Two bodies of mass 2g and 10g have position vectors \(3i+2j-u\) and \(

1.	Two bodies of mass 2g and 10g have position vectors \(3i+2j-u\) and \(3i-j+3u\)respectively. Find the position vector of centre of mass.
Answer» \(M_1=2g,\, \vec r_1=3\hat i+2\hat j -u\) \(M_2=10g,\, \vec r=3\hat i -\hat j+3u\) Position vector of centre of mass = \(\frac{M_1r_1+M_2r_2}{M_1+M_2}\) = \(\frac{2(3i+2j-u)+10(3i-j+3u)}{10+2}\)= \(3i-\frac{1}{2}j+\frac{7}{3}u\)

Two bodies of mass 2g and 10g have position vectors \(3i+2j-u\) and \(3i-j+3u\)respectively. Find the position vector of centre of mass.

Answer»

\(M_1=2g,\, \vec r_1=3\hat i+2\hat j -u\)

\(M_2=10g,\, \vec r=3\hat i -\hat j+3u\)

Position vector of centre of mass = \(\frac{M_1r_1+M_2r_2}{M_1+M_2}\)

= \(\frac{2(3i+2j-u)+10(3i-j+3u)}{10+2}\)= \(3i-\frac{1}{2}j+\frac{7}{3}u\)