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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 20:34:56 IST
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is the ans 3
if not, then what is the ans
pls post the diagram, if possible
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 20:36:33 IST
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sorry the answer is not 3 also.....well wat diag r u askin of??
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The only place where you will find success before hard work is in the dictionary
A winner never quits.....a quitter never wins......
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 20:37:42 IST
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well i have tried my best....plz solve it now....
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The only place where you will find success before hard work is in the dictionary
A winner never quits.....a quitter never wins......
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 20:37:54 IST
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i am not getting a clear picture of this ques
i mean where are A and B standing. how are P and Q arranged..
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 20:38:17 IST
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see the diag... edited....
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The only place where you will find success before hard work is in the dictionary
A winner never quits.....a quitter never wins......
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 20:44:39 IST
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some one plz solve......
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The only place where you will find success before hard work is in the dictionary
A winner never quits.....a quitter never wins......
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 20:59:29 IST
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I will use the CoM method here. Initially center of mass was at rest, and since there is no external force it is still at rest. Let the velocity of platform P be x m/s wrt ground.
So velocity of A wrt ground = 5+x
0 = 50*(5+x) + 50x +10x+ 20*5/130
0 = 250 + 110x + 100
-350/110 = x
-35/11 =x
So P moves with a velcity 3.2 m/x app towards the left.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 21:08:03 IST
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Is the answer 5 m/s towards right ?
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______________________________________________
Siddhant Shah
The trouble with doing something right the first time is that nobody appreciates how difficult it is.
You can't control the wind, but you can adjust your sails.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 21:18:41 IST
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ananth, whatb is the ans given in the brilliant ans sheet
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 21:31:18 IST
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Velocity of B w.r.t P = 0 => velocity of B = velocity of P (w.r.t ground) let velocity of P be v towards left w.r.t ground = velocity of B w.r.t ground so velocity of A w.r.t ground = 5 - v conserving momentum we get 50*(5 - v) + 20*5 - 50v - 10v = 0 => v = 35/11 hence velocity of P = velocity of B w.r.t ground is 35/11 m/sec
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 21:36:15 IST
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priyesh
why is velocity of B = velocity of P (w.r.t ground)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 22:03:13 IST
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is it 2.27 m/s (25/11)in the direction opposite that of A's motion
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 22:03:14 IST
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i too think the same as @ sid.shah.90
is it 5m/s
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 22:08:35 IST
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this ques has been tried by a gud number of people
btw, it is a gud one and has been challenging our concepts
so it would be better if an expert solves it
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 22:18:44 IST
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