Two blocks of equal masses are tied with a light string passing over a massless pulley (Assuming frictionless surfaces) acceleration of centre of mass of the two blocks is $\left(g=10 \mathrm{~ms}^{-2}\right)$

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Aicharya · Accepted Answer

The acceleration of the block is $a=\frac{(5 \sqrt{3}-5) M}{(M+m)}=\frac{5}{2}(\sqrt{3}-1) \mathrm{m} / \mathrm{s}$ $\therefore$ Acceleration of centre of mass is $\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{\mathrm{m}_1 \overrightarrow{\mathrm{a}}_1+\mathrm{m}_2 \overrightarrow{\mathrm{a}}_2}{\mathrm{~m}_1+\mathrm{m}_2}=\frac{\mathrm{M}\left(\overrightarrow{\mathrm{a}}_1+\overrightarrow{\mathrm{a}}_2\right)}{2 \mathrm{M}}=\frac{\overrightarrow{\mathrm{a}}_1+\overrightarrow{\mathrm{a}}_2}{2}$
Also, $\vec{a}_1=$ a $\hat{i}, \vec{a}_2=a \hat{j}$ $\begin{aligned} & \therefore \vec{a}_{c m}=\frac{a}{2}(\hat{i}+\hat{j}) \\ & \therefore a_{c m}=\frac{a \sqrt{2}}{2}=\frac{1}{\sqrt{2}} \cdot \frac{5}{2}(\sqrt{3}-1)=\frac{5(\sqrt{3}-1)}{2 \sqrt{2}} \end{aligned}$

Two blocks of equal masses are tied with a light string passing over a massless pulley (Assuming…

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