A system consists of two particles. At t = 0, one particle is at the origin and the other particle having a mass of 600g, is on the y-axis at $y = 80 m$ . At t = 0 the centre of mass of the system is on the y axis at y = 24 m and has a velocity given by $(6 {ms}^{- 3}) t^{2} \hat{j}$ . Find the acceleration of the centre of mass, in ${ms}^{- 2}$ at time t = 2 s.

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answer is 24.

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Detailed Solution

$(a) y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}$

$24 = \frac{(m_{1}) (0) + (\frac{600}{1000}) (80)}{m_{1} + 0.6} = \frac{48}{m_{1} + 0.6}$

$\Rightarrow m_{1} = 1.4 kg$

$So, total mass = (1.4 + 0.6) kg = 2 kg$

$(b) {\bar{v}}_{cm} = (6 {ms}^{- 3}) t^{2} \hat{j}$

$\Rightarrow {\bar{a}}_{cm} = \frac{d {\bar{v}}_{cm}}{dt} = (12 {ms}^{- 3}) t \hat{j}$

${{\bar{a}}_{cm}|}_{t = 2 s} = (12) (2) = 24 {ms}^{- 2}$

$(c) {\bar{a}}_{cm} at t = 3 s = (36 {ms}^{- 2}) \hat{j}$

$\Rightarrow {\bar{F}}_{cm} = (m_{1} + m_{2}) {\bar{a}}_{cm} = (2) (36) \hat{j} = (72 N) \hat{j}$

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