Question

Two particles whose masses are 10 kg and 30 kg
and their position vectors are i +j+k and -1-1-K
respectively would have the centre of mass at -
(A) - ( + J + k)
(B) (1 + i + k)
(0) (1 +j+k)
(0) (i+j+k)​

Answer 1

Two particles whose masses are 10 kg and 30 kg and their position vectors are $\hat i+\hat j+\hat k$ and $-\hat i-\hat j-\hat k$  respectively would have the centre of mass at

$\boxed{\frac{-(\hat i+\hat j+\hat k)}{2}}$

Explanation:

If the position vector of two particles of masses $m_1$ and $m_2$  are  $\vec r_1$ and $\vec r_2$ respectively then

The position of centre of mass is given by

$\boxed{\vec r=\frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2}}$

Here

$\vec r_1=\hat i+\hat j+\hat k$

$\vec r_2=-\hat i-\hat j-\hat k$

Masses of the particles

$m_1=10$ kg

$m_2=30$ kg

Thus, the position vector is

$\vec r=\frac{10(\hat i+\hat j+\hat k)+30(-\hat i-\hat j-\hat k)}{10+30}$

$\implies \vec r=\frac{-20(\hat i+\hat j+\hat k)}{40}$

$\implies \vec r=\frac{-(\hat i+\hat j+\hat k)}{2}$

Hope this answer is helpful.

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Answer 2

Explanation:

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