Author Topic: All Mechanics DOUBTS HERE!!!  (Read 88417 times)

Offline Saladin

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Re: All Mechanics DOUBTS HERE!!!
« Reply #45 on: May 10, 2010, 11:05:54 am »
2007 january paper question 7 last part

and thanks

I am quite sure its Edexcel.

s=0.8
a=0.98
u=0<br />
We need to find out the final velocity that the particle reaches after the other particle has fallen.

v^2=u^2+2as
v=1.25

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.

mgsin(\theta)=ma
3gsin(30)=3a
-\frac{1}{2}g=a

Using this, calculate the time taken for the particle to come to rest.

v=0
u=1.25
a=\frac{1}{2}g

<br />v=u+at
0=1.25-\frac{1}{2}gt
t=0.255
The time taken to go up, is the time taken to go down, as there is same acceleration.
Thus the total time taken is 2t=0.51
Thus the total time taken is 0.51 seconds,

Offline Uchia

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Re: All Mechanics DOUBTS HERE!!!
« Reply #46 on: May 10, 2010, 11:46:08 am »
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??

Offline Saladin

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Re: All Mechanics DOUBTS HERE!!!
« Reply #47 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.

Offline Phosu

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Re: All Mechanics DOUBTS HERE!!!
« Reply #48 on: May 10, 2010, 01:37:20 pm »
jun 02 ques 4a
i just straight off rote R=6gCos30+PSin30
but marking scheme has some other calculations, can someone plz explain how to get that.

nid404

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Re: All Mechanics DOUBTS HERE!!!
« Reply #49 on: May 10, 2010, 02:17:52 pm »
must be edexcel...i don't have the paper

Offline astarmathsandphysics

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Re: All Mechanics DOUBTS HERE!!!
« Reply #50 on: May 10, 2010, 02:19:17 pm »
They resolved vertically. The answer is a the same

Offline melony

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Re: All Mechanics DOUBTS HERE!!!
« Reply #51 on: May 14, 2010, 10:49:55 am »
help please, how do u determine the direction of the resultant of three coplanar forces. i.e mechanics paper 04/m/j/05
 ???
thank yu!  :)
xOx

nid404

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Re: All Mechanics DOUBTS HERE!!!
« Reply #52 on: May 14, 2010, 11:40:36 am »
I assume it's the second question

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= \sqrt{(5)^2+(0.83)^2}

R=5.09N

tan \theta= 0.83/5
\theta=9.43o anticlockwise from force of 7N




Offline melony

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Re: All Mechanics DOUBTS HERE!!!
« Reply #53 on: May 14, 2010, 08:15:21 pm »
wow!! ofcourse!! thanks soo much nid!!.. +rp!
 ;)
xOx

Offline Fenomenoe9

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Re: All Mechanics DOUBTS HERE!!!
« Reply #54 on: May 15, 2010, 09:04:46 am »
Can any1 help in this question... please.

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 ^^
« Last Edit: May 15, 2010, 09:07:15 am by Fenomenoe9 »

Offline astarmathsandphysics

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Re: All Mechanics DOUBTS HERE!!!
« Reply #55 on: May 15, 2010, 09:22:56 am »
here

Offline Fenomenoe9

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Re: All Mechanics DOUBTS HERE!!!
« Reply #56 on: May 15, 2010, 10:00:13 am »
wow m8 thanks soo much!!! and good luck with ur website :D

Offline wakemeup

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Offline astarmathsandphysics

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Offline melony

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Re: All Mechanics DOUBTS HERE!!!
« Reply #59 on: May 15, 2010, 09:53:09 pm »
Quote
Can any1 help in this question... please.

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 ^^
Hey! i knw astars alredy dunt buh i gav it a shot usin a diff method if it helps.. :)
xOx