Diagram shows two identical vessels X and Y connected by short pipe with a tap. Initially, X is filled

 Q#64: [Energy > Potential energy](Past Exam Paper – June 2009 Paper 1 Q15)

Diagram shows two identical vessels X and Y connected by short pipe with a tap.

Initially, X is filled with water of mass m to depth h, and Y is empty. When tap is opened, water flows from X to Y until depths of water in both vessels are equal.

How much potential energy is lost by water during this process? (g = acceleration of free fall)

Solution 64:

Answer: B.

Consider the water in both vessels as one body with one mass.

Since the mass of the water is spread all over the volume it occupies, the centre of mass should be considered when calculating the (gravitational) potential energy.

Before the tap is opened, the centre of mass of the water in vessel X may be considered to be at its centre (that is, at half the water level [h/2]).

Potential energy of water before tap is opened = m g (h/2) = mgh / 2

When the tap is opened, the total mass of water in the whole body is the same, but the water level is now at a height of h/2 in both vessels.

The centre of mass may be considered to be at the centre of the volume of water, that is, at a height of half the water level [water level = h/2, so centre of gravity is now at a height of h/4]

Potential energy of water after tap has been opened = m g (h/4) = mgh / 4

Lost in potential energy = (mgh / 2) – (mgh / 4) = mgh / 4

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