Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Particles of masses $100\;g$ and $300\;g$ have position vectors $(2 \hat {i}+5 \hat {j}+13 \hat {k})$ and $(-6 \hat i +4 \hat j+2 \hat k).$ Position vector of their center of Mass is

\[\begin {array} {1 1} (a)\;\frac{-16}{4}\hat{i}+\frac{17}{4} \hat{j}+\frac{7}{4} \hat {k} \\ (b)\;\frac{20}{4} \hat{i}+\frac{17}{4} \hat {j}+\frac{7}{4} \hat{k} \\ (c)\;\frac{-16}{4} \hat{i}+\frac{17}{4} \hat {j}+\frac{19}{4} \hat{k} \\ (d)\;\frac{-16}{4} \hat{i}+\frac{13}{4} \hat {j}+\frac{19}{4} \hat{k} \end {array}\]

Can you answer this question?

1 Answer

0 votes
$\overrightarrow{r}_{cm}=\large\frac{m_1\overrightarrow{r_1}+m_2 \overrightarrow {r_2}}{(m_1+m_2)}$
So the correct answer is $\large\frac{-16}{4} \hat{i}+\frac{17}{4} \hat {j}+\frac{19}{4} \hat{k}$
answered Nov 20, 2013 by meena.p
edited Jun 16, 2014 by lmohan717

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App