2023 A/L Physics Past Paper · Question 28 (English Medium)

Answer: (5) $P_0 - \dfrac{15}{2}\rho v_0^2$

Given information

What we know:

The fluid flows through a horizontal pipe, so both parts are at the same height.

The wide part has the inner radius $r$, the pressure $P_0$ and the speed $v_0$.

The narrow part has the inner radius $\dfrac{r}{2}$.

The fluid has the density $\rho$. It cannot be squeezed, and it has no viscosity.

What the question asks

We want: the pressure $P$ of the fluid in the narrow part.

What each letter means:

$\rho$ ($\text{kg m}^{-3}$): the density of the fluid.

$r$ ($\text{m}$): the inner radius of the wide part.

$A_1$, $A_2$ ($\text{m}^2$): the cross-sectional areas of the wide and of the narrow part.

$P_0$, $v_0$ ($\text{Pa}$, $\text{m s}^{-1}$): the pressure and the speed in the wide part.

$P$, $v$ ($\text{Pa}$, $\text{m s}^{-1}$): the pressure and the speed in the narrow part.

Relevant theory

$$\therefore A_1v_1 = A_2v_2$$

On any point along the same streamline of a steady flow of an incompressible fluid of negligible viscosity, the sum of the pressure, the potential energy per unit volume and the kinetic energy per unit volume is a constant.

$$p_1 - p_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$

The first quote says the volume that goes in each second must come out each second, so a narrower part means a faster flow. The second quote says the pressure, together with the energy of motion and the energy of height, adds up to a constant along the flow. The third quote is that sum for two places at the same height.

For example, think of a garden hose. When you squeeze the end with your thumb, the water shoots out faster. The same water has to get through a smaller opening each second.

The water cannot speed up for free. It spends some of its pressure to do it. So the faster part has the smaller pressure.

Required equations

$A = \pi r^2$

$A_1 v_0 = A_2 v$

$P_0 - P = \dfrac{1}{2}\rho (v^2 - v_0^2)$

$\rho v^2$ is in $\text{kg m}^{-3} \times \text{m}^2\text{s}^{-2} = \text{kg m}^{-1}\text{s}^{-2}$, which is $\text{Pa}$, the same unit as the pressure. So the terms can be added and taken away.

Applying it to this question

$\pi r^2$

$A_1 = \pi r^2$

$\dfrac{A_1}{4}$

$A_2 = \pi\left(\dfrac{r}{2}\right)^2 = \dfrac{\pi r^2}{4}$

$4v_0$

$v = \dfrac{A_1}{A_2}\,v_0 = 4\,v_0$

$15v_0^2$

$v^2 - v_0^2 = 16v_0^2 - v_0^2$

$\dfrac{15}{2}\rho v_0^2$

$P_0 - P = \dfrac{1}{2}\rho \times 15v_0^2$

$P_0 - \dfrac{15}{2}\rho v_0^2$

$P = P_0 - \dfrac{15}{2}\rho v_0^2$

$s * s - 1$

Reasoning and calculation

Now match it with the five options.

(1) $\dfrac{P_0}{4}$ has no $\rho$ and no $v_0$. It does not match.

(2) $\dfrac{P_0}{2}$ has no $\rho$ and no $v_0$. It does not match.

(3) $P_0 - \dfrac{1}{2}\rho v_0^2$ takes away too little. It does not match.

(4) $P_0 - \dfrac{3}{2}\rho v_0^2$ takes away too little. It does not match.

(5) $P_0 - \dfrac{15}{2}\rho v_0^2$ is what we found. This matches.

Checks

Do it a second way. The sum of the pressure and $\dfrac{1}{2}\rho v^2$ stays the same, as the second quote says.

$$P_0 + \dfrac{1}{2}\rho v_0^2 = P + \dfrac{1}{2}\rho (4v_0)^2$$

The wide part has $\dfrac{1}{2}\rho v_0^2$ and the narrow part has $8\rho v_0^2$. So the narrow part has $\dfrac{15}{2}\rho v_0^2$ more of the energy of motion.

$$P = P_0 + \dfrac{1}{2}\rho v_0^2 - 8\rho v_0^2 = P_0 - \dfrac{15}{2}\rho v_0^2$$

The same answer. The pressure is lower in the narrow part, where the fluid is faster, as the third quote says.

Conclusion

$P = P_0 - \dfrac{15}{2}\rho v_0^2$

Conclusion

First use the equation of continuity to find the speed in the narrow part. The area goes with the radius squared.

Then use Bernoulli's principle for two places at the same height. The pressure falls where the speed rises.

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