AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
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 \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
- A \(\frac{5(\sqrt{3}-1)}{2}\)
- B \(\frac{5(\sqrt{3}-1)}{2 \sqrt{2}}\)
- C \(\frac{5(\sqrt{3}+1)}{2 \sqrt{2}}\)
- D \(\frac{5(\sqrt{3}-1)}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{5(\sqrt{3}-1)}{2 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
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…
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