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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 May 2008 17:09:10 IST
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A particle of mass m ia projected with veolicity v making an angle of 45 0 with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentumwill be ?
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sahya.. |
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17 May 2008 17:16:51 IST
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the answer is 2mvsin(theta)
in projectile motion because no accleration acts in horizontal direction so horizontal component of velocity remain constant.so change in linear momentum take place due to change in vertical component of vel..i.e. vsin(theta)
change in momentum = mvsin(theta)-mvsin(-theta)
=2mvsin(theta)
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17 May 2008 17:17:13 IST
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17 May 2008 17:21:22 IST
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ya that's right i forgot to take angle...sorry....thanks to ryuamakusa....
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17 May 2008 17:22:30 IST
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nooooo..no problem even i make a lot of careless mistakes.
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17 May 2008 17:28:35 IST
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mvcos@ - (-mvcos@)
= mv/root2 - (-mv/root2)
= mv(root2)
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---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
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18 May 2008 20:14:21 IST
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i think spidey...u r wrong....its sin@ not cos@,,,,
i explained above,,,,,if u can explain how,,,,plzzzz
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18 May 2008 20:17:19 IST
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well see it all depends on which angle they take as @
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18 May 2008 20:22:01 IST
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PLS IGNORE THE DOWN DCIRCLE.
THE INITIAL HORIZONTAL MOMENTUM IS CONSTANT
NOW FOR VERTICAL MOTION
CHANGE=MvSIN@-(-MvSIN@)
=2MVSIN@=2MV/ROOT
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18 May 2008 20:23:23 IST
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ya varun is right,,,,,
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18 May 2008 20:24:31 IST
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ok ok ok thanks for correction,,,,thanks RyuAmakusa....
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