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Four particles A, B, C and D with masses m_{A}=m, m_{B}=2m, m_{C}=3m\; and\; m_{D}=4m are the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is :

 

 

  • Option 1)

    \frac{a}{5}(\widehat{i}-\widehat{j})
     

  • Option 2)

    a(\widehat{i}+\widehat{j})

  • Option 3)

    Zero

     

  • Option 4)

    \frac{a}{5}(\widehat{i}+\widehat{j})

 

Answers (1)

best_answer

\vec{a}_{cm}=\frac{ \sum m\vec{a}}{ \sum m}=\frac{m(-a\hat{i}\: )+(2m)(a\hat{j}\: )+(3m)(a\hat{i}\: )+4m(-a\hat{j}\: )}{m+2m+3m+4m}

\vec{a}_{cm}=\frac{2(\; \hat{i}-\hat{j}\: )a}{10}=\frac{1}{5}(\: \hat{i}-\hat{j}\: )a

\vec{a}_{cm}=\frac{a}{5}(\: \hat{i}-\hat{j}\; )


Option 1)

\frac{a}{5}(\widehat{i}-\widehat{j})
 

Option 2)

a(\widehat{i}+\widehat{j})

Option 3)

Zero

 

Option 4)

\frac{a}{5}(\widehat{i}+\widehat{j})

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