Question

The coordinates of the positions of particles of mass $7,4\text{ and 10}gm$ are $(\text{1,5,} - 3),(\text{2,}5,7)\$and $(\text{3, 3, } - 1)cm$ respectively. The position of the centre of mass of the system would be

• A. $\left( - \frac{15}{7},\frac{85}{17},\frac{1}{7} \right)\text{ }cm$
• B. $\left( \frac{15}{7}, - \frac{85}{17},\frac{1}{7} \right)\text{ }cm$
• C. $\left( \frac{15}{7},\frac{85}{21}, - \frac{1}{7} \right)\text{ }cm$
• D. $\left( \frac{15}{7},\frac{85}{21},\frac{7}{3} \right)\text{ }cm$

Answer 1

Answer: (C)

$\left( \frac{15}{7},\frac{85}{21}, - \frac{1}{7} \right)\text{ }cm$

Answer 2

$m_{1} = 7gm$, $m_{2} = 4gm$, $m_{3} = 10gm$

and $\overset{\rightarrow}{r_{1}} = (\widehat{i} + 5\widehat{j} - 3\widehat{k}),r_{2} = (2i + 5j + 7k),$ $r_{3} = (3\widehat{i} + 3\widehat{j} - \widehat{k})$

Position vector of center mass

$\overset{\rightarrow}{r} = \frac{7(\widehat{i} + 5\widehat{j} - 3\widehat{k}) + 4(2\widehat{i} + 5\widehat{j} + 7\widehat{k}) + 10(3\widehat{i} + 3\widehat{j} - \widehat{k})}{7 + 4 + 10} = \frac{(45\widehat{i} + 85\widehat{j} - 3\widehat{k})}{21}$

⇒ $\overset{\rightarrow}{r} = \frac{15}{7}\widehat{i} + \frac{85}{21}\widehat{j} - \frac{1}{7}\widehat{k}$. So coordinates of centre of mass $\left\lbrack \frac{15}{7},\frac{85}{21},\frac{- 1}{7} \right\rbrack$.