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 $=10 \mathrm{~ms}^{-2}$ )

Question

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{a}=\left(\frac{\mathrm{m}_2-\mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right) \mathrm{g}=\left(\frac{2 \mathrm{~m}-\mathrm{m}}{2 \mathrm{~m}+\mathrm{m}}\right) \times 10 \\ & =\frac{10}{3} \mathrm{~ms}^{-2} \end{aligned}$ After time, $\mathrm{t}=5.45$ $\begin{aligned} & \mathrm{v}_1=\mathrm{v}_2=\mathrm{at}=\frac{10}{3} \times 5.4=18 \mathrm{~ms}^{-1} \\ & \therefore \mathrm{v}_{\mathrm{cm}}=\frac{\mathrm{m}_2 \mathrm{v}_2-\mathrm{m}_1 \mathrm{v}_1}{\mathrm{~m}_1+\mathrm{m}_2}=\frac{18(2 \mathrm{~m}-\mathrm{m})}{(\mathrm{m}+2 \mathrm{~m})} \\ & =6 \mathrm{~ms}^{-1} \end{aligned}$

Two blocks of masses m and 2 m are connected by a massless string which passes over a fixed frictionless…

Solution