Solved problem: Momentum conservation - variable mass.

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Statement

difficulty

A truck moves 40 Km/hr in a straight line over a conventional road. An open container is placed on the truck. The total weight is 1800 Kg. When it starts to rain, the container got filled with water at 6 L/ min. Paying no heed to friction , what is the speed of the truck after one hour of rain?

Solution

Data

Initial velocity of the truck v_i = 40 km/h = 40·1000/3600 = 11.11 m/s
Mass of the set m = 1800 kg
Filling speed: 6 L/min
Final time considered t = 1 h = 60 min = 3600 s

Resolution

We can solve the prpblem applying the linear momentum conservation principle: in absence of outer forces the momentum remains constant:

$\sum \vec{F} = 0 \Rightarrow \vec{p} = constant \Rightarrow {\vec{p}}_{i} = {\vec{p}}_{f} \Rightarrow m_{i} \cdot v_{i} = m_{f} \cdot v_{f}$

Watch that, on one side, we have simply considered the magnitudes, and on the other side, we have distinguish an initial mass m_i of a final mass m_f. Indeed, while the container is being filled, the set is increasing, approximatly 6 litres per minute. The litre is a measure of volume equivalent to 1dm³. If we suppose the densit...

$1 \frac{g}{c m^{3}} \cdot \underset{conversion factor}{\underset{⏟}{\frac{1 k g}{1000 g} \cdot \frac{1000 c m^{3}}{1 d m^{3}}}} = 1 k g / d m^{3} = 1 k g / L$

...then we have every minute the mass of the set increases 6 kg. A way to express this mathematically is by using the continuous equation of a straight line, assuming the straight line is the mass and the variable from which time depends. We know two points of the given straight line:

For t = 0, the mass is 1800 kg
For t = 60 min the mass is 1800 + 60·6 = 2160

So, the equation of the mass according to the time is left is:

$\frac{x - a_{1}}{b_{1} - a_{1}} = \frac{y - a_{2}}{b_{2} - a_{2}} \Rightarrow \frac{t - t_{1}}{t_{2} - t_{1}} = \frac{m - m_{1}}{m_{2} - m_{1}} \Rightarrow \frac{t - 0}{60} = \frac{m - 1800}{2160 - 1800} \Rightarrow m = 1800 + \frac{360}{60} \cdot t$

Where t is expressed in minutes. This way, we are ready to calculate the velocity required. For = 60 min we have m_f = 2160 kg and coming back to the conservation principle:

$m_{i} \cdot v_{i} = m_{f} \cdot v_{f} \Rightarrow v_{f} = \frac{m_{i} \cdot v_{i}}{m_{f}} = \frac{1800 \cdot 11.11}{2160} = 9.25 m / s$

For the calculation of the general expression of the velocity based on the time, we simply replace the expression of the mass according to the time in the expression of the conservation principle of momentum:

$m_{i} \cdot v_{i} = m_{f} \cdot v_{f} \Rightarrow v_{f} = \frac{1800 \cdot 11.11}{1800 + \frac{360}{60} \cdot t} = \frac{19998}{1800 + 6 \cdot t} m / s$

Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.

Formulas

Related sections

\vec{p} = m \cdot \vec{v}

Linear Momentum

\sum \vec{F} = 0 \Leftrightarrow \vec{p} = constant \Leftrightarrow \frac{d \vec{p}}{d t} = 0

Newton´s First Law

Momentum Conservation Principle

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