Explanation:
The given data is as follows.
Mass of small bucket (m) = 4 kg
Mass of big bucket (M) = 12 kg
Initial velocity ([tex]v_{o}[/tex]) = 0 m/s
Final velocity ([tex]v_{f}[/tex]) = ?
Height [tex]H_{o} = h_{f}[/tex] = 2 m
and, [tex]H_{f} = h_{o}[/tex] = 0 m
Now, according to the law of conservation of energy
starting conditions = final conditions
[tex]\frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}[/tex]
[tex]\frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)[/tex]
235.44 = [tex]8V^{2}_{f}[/tex] + 78.48
[tex]V_{f}[/tex] = 4.43 m/s
Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.
