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Centre of mass is at point X (1,1,1) when system consists of particles of masses 2, 3, 4 and 5 kg. If the centre of mass shifts to point Y (2, 2, 2) on removal of the mass of 5 kg. What was its position ? |
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Answer» Solution :Let `VEC(r )_(1), vec(r )_(2), vec(r )_(3)`, are the position vectors of the PARTICLES of mass `m_(1)=2KG, m_(2)=3 KG, m_(3)=4 kg`and `m_(4)=5 kg` respectively. Initially, the centre of mass was at the point `xx(1,1,1)`, then `hat(i)+hat(J)+hat(k)=(m_(1)vec(r )_(1)+m_(2)vec(r )_(2)+m_(3)vec(r )_(3)+m_(4)vec(r )_(4))/(m_(1)+m_(2)+m_(3)+m_(4))` `rArr (2vec(r )_(1)+3vec(r )_(2)+4vec(r )_(3)+5vec(r )_(4))/(2+3+4+5)=hat(i)+hat(j)+hat(k)` `rArr 2vec(r )_(1)+3vec(r )_(2)+4vec(r )_(3)+5vec(r )_(4)=14hat(i)+14hat(j)+14hat(k)` ....(1) On removing the mass `m_(4)=5 kg`, the centre of mass shifts to the point Y (2, 2, 2), then `2hat(i)+2hat(j)+2hat(k)=(m_(1)vec(r )_(1)+m_(2)vec(r )_(2)+m_(3)vec(r )_(3))/(m_(1)+m_(2)+m_(3))` `rArr 2vec(r )_(1)+3vec(r )_(2)+4vec(r )_(3)=18hat(i)+18hat(j)+18hat(k)`....(2) By subtracting the equation (2) from (1). `vec(r )_(4)=-(4)/(5)(hat(i)+hat(j)+hat(k))` |
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