A) \[(-1,-1,-1)\]
B) \[(-2,-2,-2)\]
C) \[(2,2,2)\]
D) \[(1,1,1)\]
Correct Answer: B
Solution :
\[{{x}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[\Rightarrow \,\,\,{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}={{x}_{cm}}({{m}_{1}}+{{m}_{2}}+{{m}_{3}})\] Similarly, \[{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}={{y}_{cm}}({{m}_{1}}+{{m}_{2}}+{{m}_{3}})\] \[{{m}_{1}}{{z}_{1}}+{{m}_{2}}{{z}_{2}}+{{m}_{3}}{{z}_{3}}={{z}_{cm}}({{m}_{1}}+{{m}_{2}}+{{m}_{3}})\] Given, \[{{m}_{1}}=1kg,\,\,{{m}_{2}}=2kg,\,\,{{m}_{3}}=3kg\] \[{{x}_{cm}}={{y}_{cm}}={{z}_{cm}}=3m\] \[\therefore \,\,\,\,\,{{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}=18\] \[{{y}_{1}}+2{{y}_{2}}+3{{y}_{3}}=18\] \[{{z}_{1}}+2{{z}_{2}}+3{{z}_{3}}=18\] Now, if \[{{m}_{4}}=4\,kg\] is introduced in the system, \[{{x}_{cm}}=\frac{{{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}+4{{x}_{4}}}{1+2+3+4}=1\] \[\Rightarrow \,\,\frac{18+4{{x}_{4}}}{10}=1\Rightarrow {{x}_{4}}=-2\] Similarly, \[{{y}_{4}}=-2;\] \[{{z}_{4}}=-2\]
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