Question

Particles of masses 1kg, 1kg are at the points whose position vectors are $i+j$, $3i+j$. The position vector of their center of mass is

& \text{A)}\dfrac{1}{2}m \\\ & B)2i+\text{j} \\\ & C)\dfrac{1}{3}\left( 6i+\text{j} \right) \\\ & D)8i \\\ \end{aligned}$$

Answer 1

Masses and the position vectors of two particles are given in the question. For a system of n number of particles, the position vector of their center of mass is given by, the ratio of sum of their individual masses multiplied by their individual position vectors to the sum of their masses. To find the position vector of the center of mass of these two particles, we can use the above relation.
Formula used:
${{\vec{r}}_{cm}}=\dfrac{{{m}_{1}}{{{\vec{r}}}_{1}}+{{m}_{2}}{{{\vec{r}}}_{2}}}{{{m}_{1}}+{{m}_{2}}}$

Complete step by step solution:
Given,
${{m}_{1}}=1kg$
${{m}_{2}}=1kg$
${{\vec{r}}_{1}}=i+j$
${{\vec{r}}_{2}}=3i+j$
We have,
Position vector of the center of mass,
${{\vec{r}}_{cm}}=\dfrac{{{m}_{1}}{{{\vec{r}}}_{1}}+{{m}_{2}}{{{\vec{r}}}_{2}}}{{{m}_{1}}+{{m}_{2}}}$
Where,
${{\vec{r}}_{1}}$is the position vector of particle 1
${{\vec{r}}_{2}}$is the position vector particle 2
${{m}_{1}}$is the mass of particle 1
${{m}_{2}}$is the mass of particle 2
Substitute the given values in equation 1, we get,
${{\vec{r}}_{cm}}=\dfrac{1\left( i+j \right)+1\left( 3i+j \right)}{1+1}=\dfrac{4i+2j}{2}=\left( 2i+\text{j } \right)\text{m}$

Therefore, the answer is option B.

Note:
A point at which all the masses of a system of particles or the entire mass of the body appears to be concentrated is known as the centre of mass. The terms center of gravity and center of mass are used in a uniform gravity field to represent a unique point in a system or object that can be used to describe its response to external torques and forces. Mechanics deal with point objects and all the theorems and laws and of mechanics are valid with point objects only. Center of mass is a point which can represent the whole body as a point object.