2021 A/L Physics Past Paper · Question 16 (English Medium)

Answer: (2) $9\rho Av^2$

What the question asks

Find the force $F$ that the liquid exerts on the wall.

The answer is an expression in $\rho$, $A$ and $v$.

Relevant theory

According to Newton's second law resultant force is proportional to the rate of change of momentum.

momentum = mass × velocity

Required equations

$F = \dfrac{\Delta p}{\Delta t}$

$p = mv$

$m = \rho A (3v) t$

Reasoning and calculation

The symbols:

$\rho$ is the density of the liquid.

$A$ is the cross-sectional area of the tube.

$3v$ is the speed of the liquid before it hits the wall.

$F$ is the force on the wall.

In a time $t$ a length $3vt$ of liquid reaches the wall.

Its mass is:

$$m = \rho A (3v) t$$

It arrives at $3v$ and then runs down the wall with no velocity toward the wall.

So it loses a momentum:

$$\Delta p = m \times 3v = 9\rho A v^2 t$$

The force is the loss of momentum in each second:

$$F = \dfrac{\Delta p}{t} = \dfrac{9\rho A v^2 t}{t} = 9\rho A v^2$$

Reading the options

The result is $9\rho Av^2$. That is option (2).

Options (4) and (5) have $A^2$, so they are not forces.

Options (1) and (3) have the wrong change of velocity.

Checks

Dimension check:

$$[\rho A v^2] = ML^{-3} \times L^2 \times L^2T^{-2} = MLT^{-2}$$

This is the dimension of a force. It matches.

Limiting case. If $v \to 0$, no liquid arrives, and $F \to 0$.

Also $F$ grows with the square of the speed. A fast jet pushes much harder.

Conclusion

$F = \rho A(3v) \times 3v = 9\rho A v^2$

The liquid loses its momentum $3v$ per unit mass, and $\rho A(3v)$ is the mass reaching the wall each second.

So $F = 9\rho Av^2$. The answer is (2).

$$For\,a\,jet\,that\,hits\,a\,wall\,and\,stops,\; use\,force = mass\,per\,second\,times\,the\,change\,of\,velocity$$

Mass per second is density times area times speed.

Exam tip