KCET · Physics · Center of Mass Momentum and Collision
Three particles of mass \(1 \mathrm{~kg}, 2 \mathrm{~kg}\) and 3 kg are placed at the vertices \(\mathrm{A}, \mathrm{B}\) and C respectively of an equilateral triangle ABC of side 1 m . The centre of mass of the system from vertex A (located at origin) is
- A \(\left(\frac{7}{12}, \frac{3 \sqrt{3}}{12}\right)\)
- B \(\left(\frac{9}{12}, \frac{3 \sqrt{3}}{12}\right)\)
- C \(\left(\frac{7}{12}, \frac{6+3 \sqrt{3}}{12}\right)\)
- D \((0,0)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{7}{12}, \frac{3 \sqrt{3}}{12}\right)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{x}_{\mathrm{cm}}=\frac{\sum \mathrm{mixi}}{\sum \mathrm{mi}} \)
\( =\frac{1 \times 0+2 \times 1+3 \times \frac{1}{2}}{6}=\frac{7}{12} \)
\( \mathrm{y}_{\mathrm{cm}}=\frac{\sum \mathrm{miyi}}{\sum \mathrm{mi}}=\frac{1 \times 0+2 \times 0+3 \times \frac{\sqrt{3}}{2}}{6}=\frac{3 \sqrt{3}}{12}\)
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