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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
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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Yahoo!Best ResponseYou've already chosen the best response.0
Position of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3)
 one year ago

DLSBest ResponseYou've already chosen the best response.0
I didn't understand the coordinates,is it 1,0 2,0 & 3,0?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
There are three Coordinates....For the 1 St one..CM is @ (1,2,3)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
First Find the coordinates of m1 m2 and m3
 one year ago

Yahoo!Best 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

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

MashyBest 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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