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.

Answer 1

Given that a = $3.26m/s^2$

T = 3.9 N

After 2 sec, mass $m_1$ has velocity

v = u + at = 0 +3.26 $\times$ 2

= 6.52 m/s upward

At this time $m_2$ is moving 6.52 m/s downward.

At time 2 sec, $m_2$ stops for a moment.

But $m_1$ is moving upward with velocity 6.52 m/s. It will continue to move till final velocity (at highest point) becomes zero.

Here, v = 0, u = 6.52

a = -g = - 9.8 $m/s^2$

[since $m_1$ is moving upward]

v = u+ at = 6.52 +(-9.8) t

$\Rightarrow t=\frac{6.52}{9.8}=0.66=\frac{2}{3}sec.$

During this period of $\frac{2}{3}$ sec, mass $m_2$ also starts moving downward. So the string becomes tight again after a time of $\frac{2}{3}$ sec.