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27 Mar 2008 08:03:55 IST
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Three particles each of mass m are connected with each other by a massless rod of length 2a to form an equilateral triangle.The system is placed on a smooth horizontal surface.A particle of mass m moving with speed V along the line BC
collides with particle B and comes to rest..Find tension in rod after collision..
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IMPOSSIBLES ARE OFTEN UNTRIED... |
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27 Mar 2008 10:25:18 IST
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really nice!!!!!!!!!!!!!
someone plz answer.......
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I am the BEST
rest are GOOD !!! |
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27 Mar 2008 11:36:22 IST
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someone solve it!!!!!!!
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27 Mar 2008 11:48:09 IST
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i'm not sure
i'll try
now consider the whole triangle as one system
conservation of momentum
mv + 3m(0) = 3m(v') + m (0)
v' = v/3
s= 2a
v^2 +u^2 = 2As
u =0
s = 2a
Accelration A = (v/3)^2 / 2 (2a) = v^2 / 36a
now
T + T cos60 = ma
T ( 3/2) = mv^2/36a
T = mv^2 /54a
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27 Mar 2008 12:14:19 IST
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ans is given mv^2/72a
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conserving momentum along the line of action of force ..
mv = m(0) + 3m(v2)
v2 = v / 3
tensionn in which rod is required dude
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Tarun Sharma
BITS - PILANI |
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27 Mar 2008 12:22:14 IST
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we can use the concept that change in momentum which is
mv / 3 is equal to the NET force
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Tarun Sharma
BITS - PILANI |
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27 Mar 2008 12:31:00 IST
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@ tarun:
how is it possible for change in momentum be equal to force????
thts wrong from the roots of units n dimensions itself
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Nitwit Blubber Odment Tweak
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