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DLS
 3 years 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)
DLS
 3 years 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)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Position of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3)

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First Find the coordinates of m1 m2 and m3

anonymous
 3 years ago
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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait .. i know.. on call with gf :P.. 2 mins

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok.. 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
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