Solve the following :


Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment $2.0$ $s$ after the system is set into motion. Find the time elapsed before the string is tight again.


Use law of kinematics $v=u+a t$

$\mathrm{V}=0+3.26 \times 2$

$=6.52 \mathrm{~m} / \mathrm{s}$

$\mathrm{m}_{2}$ is moving down with $\mathrm{v}$

$m_{1}$ is moving down with $v$

at $\mathrm{t}=25, \mathrm{~m}_{2}$ stops

$\mathrm{m}_{1}$ moves upward and reaches $\mathrm{v}=0$

$v=0, u=6.52$

$0=u+a t=6.52+(-9.8) t$

$t=\frac{6.52}{9.8}=0.66 s$

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