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[Post New]7 Apr 2007 21:36:49 IST
    Subject: work power energy
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vijivarun1965 (0)

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 Two identical bodies each of mass m are moving with velocities v1 and v2 respectively in same direction.  The kinetic energy of the system in the frame of reference attached to the centre of mass is (v is the relative velocity between the two bodies)
1) 1/4 mv2   2) 1/2 mv2    3)mv2      4) 1/8 mv2
    
[Post New]7 Apr 2007 22:05:14 IST
    Subject: work power energy
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Avinash_Bhat (665)

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Is it not 3) mv^2 ?? Let me know if am right or wrong...
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[Post New]7 Apr 2007 22:35:35 IST
    Subject: Re:work power energy
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iitkgp_bipin (6498)

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Velocity of centre of mass = (mv1+mv2)/(m+m) = (v1+v2)/2

From COM frame velocity of 1st mass = v1 - (v1+v2)/2
From COM frame velocity of 2nd mass = v2 - (v1+v2)/2

Hence KE = (m/2){v1 - (v1+v2)/2}2 + (m/2){v2 - (v1+v2)/2}2

                = m(v2-v1)2 = mv2
Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur

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[Post New]7 Apr 2007 23:47:15 IST
    Subject: work power energy
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INDIAN_ARMY19890 (1284)

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the ans is 1/4mv2
bcoz velocity of centre of mass will be =V/2 WHERE v is relative velocity so KE=1/2*2M*(V/2)2=1/4MV2
Beat others otherwise they will beat u
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[Post New]7 Apr 2007 23:48:27 IST
    Subject: work power energy
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INDIAN_ARMY19890 (1284)

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GUYS HOW YOU WRITE POWER ?
Beat others otherwise they will beat u
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