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Two particles A and B of mass 5 kg and 2 kg are projected simultaneously from same ground level as shown in Figure. Find the maximum height (in cm) reached by the center of mass. Assume that only gravitational force of attraction by the earth is present and take g =10ms^-2

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Detailed Solution

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Since both particles are projected in the air, so both particles have acceleration g downwards
$\Rightarrow {\vec{a}}_{CM} = - g \hat{j}$
$\begin{array}{l} {(v_{CM})}_{y} = \frac{5 (10 \sin 30^{\circ}) + 2 (17 \sin 30^{\circ})}{5 + 2} \\ \Rightarrow {(v_{CM})}_{y} \frac{25 + 17}{7} = \frac{42}{7} = 6 {ms}^{- 1} \end{array}$

$\begin{array}{l} \Rightarrow {(h_{CM})}_{max} = \frac{{[{(v_{CM})}_{y}]}^{2}}{2 {(a_{CM})}_{y}} = \frac{36}{2 \times 10} = 1 .8 m \\ \Rightarrow {(h_{CM})}_{y} = 180 cm \end{array}$

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Two particles A and B of mass 5 kg and 2 kg are projected simultaneously from same ground level as shown in Figure. Find the maximum height (in cm) reached by the center of mass. Assume that only gravitational force of attraction by the earth is present and take g =10ms-2