Question

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

• A. $-\frac{\left(\hat{i} + \hat{j} + \hat{k}\right)}{2}$
• B. $\frac{\left(\hat{i} + \hat{j} + \hat{k}\right)}{2}$
• C. $-\frac{\left(\hat{i} + \hat{j} + \hat{k}\right)}{4}$
• D. $\frac{\left(\hat{i} + \hat{j} + \hat{k}\right)}{4}$

Answer 1

Answer: (A)

$\overset{ \rightarrow }{r_{c m}}=\frac{m_{1} \overset{ \rightarrow }{r_{1}} + m_{2} \overset{ \rightarrow }{r_{2}}}{m_{1} m_{2}}$
$\Rightarrow \, \overset{ \rightarrow }{r_{c m}}=\frac{10 \left(\hat{i} + \hat{j} + \hat{k}\right) + 30 \left(- \hat{i} - \hat{j} - \hat{k}\right)}{40}$
$\Rightarrow \, \, \, \overset{ \rightarrow }{r_{c m}}=\frac{- 20 \left(\hat{i} + \hat{j} + \hat{k}\right)}{40}$
$\Rightarrow \, \, \, \overset{ \rightarrow }{r_{c m}}=-\frac{\left(\hat{i} + \hat{j} + \hat{k}\right)}{2}$