Question

particles of masses 100 and 300 gram have position vectors (2î+5j+13k) and (-6î+4j+2k) position vector of centre of mass is

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Answer 1

Answer:

$\bf{(4 \hat{i} +\frac{17}{4} \hat{j} + \frac{19}{4} \hat{k})$

Explanation:

Let the position vector of centre of mass be, $(x_{cm},y_{cm},z_{cm})$

=> particles of masses 100 and 300 gram have position vectors (2î+5j+13k) and (-6î+4j+2k)

   

  $m_1=100 \ grams \\ m_2=300 \ grams \\\\ x_1=2\\x_2=-6 \\\\ y_1=5\\y_2=4\\\\z_1=13\\z_2=2$

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$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2} \\\\ x_{cm}=\frac{100(2)+300(-6)}{100+300} \\\\ x_{cm}=\frac{200-1800}{400} \\\\ x_{cm}=\frac{-1600}{400} \\\\ x_{cm}=-4$

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$y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2} \\\\ y_{cm}=\frac{100(5)+300(4)}{100+300} \\\\ y_{cm}=\frac{500+1200}{400} \\\\ y_{cm}=\frac{1700}{400} \\\\ y_{cm}=\frac{17}{4}$

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$z_{cm}=\frac{m_1z_1+m_2z_2}{m_1+m_2} \\\\ z_{cm}=\frac{100(13)+300(2)}{100+300} \\\\ z_{cm}=\frac{1300+600}{400} \\\\ z_{cm}=\frac{1900}{400} \\\\ z_{cm}=\frac{19}{4}$

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⇒ The position vector of centre of mass is $\bf{(4 \hat{i} +\frac{17}{4} \hat{j} + \frac{19}{4} \hat{k})$