Two particles A and B of equal mass are at rest at position (2m, 3m) and (4m,– 1m) respectively. They start moving with velocities \left(\right. \hat{i} + 2 \hat{j} \left.\right) \left(ms\right)^{- 1} and \left(\right. - 2 \hat{i} + \hat{j} \left.\right) \left(ms\right)^{- 1} respectively. At the end of 3s, the position of their centre of mass will be

Engineering

Physics

Center of Mass

Question

Two particles A and B of equal mass are at rest at position (2m, 3m) and (4m,– 1m) respectively. They start moving with velocities $(\hat{i} + 2 \hat{j}) {ms}^{- 1}$ and $(- 2 \hat{i} + \hat{j}) {ms}^{- 1}$ respectively. At the end of 3s, the position of their centre of mass will be

(1.5 m, 5.5m)

(– 3.0m, 9.0 m)

(4.5m, 3.0 m)

(3.0, 1.0m)

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The position of a particle at the end of 3s $\vec{R} = {\vec{R}}_{0} + \vec{v} t$

$After 3 s, R_{A} = (2 \hat{i} + 3 \hat{j}) + 3 (\hat{i} + 2 \hat{j}) = 5 \hat{i} + 9 \hat{j} and$

$R_{B} = (4 \hat{i} - 1 \hat{j}) + 3 (- 2 \hat{i} + \hat{j}) = - 2 \hat{i} + 2 \hat{j}$

The co-ordinates of centre of mass, $x = \frac{5 - 2}{2} = 1.5, y = \frac{9 + 2}{2} = 5.5 m$ .

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