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DLS
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If the centre of mass of three particles of masses 10,20 and 30g is at the point(1,2,3)then where should a fourth particle of mass 40g be placed,so that the combined centre of mass of the system is (1,1,1)
 one year ago
 one year ago
DLS Group Title
If the centre of mass of three particles of masses 10,20 and 30g is at the point(1,2,3)then where should a fourth particle of mass 40g be placed,so that the combined centre of mass of the system is (1,1,1)
 one year ago
 one year ago

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@mayankdevnani
 one year ago

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Position of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3)
 one year ago

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I didn't understand the coordinates,is it 1,0 2,0 & 3,0?
 one year ago

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There are three Coordinates....For the 1 St one..CM is @ (1,2,3)
 one year ago

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1040+90/60 =1?
 one year ago

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i didnt get it what to do
 one year ago

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First Find the coordinates of m1 m2 and m3
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
(1,2,3 ) x , y , z x axis of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3) y axis of Cm = m1y1 + m2y2 + m3y3 / (m1+m2+m3) etc...
 one year ago

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u got it..)
 one year ago

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thanks :D then?
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
wait .. i know.. on call with gf :P.. 2 mins
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
ok.. so let xcm be the centre of mass of the three particle system then \[Mxcm = m1x1+ m2x2 + m3x3 \] now .. when the fourth particle is added.. the centre of mass shifts hence \[M'xcm' = m1x1+ m2x2 + m3x3+ m4x4\] or \[M'xcm' = Mxcm + m4x4\] the only thing you have to find now is x4.. so calculate.. do the same in other 2 axes
 one year ago
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