Centre of mass of 3 particles 10 kg , 20 kg and 30 kg is at (0, 0, 0). Where should a particle of mass 40 kg be placed so that the combination centre of mass will be at (3 , 3 , 3)

Ans: (b)

sol: Let co-ordinate of Center of mass is (a , b , c)

Position vector of COM , $\large \vec{r} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2} + m_3 \vec{r_3} + m_4 \vec{r_4}}{m_1 + m_2 + m_3 + m_4}$

$\large \vec{r} = \frac{10 \times 0 + 20 \times 0 + 30 \times 0 + 40 \times (a \hat{i} + b\hat{j} + c \hat{k})}{10 + 20 + 30 + 40} $

$\large (3 \hat{i} + 3\hat{j} + 3 \hat{k}) = \frac{ 40 \times (a \hat{i} + b\hat{j} + c \hat{k})}{100} $

$\large 3( \hat{i} + \hat{j} + \hat{k}) = \frac{2(a \hat{i} + b\hat{j} + c \hat{k})}{5} $

$\large 15( \hat{i} + \hat{j} + \hat{k}) = 2 (a \hat{i} + b\hat{j} + c \hat{k})$

2 a = 15 , 2b = 15 , 2 c = 15

a = 7.5 , b= 7.5 , c = 7.5