AP EAMCET · PHYSICS · Laws of Motion
Two blocks of masses m and 2 m are connected by a massless string which passes over a fixed frictionless pulley. If the system of blocks is released from rest, the speed of the centre of mass of the system of two blocks after a time of 5.4 s is
(Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(6 \mathrm{~ms}^{-1}\)
- B \(8 \mathrm{~ms}^{-1}\)
- C \(4 \mathrm{~ms}^{-1}\)
- D \(12 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(A) \(6 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{a}=\left(\frac{\mathrm{m}_2-\mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right) \mathrm{g}=\left(\frac{2 \mathrm{~m}-\mathrm{m}}{2 \mathrm{~m}+\mathrm{m}}\right) \times 10 \\ & =\frac{10}{3} \mathrm{~ms}^{-2} \end{aligned}\) After time,…
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