From the conservation of energy, the energies can be expressed as:
\({m_{\rm{B}}}gh = \frac{1}{2}{m_{\rm{A}}}{v^2} + \frac{1}{2}{m_{\rm{B}}}{v^2} + \frac{1}{2}I{\omega ^2} + {m_{\rm{A}}}gh\) … (i)
Here, v is the speed of mass B before it hits the ground,\(\omega \)is the angular speed of the pulley, and h is the height of mass B above the ground.
The moment of inertia of the cylindrical pulley can be expressed as:
\(I = \frac{{M{R^2}}}{2}\)
The angular velocity of the pulley can be expressed as:
\(\omega = \frac{v}{R}\)
Substitute the values in equation (i).
\(\begin{aligned}{c}{m_{\rm{B}}}gh &= \frac{1}{2}{m_{\rm{A}}}{v^2} + \frac{1}{2}{m_{\rm{B}}}{v^2} + \frac{1}{2} \times \frac{{M{R^2}}}{2} \times {\left( {\frac{v}{R}} \right)^2} + {m_{\rm{A}}}gh\\{m_{\rm{B}}}gh - {m_{\rm{A}}}gh &= \frac{1}{2}{m_A}{v^2} + \frac{1}{2}{m_{\rm{B}}}{v^2} + \frac{1}{2} \times \frac{{M{v^2}}}{2}\\\left( {{m_B} - {m_{\rm{A}}}} \right)gh &= \left( {\frac{1}{2}{m_A} + \frac{1}{2}{m_{\rm{B}}} + \frac{1}{4} \times M} \right){v^2}\\{v^2} &= \frac{{2\left( {{m_B} - {m_{\rm{A}}}} \right)gh}}{{\left( {{m_A} + {m_{\rm{B}}} + \frac{M}{2}} \right)}}\end{aligned}\)
Substitute the values in the above equation.
\(\begin{aligned}{c}{v^2} &= \frac{{2\left( {38{\bf{ }}{\rm{kg}} - 32{\bf{ }}{\rm{kg}}} \right) \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 2.5{\rm{ m}}}}{{\left( {38{\bf{ }}{\rm{kg}} + 32{\bf{ }}{\rm{kg}} + \frac{{3.1{\bf{ }}{\rm{kg}}}}{2}} \right)}}\\{v^2} &= 4.113{\rm{ }}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}\\v &= \sqrt {4.113} {\rm{ m/s}}\\ &= 2.03{\rm{ m/s}}\end{aligned}\)
Thus, the speed of mass B just before it strikes the ground is 2.03 m/s.
