IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: nid404 on November 02, 2009, 07:02:46 am
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In fig m1=4m2. Pulleys are smooth and the thread is light. At t=0 system is released from rest. Find the maximum height reached by m2
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Will answer when i get home
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Will answer when i get home
k thank you
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In fig m1=4m2. Pulleys are smooth and the thread is light. At t=0 system is released from rest. Find the maximum height reached by m2
since m1 has two strings attached it has half the acc
4mg-2T=2ma (1) for m1
T-mg=ma (2) for m2
(1)+2*(2)
2mg=4ma a=g/2
fpor m1 a=g/4=2.45 v^2=u^2+2as v2=0+2*9.8/4*0.2=0.98 v=sqrt(0.98)=0.99
The speed of m2 is twice the speed of m1 ie 1.98
after m1 stops m2 takes on acceleration -g v^2=u^2+2as s=(0-3.96)/2*9.8=0.2
height reached=0.4+0.2=0.6
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In fig m1=4m2. Pulleys are smooth and the thread is light. At t=0 system is released from rest. Find the maximum height reached by m2
since m1 has two strings attached it has half the acc
4mg-2T=2ma (1) for m1
T-mg=ma (2) for m2
(1)+2*(2)
2mg=4ma a=g/2
fpor m1 v^2=u^2+2as v2=0+2*9.8/2*0.2=1.96 v=sqrt(1.96)=1.4
after m1 stops m2 takes on acceleration -g v^2=u^2+2as s=(0-1.96)/2*9.8=0.1
height reached=0.2+0.1=0.3
answer is 6.6 cm ???
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I looked at my post again. Did you mean 0.6m?
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I looked at my post again. Did you mean 0.6m?
yup sorry...0.6 cm
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I corrected my post.
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thank you sir
I have some more comin :(
I'm bad at this.....i hate constraints >:(
Please help me in this one too
In the arrangement shown, the mass of the bar and wedge is same and wedge angle is 37 degrees. The masses of the pulley and thread r negligible. Friction is absent. Find the acceleration of the wedge
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Will answer when i get home
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Will answer when i get home
cool
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Resolving horizontally for the wedge
T+mgsin37-Tcos37=ma ma=mgsin37+T(1-cos37) (1)
resolving down the slope for the block
mgsin37-T=ma/cos37 so T=mgsin37-ma/cos37 sub into (1)
ma=mgsin37+(mgsin37-ma/cos37)(1-cos37)
then divide by m
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Resolving horizontally for the wedge
T+mgsin37-Tcos37=ma ma=mgsin37+T(1-cos37) (1)
resolving down the slope for the block
mgsin37-T=ma/cos37 so T=mgsin37-ma/cos37 sub into (1)
ma=mgsin37+(mgsin37-ma/cos37)(1-cos37)
then divide by m
I know this may kinda irritate you......but when I solve this...I get 5g/13... :-[........the answer is 13mg/18..damn!!!
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could u draw a free body diagram please....I won't understand this otherwise
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Resolving horizontally for the wedge
T+mgsin37-Tcos37=ma ma=mgsin37+T(1-cos37) (1)
resolving down the slope for the block
mgsin37-T=ma so T=mgsin37-ma sub into (1)
ma=mgsin37+(mgsin37-ma)(1-cos37)
ma+ma(1-cos37)=mgsin37+mg(1-cos37) so a=(mgsin37+mg(1-cos37) )/(m+m(1-cos37))
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It works now
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baapre is this a levels physics???
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will try and draw a diagram
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I think it is iit
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I think it is iit
yup and A level is nuin compared to this crap.....I'm still tryin to figure that out. I feel dumb ::)
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ohh!!
so this is iit!!
thatswhat iwas thining how can A levels be so tough!!!
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here
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here
that makes it easier. thanks :)
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There r a few more. Please answer them asap....I'm so screwed for my major test :(
Q. A steel ball is suspended from the ceiling of an accelerating carriage by means of two cords A and B. Determine the acceleration- a of the carriage which will cause the tension in A to be twice that in B
Ans-g/3 X root3
I don't even know where to begin from in this one. I'm flunking this test or sure
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In the figure shown, friction force between the bead and the thread is mg/4. Find the time in which the bead loses contact with the string after the system is released from rest.
Ans- t= root of (8l/7g)
I'm getting root (8l/3g) ???
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I neeeeed Helllllllpppppppppppppppp....................any1
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I will be up in 2 few minutes. When is your test
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I will be up in 2 few minutes. When is your test
test on sunday.....but I have my class in like.......1hr....I am supposed to submit these questions to my sir......I dunno friction......I suck at it big time
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What is the mass of the ball and carriage
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What is the mass of the ball and carriage
nuin mentioned
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for the 1st
>2T root 3 /2-Troot 3 /2=ma so Troot 3 /2=ma m=maa osf ball
res vertically 2T/2+T=mg so T=2mg/3
then from first eq a=2mg/3 *root 3 /2m=groot 3 /3 =
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working back for the secon one I get the acceleration of the bead is 7g/4 which is impossible
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I can see the problem now. As the bead is moving down the string., the string is accelerating upwards so the length of time will be shorter. I have to find the acceleration of the m particle
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I think I have it. Solve the equations a1+a2=g
and a2=g/4 to get a1=3g/4 then string is passing through bead at 2a1+a2=7g/4 and use s=1/2at^2 with a =7g/4
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THank YoU Very MucH Sir
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Need help again :(
Find the acceleration of the wedge shown in the figure if the mass of the wedge is 4 times the mass of the block. All surfaces are frictionless. The angle of the inclined surface of the wedge is 37o with the horizontal.
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It will be late tonight when i answer this.
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It will be late tonight when i answer this.
np :)
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Well I'd have to add a few more to this ::) I'm dumb u c :D
In the figure shown,pulley and strings are light and inextensible.Initially all the bodies are at rest when string connecting A and B is cut. Find the initial acceleration of A,C and the pulley
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Anybody give it try plzz.....
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hey nid about that wedge n block.....heres what i did wit my m1 knowledge n a lil thinkin :P
(http://img526.imageshack.us/img526/1271/mechwedge.jpg)
Since all surfaces r frictionless, the wedge resting on ground would slip as the block moves down. So i guess v need to split the component of block's weight (4mg) into a force pushing the wedge... so that it slips left
I don't think this is the right way...mayb astar can help
Interestin ques...even i wanna know the answer :D
did u ask in goiit.com?
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Thanks for givin it a try :)
I saw that site. I registered but I'm still to ask questions there....thanks a lot tho :D
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For the pulley when the string is cut there is a resultant for ce mg due to the tension in the spring acting on a total mass of 3m hence using F=ma
mg=3ma so a =g/3
(The tension initially is 4mg becasue`the system is in equilibrium)
resolving vertically for A
T-mg=ma
and for B
2mg-T=2ma
solve these to get a=g/3
so the acceleration of A is g/3+g/3=2g/3 and the acceleration of B is g/3-g/3=0
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For the black and wedge problem try this attachement
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thanks sir... :)