Four particles A, B, C and D with masses mA = m, mB = 2m, mC = 3m and… | JEE Main 2019 (8 Apr, Shift 1)

Q7JEE Main 2019 · 8 APR · SHIFT 1Medium

Motion of Centre of Mass

Four particles A, B, C and D with masses mA = m, mB = 2m, mC = 3m and mD = 4m are at 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:

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Answer: (b)

From the figure: $\vec{a}_A = -a\hat{i}$ , $\vec{a}_B = a\hat{j}$ , $\vec{a}_C = a\hat{i}$ , $\vec{a}_D = -a\hat{j}$ .
$\sum m\vec{a} = m(-a\hat{i}) + 2m(a\hat{j}) + 3m(a\hat{i}) + 4m(-a\hat{j}) = 2ma\hat{i} - 2ma\hat{j}$
$\vec{a}_{cm} = \frac{2ma(\hat{i}-\hat{j})}{10m} = \frac{a}{5}(\hat{i}-\hat{j})$
So the answer is (b).

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