The co-ordinates of centre of mass of particles mass 10, 20 and 30 gm are (1, 1, 1) cm. The position co-ordinates of mass 40 gm which when added to the system, the position of combined centre of mass be at (0, 0, 0) are:-

\frac{3}{2}, \frac{3}{2}, \frac{3}{2}

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- \frac{3}{2}, - \frac{3}{2}, - \frac{3}{2}

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\frac{3}{4}, \frac{3}{4}, \frac{3}{4}

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- \frac{3}{4}, - \frac{3}{4}, - \frac{3}{4}

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Solution

The correct option is B $- \frac{3}{2}, - \frac{3}{2}, - \frac{3}{2}$
$\begin{matrix} i n i t i a l y : i n x d i r e c t i o n x_{c m} = \frac{10 (x_{1}) + 20 (x_{2}) + 30 (x_{3})}{60} = 1 \Rightarrow 10 (x_{1}) + 20 (x_{2}) + 30 (x_{3}) = 60 f i n a l l y : i n x d i r e c t i o n x_{c m} = \frac{10 (x_{1}) + 20 (x_{2}) + 30 (x_{3})}{100} = 0 \Rightarrow 60 + 40 (x_{4}) = 0 \Rightarrow x_{4} = \frac{- 6}{4} = \frac{- 3}{2} s i m i l a r l y y_{4} = \frac{- 3}{2} z_{4} = \frac{- 3}{2} ∴ (\frac{- 3}{2}, \frac{- 3}{2}, \frac{- 3}{2}) s h o u l d b e p o s i t i o n o f 40 g m H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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The co-ordinates of centre of mass of particles mass 10, 20 and 30 gm are (1, 1, 1) cm. The position co-ordinates of mass 40 gm which when added to the system, the position of combined centre of mass be at (0, 0, 0) are:-

The correct option is B −32,−32,−32initialy:inxdirectionxcm=10(x1)+20(x2)+30(x3)60=1⇒10(x1)+20(x2)+30(x3)=60finally:inxdirectionxcm=10(x1)+20(x2)+30(x3)100=0⇒60+40(x4)=0⇒x4=−64=−32similarlyy4=−32z4=−32∴(−32,−32,−32)shouldbepositionof40gmHence,optionBiscorrectanswer.