TS EAMCET · Physics · Center of Mass Momentum and Collision
Two blocks of equal masses are tied to the ends of a light string. The string passes over a mass less pulley fixed on frictionless surface as shown in the figure. The acceleration of the centre of mass of the blocks is ( \(g\) - acceleration due to gravity)
- A \(\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) \mathrm{g}\)
- B \(\left(\frac{\sqrt{3}+1}{4 \sqrt{2}}\right) \mathrm{g}\)
- C \(\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right) \mathrm{g}\)
- D \(\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) \mathrm{g}\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) \mathrm{g}\)
Step-by-step Solution
Detailed explanation
Since, \(\mathrm{mg} \sin 60>\mathrm{mg} \sin 30\). So, block will go left side. \(\mathrm{T}-\mathrm{mg} \sin 30=\mathrm{m}\) a \(\mathrm{T}-\frac{\mathrm{mg}}{2}=\mathrm{m}\) a \(\mathrm{mg} \sin 60-\mathrm{T}=\mathrm{m}\) a…
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