Question

A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the floor (Fig. P7.55). Use the principle of conservation of energy to find the speed with which this bucket strikes the floor. You can ignore friction and the mass of the pulley.
Two paint buckets connected

Answer 1

PE lost by M2 =12kg bucket will be M2gh. PE gained by M1 = 4kg will be M1gh. Net loss of PE will be:

(M2-M1)gh. Total KE as it hits the floor will be: (1/2)(M1 + M2)v^2

(Note they both will be moving at the same speed because the rope doesn't stretch.)

Thus conserv. of Energy gives us:

(M2-M1)gh = (1/2)(M1 + M2)v^2 or

v = sqrt[2(M2-M1)gh / (M1 + M2)].

Plug in M1 = 4, M2 = 12, g=9.8, h=2 and you will get

v = 4.43 m/s

Answer 2

SOLUTIION : 

Let v be speed when 12 kg bucket  hits the floor. 

Net P. E. Lost = K.E gained by both the masses.

(12 - 4) * g * 2 = 1/2 (12 + 4) v^2

=> 16 g = 8 v^2

=> v = sqrt(2g) 

=> v = sqrt(2 * 9.8)

=> v = speed at which bucket hits the floor =  4.427 m/s (ANSWER)