Question:
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.
Solution:
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 $\mathrm{v}=0$
$v=0, u=6.52$
$0=u+a t=6.52+(-9.8) t$
$t=\frac{6.52}{9.8}=0.66 s$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.