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Units And Measurement

Question
CBSEENPH11017404

Two particles of masses 100gm and 300 gm have position coordinates (2,5,13) and (-6,4,-2) respectively. Find the position coordinates of centre of mass.

Solution
The position coordinates of centre of mass are given by, 
X = fraction numerator straight m subscript 1 straight x subscript 1 space plus space straight m subscript 2 straight x subscript 2 over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction 
Y = fraction numerator straight m subscript 1 straight y subscript 1 space plus space straight m subscript 2 straight y subscript 2 over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction 
Z = fraction numerator straight m subscript 1 straight z subscript 1 space plus space straight m subscript 2 straight z subscript 2 over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
Substituting the respective values, we get 
straight X space equals space fraction numerator 100 space cross times space 2 space plus space 300 space cross times space left parenthesis negative 6 right parenthesis over denominator 100 plus 30 end fraction space equals space minus 4 space

straight Y space equals space fraction numerator 100 space cross times space 5 space plus space 300 space cross times space 4 over denominator 100 space plus space 300 end fraction space equals space 17 over 4

straight Z space equals space fraction numerator 100 space cross times space 13 space plus space 300 space cross times space left parenthesis negative 2 right parenthesis over denominator 100 space plus space 300 end fraction space equals space 7 over 4 
Therefore, the position coordinates of centre of mass are (negative 4 comma space 17 over 4 comma space 7 over 4