AP EAPCET 2024 - 22th May Evening Shift Physics - Center of Mass and Collision Question 17 English Explanation

From the figure,

$$ \begin{aligned} & m g \sin 60^{\circ}-T=m a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & T-m g \sin 30^{\circ}=m a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $$

Adding both Eq. (i) and Eq. (ii), we get

$$ m g\left(\sin 60^{\circ}-\sin 30^{\circ}\right)=2 m a $$

$$ \begin{aligned} &\begin{aligned} & a=\frac{g}{2}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right) \\ & a=\frac{g}{4}(\sqrt{3}-1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned}\\ &\begin{aligned} & \text { Given, } m_1=m_2=m \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & \mathbf{a}_1=-a \cos 60^{\circ} \hat{\mathbf{i}}-a \sin 60 \hat{\mathbf{j}}=\frac{a}{2}(-\hat{\mathbf{i}}-\sqrt{3 \mathbf{j}}) \\ & \mathbf{a}_2=-a \cos 30^{\circ} \hat{\mathbf{i}}+a \sin 30 \hat{\mathbf{j}} \\ & \\ & =\frac{a}{2}(-\sqrt{3 \hat{\mathbf{i}}}+1 \hat{\mathbf{j}}) \end{aligned} \end{aligned} $$

Acceleration of the centre of mass

$$ \mathbf{a}_{\mathrm{com}}=\frac{m_1 \mathbf{a}_1+m_2 \mathbf{a}_2}{m_1+m_2} $$

From Eq. (iv), $\mathbf{a}_{\text {com }}=\frac{\mathbf{a}_1+\mathbf{a}_2}{2}$

$$ \begin{gathered} \mathbf{a}_{\text {com }}=\frac{\frac{a}{2}(-\hat{\mathbf{i}}-3 \sqrt{3 \hat{\mathbf{j}}})+\frac{a}{2}(-\sqrt{3 \hat{\mathbf{i}}}+\hat{\mathbf{j}})}{2} \\ \left|\mathbf{a}_{\text {com }}\right|=\left\lvert\, \frac{a}{4}[(-1-\sqrt{3}) \hat{\mathbf{i}}+(1-\sqrt{3 \hat{\mathbf{j}})] \mid}\right. \\ \quad\left[\because|a \hat{\mathbf{i}}+b \hat{\mathbf{j}}|=\sqrt{a^2+b^2}\right] \end{gathered} $$

$$ \begin{aligned} &\begin{aligned} \left|\mathbf{a}_{\mathrm{com}}\right| & =\frac{a}{4}\left[\sqrt{(-1-\sqrt{3})^2+(1-\sqrt{3})^2}\right] \\ & =\frac{a}{4}[\sqrt{1+3+2 \sqrt{3}+1+3-2 \sqrt{3}}] \\ & =\frac{a}{4} 2 \sqrt{2} \end{aligned}\\ &\text { Putting value of a from Eq. (iii), }\\ &\begin{aligned} & =\frac{g}{4 \cdot 4} \cdot 2 \sqrt{2}(\sqrt{3}-1) \\ & =\frac{10 \times 2 \sqrt{2} \times(\sqrt{3}-1)}{4 \times \sqrt{2} \times 2 \sqrt{2}} \\ & =\frac{10}{4 \sqrt{2}}(\sqrt{3}-1) \\ \left|a_{\mathrm{com}}\right| & =\frac{5(\sqrt{3}-1)}{2 \sqrt{2}} \mathrm{~m} / \mathrm{s}^2 \end{aligned} \end{aligned} $$

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