Post by: Dania on October 09, 2010, 06:48:46 pm
Here is the question below, and the extract from the mark scheme.
Post by: S.M.A.T on October 09, 2010, 08:18:32 pm
Post by: Dania on October 09, 2010, 08:23:30 pm
Post by: S.M.A.T on October 09, 2010, 08:31:54 pm
U welcome :)
Post by: Dania on October 09, 2010, 08:50:20 pm
Post by: Dania on October 10, 2010, 07:34:34 pm
Post by: S.M.A.T on October 10, 2010, 10:52:46 pm
Post by: Dania on October 14, 2010, 05:41:28 pm
I have another question. Take a look at the attached file, thanks :)
P.S Don't bother yourself answering (i) for me. I'm only interested in (ii), but you'll have to do the whole question to effectively help me out.
Post by: nid404 on October 14, 2010, 06:15:12 pm
Post by: S.M.A.T on October 14, 2010, 08:45:18 pm
The frictional force provides forward acceleration since there is no more tension(once block B is on the ground)Garfield u are wrong ::).....frictional force is an opposing force so it always try to decelerate an object :)
When particle B is about to hit the floor the speed of A is also 3.95ms^-2,this is because the acceleration of A&B are same before B hit the ground and they both travel the same distance(1.3m).
Once particle B is on the ground the tension in the string become zero and the only force acting on A is frictional force and it is decelerating particle A.
therefore
u=3.95ms^-1
frictional force=ma
-0.06g=0.2a
a=-0.3g
s=2.1-1.3=0.8m(1.3m is already travelled by A when B was about to hit the ground)
(v^2)=(u^2)+2as
(v^2)=(3.95^2)+(2*-0.3g*0.8 )
v=3.29
Post by: thecandydoll on October 15, 2010, 03:04:53 am
i think i land up with 54/75 or something
so doing well in mechanics,is there a chance i can get a b?
I have a doubt.
Mechanics Paper 4.M/J 2003.
q2
Post by: elemis on October 15, 2010, 04:44:51 am
As they are asking for the resultant force in the direction PQ we must resolve for forces in the y axis.
i.e. : 10 cos 30 + 10 cos 30 - 6 cos 60= 14.3 N
The direction 'perpendicular to PQ' is basically the x axis. So we resolve forces in the x axis.
10 cos 60 + 6 cos 30 - 10 cos 60 = 5.20 N
For the last part we know the magnitude of the horizontal and vertical forces. The resultant is the red line. (see 2nd pic)
Using Pythagoras Theorem : 5.22 + 14.3 2 = 231.53
Hence,
Post by: nid404 on October 15, 2010, 04:46:22 am
Garfield u are wrong ::).....frictional force is an opposing force so it always try to decelerate an object :)
negative acceleration...my bad -_-
Post by: Deadly_king on October 15, 2010, 05:11:46 am
I feel so upset screwing up my p1 math.
i think i land up with 54/75 or something
so doing well in mechanics,is there a chance i can get a b?
I have a doubt.
Mechanics Paper 4.M/J 2003.
q2
Hmm........I guess it is possible :)
You need to catch up in p4 though!
Post by: thecandydoll on October 15, 2010, 12:31:00 pm
Where does the 6cos60 come from :(
OMGGGG I DONT GETTT IT.
brain has shutdown.
First we draw a set of axis. Then we place the forces on them as shown in the first pciture.
As they are asking for the resultant force in the direction PQ we must resolve for forces in the y axis.
i.e. : 10 cos 30 + 10 cos 30 - 6 cos 60= 14.3 N
The direction 'perpendicular to PQ' is basically the x axis. So we resolve forces in the x axis.
10 cos 60 + 6 cos 30 - 10 cos 60 = 5.20 N
For the last part we know the magnitude of the horizontal and vertical forces. The resultant is the red line. (see 2nd pic)
Using Pythagoras Theorem : 5.22 + 14.3 2 = 231.53
Hence,= 15.2 N (3 s.f.)
Post by: nid404 on October 15, 2010, 01:56:23 pm
Where does the 6cos60 come from :(
OMGGGG I DONT GETTT IT.
brain has shutdown.
6cos60 is the same as 6sin30
Post by: elemis on October 15, 2010, 02:21:46 pm
Post by: Dania on October 15, 2010, 02:50:53 pm
I keep getting a tension of 8.32N when in fact I need to get 3N. I think I'm resolving the forces incorrectly, can someone please draw it out for me, and solve the question.
Post by: nid404 on October 15, 2010, 03:12:22 pm
Post by: nid404 on October 15, 2010, 03:27:20 pm
Post by: elemis on October 15, 2010, 03:29:52 pm
6cos60 is the same as 6sin30
Yeah, I prefer using cosine instead of sine. I'm biased that way :P
Post by: nid404 on October 15, 2010, 03:30:47 pm
Yeah, I prefer using cosine instead of sine. I'm biased that way :P
Racist :P
Post by: elemis on October 15, 2010, 03:36:27 pm
Racist :P
wtv :P
Post by: Deadly_king on October 15, 2010, 03:37:48 pm
attached
Quite long but good ;)
You could have resolved the weight P along the line of the string PS. But first you need to find the angles of the triangle APS.
T = 5cos 53.1 = 3.0N :)
Am sure this method will make Ari happy :P
Post by: nid404 on October 15, 2010, 03:41:12 pm
yup deadly king, that's one way to go abt it :)
Post by: elemis on October 15, 2010, 03:49:47 pm
Am sure this method will make Ari happy :P
This wasnt my question smarty pants ::)
Post by: Deadly_king on October 15, 2010, 04:06:09 pm
hahaha :PHats off to you though.....you were faster since you started earlier :P
yup deadly king, that's one way to go abt it :)
@ Ari : I know .......but I used cosine and that's what you prefer, right?
Post by: Dania on October 16, 2010, 01:06:38 pm
Post by: nid404 on October 16, 2010, 01:35:57 pm
for the second part, acceleration, t seconds after passing through A is given by (10-3t)
at t=0, v=5m/s
v=10t-0.15t2+C
substituting t=0 and v=5
you get c=+5
Post by: Dania on October 16, 2010, 11:45:27 pm
(i) Express x and y in terms of t, and hence show that y= 3x- 0.1x2
The ball hits the sea at a point which is 25m below the level of O.
(ii) Find the horizontal distance between the cliff and the point where the ball hits the sea.
Post by: S.M.A.T on October 17, 2010, 02:48:12 am
A ball is projected from a point O on the edge of a vertical cliff. The horizontal and vertically upward components of the initial velocity are 7ms-1 and 21ms-1 respectively. At the time t seconds after the projection the ball is at the point (x,y) referred to horizontal and vertically upward axes through O. Air resistance may be neglected.
(i) Express x and y in terms of t, and hence show that y= 3x- 0.1x2
The ball hits the sea at a point which is 25m below the level of O.
(ii) Find the horizontal distance between the cliff and the point where the ball hits the sea.
(i) x=7t(horizontal velocity always remain constant so the formula used is s=vt)
y=21t-0.5gt2(s=ut+0.5at2)
t=(x/7)
y=21t-0.5gt2
y=21(x/7) -0.5g(x/7) 2
y=3x-0.1x2
(ii) y=-25(because the question said y is vertically upward axes through O the ball hits below O so negative 25)
3x-0.1x2=-25
0.1x2-3x-25=0
solving this equation we get
x=36.8m
Post by: Arthur Bon Zavi on October 17, 2010, 07:25:34 am
(i) x=7t(horizontal velocity always remain constant so the formula used is s=vt)
y=21t-0.5gt2(s=ut+0.5at2)
t=(x/7)
y=21t-0.5gt2
y=21(x/7) -0.5g(x/7) 2
y=3x-0.1x2
(ii) y=-25(because the question said y is vertically upward axes through O the ball hits below O so negative 25)
3x-0.1x2=-25
0.1x2-3x-25=0
solving this equation we get
x=36.8m
Have You Completed yous A Levels? :D
Post by: S.M.A.T on October 18, 2010, 10:55:13 am
Have You Completed yous A Levels? :D
no,still doing my A2.:)
Post by: Arthur Bon Zavi on October 18, 2010, 01:22:20 pm
no,still doing my A2.:)
Well Then Go Ahead! 8)
Post by: astarmathsandphysics on October 20, 2010, 09:03:36 am
Post by: Ghost Of Highbury on October 20, 2010, 01:24:08 pm
attached
Can u help with the second subpart too plz?
Y do they use F = T cos (sin-1 0.6)
Post by: elemis on October 20, 2010, 02:07:43 pm
Can u help with the second subpart too plz?
Y do they use F = T cos (sin-1 0.6)
What question are you referring to ?
Post by: Ghost Of Highbury on October 20, 2010, 02:20:45 pm
Post by: elemis on October 20, 2010, 03:02:26 pm
iVE quoted it.
But whats the ORIGINAL question ? All you've quoted is the answers Nid gave you - I cant find your question anywhere.
Post by: Ghost Of Highbury on October 20, 2010, 03:27:40 pm
But whats the ORIGINAL question ? All you've quoted is the answers Nid gave you - I cant find your question anywhere.
Click on the "Quote from...." mate.
Post by: Ghost Of Highbury on October 20, 2010, 03:33:03 pm
I think it should be F = T + Wsin53.1 (weight component)
But the answer seems to ignore the W, and instead give the equation F = Tsin53.1
Post by: elemis on October 20, 2010, 03:41:24 pm
Click on the "Quote from...." mate.
I did that. I'm not dumb you know.... :P
I figured out you were talking about Dania's question.
This whole time I was looking for a question posted by YOU.
Post by: Ghost Of Highbury on October 20, 2010, 03:56:38 pm
I did that. I'm not dumb you know.... :P
I figured out you were talking about Dania's question.
This whole time I was looking for a question posted by YOU.
Ahh ok.
Post by: Deadly_king on October 20, 2010, 04:29:29 pm
I know that the Friction force is stopping the ring to slip and opposing T too.
I think it should be F = T + Wsin53.1 (weight component)
But the answer seems to ignore the W, and instead give the equation F = Tsin53.1
W acts perpendicular to the ring...............hence it has no component!
Post by: Ghost Of Highbury on October 20, 2010, 05:21:32 pm
W acts perpendicular to the ring...............hence it has no component!
Could you elaborate that plz, and sry for the trouble.
Post by: Deadly_king on October 20, 2010, 05:27:06 pm
Could you elaborate that plz, and sry for the trouble.
If you resolve the weight, it will be Wsin90 which is zero!
This is because sin 90 = 0.
Hence we omit the component of the weight ;)
Post by: Ghost Of Highbury on October 20, 2010, 05:34:09 pm
Post by: Deadly_king on October 20, 2010, 05:47:20 pm
Thank you man.
Anytime dude ;)
Post by: cooldude on October 21, 2010, 10:38:26 am
If you resolve the weight, it will be Wsin90 which is zero!
This is because sin 90 = 0.
Hence we omit the component of the weight ;)
actually it will because Wcos90 is 0 as cos 90=0 while sin 90=1
Post by: Deadly_king on October 22, 2010, 06:05:43 am
actually it will because Wcos90 is 0 as cos 90=0 while sin 90=1
Rightly corrected dude ;)
My bad :-[