question_answer

Three particles each of mass 1 kg are placed at the corners of a right angled triangle AOB, O being the origin of the coordinate system (OA and OB along positive X- direction and positive Y- direction). If OA=OB=1m, the positive vector of the centre of mass (in metres) is:

A)  \[\frac{\widehat{i}+\widehat{j}}{3}\]                                   

B)  \[\frac{2\left( \widehat{i}+\widehat{j} \right)}{3}\]

C)  \[\frac{2\left( \widehat{i}+\widehat{j} \right)}{3}\]                       

D)  \[\left( \widehat{i}-\widehat{j} \right)\]

Correct Answer: A

Solution :

                 Three particles each of mass 1 kg are placed the three corners of a right angled triangle. Here, \[{{m}_{1}}={{m}_{2}}={{m}_{3}}=1\] \[({{x}_{1}},{{y}_{1}})=(0,0)\] \[({{x}_{3}},{{y}_{3}})=(0,1)\]                 \[({{x}_{CM}},{{y}_{CM}})=?\] \[{{x}_{CM}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[{{x}_{CM}}=\frac{1.0+1.1+1.0}{1+1+1}\] \[{{x}_{CM}}=\frac{1}{3}\] \[{{y}_{CM}}=\frac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[{{y}_{CM}}=\frac{1.0+1.0+1.1}{1+1+1}\] \[{{y}_{CM}}=\frac{1}{3}\] \[\therefore \] \[\vec{r}\,CM={{x}_{CM}}\hat{i}+{{y}_{CM}}\hat{j}=\frac{1}{3}\hat{i}+\frac{1}{3}\hat{j}\] \[{{\vec{r}}_{CM}}=\frac{\hat{i}+\hat{j}}{3}\]