Solved Problem on Impulse and Linear Momentum

Solved Problem on Impulse

A tank truck is moving at a constant speed of 20 m/s. Noticing an obstacle, the driver brakes sharply, and the vehicle takes 8 s to stop. Assuming the tank has a shape of a horizontal cylinder 3 m long and is filled with oil with a density equal to 0.8 g/cm³, calculate the pressure exerted by the oil on the front wall of the tank during braking.

Problem data:

Initial speed of the truck: v_i = 20 m/s;
Final speed of the truck: v_f = 0;
Time the truck takes to stop: Δt = 8 s;
Oil tank length: L = 3 m;
Oil density: ρ = 0.8 g/cm³.

Problem diagram:

Figure 1

During the movement, the oil moves with the truck at the same speed. When the truck brakes the oil tends to continue the movement, but as it is limited by the walls of the tank it will exert pressure on the front wall of the tank (Figure 1).

Solution

First, let's convert the units of the density of oil given in grams per cubic centimeter (g/cm³) to kilograms per cubic meter (kg/m³) used in the International System of Units (SI)

\[ \begin{gather} \rho=0.8\;\frac{\cancel{\text{g}}}{\text{cm}^{3}}\times\frac{10^{-3}\;\text{kg}}{1\;\cancel{\text{g}}}\times\frac{(10^{2}\;\text{cm})^{3}}{1\;\text{m}^{3}}=0.8\;\frac{1}{\cancel{\text{cm}^{3}}}\times 10^{-3}\;\text{kg}\times\frac{10^{6}\;\cancel{\text{cm}^{3}}}{1\;\text{m}^{3}}=0.8\times 10^{3}\;\frac{\text{kg}}{\text{m}^{3}}=800\;\frac{\text{kg}}{\text{m}^{3}} \end{gather} \]

The pressure P that the oil will exert on the front wall of the tank will be given by

\[ \begin{gather} \bbox[#99CCFF,10px] {P=\frac{F_{o}}{A}} \tag{I} \end{gather} \]

where F_o is the force exerted by the oil against the tank wall during the braking time, and A is the area of the tank wall.
The impulse of the braking force will be

\[ \begin{gather} \bbox[#99CCFF,10px] {I=F_{b}\Delta t} \tag{II} \end{gather} \]

where F_b is the braking force and Δt is the time it acts on the truck.

The oil force on the tank wall and braking force are represented in Figure 2, they have the same magnitude, and opposite directions

\[ \begin{gather} F_{o}=F_{b} \tag{III} \end{gather} \]

Figure 2

Using the Impulse-Momentum Theorem, we have that the impulse is equal to the change in momentum

\[ \begin{gather} \bbox[#99CCFF,10px] {I=\Delta Q=Q_{f}-Q_{i}} \tag{IV} \end{gather} \]

equating expressions (II) and (IV) and using expression (III)

\[ \begin{gather} Q_{f}-Q_{i}=-F_{o}\Delta t \tag{V} \end{gather} \]

The momentum is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mv} \tag{VI} \end{gather} \]

substituting expression (VI) into expression (V)

\[ \begin{gather} mv_{f}-mv_{i}=-F_{o}\Delta t\\[5pt] m\times 0-mv_{i}=-F_{o}\Delta t\\[5pt] -mv_{i}=-F_{o}\Delta t\\[5pt] mv_{i}=F_{o}\Delta t \tag{VII} \end{gather} \]

Here the mass considered is the oil that presses against the tank wall, the oil mass as a function of the density given in the problem is calculated by

\[ \begin{gather} \bbox[#99CCFF,10px] {\rho =\frac{m}{V}} \end{gather} \]

\[ \begin{gather} \rho =\frac{m}{V} \tag{VIII} \end{gather} \]

where V is the volume of the cylindrical tank. The volume of a cylinder will be the area of the base multiplied by the height (Figure 3)

\[ \begin{gather} \bbox[#99CCFF,10px] {V=AL} \tag{IX} \end{gather} \]

Figure 3

substituting expression (IX) into expression (VIII)

\[ \begin{gather} m=\rho AL \tag{X} \end{gather} \]

substituting the expression (X) into the expression (VII)

\[ \begin{gather} F_{o}\Delta t=\rho ALv_{i}\\[5pt] \frac{F_{o}}{A}=\frac{\rho Lv_{i}}{\Delta t} \tag{XI} \end{gather} \]

Now we can see that the left-hand side of expression (XI) is equal to the right-hand side of expression (I), which gives us the pressure, so equating these values

\[ \begin{gather} P=\frac{\rho Lv_{i}}{\Delta t} \end{gather} \]

Substituting the numerical values given in the problem

\[ \begin{gather} P=\frac{800\times 3\times 20}{8} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {P=6000\;\text{Pa}} \end{gather} \]

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