Let
be the acceleration of the masses. By Newton's second law, we have
• for the masses on the left,
![1.3mg - T = 1.3ma](https://img.qammunity.org/2023/formulas/physics/college/1t6onmce8l3quks13c1uhwsbsapncioib3.png)
where
is the magnitude of tension in the pulley cord, and
• for the mass on the right,
![T - mg = ma](https://img.qammunity.org/2023/formulas/physics/college/k29oxcy5uefw2rbrxl9scvqmmxcz8xwppu.png)
Eliminate
to get
![(1.3mg - T) + (T - mg) = 1.3ma + ma](https://img.qammunity.org/2023/formulas/physics/college/4n812cm0yogupdru6u4jtsv6u13y322328.png)
![0.3mg = 2.3ma](https://img.qammunity.org/2023/formulas/physics/college/dn3lgp2mme03gukgqfskndt3d69yks0z3d.png)
![\implies a = (0.3)/(2.3)g \approx 0.13g \approx 1.3 (\rm m)/(\mathrm s^2)](https://img.qammunity.org/2023/formulas/physics/college/20is03abt2oyjr0ucyp0fcmmmbf6yavill.png)
Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity
such that
![v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3(\rm m)/(\mathrm s^2)\right)(0.75\,\mathrm m)} \approx 1.4(\rm m)/(\rm s)](https://img.qammunity.org/2023/formulas/physics/college/j5wkfqjs2b10slisg7woesf86i3xtstqc4.png)
When the 0.5m mass is released, the new net force equations change to
• for the mass on the right,
![mg - T' = ma'](https://img.qammunity.org/2023/formulas/physics/college/fmmsmi0u927xican4tchdljoiayg7bqbnc.png)
where
and
are still tension and acceleration, but not having the same magnitude as before the mass was removed; and
• for the mass on the left,
![T' - 0.8mg = 0.8ma'](https://img.qammunity.org/2023/formulas/physics/college/uogmb37cuul0za2s2u1oioev5qju4eeieh.png)
Eliminate
.
![(mg - T') + (T' - 0.8mg) = ma' + 0.8ma'](https://img.qammunity.org/2023/formulas/physics/college/vfolcg225xl3pv6w42gm2vjpd0le8vives.png)
![0.2mg = 1.8 ma'](https://img.qammunity.org/2023/formulas/physics/college/ev2s5hs9m1ln170hl8b0qegowh2xdfeagq.png)
![\implies a' = (0.2)/(1.8)g = \frac19 g \approx 1.1(\rm m)/(\mathrm s^2)](https://img.qammunity.org/2023/formulas/physics/college/l8cr5zpylrpaovbhk0opeq5kcebvi79lzs.png)
Now, the right-hand mass has an initial upward velocity of
, but we're now treating down as the positive direction. As it returns to its starting position, its speed
at that point is such that
![{v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4(\rm m)/(\rm s)\right)^2 + 2\left(1.1(\rm m)/(\mathrm s^2)\right)(0.75\,\mathrm m)} \approx \boxed{1.9(\rm m)/(\rm s)}](https://img.qammunity.org/2023/formulas/physics/college/hizw0ba61nugjma15nx3pg3ewxfid0370x.png)