determine final speed of first mass after falling using conservation of energy?

Question

A 5.9 kg mass is attached to a light cord that passes over a massless, frictionless pulley. The other end of the cord is attached to a 3 kg mass.

The acceleration of gravity is 9.8 m/s^2.

Use conservation of energy to determine the final speed of the first mass after it has fallen (starting from rest) 5.3 m: Answer in units of m/s.

Answer 1

Use the subscripts l and s for large and small. Let mass Ms start at its 0 Potential energy position. Let mass Ml start with potential energy from its position 5.3 m over final position. And let subscripts 1 and 2 indicate before and after.

Total Energy at start = Total Energy after larger mass falls 5.3 m.
KEs1 + PEs1 + KEl1 + PEl1 = KEs2 + PEs2 + KEl2 + PEl2
0 + Ms*g*0 + 0 + 5.9 kg*g*5.3 m = (1/2)*Ms*v^2 + 3 kg*g*5.3 m + (1/2)*Ml*v^2 + 5.9 kg*g*0

Simplifying
5.9 kg*g*5.3 m = (1/2)*Ms*v^2 + 3 kg*g*5.3 m + (1/2)*Ml*v^2
5.9 kg*g*5.3 m - 3 kg*g*5.3 m = (1/2)*v^2(Ms + Ml)
2.9 kg*g*5.3 m = (1/2)*v^2*8.9 kg
Solve for v^2 and take the sqrt.

Answer 2

5.9 kg minus 3 kg = 2.9 kg so work out the problem as if it was just a 2.9 kg mass & disregard the rest...