Question

Two particles whose masses are 10 kg and 30 kg and their position vectors are i + i + K and -i-j-k
respectively would have the centre of mass at -
(1) (1 + i + k)
(2) (1 + i + k)
(3) - (1 + i +k)
(4) (1 + i + 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

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

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

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

$\vec r=\frac{-20\hat i-20\hat j-20\hat k}{40}$

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

Hope this answer is helpful.

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Answer 2

Answer:

Hope this answer is helpful.