Select Page

# Physics HL

## Paper 2 – Exam 3

#### Question 1

A space station supply ship is launched vertically upwards from a launchpad. When it reaches a height of 23 km, it releases one of its external fuel tanks which falls to the ground. As the fuel tank falls, it reaches terminal velocity at a height of 11 km.

a.

i. Outline, with reference to Newton’s first law of motion, why the resultant force on the fuel tank after it has reached terminal velocity is zero.

[2 marks]

ii. Draw a fully-labelled free-body diagram to represent the forces acting on the fuel tank at the moment when it reaches a height of 8 km.

[2 marks]

b. Sketch on the axes below how the height of the fuel tank varies with time from the moment it is released by the rocket until it reaches the surface of the Earth. On your sketch, clearly show the height where the fuel tank was released by the rocket and the height where the fuel tank reached terminal velocity.

[3 marks]

c. After the supply ship leaves its cargo at the space station, it returns to Earth. Upon reaching the surface of the Earth, it lands on a landing pad floating in the ocean.

i. Outline the law of conservation of linear momentum.

[2 mark]

ii. Outline, with reference to change in momentum, why landing on a floating landing pad as opposed to one on solid ground makes it less likely that the rocket will be damaged.

[2 marks]

Newton’s first law states that a body remains at rest or moves with constant velocity unless an external force acts on it. After the fuel tank reaches terminal velocity, it moves with constant velocity, so the resultant force on the fuel tank must be zero.

Note: the two forces must have the same magnitude.

Note: After the tank reaches a height of 11 km, the graph must be a straight line.

If no external forces act on a system, total linear momentum remains constant.

Landing in water increases the time it takes the rocket to come to rest.

Since F = $\frac{\Delta p}{\Delta t}$, the force acting on the rocket while it is coming to rest is smaller, hence it is less likely that the rocket will be damaged.

#### Question 2

A third-harmonic standing sound wave is formed in a pipe that is closed at one end and open at the other. The diagram shows the displacement of the wave at time t. The speed of sound is 340 m s$^{-1}$.

a.

i. Outline how a standing wave is formed.

[1 mark]

ii. Outline two differences between a standing wave in a pipe that is open at both ends and a standing wave in a pipe that is open at one end and closed at the other.

[2 marks]

b. On the diagram, label a point with the letter A that shows the position of an antinode.

[1 mark]

c. The diagram shows two air molecules, P and Q, on the standing wave at time t

i. State and explain which particle has a larger magnitude of acceleration at time t.

[2 marks]

ii. Explain whether Q is at the centre of a compression or a rarefaction.

[2 marks]

The sound wave incident on the pipe and the sound wave reflected at the closed end of the pipe superpose and form the standing wave.

In a pipe that is open at both ends, all integer harmonics (1, 2, 3, etc.) are available. In a pipe that is open at one end and closed at the other, only odd integer harmonics (1, 3, 5, etc.) are available.

In a pipe that is open at both ends, there is an antinode at each end. In a pipe that is open at one end and closed at the other, there is an antinode at the open end and a node at the closed end.

The air molecules vibrate with simple harmonic motion (SHM), so the magnitude of their acceleration is proportional to their displacement.

Since P has a larger displacement, it has a larger magnitude of acceleration.

The air molecules to the left of Q are displaced to the right and the air molecules to the right of Q are displaced to the left.

Hence Q is at the centre of a compression.

#### Question 3

A student builds a simple circuit using copper wire that includes a heater and a cell.  The following data is available for the circuit:

Emf of the cell $12$ V

Internal resistance of the cell $1.5 \Omega$

Diameter of copper wire $0.92$ mm

Resistivity of copper $1.7 \times 10^{-8} \Omega$ m

a. The student would like to build the circuit so that the maximum internal resistance of the wiring is 62 m$\Omega$. Find the maximum length of the wire the student can use. Give your answer to an appropriate number of significant figures.

[3 marks]

b. Outline what happens to the resistance of the wiring when, instead of the original wire, the student uses a wire with a square-shaped cross sectional area where the side length of the square is 0.92 mm.

[2 marks]

c. The student uses an ammeter and measures the current to be 3.0 A in the circuit. The student assumes that the ammeter is ideal.

i. Outline what is meant by an ideal ammeter.

[1 mark]

ii. Calculate the power dissipated in the heater. Assume that the wiring has negligible internal resistance.

[2 marks]

d. The student would like to measure the resistance of the heater. Draw a circuit diagram to show how she can do this by using a potential divider arrangement.

[2 marks]

$\frac{(62 \times 10^{-3}) \times (0.46 \times 10^{-3})^{2} \times \pi}{1.7 \times 10^{-8}} \approx 2.4$ m

Answer must be given to 2 s. f. in order to earn final mark.

Cross-sectional area of circular wire: $\pi \times 0.46^{2} \approx 0.665$ mm$^{2}$

Cross-sectional area of square-shaped wire: $0.92^{2} \approx 0.846$ mm$^{2}$

The cross-sectional area of the square-shaped wire is larger, so resistance decreases.

An ideal (am)meter does not affect the circuit, so the ammeter should have no resistance.

$12 = 3(R + 1.5)$

$R = 2.5 \Omega$

P = $3^{2} \times 2.5$ = 22.5 W $\approx$ 23 W

#### Question 4

During the process of electrical energy production in some nuclear power plants, uranium-239 (U-239) decays into neptunium-239 (Np-239) through $\beta^{-}$decay. The decay is shown on the following Feynman diagram.

a. Complete the Feynman diagram by inserting the four missing particles.

[2 marks]

b. The Rutherford-Geiger-Mardsen scattering experiment provided evidence for a new model of the atom.

i. Describe this model.

[2 marks]

ii. Explain why, based on classical physics, electrons would not orbit the nucleus.

[3 marks]

iii. State one reason why scientists use models in their work.

[1 mark]

The atom has a very small positively charged nucleus that contains almost all the mass of the atom.

Negatively charged electrons orbit the nucleus.

An electron in orbit performs circular motion.

During circular motion, the electron is accelerated.

When an electron is accelerated, it loses energy in the form of electromagnetic radiation.

As a result, the electron would get closer and closer to the nucleus and eventually spiral into the nucleus.

Models help simulate ideas.

Models provide a simplified explanation of reality.

Models help in the development of new hypotheses.

Models help when explaining concepts.

#### Question 5

A moon has a circular orbit around a planet. The radius of the planet is R, the radius of the moon is r, the time it takes for the moon to complete one full orbit around the planet is T and the mass of the planet is M and the mass of the moon is m.

a. Show that the shortest distance between the surface of the moon and the surface of the planet is $\sqrt[3]{\frac{T^{2}GM}{4 \pi^{2}}} - (R+r)$

[3 marks]

b. The following data is known about the moon and the planet:

Mass of the planet = $8.3 \times 10^{25}$ kg

Diameter of the planet = 10500 km

Ratio of $\frac{r}{R} = \frac{1}{4}$

Shortest distance between the surface of the planet and the surface of the moon = 75R

Find, in days, the orbital period of the moon.

[4 marks]

c. The ratio of $\frac{m}{M} = \frac{1}{80}$.

i. Show that the orbital speed of the moon is 3.7 km s$^{-1}$.

[3 marks]

ii. Find the gravitational force exerted by the moon on the planet.

[2 marks]

Equating centripetal acceleration of the moon and gravitational field strength:: $\frac{4 \pi^{2} r}{T^{2}} = \frac{GM}{r^{2}}$

$R_{orbital} = \sqrt[3]{\frac{T^{2}GM}{4 \pi^{2}}}$

Shortest distance between the surface of the moon and the surface of the planet:

$R_{orbital}-(R+r) = \sqrt[3]{\frac{T^{2}GM}{4 \pi^{2}}}-(R+r)$

$75R = \sqrt[3]{\frac{T^{2}GM}{4 \pi^{2}}}-1.25R$

$T = \sqrt{\frac{(76.25R)^{3} \times 4 \pi^{2}}{GM}}$

$T = \sqrt{\frac{(76.25 \times 5250 \times 10^{3})^{3} \times 4 \pi^{2}}{(6.67 \times 10^{-11}) \times (8.3 \times 10^{25})}} \approx 676358$ s

$T = \frac{676358}{60 \times 60 \times 24} \approx 7.8$ days

$R_{orbital} = \sqrt[3]{\frac{676358^{2} \times (6.67 \times 10^{-11}) \times (8.3 \times 10^{25})}{4 \pi^{2}}} \approx 4 \times 10^{8}$ m

v = $\frac{2 \pi \times (4 \times 10^{8})}{676358} \approx 3716$ m s$^{-1} \approx 3.7$ km s$^{-1}$

(Calculated answer must be given to at least 3 s. f. to earn final mark)

M = $\frac{8.3 \times 10^{25}}{80} \approx 1.04 \times 10^{24}$ kg

F = $\frac{1.04 \times 10^{24} \times 3716^{2}}{4 \times 10^{8}} \approx 3.6 \times 10^{22}$ N

#### Question 6

A student uses a heater in an experiment to estimate the specific latent heat of fusion of potassium. The following data is available:

Power output of the heater: 23 W

Mass of potassium: 150 grams

Specific heat capacity of potassium: 770 J kg$^{-1}$ K$^{-1}$

Initial temperature of potassium: 5.0 $^{o}$

Melting point of potassium: 63.2 $^{o}$

It takes the student 11.5 minutes to completely melt the potassium.

a. Outline the difference between the molecular structure of potassium at 60 $^{o}$C and 70 $^{o}$C .

[2 marks]

b. Estimate the specific latent heat of fusion of potassium.

[3 marks]

c. Suggest one reason why your answer to part (c) is an estimate and how this affects the value of specific latent heat of fusion.

[2 marks]

d. When scientists carry out experiments, the results are often reported in peer-reviewed scientific journals. Outline why scientists choose to share their work in this manner.

[1 mark]

At 60 $^{o}$C potassium is a solid, so molecules are closer to each other and have fixed positions.

At 70 $^{o}$C potassium is a liquid, so bonds between molecules are broken and they can move more freely than in a solid.

(23)(11.5)(60) = (0.15)(770)(58.2) + 0.15L

L = 60986 J kg$^{-1}$ $\approx$ 61 kJ kg$^{-1}$

Some thermal energy from the heater is transferred to the surroundings (instead of the potassium), so the estimated value is larger than the actual value of the specific latent heat.

Sharing work in this manner ensures the validity of the work.

#### Question 7

A pumped storage hydroelectric system consists of an upper lake, a connecting pipe leading from the upper lake to the lower lake and a turbine at the end of the connecting pipe. The following data is available for a pumped storage hydroelectric system:

Depth of upper lake: $28$ m

Length of pipe: $144$ m

Angle that the pipe makes with the horizontal: $62^{o}$

Base area of upper lake: $2.72$ km$^{2}$

Density of water: $997$ kg m$^{-3}$

a. Outline, with reference to energy changes, how electrical energy is produced in a pumped storage hydroelectric system.

[2 marks]

b. The system has six identical generators. It takes $5.47 \times 10^{4}$ seconds for the upper lake to empty completely.

i. Show that the total mass of the water in the upper lake when it is full is $7.6 \times 10^{10}$ kg.

[3 marks]

ii. Calculate the maximum power output of one generator.

[4 marks]

iii. Suggest why your answer to part (b) (ii) is a maximum.

[1 mark]

The gravitational potential energy of the water is transferred to kinetic energy of the turbines.

The kinetic energy of the turbines is transferred to kinetic energy of the generators and the kinetic energy of the generators is transferred to electrical energy.

Volume of water in upper lake = $2.72 \times 10^{6} \times 28 \approx 7.62 \times 10^{7}$ m$^{3}$

Mass = $(7.62 \times 10^{7}) \times 997 \approx 7.59 \times 10^{10}$ kg $\approx 7.6 \times 10^{10}$ kg

Calculated answer must be given to at least 3 s. f. in order to earn final mark.

Vertical height between the bottom of upper lake and turbine:

$144 \times sin(62^{o}) \approx 127$ m

Average height through which water falls to reach turbine:

$127 + \frac{28}{2} = 141$ m

Total energy generated by all water in upper lake:

$E_{total} = (7.6 \times 10^{10}) \times 9.81 \times 141 \approx 1.05 \times 10^{14}$ J

Total power output = $\frac{1.05 \times 10^{14}}{5.47 \times 10^{4}} \approx 1.92 \times 10^{9}$ W

Power output of one generator = $\frac{1.92 \times 10^{9}}{6} \approx 3.2 \times 10^{8}$ W or $320$ MW

There are energy/power losses, because the system is not fully efficient.

OR

Energy/power is lost to the surroundings during the process.

#### Question 8

A mass of 0.48 kg is attached to a spring and the system is made to oscillate with simple harmonic motion (SHM) on a horizontal, frictionless surface. The mass passes through the equilibrium position $240$ times per minute.

a. The kinetic energy of the mass as it passes through the equilibrium position is $400$ mJ.

i. There are two points where the restoring force acting on the block is at its maximum. Show that the distance between these points is approximately $21$ cm.

[4 marks]

ii. Find the distance of the block from the equilibrium position when the speed of the block is $0.8$ m s$^{-1}$

[2 marks]

iii. Find the magnitude of the net force acting the block when its speed is $0.8$ m s$^{-1}$

[2 marks]

b. The experiment is moved to planet M. The gravitational acceleration on planet M is $g_{M}$. It is known that $\frac{g_{M}}{g} = 3$. Another change is that three more identical springs are placed parallel to the original spring. Outline, without calculations, how these changes affect the frequency with which the block oscillates.

[4 marks]

$T = \frac{2 \times 60}{240} = 0.5$ s

$\omega = \frac{2 \pi}{0.5} = 4 \pi$ s$^{-1}$

Restoring force is maximum when displacement is maximum.

$0.4 = \frac{1}{2} \times 0.48 \times (4 \pi)^{2} \times x_{0}^{2}$

$x_{0} \approx 0.103$ m $\approx 10.3$ cm

Distance between two maximum displacement positions:

$2 \times 10.3 \approx 20.6 \approx 21$ cm

Calculated answer must be given to at least 3 s. f. in order to earn final mark.

$0.8 = 4 \pi \times \sqrt{0.103^{2} - x^{2}}$

$x \approx 0.081$ m or $8.1$ cm

If $0.105$ m is used for $x_{0}$, the answer is $0.083$ m or $8.3$ cm

$a = -(4 \pi)^{2} \times 0.081 \approx -13.3$ m s$^{-2}$

$F = 0.48 \times 13.3 \approx 6.4$ N

Period of oscillation for a mass-spring system can be calculated using $T = 2 \pi \sqrt{\frac{m}{k}}$

When three additional springs are placed parallel to the original spring, the value of $k$ is quadrupled.

As a result, $T$ is halved. Since frequency is inversely proportional to the period, this change causes the frequency to double.

Based on the equation above, the time period of a mass-spring system is not affected by the gravitational acceleration so moving the experiment to another planet does not affect the frequency.

This means that after the changes the frequency of the oscillations doubles.

#### Question 9

Two capacitors and a resistor of resistance $2.1 \times 10^{6}$ $\Omega$ are connected to a cell of emf $24$ V and negligible internal resistance. The circuit is in vacuum.

The switch is connected to X and the capacitors are fully charged. The energy stored in $C_{1}$ $2.41 \times 10^{-4}$ J, the potential difference across  $C_{2}$ is $15$ V and the combined capacitance of $C_{1}$ and $C_{2}$ is $3.8$ $\mu$F.

a.

i. Show that the capacitance of $C_{1}$ is $6.0$ $\mu$F.

[3 marks]

ii. Find the energy stored in $C_{2}$.

[2 marks]

b. The switch is moved to position Y and the capacitors discharge. The initial current in the circuit is $I_{0}$

i. Find the value of $\tau$, the time constant of the circuit.

[1 mark]

ii. Calculate the time it takes for the initial current in the circuit to fall to $38$% of its initial value.

[2 marks]

The capacitors are connected in series, so $V_{1} = 24 - 15 = 9$ V

$C_{1} = \frac{2 \times (2.41 \times 10^{-4})}{9^{2}} \approx 5.95 \times 10^{-6}$ F $\approx 6.0$ $\mu$F

Calculated answer must be given to at least 3 s. f. in order to earn final mark.

The capacitors are connected in series, so $\frac{1}{3.8} = \frac{1}{6} + \frac{1}{C_{2}}$

$C_{2} \approx 10.4 \times 10^{-6}$ F

$E = \frac{1}{2} \times (10.4 \times 10^{-6}) \times 15^{2} \approx 1.2 \times 10^{-3})$ J

$\tau = (2.1 \times 10^{6}) \times (3.8 \times 10^{-6}) \approx 8.0$ s

$0.38I_{0} = I_{0}e^{-\frac{t}{8.0}}$

$t = -8 \times ln(0.38) \approx 7.7$ s

#### Question 10

In an experiment electrons are passed through a thin slice of crystal. A series of bright and dark rings are observed on a screen behind the crystal.

a. The electrons are accelerated through a potential difference of $V$. The de Broglie wavelength of the electrons is given by the equation $\lambda = \frac{h}{p}$. Show that $\lambda = \frac{h}{\sqrt{2em_{e}V}}$

[3 marks]

b. When $V = 4.3$ kV, the electrons are accelerated to $X$% of the speed of light in vacuum.

i. Find the de Broglie wavelength of the electrons.

[1 mark]

ii. Find the value of $X$.

[2 marks]

c. The uncertainty in the position of some of the electrons that reach the screen is small. The Heisenberg uncertainty principle makes a prediction about another property of these electrons. Outline this prediction.

[2 marks]

Kinetic energy of electrons = Work done on the electrons by the electric field

$\frac{1}{2}m_{e}v^{2} = eV$

There are different ways to combine this equation with the equation for the de Broglie wavelength. An example is:

Multiplying both sides by $2m_{e}$:

$(m_{e}v)^{2} = 2em_{e}V$

$m_{e}v = p = \sqrt{2em_{e}V}$

Substituting into the equation for the de Broglie wavelength:

$\lambda = \frac{h}{\sqrt{2em_{e}V}}$

$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (1.60 \times 10^{-19}) \times (9.11 \times 10^{-31}) \times (4.3 \times 10^{3})}}$

$\lambda \approx 1.87 \times 10^{-11}$ m $\approx 1.9 \times 10^{-11}$ m

Using $\lambda = \frac{h}{mv}$
$v = \frac{6.63 \times 10^{-34}}{(1.87 \times 10^{-11}) \times (9.11 \times 10^{-31})} \approx 3.89 \times 10^{7}$ m s$^{-1}$
$X = \frac{3.89 \times 10^{7}}{3 \times 10^{8}} \times 100$% $\approx 13$%
Based on the equation describing the Heisenberg uncertainty principle, $\Delta x \Delta p \geq \frac{h}{4 \pi}$, small uncertainty in the position of the electrons means that there is a large uncertainty in the momentum of the electrons.