No4
Post by: The Golden Girl =D on October 19, 2010, 04:39:14 pm
@GG are you doing Edexcel or CIE M1 ?
Edexcel
Post by: Deadly_king on October 19, 2010, 05:29:53 pm
Hey ,I just wanted to know How Am i supposed to study for M1 , I want to know what is the best way to study for this Subject ?
Because i got this really Low grade in the Quiz and I don't want to show my Dad :S Since I'm dead meat for sure :-X :-X
Thank You In Advance .
M1 is supposed to be easy especially for those doing physics.
Hmm............studying M1 does not actually require much knowledge. You just need to understand the basic principles. Then only practice might help you perform well ;)
Post by: The Golden Girl =D on October 19, 2010, 07:36:05 pm
M1 is supposed to be easy especially for those doing physics.
Hmm............studying M1 does not actually require much knowledge. You just need to understand the basic principles. Then only practice might help you perform well ;)
Practice ..Ohh Ya :S :S
Thanks :)
Post by: **RoRo** on November 26, 2010, 04:09:31 pm
Question, from the chapter vectors, I came across this question and I got the answer from the CD that came with the book, but I can't understand
Q. An aeroplane flies from airport A to airport B 80 km away on a bearing of 070?. From B the aeroplane flies to airport C, 60 km from B. Airport C is 90 km from A. Find the two possible directions for the course set by the aeroplane on the second stage of its journey.
Ans. ?cos?? =(62+82?92) / (2×6×8) =0.19791…
? =78.6°(3 s.f.)
?bearing ofC from B is
180°+70°??=171.4°(1 d.p.) or 180°+70°+?=323.6°(1 d.p.)
& the diagram that they showed is attached below.
I don't understand how did they reach to this diagram and hence the solution!
Any help will be much appreciated!
Thanks a lot in advance! :)
Post by: ashish on November 27, 2010, 06:41:24 am
Mechanics shall drive me nuts soon!! :/
Question, from the chapter vectors, I came across this question and I got the answer from the CD that came with the book, but I can't understand
Q. An aeroplane flies from airport A to airport B 80 km away on a bearing of 070?. From B the aeroplane flies to airport C, 60 km from B. Airport C is 90 km from A. Find the two possible directions for the course set by the aeroplane on the second stage of its journey.
Ans. ?cos?? =(62+82?92) / (2×6×8) =0.19791…
? =78.6°(3 s.f.)
?bearing ofC from B is
180°+70°??=171.4°(1 d.p.) or 180°+70°+?=323.6°(1 d.p.)
& the diagram that they showed is attached below.
I don't understand how did they reach to this diagram and hence the solution!
Any help will be much appreciated!
Thanks a lot in advance! :)
it's like or it's a locus question.
locus of a point of r cm is a circle with radius r cm
and in this particular question C will be where the two circle intersect each other
sorry for the handwriting
Post by: **RoRo** on November 27, 2010, 07:44:40 am
it's like or it's a locus question.
locus of a point of r cm is a circle with radius r cm
and in this particular question C will be where the two circle intersect each other
sorry for the handwriting
Ohhh, now that makes sense!
Thanks a lot! ;D
Post by: **RoRo** on November 27, 2010, 12:18:15 pm
A particle P is moving along the x-axis with constant deceleration 2.5ms?2. At time t=0s, P passes through the origin with velocity 20ms?1 in the direction of x increasing. At time t=12s, P is at the point A. Find
(a) the distance OA,
(b) the total distance P travels in 12 s.
The answer for (a) is 60m and for (b) it's 100m.
HOW?
Post by: elemis on November 27, 2010, 12:22:58 pm
Post by: elemis on November 27, 2010, 12:27:45 pm
s = 20*12 + 0.5*-2.5*122
s = 60m
--------------------------------------------
I cant answer the second part since you must have left some information out or maybe the question is wrong.
Post by: **RoRo** on November 27, 2010, 12:42:17 pm
s = ut + 0.5at2
s = 20*12 + 0.5*-2.5*122
s = 60m
--------------------------------------------
I cant answer the second part since you must have left some information out or maybe the question is wrong.
Thank you!
Nope, the question is complete from the textbook.
According to the CD that I have, this is the working of the answer.
The particle will turn round when v=0.
a =?2.5, u=20, v=0, s=?
v2 =u2+2a?s
02 =202?5s
?s=80
The total distance P travels is (80+20)m=100m.
& the photo attached is the photo that they have shown.
It's alright, I figured it out now, I copied a wrong number into my calculator and hence my entire working was wrong! -.-'
Thank you! ;D
Post by: elemis on November 27, 2010, 12:44:30 pm
Post by: **RoRo** on November 27, 2010, 12:59:37 pm
1. How many s.f. are we supposed to leave our answers to? 2 or 3?
2. For questions on bearings, do we leave our angles to only 1 d.p.?
3. If a bearing is 42.5, do we leave the answer as 42.5 or 042.5? [since bearings are supposed to consist of 3 digits]
4. If an answer to a question is a long decimal, for example, an answer is 0.714285714 on the calculator, do I leave the answer as 0.71 or (5/7)?
Post by: elemis on November 27, 2010, 01:08:34 pm
Angles to 1 d.p.
Giving it as either will get the mark, however, to be safe stick to 042.5
State the answer exactly i.e. in fractions or surds whenever possible.
Post by: **RoRo** on November 27, 2010, 01:09:52 pm
Post by: elemis on November 27, 2010, 01:14:39 pm
Okie! Thanks a lot! ;D
Good luck ;)
When are you doing your M1 exam ? CIE or Edexcel ?
Post by: **RoRo** on November 27, 2010, 01:19:02 pm
I'm doing Edexcel, in the May/June 2011 session.
But I've got my first term school exams in a week, so yea, I'm preparing for them now!
What about you?
Post by: elemis on November 27, 2010, 01:23:10 pm
Thank you! :D
I'm doing Edexcel, in the May/June 2011 session.
But I've got my first term school exams in a week, so yea, I'm preparing for them now!
What about you?
I'm doing M1 this coming January.
Post by: **RoRo** on November 27, 2010, 01:25:47 pm
My school doesn't allow us to take any exams in November [for CIE] and January [for Edexcel], just the May/June session! :/
Post by: mdwael on November 27, 2010, 04:50:53 pm
A particle P is moving on the x-axis with constant deceleration of 4 (ms^-2). At time t= 0, P passes through the origin O with velocity 14 (ms-1) in the positive direction. The point A lies on the axis and OA = 22.5 m.
a) Find the difference between the times when P passes through A. (ans 2 sec)
b) Find the total distance travelled by P during the interval between these times. (ans 4m)
I got the first part as 2 sec which is correct, the second part is 4m but I dont understand how do you reach this answer?
Thanks in advance
Post by: astarmathsandphysics on November 27, 2010, 11:57:22 pm
u=
v=
a=-4
t=2
by symmetry v=-u
a=(v-u)/t
-4=2v/2
v=-4
s=
u=4
v=0
a=-4
t=1
s=ut+1/2at^2=4*1+1/2*-4*1^2=2
2m to stop and 2m back=4m
Post by: astarmathsandphysics on December 06, 2010, 10:02:17 pm
jaidee drove in the opposite direction going 6mph slower then perry for one hour after which time they
were 174 min apart. What was perry's speed.
I assume you mean 174 miles apart.
Perry's speed is v. After t hours Perry's distance from start point is vt.
Jaidee's distqance from start point is (t-1)(v-6)
vt+ (t-1)(v-6)=174
t=2 cos jaidee drove for 1 hour, 1 hour less than Perry.
2v+v-6=174
v=30
Post by: S.M.A.T on December 12, 2010, 08:31:36 pm
someone please help.
thanks in advance :D
Post by: The Golden Girl =D on December 15, 2010, 07:09:20 pm
Hmm in this Equation :
S=ut+0.5a(time Square)
When do we take the S as Displacement and when do we take it as Distance :'(
Please help
thx =]
Post by: iluvme on December 15, 2010, 07:17:49 pm
Post by: The Golden Girl =D on December 15, 2010, 07:25:07 pm
S is always the displacement and never the distance.
Hmm well here is a Question that says :
a book falls off the top of a bookcase.the shelf is 1.4m above the a wooden floor . (a) find the time the book takes to reach the floor , (b) the speed with which the book strikes the floor .
here we used the 1.4m as distance in the Equation I mentioned Prior , while in another Question :
A ball is projected vertically upwards ,from a point X which is 7m above the ground , with speed 21 m/s .Find (a) the greatest height above the ground reached by the ball , (b) the time of flight of the ball
here we used displcement as -7 , hence in the Equation S was used as Discplacement .
y there as distance and here as dicplacement :-\
Post by: The Golden Girl =D on December 15, 2010, 07:35:30 pm
Hmm well here is a Question that says :
a book falls off the top of a bookcase.the shelf is 1.4m above the a wooden floor . (a) find the time the book takes to reach the floor , (b) the speed with which the book strikes the floor .
here we used the 1.4m as distance in the Equation I mentioned Prior , while in another Question :
A ball is projected vertically upwards ,from a point X which is 7m above the ground , with speed 21 m/s .Find (a) the greatest height above the ground reached by the ball , (b) the time of flight of the ball
here we used displcement as -7 , hence in the Equation S was used as Discplacement .
y there as distance and here as dicplacement :-\
I'll go study since I still got some stuff to go over , So I'll see this thread in the morning iA =]
Please DO Help , I really need this :-\
thx =D
Post by: iluvme on December 15, 2010, 08:38:19 pm
Post by: The Golden Girl =D on December 15, 2010, 08:39:38 pm
Post by: elemis on December 16, 2010, 06:21:20 am
However, the second question states the object falls with constant velocity indicating no acceleration hence we can use the simple equation :
speed = distance/time
Post by: astarmathsandphysics on December 21, 2010, 03:51:07 pm
Post by: @d!_†oX!© on January 20, 2011, 04:26:15 pm
I have an exam in 3 days from now, and I feel so dumb cuz I don't know one whole chapter :P
Thought the sf people would be my last bet, so here it goes.
The problem is "Centre of Mass". Basically, the whole topic is a problem, but I guess posting all the questions that I have will be too much :P
So, I would like if someone could explain my the basic concept behind it, using the following two types of questions I have.
Post by: astarmathsandphysics on January 20, 2011, 10:42:22 pm
Post by: @d!_†oX!© on January 21, 2011, 09:45:40 am
hereThanks astar! You're always the rescue sir :)
Post by: @d!_†oX!© on January 22, 2011, 01:09:38 pm
Thanks in advance
Post by: astarmathsandphysics on January 22, 2011, 07:59:16 pm
Post by: The Golden Girl =D on January 23, 2011, 05:39:37 pm
A particle of mass 6 kg hangs in eguilibrium , suspended by two light extensible strings ,inclined at 60 degree and 45 degree to the horizontal ,as shown. Find the tension in each of the strings.
Answer is : S = 30.4 or 30.5 and T = 43.0 ...
what the hell is S o.0 ?
Post by: astarmathsandphysics on January 23, 2011, 07:59:16 pm
Post by: Zishi on February 13, 2011, 07:21:45 am
Thanks~
Post by: astarmathsandphysics on February 15, 2011, 01:08:56 pm
Post by: **RoRo** on February 17, 2011, 12:43:37 pm
Guys, I need help in Dynamics, section 5.3, which is about particles inclined on a plane where you need to resolve the forces parallel and perpendicular to the plane.
I don't understand how the forces are resolved, how do they know which angle is it and when to use sin[theta] and cos[theta].
Can someone please explain this to me?
Thanks a lot! :)
& here's an example of a question, in case anyone would need it. :)
A particle of mass 5 kg is at limiting equilibrium on a rough plane which is inclined at an angle of 30 degrees to the horizontal. Find the coefficient of friction between the particle and the plane.
Post by: The Golden Girl =D on February 17, 2011, 01:18:45 pm
Mechanics is on it's way to drive me insane! -.-"
Guys, I need help in Dynamics, section 5.3, which is about particles inclined on a plane where you need to resolve the forces parallel and perpendicular to the plane.
I don't understand how the forces are resolved, how do they know which angle is it and when to use sin[theta] and cos[theta].
Can someone please explain this to me?
Thanks a lot! :)
& here's an example of a question, in case anyone would need it. :)
A particle of mass 5 kg is at limiting equilibrium on a rough plane which is inclined at an angle of 30 degrees to the horizontal. Find the coefficient of friction between the particle and the plane.
I have two pics that I hope will explain the general idea to you.
look at the worked solution for the Question on this page if you need a better idea of how to do it:
http://www.examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/M1/particles-in-equilibrium/revision-guide.php
=]
Post by: The Golden Girl =D on February 17, 2011, 01:39:41 pm
Quote
& here's an example of a question, in case anyone would need it. :)
A particle of mass 5 kg is at limiting equilibrium on a rough plane which is inclined at an angle of 30 degrees to the horizontal. Find the coefficient of friction between the particle and the plane.
Post by: **RoRo** on February 17, 2011, 06:03:55 pm
I have two pics that I hope will explain the general idea to you.
look at the worked solution for the Question on this page if you need a better idea of how to do it:
http://www.examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/M1/particles-in-equilibrium/revision-guide.php
=]
I UNDERSTOOD IT!! =D
Thanks ALOTT!! :D
Now a question, check the photo attached for the question, the diagram and the answer from the CD that came with the textbook.
How is the acceleration = 5/7, when in part a) it was 1.4?
Post by: The Golden Girl =D on February 17, 2011, 08:12:32 pm
I UNDERSTOOD IT!! =D
Thanks ALOTT!! :D
Now a question, check the photo attached for the question, the diagram and the answer from the CD that came with the textbook.
How is the acceleration = 5/7, when in part a) it was 1.4?
I'm sorry but can you tell me the exact exercise and the number of the Question please.
Post by: **RoRo** on February 17, 2011, 09:07:38 pm
I'm sorry but can you tell me the exact exercise and the number of the Question please.
I'm sorry, I forgot to attach the file! :/
There you go with it, it's Exercise 3F Q4b
Post by: elemis on February 18, 2011, 09:46:34 am
I'm sorry, I forgot to attach the file! :/
There you go with it, it's Exercise 3F Q4b
After A hits the ground the string goes slack and there is no tension. Hence, B continues to move upwards but under the force of gravity.
That is, the acceleration of B will be = -9.8 ms-2
Post by: The Golden Girl =D on February 18, 2011, 10:05:54 am
I'm sorry, I forgot to attach the file! :/
There you go with it, it's Exercise 3F Q4b
my teacher taught us a different way of doing this :
we want the GREATEST height and so we have to do TWO steps.
Step 1;
the Speed with which the Particle A strikes the ground
u = o , v = ? ,a = 1.4 ms-1 , s = 2
v2 = u2 + 2as
v2 = 0 +( 2 * 1.4 * 2 )
v2 = 5.6
v = 2.37 ms-1
Step 2
the distance traveled by B after A has reached the ground
u = 2.37 , v = 0 , a = - 9.8 (going up hence against gravity ) , s = ?
v2 = u2 + 2as
0 = (2.37)2 + (2*-9.8* s )
{19.6 s = (2.37)2 } / 19.6
s = 0.2865765
MAX height = 2 + 0.2865765
= 16/7 m
=]
Post by: elemis on February 18, 2011, 10:09:22 am
my teacher taught us a different way of doing this :
we want the GREATEST height and so we have to do TWO steps.
Step 1;
the Speed with which the Particle A strikes the ground
u = o , v = ? ,a = 1.4 ms-1 , s = 2
v2 = u2 + 2as
v2 = 0 +( 2 * 1.4 * 2 )
v2 = 5.6
v = 2.37 ms-1
Step 2
the distance traveled by B after A has reached the ground
u = 2.37 , v = 0 , a = - 9.8 (going up hence against gravity ) , s = ?
v2 = u2 + 2as
0 = (2.37)2 + (2*-9.8* s )
{19.6 s = (2.37)2 } / 19.6
s = 0.2865765
MAX height = 2 + 0.2865765
= 16/7 m
=]
LOL. You didnt answer her question ::)
Post by: Zishi on February 18, 2011, 12:06:51 pm
here
Thanks for that, but why the two tensions are same? I just can't understand that; which word or thing in the question makes us understand that the tensions will be same. I've solved a lot of questions of like this before, but the tensions in the two strings were not same. Plus, please answer the third part too now!(Attached)
Post by: **RoRo** on February 18, 2011, 12:31:23 pm
After A hits the ground the string goes slack and there is no tension. Hence, B continues to move upwards but under the force of gravity.
That is, the acceleration of B will be = -9.8 ms-2
mmm, alright, that makes sense too!
My question was, when calculating the acceleration for A, it was taken as 5/7 and not 1.4 like we had found in part a) of the question, why is that?
Post by: **RoRo** on February 18, 2011, 12:32:24 pm
my teacher taught us a different way of doing this :
we want the GREATEST height and so we have to do TWO steps.
Step 1;
the Speed with which the Particle A strikes the ground
u = o , v = ? ,a = 1.4 ms-1 , s = 2
v2 = u2 + 2as
v2 = 0 +( 2 * 1.4 * 2 )
v2 = 5.6
v = 2.37 ms-1
Step 2
the distance traveled by B after A has reached the ground
u = 2.37 , v = 0 , a = - 9.8 (going up hence against gravity ) , s = ?
v2 = u2 + 2as
0 = (2.37)2 + (2*-9.8* s )
{19.6 s = (2.37)2 } / 19.6
s = 0.2865765
MAX height = 2 + 0.2865765
= 16/7 m
=]
This makes more sense to me than the working on the CD, thanks a lot and sorry for all the trouble! :)
Post by: **RoRo** on February 18, 2011, 01:05:18 pm
I don't understand how part c) is solved, the working is given below!
Check the attachment, it has the question and the answer!
Note:
Acceleration of the system = 0.613
Tension in the string = 27.6
Post by: elemis on February 18, 2011, 01:20:11 pm
mmm, alright, that makes sense too!
My question was, when calculating the acceleration for A, it was taken as 5/7 and not 1.4 like we had found in part a) of the question, why is that?
In part a tension was an added factor, when the tension disappears the acceleration changes.
M1 is about common sense and spatial awareness.
Post by: elemis on February 18, 2011, 01:21:44 pm
Okay, another question:
I don't understand how part c) is solved, the working is given below!
Check the attachment, it has the question and the answer!
Note:
Acceleration of the system = 0.613
Tension in the string = 27.6
Consider the two tensions as forces (never mind the direction they act in) and find the resultant force.
Post by: astarmathsandphysics on February 18, 2011, 01:27:03 pm
Post by: astarmathsandphysics on February 18, 2011, 01:28:54 pm
Post by: **RoRo** on February 18, 2011, 01:37:54 pm
Consider the two tensions as forces (never mind the direction they act in) and find the resultant force.
Alright, thank you very much! :D
Post by: Zishi on February 19, 2011, 04:39:48 am
The two tensions are the same because the particle in free to move.Alright, thank you! And what about the third part which I attached in my previous post?
EDIT: By the way I've attached my diagram showing what I did, and how I resolved forces, etc. Please tell me why I can't find R, and if is any direction wrong?!
Post by: Sue T on February 21, 2011, 01:50:02 pm
7. Two small spheres P and Q of equal radius have masses m and 5m respectively. They lie
on a smooth horizontal table. Sphere P is moving with speed u when it collides directly
with sphere Q which is at rest. The coefficient of restitution between the spheres is e,
where e > 1/5
(a) (i) Show that the speed of P immediately after the collision is u/6(5e-1)
(ii) Find an expression for the speed of Q immediately after the collision, giving your
answer in the form ?u, where ? is in terms of e.
the marking scheme dint say nything bout this one - wat is it??!
Three small spheres A, B and C of equal radius lie at rest in a straight line on a smooth
horizontal table, with B between A and C. The spheres A and C each have mass 5m, and the
mass of B is m. Sphere B is projected towards C with speed u. The coefficient of restitution
between each pair of spheres is .
(b) Show that, after B and C have collided, there is a collision between B and A.
th marking scheme said:
After B hits C, velocity of B = “v” = 1/6 (1 – 5*4/5 )u = – ½u
velocity < 0 ? change of direction ? B hits A
but were did all the negative come from - wasn't it u/6(5e-1)? why'd they reverse the signs out of no where??! is it jus 4 the sake of proving tht there'll be a collision?!
Post by: **RoRo** on February 21, 2011, 04:19:00 pm
Check the attachment below for the question.
Post by: elemis on February 21, 2011, 04:20:40 pm
How do you solve part c) ?
Check the attachment below for the question.
2 minutes.
Post by: **RoRo** on February 21, 2011, 04:35:11 pm
(a) -2g-0.4(2g*0.8 ) = 2*a
Therefore a = -12.936 ms-2
Using V^2-U^2=2as we find distance = 0.154607 m ?
Correct ?
Nope, the correct answer is 1.1ms-1 (2 s.f.)
& the working on the CD is attached, but I don't understand how did the get that 2a = 14g/25?
In part b) we had to show that the box will slide down the plane, so I took the net force down the plane and that was the answer [14g/25], how did that change to acceleration all of a sudden?
Post by: elemis on February 21, 2011, 04:40:40 pm
Nope, the correct answer is 1.1ms-1 (2 s.f.)
& the working on the CD is attached, but I don't understand how did the get that 2a = 14g/25?
In part b) we had to show that the box will slide down the plane, so I took the net force down the plane and that was the answer [14g/25], how did that change to acceleration all of a sudden?
I made a mistake using the wrong initial speed.
When the diagram slides back down the plane it will accelerate obviously.
Using the below diagram :
We can see the friction acts in the opposite direction to motion. Resolving the forces :
Hence, 2g*sin alpha -0.4*2g cos alpha = 2*a
a = 2.744 ms^-2
Since the velocity at the top of the slope will be zero :
v^2 = 0^2 + 2*a*0.22
V = 1.10 ms^-1
Post by: elemis on February 21, 2011, 04:44:18 pm
Considering the frictional force only : 0.4*R = 6.272 N
Comparing this to the weight : 2g*0.6 = 11.76 N
We can see that friction cannot prevent the weight from pulling the object down.
Post by: **RoRo** on February 21, 2011, 05:06:24 pm
I made a mistake using the wrong initial speed.
When the diagram slides back down the plane it will accelerate obviously.
Using the below diagram :
We can see the friction acts in the opposite direction to motion. Resolving the forces :
Hence, 2g*sin alpha -0.4*2g cos alpha = 2*a
a = 2.744 ms^-2
Since the velocity at the top of the slope will be zero :
v^2 = 0^2 + 2*a*0.22
V = 1.10 ms^-1
How is s=0.22?
And the acceleration will not be taken as a negative value because when we resolved the forces and found the acceleration, the resultant was in the downwards direction?
Post by: **RoRo** on March 20, 2011, 04:54:34 pm
The answer for a) 11.4N and for b) is 13.9N
I found that I can get the answer by doing the following:
R= 11.4Cos35 + 8Cos55 = 13.9
But I don't understand why is it an addition sign?
The horizontal 8N force, when resolved, should have 2 values, one that acts up the plane and the other one is perpendicular to it, why is it pointing downwards? [to get the answer 13.9]
I'm hoping my question is clear, if not, please let me know and I'll clarify further! :)
Post by: astarmathsandphysics on March 20, 2011, 11:42:00 pm
Post by: **RoRo** on March 21, 2011, 11:57:36 am
look at the arrows on the 8N force on the diagram below. The force is pushing the block into the plane
Oh, okay, I got it, thanks a lot! :)
Post by: The Golden Girl =D on March 22, 2011, 04:03:54 pm
I NEED this URGENTLY :'(
Post by: elemis on March 22, 2011, 04:05:54 pm
Chapter 6 , Page 146 , Q1}a] ..... Q2}a] ....and Q4}a]
I NEED this URGENTLY :'(
10 minutes.
Post by: elemis on March 22, 2011, 04:10:45 pm
Hence,
Q1
r = 3j +8i
= 8i +3j
Q2
6i+13j = 2i+3j + 2v
4i +10j = 2v
2i +5j= v
Can you apply the same method for Question 4 ?
Post by: The Golden Girl =D on March 22, 2011, 04:14:21 pm
Position vector after a time = Initial Position vector + velocity*time
Hence,
Q1
r = 3j +8i
= 8i +3j
Q2
6i+13j = 2i+3j + 2v
4i +10j = 2v
2i +5j= v
Can you apply the same method for Question 4 ?
Thanks mate :)
Post by: **RoRo** on April 07, 2011, 07:25:36 pm
Check the attachment for the questions and the answers [at the bottom]
I need the steps for the following questions:
4b)
5a)b)
7c)d)
8c)
Thanks A LOT in advance! :)
Post by: iluvme on April 07, 2011, 08:15:56 pm
Not sure about the others :S
Post by: **RoRo** on April 07, 2011, 08:32:56 pm
4 b)
Not sure about the others :S
Thank you so much!
I did a super-stupid mistake, instead of adding 4 to 3 and writing the answer as 7, I wrote it as 1! -.-"
Post by: Darkstar3000 on April 09, 2011, 03:53:29 pm
Mark scheme : http://www.carlgauss.co.uk/PDFs/A-Level%20Papers/M2/M2%20-%202008%20Jan%20Mark%20Scheme.pdf
Am I wrong or is the mark scheme wrong because when I calculate the answer I'm getting 0.72 but they get 0.18
----------------------------------------------------------
How do I do question number 6, the whole thing ?
Thanks
Post by: elemis on April 09, 2011, 07:10:41 pm
Post by: elemis on April 10, 2011, 06:20:02 am
Considering vertical forces you will find that R = 4mg
Taking moments about point B :
2a*cos60*mg + 3a*cos60*3mg + 4a*cos30*4mg
Solve this and you will get an answer of : 0.180
Question 6
Attempt the question and show me your workings. I'll see where you've gone wrong and correct you.
Post by: Darkstar3000 on April 10, 2011, 11:59:08 am
Question 5
Considering vertical forces you will find that R = 4mg
Taking moments about point B :
2a*cos60*mg + 3a*cos60*3mg + 4a*cos30*4mg= 4a*cos60*4mg
Solve this and you will get an answer of : 0.180
Question 6
Attempt the question and show me your workings. I'll see where you've gone wrong and correct you.
Thank you for question 5.
As for 6, I have no idea where to start it
Post by: jellybeans on April 11, 2011, 08:01:44 am
For (a) I equated j component = 0
-> (UsinthetaT - ½gcosthetaT2) = 0
-> (Usin60T - ½gcos30T2) = 0
-> T=0, (root3)U/2 - (root3)g/4(T))T=0
-> (root3)g/4(T) = (root3)U/2
so i got -> T=2U/g
but it says 2U/g(root3)
could somebody help me on this question please.. thanks! :)
Post by: **RoRo** on April 12, 2011, 12:40:02 pm
Thank you! :)
Post by: HUSH1994 on April 17, 2011, 12:55:21 pm
Post by: iluvme on April 17, 2011, 01:18:49 pm
MJ 2003 Q4 part 2,i did it wrong and then the mark scheme had 2 dots above x,anyone can help me with this
CIE?
Post by: HUSH1994 on April 17, 2011, 02:46:46 pm
Post by: iluvme on April 17, 2011, 06:14:15 pm
Here you go,
Post by: HUSH1994 on April 19, 2011, 04:51:40 pm
Post by: **RoRo** on April 20, 2011, 11:46:14 am
a) the value of u
Ans is 28
b) the total time for which the particle is more than 17 1/2m above A.
Ans is 4 2/7
My question is, when doing part b) I get 2 answers 5 and 5/7, why do we subtract them to get 4 2/7?
Post by: astarmathsandphysics on April 25, 2011, 03:28:25 pm
Post by: **RoRo** on April 25, 2011, 08:36:49 pm
The particle rises above 17 1/2 m at t=5/7 and falls below 17 1/2 m at t=5 so is above 17 1/2 m for 5-2/7 =4 2/7 s
Ohh, okay, thank you! :)
Can you please answer my questions in my previous post here? I uploaded a picture with a couple of questions that I needed help on, thanks a lot! :)
Post by: astarmathsandphysics on April 25, 2011, 09:05:04 pm
Post by: astarmathsandphysics on April 25, 2011, 10:56:34 pm
did yo use 30 or 60?
Post by: astarmathsandphysics on April 26, 2011, 01:48:34 pm
Post by: **RoRo** on April 29, 2011, 09:21:12 am
those questions answered in attachments
Thank you very much for your help Sir, I'll go through them once I start revising Mechanics again, because right now I've got my A2 math mocks coming up next week!
Post by: doubtigetastar on April 29, 2011, 09:02:14 pm
a uniform lamina ABCDEF is formed by removing a rectangle of dimensions 8 cm and 12 cm from a square of side length 20 cm , find the distance of the centre of mass of the lamina from its edges : a) AF b) FE
i made a diagram it's in the attachments .
Post by: astarmathsandphysics on April 29, 2011, 10:28:23 pm
Post by: astarmathsandphysics on April 30, 2011, 07:53:00 am
Post by: doubtigetastar on April 30, 2011, 11:24:27 am
one last question , how did you get the distance of the big rectangle (area 400) from the line FE :D
Post by: astarmathsandphysics on April 30, 2011, 03:04:11 pm
Post by: Weaam on May 03, 2011, 02:37:41 pm
Most importantly M1 Jan 09 mark schemee
Post by: astarmathsandphysics on May 04, 2011, 07:41:33 am
Post by: wakemeup on May 07, 2011, 09:05:20 pm
Post by: elemis on May 08, 2011, 08:33:57 am
Can someone please clearly explain 8d of this question paper to me? The mark scheme isn't making much sense at all.
The displacement equation for 0<t<4 :
4t2 -0.5t3
From t>4 :
16t - t2 -16
In the first 4 seconds the displacement is : 32m
It is at instantaneous rest at 8 seconds. At this point it changes direction such that its displacement from the origin will begin to DECREASE.
At 8 seconds displacement is : 48
At 10 seconds : 44 m NOTE : The displacement has decreased as previously stated.
Hence, it travels 48m FORWARDS and 4 m BACKWARDS.
Thus, 48 + 4 = 52m
Post by: wakemeup on May 08, 2011, 12:16:50 pm
Post by: iluvme on May 11, 2011, 12:23:59 pm
I got a=6ms-2
How do you show the speed as 3.95 ms-1?
Post by: elemis on May 11, 2011, 01:01:56 pm
June 2010 Paper 41 Question 6 i)
I got a=6ms-2
How do you show the speed as 3.95 ms-1?
CIE ? Give me a minute.
Post by: elemis on May 11, 2011, 01:11:37 pm
T - 0.3*0.2*10 = 0.2a
Adding the two equation to eliminate T gives :
0.45*10 - 0.3*0.2*10 = 0.65a
Solving for a :
a = 6
Now. If you read the first part of the question you'll notice a very important piece of information which I think you've missed :
Quote
Particles A and B, of masses 0.2 kg and 0.45 kg respectively, are connected by a light inextensible
string of length 2.8m. The string passes over a small smooth pulley at the edge of a rough horizontal
surface, which is 2m above the floor. Particle A is held in contact with the surface at a distance of
2.1m from the pulley and particle B hangs freely (see diagram). The coefficient of friction between
A and the surface is 0.3. Particle A is released and the system begins to move.
That means the distance from the pulley to B is 0.7 m. Hence, B is 1.3 m above the ground.
Use v^2=u^2 +2as to find the final speed.
Post by: iluvme on May 11, 2011, 01:16:44 pm
Could you help me with second part too?
Post by: elemis on May 11, 2011, 01:17:18 pm
Thanks Ari.
Could you help me with second part too?
Okay. Hang on.
Post by: elemis on May 11, 2011, 01:25:46 pm
Hence, for A :
-0.3*0.2*10 = 0.2*a
a = -3
However, by this time A has moved 1.3 metres towards the pulley (since B fell 1.3 m) and A has a speed of 3.95.
So using this data you should be able to calculate the final speed.
Post by: iluvme on May 11, 2011, 01:31:40 pm
Post by: elemis on May 11, 2011, 01:38:51 pm
Why is it -0.3?
Since friction is acting in the direction OPPOSITE to motion.
Post by: iluvme on May 11, 2011, 01:41:48 pm
Thanks Ari. :)
Post by: **RoRo** on May 12, 2011, 02:52:11 pm
The secret are you edexcel?
and By the way in MJ 2011 P41 the meu(u) was over one, my teacher told me its correct but it can never go beyond 1,so this was a trick?
Last thing is that how do u calculate squared trigonometry fuction,eg:
tan^2(Theta)= -0.56
Yupp, I am Edexcel and I assume the paper you're talking about is CIE!
Post by: **RoRo** on May 12, 2011, 03:01:23 pm
Thank you ;D
Post by: cashem'up on May 12, 2011, 04:56:53 pm
is attached to a fixed point. A particle of mass 4 kg is attached to the other end. Find the extension,
e metres, of the string when the particle hangs freely in equilibrium at E.[2]
When the particle is in equilibrium at E, it is given a blow of impulse 20Ns vertically downwards.
Show that, while the string remains taut, the motion of the particle is simple harmonic, and find the
period and amplitude of this simple harmonic motion. [8]
Find also the time that elapses between the blow and the instant when the string first becomes slack.
[4]
The first part is easy to get... but i am having difficulties in the subsequent parts... please help
Post by: leebux101 on May 12, 2011, 05:37:55 pm
Post by: elemis on May 13, 2011, 11:01:40 am
MJ 2003 P4, q 5 ii . PLEASE ASAP!
Consider the two particles to be one single particle of mass 0.4 kg.
Hence,
0.4*10 - (0.4+0.2) = 0.4*a
a = 8.5
Now consider only one of the particles (I will use particle B) :
B has its weight acting downwards, tension upwards and air resistance of force 0.2 N.
Keeping in mind that B is accelerating at 8.5 m/s2 can you calculate the tension ?
Post by: HUSH1994 on May 13, 2011, 11:41:19 am
Last thing is that how do u calculate squared trigonometry fuction,eg:
tan^2(Theta)= -0.56
Post by: The Golden Girl =D on May 13, 2011, 12:19:33 pm
Can someone please explain to me how do you solve Q7d in January 2010, Q7c in January 2008, Q7f in January 2007 [Edexcel papers]?
Thank you ;D
Jan 08 ; Q.7]c) the fact that you ignored the mass/weight of the particles A and B.
for such Questions I suggest you read chapter one in the book *the ones with only words and definitions*
Jan 07 : Q.7]f)
16:00 - 14:00 = 2 hrs
at 16:00 , t = 2hrs
11i + (17+10) j
rShip = 11i = 27j
S from R i.e. RS
RS = rShip - rrock
= (11i + 27j ) - (8.5 + 23 j )
= (2.5 + 4j )
distance RS = square root of (2.52) + (42)
Distance RS = 4.72 Km
Post by: **RoRo** on May 13, 2011, 12:54:42 pm
Jan 08 ; Q.7]c) the fact that you ignored the mass/weight of the particles A and B.
for such Questions I suggest you read chapter one in the book *the ones with only words and definitions*
Jan 07 : Q.7]f)
16:00 - 14:00 = 2 hrs
at 16:00 , t = 2hrs
11i + (17+10) j
rShip = 11i = 27j
S from R i.e. RS
RS = rShip - rrock
= (11i + 27j ) - (8.5 + 23 j )
= (2.5 + 4j )
distance RS = square root of (2.52) + (42)
Distance RS = 4.72 Km
I think you've solved the wrong questions,
For Q7c in Jan 2008, the question was find the speed of A as it reaches P!
& for 7f in Jan 2007, it's a question from Dynamics not vectors
Post by: leebux101 on May 13, 2011, 04:27:31 pm
Consider the two particles to be one single particle of mass 0.4 kg.yes thankyou sooooo much! =D =D
Hence,
0.4*10 - (0.4+0.2) = 0.4*a
a = 8.5
Now consider only one of the particles (I will use particle B) :
B has its weight acting downwards, tension upwards and air resistance of force 0.2 N.
Keeping in mind that B is accelerating at 8.5 m/s2 can you calculate the tension ?
Post by: leebux101 on May 13, 2011, 04:29:24 pm
Post by: 8T on May 13, 2011, 06:40:06 pm
Could someone help out in the following questions ?
M1 Jan O8
4 C
6 b c d
7 all of it
I know they are many but please if anyone could explain how we solve them ?
Thanks in advance :D
Post by: 8T on May 13, 2011, 07:02:46 pm
ON 2009 P41 q4 , all of it, and can u please explain with a diagram of resolved forces please? =d
This edexcel right ?
Do you mean May or Jan 2009 ???
Post by: leebux101 on May 13, 2011, 07:14:59 pm
Post by: HUSH1994 on May 14, 2011, 06:39:49 am
https://studentforums.biz/index.php?action=dlattach;topic=7281.0;attach=13300;image (By iluvme)
Post by: Naruto123 on May 14, 2011, 02:11:42 pm
like i knw the forces tension nd recently thrust arre there any oder such forces found in connected particles ?
and plz wid a diagram can som1 explain thrust (pushing force) clearly
thanku
Post by: The Golden Girl =D on May 14, 2011, 05:13:05 pm
Hello :)
Could someone help out in the following questions ?
M1 Jan O8
4 C
6 b c d
7 all of it
I know they are many but please if anyone could explain how we solve them ?
Thanks in advance :D
check this website => http://www.examsolutions.co.uk/
the section that has the Edexcel papers :)
Post by: leebux101 on May 14, 2011, 05:23:41 pm
ON 2009 P41 q4 , all of it, and can u please explain with a diagram of resolved forces please? =dcan someone please explain this?? :(
Post by: iluvme on May 14, 2011, 06:01:38 pm
can someone please explain this?? :(
Here you go, and the iii) sorry its not that legible.
Post by: leebux101 on May 14, 2011, 07:40:38 pm
Here you go, and the iii) sorry its not that legible.u are awesome .
Post by: iluvme on May 14, 2011, 08:49:19 pm
Post by: leebux101 on May 14, 2011, 08:50:43 pm
Did you get part iii)?yup! thanks ALOT!
Post by: SkyPilotage on May 14, 2011, 09:11:32 pm
I use Work by Driving force=change in p.e+change in k.e+work agnst resistance
Others use Work by D - Work agnst R = Final M.E - Initial M.E???
I know both are the same but which is prefered to be used as A RULE before substituting..???
Another question..
When they ask for Loss in K.e..Do we give it as +ve loss or -ve(change) Do you get what I mean?? For ex: if they ask for change answer wod be -ve what if they ask for Loss we give as +ve right?? What abt the equation (rule( do we give it as 0.5m(u^2-v^2) or 0.5(v^2-u^2) then state its -ve because its loss??)
Plz answer me If you are sure ;)
Thank You ..
Post by: Blizz_rb93 on May 14, 2011, 09:33:37 pm
help please
Post by: iluvme on May 14, 2011, 10:10:36 pm
yup! thanks ALOT!
Welcome :)
Post by: leebux101 on May 14, 2011, 10:38:23 pm
Hey guys what rule do you use for work??ok for ur first question , i had the same confusion, but i have learned that logic works best in this kind of problem, For example >
I use Work by Driving force=change in p.e+change in k.e+work agnst resistance
Others use Work by D - Work agnst R = Final M.E - Initial M.E???
I know both are the same but which is prefered to be used as A RULE before substituting..???
Another question..
When they ask for Loss in K.e..Do we give it as +ve loss or -ve(change) Do you get what I mean?? For ex: if they ask for change answer wod be -ve what if they ask for Loss we give as +ve right?? What abt the equation (rule( do we give it as 0.5m(u^2-v^2) or 0.5(v^2-u^2) then state its -ve because its loss??)
Plz answer me If you are sure ;)
Thank You ..
work is done by the driving force to overcome loss in ke and work done against resistance, because they are both actually working against letting an object move ryt? cuz ke is lost bcuz there was resistance.. do u get it? so work dont by driving force = Loss in ke + work done by resistive forces... secondly this formula >> work done by DF - Work done by Resistive forces = gain in KE , apply this to the situation, if u dont get the answer then u need to think logically about the changes invloved...
if u dont understand this, then PLEASE do ASK me again and i will give u specific example from pastpapers,
for ur second problem >>
u can change the formula from 1/2m(v^2 - u^2) to 1/2m(u^2 - v^2) .. we always write the value as postive no matter if its gain or loss
the first formula is for gain because final speed will be greater than initial speed so u subtract v^2 from u^2.
the second formula is for loss cuz obviously initial speed will be greater than final speed so u subtract u^2 from v^2 ..
do u get it?
Post by: The Golden Girl =D on May 15, 2011, 10:49:36 am
I got the S= 0.025 , t = 0.0714 , as well as the v = 0.7 , but I can't seem to get the Total distance nor total time right =.=
Thank you in advance :)
Attached.
Post by: EMO123 on May 15, 2011, 11:47:36 am
URGENT June 2010 Q8]d) , Edexcel By the way..there is no Q8 d)
I got the S= 0.025 , t = 0.0714 , as well as the v = 0.7 , but I can't seem to get the Total distance nor total time right =.=
Thank you in advance :)
Attached.
Post by: The Golden Girl =D on May 15, 2011, 01:00:20 pm
Post by: The Golden Girl =D on May 15, 2011, 06:48:05 pm
URGENT June 2010 Q8]c) , Edexcel By the way..
I got the S= 0.025 , t = 0.0714 , as well as the v = 0.7 , but I can't seem to get the Total distance nor total time right =.=
Thank you in advance :)
Attached.
Anyone :-X
Post by: HUSH1994 on May 16, 2011, 04:09:50 am
Post by: your sister on May 16, 2011, 06:31:37 am
Post by: HUSH1994 on May 16, 2011, 07:05:19 am
goodluck for M1 cie-ers :)Thanks and same to you,exam in 2 hours :D
Post by: ........ on May 16, 2011, 03:23:20 pm
Hey guys what rule do you use for work??
I use Work by Driving force=change in p.e+change in k.e+work agnst resistance
Others use Work by D - Work agnst R = Final M.E - Initial M.E???
I know both are the same but which is prefered to be used as A RULE before substituting..???
Another question..
When they ask for Loss in K.e..Do we give it as +ve loss or -ve(change) Do you get what I mean?? For ex: if they ask for change answer wod be -ve what if they ask for Loss we give as +ve right?? What abt the equation (rule( do we give it as 0.5m(u^2-v^2) or 0.5(v^2-u^2) then state its -ve because its loss??)
Plz answer me If you are sure ;)
Thank You ..
Do you by any chance study from Mr Shariq in Jeddah
Post by: SkyPilotage on May 18, 2011, 02:32:43 pm
Do you by any chance study from Mr Shariq in JeddahMr. Shariq? Nope...
Post by: mdwael on May 18, 2011, 06:41:06 pm
Anyone :-X
Do you still want the answer?
Post by: SkyPilotage on May 18, 2011, 06:43:27 pm
Do you still want the answer?lol i doubt :D
Post by: The Golden Girl =D on May 18, 2011, 07:26:12 pm
Do you still want the answer?
Thx but I've already done the exam :)
Post by: username on May 19, 2011, 07:13:15 am
Post by: elemis on May 19, 2011, 07:22:43 am
so we can discuss yet???
Yes, but not here.
Discuss here :
https://studentforums.biz/gce-as-a2-level-(cie-edexcel)/2011-mayjune-examination-discussion-here-!!!!/525/ (https://studentforums.biz/gce-as-a2-level-(cie-edexcel)/2011-mayjune-examination-discussion-here-!!!!/525/)
Post by: aloha32 on May 24, 2011, 09:38:01 pm
Can anyone please explain why in June 2005 Qs 7 paper we take moments about F and not A?
and Nov 09 Q5 part one for circular motion..how is sin theta = 3/8
and sorry this one's REALLY confusing too..why isn't the speed 10m/s in Q5 Nov 2003 after rebounding at right angles? The mark scheme says "NOT 10m/s" and if you had correct method in 10m/s, there would be no marks eithers. So how did we get 10m/s?
Thank you!
Post by: yasser37 on May 24, 2011, 10:38:02 pm
These are the doubts I have so far
please help :)
November 09 , 51, Q4 (ii)
November 10 . 53 , Q4(ii)
November 09 , 52 ,Q1
June 10 , 53, Q(i) ---- A Drawing of angles taken would be greatly appreciated
June 09, 5, Q2 (i) Same as above and Q(6) (ii)
I hope that someone can help me before my exam
+rep for helper :-*
Post by: yasser37 on May 25, 2011, 07:51:52 am
Post by: yasser37 on May 25, 2011, 01:55:22 pm
Post by: elemis on May 25, 2011, 01:56:38 pm
Hi
These are the doubts I have so far
please help :)
November 09 , 51, Q4 (ii)
November 10 . 53 , Q4(ii)
November 09 , 52 ,Q1
June 10 , 53, Q(i) ---- A Drawing of angles taken would be greatly appreciated
June 09, 5, Q2 (i) Same as above and Q(6) (ii)
I hope that someone can help me before my exam
+rep for helper :-*
I'll see what I can do.
Post by: elemis on May 25, 2011, 02:07:25 pm
Question 4
Let us say that at 0.3 seconds the object is travelling at an angle alpha to the horizontal, where tan alpha = 7/24 (as the question states).
We know the horizontal component of the velocity is 24 m/s.
If tan alpha = 7/24 ..... sin alpha = 7/25
Hence, at 0.3s the projectile has a vertical component of velocity of :
25 sin alpha i.e. 7m/s
Using v = u+at you should be able to find the initial VERTICAL component of velocity to be 10m/s.
Can you proceed by yourself now ?
Post by: elemis on May 25, 2011, 02:25:31 pm
Question 4
Take moments about B.
See if the diagram helps you.
Post by: yasser37 on May 25, 2011, 03:01:24 pm
hope that you can finish what you started ::)
+rep
Post by: elemis on May 25, 2011, 03:47:32 pm
Thanks a lot mate
hope that you can finish what you started ::)
+rep
November 09 , 52 ,Q1 <----- Not in my syllabus
June 10 , 53, Q(i) ---- A Drawing of angles taken would be greatly appreciated What question ?
June 09, 5, Q2 (i) Same as above and Q(6) (ii) There is no question 2 i from June 2009 and Question 6 ii is not in my syllabus.
Post by: yasser37 on May 25, 2011, 04:30:18 pm
November 09 , 52 ,Q1 <----- Not in my syllabus
June 10 , 53, Q(i) ---- A Drawing of angles taken would be greatly appreciated What question ?
June 09, 5, Q2 (i) Same as above and Q(6) (ii) There is no question 2 i from June 2009 and Question 6 ii is not in my syllabus.
June 10 53 Question 4 part i
and I meant june 08 question 2 i
sorry for the mistake
Post by: Sue T on June 03, 2011, 08:59:51 am
Post by: elemis on June 03, 2011, 09:04:00 am
can any1 please link me to the june 2009 ms of M2 ??
Edexcel ?
Post by: elemis on June 03, 2011, 09:07:05 am
Post by: Sue T on June 03, 2011, 10:08:43 am
Here.thank you
Post by: Sue T on June 05, 2011, 10:54:28 am
Post by: elemis on June 05, 2011, 11:29:13 am
At point Q the velocity is such that it is PERPENDICULAR to the initial velocity line.
You have to find this point Q or in other words:
Find the point alone the trajectory which has a gradient equal to the gradient of the normal (dotted red line) at point x=0.
Can you do it yourself now ?
Post by: Sue T on June 05, 2011, 11:31:06 am
The initial velocity line represents the tangent to the point at x=0.yesss, got it! - thank you
At point Q the velocity is such that it is PERPENDICULAR to the initial velocity line.
You have to find this point Q or in other words:
Find the point alone the trajectory which has a gradient equal to the gradient of the normal (dotted red line) at point x=0.
Can you do it yourself now ?
Post by: elemis on June 05, 2011, 11:40:57 am
yesss, got it! - thank you
Glad to have helped.
Post by: 3bood on June 09, 2011, 07:55:07 pm
june 2010 question 3b
jan 2010 question 7, question 8c
june 2008 question 6c
plzz guys answer ASAP, this is edexcel :)
Post by: Darkstar3000 on June 11, 2011, 08:37:36 pm
(i) Speed of P after collision
(ii)Speed of Q after collision
See my situation has the balls colliding and rebounding in opposite directions but the mark scheme has them colliding and going in the same direction. How do I figure out which direction they will go in ? because I have the same answers but with different signs like my answer for (i) is u(e-1)** mark scheme has it as u(1-e). If I did this in the exam, will my answer be rejected ?
My answer for part B is correct because the direction that I chose for it is the correct one but A is wrong because of its direction
This is Jan 2010
Post by: wakemeup on June 12, 2011, 09:36:15 am
Q 2) Two particles, P, of mass 2m, and Q, of mass m, are moving along the same straight line on a smooth horizontal plane. They are moving in opposite directions towards each other and collide. Immediately before the collision the speed of P is 2u and the speed of Q is u. The coefficient of restitution between the particles is e, where e<1. Find, in terms of u and e
(i) Speed of P after collision
(ii)Speed of Q after collision
See my situation has the balls colliding and rebounding in opposite directions but the mark scheme has them colliding and going in the same direction. How do I figure out which direction they will go in ? because I have the same answers but with different signs like my answer for (i) is u(e-1)** mark scheme has it as u(1-e). If I did this in the exam, will my answer be rejected ?
My answer for part B is correct because the direction that I chose for it is the correct one but A is wrong because of its direction
This is Jan 2010
Think about it intuitively. If one ball has a bigger mass and is moving at a faster speed than the smaller one, even though in the opposite direction, which direction will they go in? In the direction of the velocity of the ball with the bigger mass. So both will move in the same direction.
Post by: Darkstar3000 on June 12, 2011, 09:59:27 am
Think about it intuitively. If one ball has a bigger mass and is moving at a faster speed than the smaller one, even though in the opposite direction, which direction will they go in? In the direction of the velocity of the ball with the bigger mass. So both will move in the same direction.
I can see how that would work but what if B had a larger mass ? Which direction would they go in ?
Post by: wakemeup on June 12, 2011, 10:48:19 am
I can see how that would work but what if B had a larger mass ? Which direction would they go in ?
Then we can't know for sure. In that case you can either consider both the final velocities in the same or opposite directions. As long as you take care with the +ve and -ve signs you should be able to get the correct answer either way. If you don't know which directions the balls will go in, I recommend considering them both in the same direction as it will make your calculations less messy.
But usually, you will be able to guess the directions of the final velocities intuitively or the question itself will say whether they're going in the same or opposing directions.
Post by: Darkstar3000 on June 12, 2011, 11:55:25 am
Then we can't know for sure. In that case you can either consider both the final velocities in the same or opposite directions. As long as you take care with the +ve and -ve signs you should be able to get the correct answer either way. If you don't know which directions the balls will go in, I recommend considering them both in the same direction as it will make your calculations less messy.
But usually, you will be able to guess the directions of the final velocities intuitively or the question itself will say whether they're going in the same or opposing directions.
Thank you for your help, I hope intuition works out :D
Post by: curiousguy on September 12, 2011, 02:34:23 pm
Post by: astarmathsandphysics on September 13, 2011, 10:05:52 am
Post by: curiousguy on September 25, 2011, 04:41:47 pm
04/0/n/01 - 2(ii) and 4(i)
04/m/j/02- 5(ii)
04/m/j/03- 4
And can any one upload the solution bank book for m1 edexcel
Post by: astarmathsandphysics on September 26, 2011, 09:19:17 am
Post by: astarmathsandphysics on September 26, 2011, 03:37:32 pm
Post by: The Golden Girl =D on September 26, 2011, 07:02:17 pm
Click HERE to view the Mentally corrupted Chapter =,= (http://www.pearsonschoolsandfecolleges.co.uk/Secondary/Mathematics/16+/EdexcelModularMathematicsforASandALevel/Resources/Mechanics2/Mechanics2_Chapter1_DRAFT.pdf)
I'm SO LOST in this subject even though I'm a Physics Student (this is my SECOND Week of School too T_T ). I guess the fact that Trignometry takes part is making me go ALL Crazy , this isn't Physics this is just A bunch of weird equations combined together => That's how I see it right now :-X :-\ :'(
I need some good demonstrations or some videos or notes that explains this chapter in a Beautiful and easy way i.e. make it user friendly to seriously confused students.
I need it before the Weekend :-[ , god Knows if my teacher will give us a Drop Quiz next week ::)
Thanks in Advance =]
Post by: astarmathsandphysics on September 26, 2011, 09:47:13 pm
Have separate columns for suvat - a vertical suvat and a horizontal suvat.
vertically horizontally
s =y s=x
u_y=u sin alpha u_x=u cos alpha
v v=u cos alpha since there are no horizontal forces so horizontal s[peed is constant and a=0
a=-9.8 a=0
t t is the same vertivcally and horizontally
Typically you find t by using suvat vertically then put into the suvat equations horizontally to find the range etc.
Post by: The Golden Girl =D on September 27, 2011, 02:13:08 pm
Post by: astarmathsandphysics on September 27, 2011, 02:22:00 pm
Post by: astarmathsandphysics on September 27, 2011, 02:33:06 pm
Post by: The Golden Girl =D on September 27, 2011, 03:08:30 pm
Thanks =]
Post by: astarmathsandphysics on September 27, 2011, 03:11:11 pm
MAKE SURE YOUR CALCULATOR IS IN DEGREES MODE!
Post by: The Golden Girl =D on September 27, 2011, 03:12:31 pm
Post by: The Golden Girl =D on October 12, 2011, 05:30:42 pm
Didn't know how to solve the Whole thing =,=
Q.6. A ball is thrown at 14m/s at an angle of elevation of 60. a) Find it's horizontal and vertical distances from the point of projection after ONE second. b) Find the direction of motion of the ball.
I can't seem to get the right answer for part b) :-\
Can someone answer it in details Please and I'd prefer it if you drew a Diagram , it'll make it easier for me to understand what you're doing =]
Post by: astarmathsandphysics on October 12, 2011, 10:20:09 pm
Post by: astarmathsandphysics on October 16, 2011, 11:27:48 pm
Post by: dk bose on October 25, 2011, 12:09:59 pm
the question is as follows-
A particle P moved in a straight line with constant retardation. At the instants when P passed through the points A,B and C it was moving with speeds 10m/s,7m/s and 3m/s respectively.
Prove that AB/BC=51/40.
Post by: astarmathsandphysics on October 25, 2011, 12:29:40 pm
Post by: dk bose on October 25, 2011, 06:06:38 pm
A block of mass 3 kg is pulled along a rough horizontal floor by a constant force of magnitude 20 N inclined at an angle 60 degrees to the upward vertical. The acceleration of the block has magnitude 2m/s.Calculate,to 2 dp the value of the coefficient of friction between the block and the floor.
Post by: astarmathsandphysics on October 25, 2011, 06:23:14 pm
Post by: astarmathsandphysics on October 25, 2011, 08:57:51 pm
Post by: dk bose on October 25, 2011, 09:10:26 pm
A particle starts at rest at O and moves along Ox.During the first 4s of its motion it accelerates uniformly to a speed of 12m/s.the particle continues at this speed for 10s.It then decelerates uniformly,coming to rest after a further T seconds.Given that the total distance travelled is 180m, calculate T.
Post by: astarmathsandphysics on October 25, 2011, 09:12:31 pm
Post by: astarmathsandphysics on October 25, 2011, 09:19:29 pm
Post by: dk bose on October 26, 2011, 10:09:00 am
Post by: astarmathsandphysics on October 26, 2011, 09:46:58 pm
1/2*4*12+10*12+1/2*T*12=180
the answer is T=6
Post by: dk bose on October 27, 2011, 12:53:48 pm
A steel girder AB has weight 210N.It is held in equilibrium in a horizontal position by two vertical cables.one cable is attached to the end A.The other cable is attached to the point C on the girder,AC=90cm.The girder is modelled as a uniform rod,and the cables as light inextensible strings.
Given that the tension in the cable at C is twice the tension in the cable at A,find (a) the tension in the cable at A, (b) show that AB = 120cm.
A small load of weight W newtons is attached to the girder at B. the load is modelled as a particle.The girder remains in equilibrium in a horizontal position.The tension in the cable at C is now three times the tension in the cable at A. (c) Find the value of W.
Post by: dk bose on October 27, 2011, 05:11:33 pm
Post by: astarmathsandphysics on October 27, 2011, 06:08:15 pm
Post by: astarmathsandphysics on October 27, 2011, 11:24:33 pm
Post by: dk bose on October 28, 2011, 12:44:14 pm
Two particles A and B of mass 8kg and 10 kg respectively,are connected by a light inextensible string which passes over a light smooth pulley P.Particle B rests on a smooth horizontal table and particle A rests on a smooth plane inclined at 300 to the horizontal with the string taut and perpendicular to the line of intersection of the table and the plane as shown in the attachment.
(a) Find the magnitude of the acceleration of particle B.
(b) Find, in newtons,the tension in the string.
(c) Find the distance covered by B in the first two seconds of motion,given B does not reach the pulley.
Post by: astarmathsandphysics on October 28, 2011, 03:04:08 pm
Post by: astarmathsandphysics on October 28, 2011, 03:06:22 pm
Post by: dk bose on October 29, 2011, 10:09:52 am
A train starts from rest at a station,accelerates uniformly to its maximum speed 15m/s,travels at this speed for a time,and then decelerates uniformly to rest at the next station.the distance from station to station is 1260m,and the time spent is three-quarters of the total journey time.
(a)iilustrate this information on a velocity time graph.
(b) By using this graph,or otherwise find the total journey time.
(c) Given also that the magnitude of the deceleration is twice the magnitude of the acceleration find in m s-2,the magnitude of the acceleration.
Post by: dk bose on October 29, 2011, 10:16:43 am
A particle A is constructed to move in a straight line. it starts from rest at X and accelerates at 2m s-2 until it reaches speed V.it then travels with constant speed V for 60s before decelerating at 1 ms-2 to come to rest at Y.The total time for the motion is 105s.Find the distance XY.
Post by: dk bose on October 29, 2011, 05:29:32 pm
Post by: astarmathsandphysics on October 29, 2011, 06:57:29 pm
Post by: dk bose on October 29, 2011, 08:52:35 pm
Post by: astarmathsandphysics on October 30, 2011, 07:47:32 am
Post by: curiousguy on October 31, 2011, 09:02:42 am
Post by: astarmathsandphysics on October 31, 2011, 03:49:03 pm
Post by: dk bose on October 31, 2011, 10:00:09 pm
A particle starting from rest at time t=0 moves is a straight line and accelerates in the following way:
a=3 when 0<t<20
a=1 when 20<t<50
a=-2 when 50<t<95
where a is the acceleration in m s-2 and t is in seconds.Find the speed of the particle when t=20,t=50 and t=95.Sketch a time-speed graph for the particle int the interval 0<t<95.Find the total distance travelled bt the particle in this interval.
Post by: astarmathsandphysics on October 31, 2011, 10:18:06 pm
Post by: dk bose on October 31, 2011, 10:34:21 pm
Two particles A and B, of masses 0.4kg and 0.3kg respectively.,are connected by a light inextensible string which passes over a smooth light fixed pulley.The particles are released from rest with the string taut and the hanging parts vertical.Calculate-
(a) the acceleration of A.
(b) the tension in the string.
The particles continue to move in this system until the instant when they have each acquired speed 3.5m/s.At this instant both A and B are 1.4m above horizontal ground and the string is cut.Give that each particle now moves freely under gravity find the difference in the times,measured from the instant when the string is cut,for A and B to reach the ground.
Post by: dk bose on October 31, 2011, 10:39:07 pm
A particle moves down a line of greatest slope of a rough plane which is inclined at 30 degress to the horizontal.The particle starts from rest and covers 3.5m in time 2s.Find the coefficient of friction between the particle and the plane.
anyone plzzzzzzzz !!!!!Try to solve my these threee doubts and post me the solutions as fast as possible!!!!!!!!!!!! itzzzzzzzzzz urgent !!!!!!!!!!
Post by: astarmathsandphysics on October 31, 2011, 11:17:31 pm
Post by: astarmathsandphysics on November 01, 2011, 06:26:10 am
Post by: dk bose on November 01, 2011, 08:49:28 am
Post by: astarmathsandphysics on November 01, 2011, 11:06:44 am
Post by: astarmathsandphysics on November 01, 2011, 11:51:46 am
Post by: dk bose on November 02, 2011, 09:04:35 am
Two joggers,A and B, are running with constant velocity on level parkland. At a certain instant, A and B have position vectors (-60i+210j)m and (30i-60j)m respectively,referred to a fixed origin O.Ninety seconds later,A and B meet at the point with position vector (210i+120j)m.
(a) Find,as a vector in terms of i and j the velocity of A relative to B.
(b) Verify that the magnitude of the velocity of A relative to B is equal to the speed of A.
Post by: dk bose on November 02, 2011, 09:17:41 am
A particle,P,of mass 0.3kg,is moving in a straight line on a rough horizontal floor.The speed of P decreases from 7.5m s-1 to 4 m s-1 in time T seconds.Given that the coefficient of friction between P and the floor is 1/7(one divided by 7) find:
(a) the magnitude of the frictional force opposing the motion of P.
(b) The value of T.
try to solve these two sums as give me the solutions as fast as possible!!!!!!! itzzzz urgent!!!!!!!!!!!!
Post by: astarmathsandphysics on November 02, 2011, 03:35:07 pm
Post by: dk bose on November 02, 2011, 08:23:38 pm
(a)A ball of mass 0.3kg is released from a point at a height of 10m above horizontal ground.After hitting the ground the ball rebounds to a height of 2.5m.calculate,in N s, the magnitude of the impulse of the force exerted by the ground during the impact.
(b) State two assumptions you have made about the ball and the forces acting on it in order to solve part(a).
Post by: astarmathsandphysics on November 03, 2011, 08:50:22 pm
Post by: dk bose on November 05, 2011, 06:57:17 am
(a)Sketch a speed-time graph for this journey.Hence,or otherwise,find
(b) the deceleration in m s-2 of the train during the final 0.6km,
(c) the value of T,
(d) the total time for the journey.
Post by: dk bose on November 05, 2011, 07:03:59 am
(a) the acceleration of the particle.
(b) the speed of the particle.
anyone plzz solve these two sums and give me the solutions as fast as possible!!!!!!!!!!!
Post by: astarmathsandphysics on November 05, 2011, 10:38:24 am
Post by: astarmathsandphysics on November 06, 2011, 07:56:45 am
Post by: dk bose on November 06, 2011, 10:40:51 pm
(a)calculate in N to 2sf the value of this constant force.
2)A straight uniform rigid beam AB,of length 6 m and mass 15 kg, is supported at A and B and rests horizontally.A load of mass 60kg is placed on the beam at the point C.Given that the magnitude of the force exerted on the support at A is twice the magnitude of the force exerted on the support at B,find the distance AC.
plzz anyone solve these two sums and post the solutions as fast as possible!!!!!!!
Post by: astarmathsandphysics on November 07, 2011, 07:03:10 am
Post by: astarmathsandphysics on November 07, 2011, 11:15:58 pm
Post by: dk bose on November 08, 2011, 10:19:20 pm
Determine the forces P and Q expressing each in terms of i and j.
2)At 11:00 hours the position vector of an aircraft relative to an airport O is (200i+30j)km,i and j being unit vectors due east and due north respectively.The velocity of the aircraft is (180i-120j)km/h.find:
(a) how far it is from O at 12:00.
3) this question is quite huge so i cant write it here.it is from heinemann modular mechanics unit 1 textbook.page no 57.review exercise 1.question no 7,part (f).
Solve all these 3 questions and give the solutions as fast as possible!!!!!
Post by: astarmathsandphysics on November 09, 2011, 08:17:11 am
Post by: Becca on November 09, 2011, 09:15:28 am
First Question:
P is parallel to (i+4j), so P= x(i+4j)
Q is parallel to (i+j), so Q= y(i+j)
x & y are to be calulated, as follows:
since R is the resultant of P and Q, R = P+Q
therefore,
R = P+Q
(7i+16j) = x(i+4j) + y(i+j)
(7i+16j) = (xi + 4xj) + (yi +yj)
(7i+16j) = [(x+y)i + (4x+y)j]
Hence, based on the above equation, you can say that:
x+y = 7
y=7-x >>>>>>>>>>>>>>>>> eqn 1
and,
4x+y = 16
y= 16-4x >>>>>>>>>>>>>>>>>>>eqn 2
Sub eqn 1 into eqn 2:
7-x=16-4x
3x=9
x=3
Sub x=3 into eqn 1
y=7-3
y=4
Therefore, P=x(i+4j)
P= 3(i+4j)
P=3i + 12j
and Q=y(i+j)
Q=4(i+j)
Q=4i+4j
Second Question:
1 hour passes from 11.00 to 12.00, so the plane moves through a vector of (180i-120j) within this time. The initial vector is (200i+30j).
Therefore, the final vector is:
(200i+30j) + (180i-120j) = (380i-90j)
However, since the question is asking for the distance of the aeroplane from the airport, we must find the magnitude of the above vector via Pythagoras' theorem:
?[380^2+ (-90)^2] = 390.5 km
^For some reason, I couldn't insert the square root sign; it became a '?' instead.
Post by: Becca on November 09, 2011, 09:32:06 am
Post by: astarmathsandphysics on November 09, 2011, 02:18:01 pm
Post by: dk bose on November 11, 2011, 11:56:40 pm
(a) the tension T in the horizontal rope giving ur ans to the nearest N.
2)A small sphere R,of mass 0.08kg,moving with speed 1.5m/s,collides directly with another small sphere S,of mass 0.12kg,moving in the same direction with speed 1m/s. Immediately after the collision R and S continue to move in the same direction with speeds U m/s and V m/s respectively.
Given that U:V=21:26,
(a) show that V = 1.3,
(b) find the magnitude of the impulse, in Ns, received by R as a result of the collision.
Post by: iluvme on November 12, 2011, 02:39:15 pm
1)A body of mass 5kg is held in equilibrium under gravity by two inextensible light ropes.One rope is horizontal,the other is at an angle teetha to the horizontal,as shown in the attachment.The tension in the rope inclined at teetha to the horizontal is 72 N.
(a) the tension T in the horizontal rope giving ur ans to the nearest N.
2)A small sphere R,of mass 0.08kg,moving with speed 1.5m/s,collides directly with another small sphere S,of mass 0.12kg,moving in the same direction with speed 1m/s. Immediately after the collision R and S continue to move in the same direction with speeds U m/s and V m/s respectively.
Given that U:V=21:26,
(a) show that V = 1.3,
(b) find the magnitude of the impulse, in Ns, received by R as a result of the collision.
Its been a long time since I did my M1, I did the first question for you, as for the second, :X
Post by: astarmathsandphysics on November 12, 2011, 09:35:59 pm
Post by: dk bose on November 13, 2011, 07:00:47 am
Post by: astarmathsandphysics on November 13, 2011, 09:48:19 pm
Post by: dk bose on November 15, 2011, 04:39:29 pm
Post by: astarmathsandphysics on November 15, 2011, 04:54:42 pm
Post by: astarmathsandphysics on November 15, 2011, 09:30:59 pm
Post by: astarmathsandphysics on November 15, 2011, 09:33:43 pm
The particle question answered on last page
Post by: dk bose on November 16, 2011, 05:38:54 am
Post by: astarmathsandphysics on November 16, 2011, 08:30:45 am
Post by: dk bose on November 16, 2011, 05:06:33 pm
Post by: dk bose on November 17, 2011, 08:05:45 am
Post by: astarmathsandphysics on November 17, 2011, 08:59:03 am
Post by: dk bose on November 20, 2011, 08:52:58 am
Post by: Malak on November 20, 2011, 03:01:59 pm
You are not making us or people who helped you do your homework, right?
Post by: dk bose on November 20, 2011, 09:23:18 pm
Post by: astarmathsandphysics on November 20, 2011, 10:48:20 pm
Post by: dk bose on November 21, 2011, 10:00:01 am
Post by: dk bose on November 21, 2011, 10:29:29 am
Post by: dk bose on November 21, 2011, 11:32:14 am
Post by: astarmathsandphysics on November 21, 2011, 02:24:10 pm
Post by: dk bose on November 22, 2011, 10:16:39 am
Post by: Arthur Bon Zavi on November 22, 2011, 10:18:01 am
Post by: astarmathsandphysics on November 22, 2011, 11:26:49 am
Post by: Malak on November 22, 2011, 11:56:27 am
hello.i dont go to school,so i have to self study for which i dont have a tuition teacher so i ask all my doubts here for help.!!! now ok??Well, there is nothing to be mad about. I just had to make sure.
Post by: dk bose on November 23, 2011, 10:55:43 am
@ D. K. Bose - did you see Delhi Belly ? ::)
@Arthur Bon Zavi - yup,obviously.didnt you like it??
Post by: dk bose on November 23, 2011, 11:12:47 am
Thanks anyway,but if u dont mind here are some more of my doubts,the circled questions!!!!!
Mr astarmathsandphysics, can u solve my this doubt at the end of page 40,my new doubts???.its in scan0008 file!!!
Post by: astarmathsandphysics on November 23, 2011, 09:48:15 pm
Post by: astarmathsandphysics on November 24, 2011, 09:19:30 am
Post by: donhassan on November 29, 2011, 07:35:56 pm
Post by: Arthur Bon Zavi on December 03, 2011, 09:39:43 am
@Arthur Bon Zavi - yup,obviously.didnt you like it??
Yes, I did.
Post by: dk bose on December 13, 2011, 02:03:22 pm
Post by: astarmathsandphysics on December 13, 2011, 03:45:15 pm
Post by: The Golden Girl =D on December 13, 2011, 04:04:54 pm
Why did we include both GPE and KE ?
My answer is 18 m/s which is WRONG . the right answer is 8.7 m/s :-\ :-X
Post by: astarmathsandphysics on December 13, 2011, 09:24:53 pm
Post by: The Golden Girl =D on December 13, 2011, 09:26:49 pm
Post by: dk bose on December 21, 2011, 02:31:01 pm
here
Mr astarmathsandphysics,the (c) part is wrong in this question.its answer should 90m,so plzz try the sum again.
Post by: astarmathsandphysics on December 21, 2011, 09:32:31 pm
Post by: dk bose on December 21, 2011, 11:07:19 pm
Post by: astarmathsandphysics on December 22, 2011, 01:29:49 pm
Post by: dk bose on December 27, 2011, 08:32:30 am
Post by: astarmathsandphysics on December 28, 2011, 09:46:50 am
Post by: ASdude on January 03, 2012, 08:23:44 pm
I am using Edexcel AS and A level Modular Mathematics M1: Unit 6 Vectors page 137-138 Example number 5:
I understood most of it: but the last part, I don't know how did they use the big expression to form 2 simultaneous equations. Please can you type in the detailed steps.
Please Reply ASAP.
Post by: astarmathsandphysics on January 04, 2012, 08:50:13 am
Post by: ASdude on January 04, 2012, 07:06:57 pm
I don't have the textbook. Can you post the question? Maybe scan it.
Please see the post above. I just uploaded the question. PLEASE REPLY ASAP!!
Post by: astarmathsandphysics on January 04, 2012, 10:48:42 pm
Post by: -Cornea on February 11, 2012, 07:59:50 pm
Post by: astarmathsandphysics on February 12, 2012, 10:59:50 pm
Post by: -Cornea on February 14, 2012, 11:03:16 am
Post by: Malak on February 16, 2012, 12:23:40 pm
The most I need is the free body diagram 'cause that is where I think I am messing up ::)
The answer for both parts is 9.8N
Post by: astarmathsandphysics on February 19, 2012, 12:40:21 pm
Post by: WARRIOR on February 22, 2012, 05:54:59 pm
EXERCISE 1 A , Q 14 . I got a) but i dont know how to get b :(
Post by: astarmathsandphysics on February 23, 2012, 09:47:47 am
Post the question.
Post by: LekhB on February 26, 2012, 01:04:45 pm
(a) the magnitude of the frictional force opposing the motion of P.
(b) The value of T.
Post by: astarmathsandphysics on February 26, 2012, 01:19:54 pm
Post by: LekhB on February 26, 2012, 02:56:48 pm
I had some problems with this one too.
A box of mass 4.5 kg is pulled at a constant speed of 2 m/s alone a rough horizontal floor by a horizontal force of magnitude 15 N.
(i) Find the coefficient of friction between the box and the floor.
The horizontal pulling force is now removed. find
(ii) the deceleration of the bx in the subsequent motion
(iii) the distance travelled by the box rom the instant the horizontal force is removed until the box comes to rest.
And this one too:
A particle of mass 0.5 kg is placed on a rough plane inclined at an angle ? to the horizontal where sin ?= 4/5. Initially, the particle is at rest under the action of a force P N applied in a upward direction parallel to the plan. Given that the coefficient of friction between the particle and the plan is 2/5, calculate the value of P when
(i) the particle is about to move up the plane,
(ii) the particle is about to move down the plan.
If the force P is now removed, calculate the acceleration of the particle down the plane.
Post by: sweetsh on February 26, 2012, 05:22:07 pm
Thank you!
I had some problems with this one too.
A box of mass 4.5 kg is pulled at a constant speed of 2 m/s alone a rough horizontal floor by a horizontal force of magnitude 15 N.
(i) Find the coefficient of friction between the box and the floor.
The horizontal pulling force is now removed. find
(ii) the deceleration of the bx in the subsequent motion
(iii) the distance travelled by the box rom the instant the horizontal force is removed until the box comes to rest.
And this one too:
A particle of mass 0.5 kg is placed on a rough plane inclined at an angle ? to the horizontal where sin ?= 4/5. Initially, the particle is at rest under the action of a force P N applied in a upward direction parallel to the plan. Given that the coefficient of friction between the particle and the plan is 2/5, calculate the value of P when
(i) the particle is about to move up the plane,
(ii) the particle is about to move down the plan.
If the force P is now removed, calculate the acceleration of the particle down the plane.
Hello,
I apologize I won't be able to answer your question i'm on my mobile now but a very good hint is in every question like that, draw the plane inclined with the particular object as a "block" or a "circle" and then put arrows for each force acting on it(i.e gravity, friction, N, .... )
Post by: astarmathsandphysics on February 26, 2012, 05:46:42 pm
Post by: WARRIOR on March 01, 2012, 07:26:26 pm
[3t(sq) -12t + 5 ] m/s
When t=0 , P is at ORIGIN O. Find
a) The velocity of p when its acceleration is zero
b)the values of t when p is again at O
c) The distance travelled by p in the interval 3=<t=<4
i got a and b but i dont get c at all.
I understand by getting the maximu displacement , when V = 0 , i get 13.12... . And at t = 3 displacement is -12 , and at t= 4 displacement is -12 again ! So displacement must be
(13.12-12)X2 ?? explain please the right answer is 48
Post by: astarmathsandphysics on March 01, 2012, 07:55:18 pm
Post by: astarmathsandphysics on March 01, 2012, 08:00:29 pm
Post by: WARRIOR on March 01, 2012, 08:26:39 pm
I got 1.926 and I cant post my working until my scanner recovers from the thumping I gave itThis is from the M2 Edexcel book . Page 18 ,Exercise 1B question 10 !
The answer at the back
a)7
b)1,5
c)48m
Thanks anyway !
Post by: astarmathsandphysics on March 01, 2012, 08:36:09 pm
Check you have the correct page
Post by: WARRIOR on March 02, 2012, 10:06:15 am
F= ( 3t ) i + (4t-5)j .
When t=1 , the velocity of P is 12 i m/s
Find velocity of P after t seconds
The angle the direction of motion of p makes with i when t=5 .
also i woudl like to know , when the question says for example
Find t when v is paraller to i . This means j = 0
but when he says i+j or i-j , what does this mean?
Gradient = 1 and gradient = -1
Post by: astarmathsandphysics on March 02, 2012, 10:21:24 am
I did for i+j at bottom of page.
Post by: WARRIOR on March 02, 2012, 10:38:51 am
Didn't read the question properly.Your answers are right sir ! but how is 12i=3i-6j + c => c= -3i+6j ?????
I did for i+j at bottom of page.
isnt is supposed to be 9i+6j?
Post by: astarmathsandphysics on March 02, 2012, 10:48:47 am
Post by: WARRIOR on March 02, 2012, 10:55:03 am
Now I get 9i+18j after checking
I got the same at first , but i checked the answer at the back it was exactly like this
( 3t(sq) + 9) i + ( 4t(cubed) -10t + 6 ) j
Very weird book !!!!
anyway thanks a million for your answers and explanations ! + rep!
Post by: astarmathsandphysics on March 02, 2012, 10:59:49 am
When I did it the first time I ignored the 12i
and the second time I got 12i+6j=18j
Post by: WARRIOR on March 02, 2012, 11:26:05 am
BOOK is correct!!!HOW :o im so slow , i still dont understand how 12i= 3i + ( 4 -10 ) j could be 12i+6j=18j
When I did it the first time I ignored the 12i
and the second time I got 12i+6j=18j
Post by: astarmathsandphysics on March 02, 2012, 11:32:16 am
Shall I make a video the this?
theeducationchannel.info
Post by: WARRIOR on March 02, 2012, 03:54:10 pm
r= [ (3t(sq) - 4) i + ( 8 - 4t(sq) ) j ]
a Show that the acceleration of P is a constant .
b Find the magnitude of the acceleration of P and the size of the angle which the acceleration makes with j .
I get magnitude of acceleration = 10
But size of angle is 143.1 HOWWWWWWWWWWWWWWW
Post by: astarmathsandphysics on March 02, 2012, 08:57:20 pm
This is at an angle tan^(-1) (8/6)=53.13 below the x axis (plot (6,-8))
add 90 cos it is the angle with the j axis
Post by: LekhB on March 03, 2012, 07:52:21 pm
1/. A car is moving along a straight road with uniform acceleration. The car passes a check-point A with a speed of 12 m/s and another check-point C with a speed of 32 m/s. The distance between A and C is 1100m.
(i) Find the time, in s, taken by the car to move from A to C.
(ii) Given that B is the mid-point of AC, find in m/s to 1 decimal place, the speed with which the car passes B.
2. A, B and C are 3 points on a straight line, in that order, and the distance AB and AC are 45m and 77m respectively. A particle moves along the straight line with constant acceleration 2 m/s². Given that it taked 5 seconds to travel from A to B, find the time taken to travel from A to C.
3. An aircraft accelerates along the runway with constant acceleration a m/s². 3 points on the runway are A, B, C, where the distances AB and AC are each 120m. The speed of the aircraft as it passes A is u m/s. The aircraft takes 6s to cover the distance AB and 4s to cover the distance BC. Find the values of a and u.
It would be awesome if anyone could help. Cheers.
Post by: astarmathsandphysics on March 03, 2012, 09:37:03 pm
Post by: astarmathsandphysics on March 04, 2012, 08:10:05 am
Post by: LekhB on March 04, 2012, 08:02:07 pm
Attached 3 more questions. The connecting particles chapter is really getting on my nerves.
(http://i.imgur.com/Hrbxz.jpg)
(http://i.imgur.com/6SLXd.jpg)
Post by: astarmathsandphysics on March 04, 2012, 09:42:30 pm
I made a video.
Post by: astarmathsandphysics on March 04, 2012, 09:45:56 pm
Post by: astarmathsandphysics on March 05, 2012, 10:31:03 am
resolving for 2kg mass
T-2g=2a
Resolving for 5 kg mass
5g-T=5a
add these
g=5a so a=g/5
sub into T-2g=2a
T=2g+2*g/5 so T=12g/5
Post by: astarmathsandphysics on March 05, 2012, 10:53:07 am
Post by: LekhB on March 10, 2012, 04:50:37 pm
From a point 120m above ground level, a stone is projected vertically upwards with a speed u m/s. if the stone rises to a height of 5m above the point of projection, calculate(using g= 10 m/s²)
(i) the value of u,
(ii) the time the stone takes from projection until it reaches the ground level,
(iii) the speed of the stone at ground level.
A particle is projected vertically upwards from ground level. Between 2 seconds and 3 seconds after leaving the ground, it rises 45 m. Calculate (using g=10 m/s²)
(i) the speed of projection
(ii) the maximum height reached
(iii) the time interval for which the particle is above a level of 165m.
Post by: LekhB on March 10, 2012, 05:09:58 pm
A particle is projected vertically upwards from the ground with a speed of 36 m/s. Calculate (using g= 10 m/s²)
(i) the time for which it is above a height of 63m
What I tried :
S: 63m
u: 36 m/s
v: ? m/s
a: 10 m/s²
t: ? s
S= ut + 0.5(at²)
63 = 36(t) + 0.5(10)(t²)
-> 5t² + 36t - 63 = 0
t = [(-36) + (36² - 4(5)(-63)^0.5]/10 or [(-36) - (36² - 4(5)(-63)^0.5]/10
t= 1.46 s.
The answer is t= 1.2s.
And I need t for doing the other parts which are;
(ii) the speed which it has at this height on its way down
(iii) the total time of flight.
Post by: LekhB on March 10, 2012, 06:03:37 pm
From the foot of a vertical cliff 28.8 high, a stone was projected vertically upwards so as just to reach the top. Find its velocity of projection.
One second after the first stone was projected, another stone was allowed to fall from rest from the top of the cliff. The stones passed one another after a further t seconds at a height h m above the ground. Calculate the value of t and h.
Answers : 24 m/s(I got 13.78), 0.7, 26.4
I used to get all of these numbers right in physics. Don't know what happened here...
Another one(in which I got the first answer though):
A particle X is projected vertically upwards from the ground with a velocity of 80 m/s. Calculate the maximum height reached by X. -> I could do this part
The other part: A particle Y is held at a height of 300m above the ground. At the moment when X has droped 80m from its max height, Y is projected downwards with a velocity of v m/s. The particles reach the ground at the same time. Calculate the value of v.
I got v= 103.4
Post by: astarmathsandphysics on March 10, 2012, 09:35:07 pm
If I used 9.8 for g forgive me
Post by: LekhB on March 11, 2012, 08:50:31 pm
(http://i.imgur.com/cc7ot.jpg)
(http://i.imgur.com/nE9k6.jpg)
Post by: astarmathsandphysics on March 12, 2012, 08:01:44 pm
Post by: astarmathsandphysics on March 12, 2012, 08:26:51 pm
Post by: astarmathsandphysics on March 12, 2012, 08:50:41 pm
Post by: a.k. on March 25, 2012, 06:40:09 pm
Question - Particle A of mass 1kg is at rest 0.2m from the edge of a smooth horizontal 0.8m high table. It is connected by a light inextensible string of length 0.7m to a partcle B of mass 0.5kg. Partcle B is intially at rest at the edge of the table closest to A but then falls off. Assuming B's initial horizontal velocity to be zero find the speed with which A begins to move?
thank you
Post by: astarmathsandphysics on March 26, 2012, 07:36:04 am
Post by: fawazomar on April 21, 2012, 06:23:25 am
"A light rectangular plate ABCD has AB=20cm and AD=50cm. Particles of mass 2kg, 3kg, 5kg and 5kg are attached to the plate at the points A,B,C,D respectively. Find the distance of the centre of mass of the loaded plane from:
a. AD
b. AB"
Can anyone explain to me how they find the distance??? I know how to find the centre of mass but it stops there.
Post by: aaakhtar19 on May 03, 2012, 07:10:29 am
can any0ne help me solving this questi0n
CIE Board :
paper 04 mechanics
0ctober n0vember 2003
q6
Post by: astarmathsandphysics on May 06, 2012, 09:37:22 am
Post by: YU-GI-OH! on May 12, 2012, 10:55:26 pm
Thanks in advance :)
Post by: xainer on May 04, 2013, 06:01:44 pm
Post by: astarmathsandphysics on October 09, 2013, 12:27:32 pm