In this post, we reveal the solutions to the 2019 HSC Physics Exam Paper!

The Matrix 2019 HSC Physics Exam Paper Solutions are here! The Matrix Academic team has been hard at work on them.

Check the solutions below and see how you went!

Question | Answer | Solution |

1 | D | At max height the vertical component of velocity is zero, so the projectile travels horizontally. |

2 | B | Spectral lines at different wavelengths indicate a different Doppler shift, and hence different velocity relative to the Earth. The same set of spectral lines indicates the presence of the same element, and hence the same composition. |

3 | C | They used alpha particles, and according to Thomson’s model initially expected them to pass through atoms. |

4 | C | B class stars have a higher temperature than M class stars. |

5 | D | An induced current will only flow through the coil on the right when it experiences a change in flux, i.e. when the current in the coil on the left decreases due to opening the switch. Hence the galvanometer will read zero to begin with, then deflect, the return to zero. |

6 | A | This is dictated by Wien’s Displacement Law: \(λ_{\textrm{max}}=\frac{b}{T}\), giving \(λ_{\textrm{max}}∝ \frac{1}{T}\). |

7 | D | There will be an opposing force on the magnet according to Lenz’s Law, requiring a North pole on the solenoid close to the South pole of the magnet. The right hand coil rule gives the direction of the current. |

8 | A | Hubble found that recessional velocity was proportional to distance, hence a straight line graph. The distances in option A are consistent with Hubble’s measurements. |

9 | D | Orbital period and total energy both increase with the radius of the orbit, hence are both greater for GEO satellites which are further away from Earth than LEO satellites. |

10 | A | Malus’s Law gives \(\frac{I_B}{I_0} =\cos^2 30° =0.75\). This supports the wave model of light. |

11 | C | Kepler’s Third Law gives \(r= \sqrt[3]{40,000^2}=1170 \textrm{ AU}\) |

12 | A | This describes beta decay where a neutron (udd) is transformed into a proton (uud), an electron, and an antineutrino. Alternatively, conservation of charge can be used. |

13 | B | The photon energy is \(E=\frac{hc}{λ}=3.1×10^{-19} \textrm{J}\). The number of photons per second is the power divided by the photon energy: \(\frac{30×10^{-3}}{3.1×10^{-19}} = 9.8×10^{16}\) |

14 | B | \(v= \sqrt{\frac{GM}{r}}\), giving \(v∝\frac{1}{\sqrt{r}}\). Doubling \(r\) would change \(v\) by a factor of \(\frac{1}{\sqrt{2}}≈0.7\). |

15 | C | \(d\) is the distance between the slits, given in the question. \(\theta\) is the angle at which the light propagates from the slits to a maximum. The path the light takes forms the hypotenuse of a triangle formed by joining the centre of the slits to the central maximum, to the upper maximum. The ratio \(\frac{0.08}{0.3}\) gives the tan of \(θ\). |

16 | D | In the second scenario the net force on the particle is upwards and equal in magnitude to its weight (so as to produce the same motion upwards). The particle has both its weight and the electric force \(qE\) acting on it: \(F=mg=qE-mg\) \(E=\frac{2mg}{q}\) |

17 | D | The current through \(QR\) would create a field to the right above \(QR\) according to the right hand grip rule. The cathode rays would experience a force upwards according to the right hand palm rule (NB cathode rays are electrons, and hence are negative). Increasing the resistance would decrease the current in \(QR\) and would reduce this effect, resulting in the cathode rays moving downwards towards \(Z\). |

18 | B | According to Faraday’s Law and Lenz’s Law, the coil experiences an anticlockwise EMF when it enters the magnetic field, and a clockwise EMF when it exits the field. Since the battery is producing a clockwise current, the EMF will initially oppose the existing current, dimming the lamp on the way in to the field, and then increased the current on the way out, making the lamp brighter. |

19 | D | Energy had to be released in the reaction regardless of the properties of \(Z\). The binding energy of \(Y\) is greater than the sum of the binding energies of the reactants \(W\) and \(Z\). Breaking \(W\) and \(Z\) into constituent particles would consume \(2.22 + 8.48 = 10.7 \ \textrm{MeV}\) of energy, but forming \(Y\) will release \(28.3 \ \textrm{MeV}\), meaning \(28.3 – 10.7 = 17.6 \ \textrm{MeV}\) is released without considering \(Z\). The binding energy of \(Z\) will be zero (if it is a single particle) or positive (if it is a nucleus consisting of at least two particles). Its binding energy will also be released, meaning the energy released is \(17.6 \ \textrm{MeV}\) + the binding energy of \(Z\). |

20 | B | This is a statement of Newton’s Third Law. The force of the blackboard on the cube provides the centripetal force required to keep the cube in uniform circular motion. This force also determines the friction through \(f=μN\). |

De Broglie suggested that electrons and other matter particles would have a wave nature in addition to their particle nature.

He proposed their wavelength would depend on Planck’s constant and their momentum, \(λ=\frac{h}{mv}=\frac{h}{p}\).

Linear velocity: The Doppler shift of spectral lines observed in the star’s light will indicate i’s linear velocity along the line of sight between the Earth and the star. A red shift indicates the star is moving away from Earth and a blue shift indicates the star is moving towards Earth. The amount of shift indicates the speed.

Rotational velocity: The broadening of spectral lines (due to Doppler broadening) indicates the rotational velocity of the star. More broadening indicates a higher rotational velocity.

Pressure: The broadening of the edges of spectral lines (due to pressure broadening) indicates the pressure in the star’s atmosphere. More broadening indicates a higher pressure.

The data given should be plotted on the graph provided. The line of best fit should be drawn.

The line of best fit should be extrapolated to find the y-intercept which is the negative of the work function.

The work function is around \(4 \ \textrm{eV}\).

The input power is given in the question \(V_P I_P=500 \ \text{W}\). The output power can be calculated from \(V_S I_S= 50×9=450 \ \textrm{W}\). Although there is a discrepancy, with the output power being \(50 \ \textrm{W}\) lower than the input power, conservation of energy is not violated.

The solid iron core used will experience a change in flux from the coils, which will result in an induced EMF and induced eddy currents. These eddy currents will generate heat.

Heat is also produced in the wires of the coils due to their resistance. This heat and the energy of the eddy currents will correspond to the \(50 \ \textrm{W}\) noted above.

Laminating the iron core will prevent eddy currents from forming large loops due to the insulating material. As a result only much smaller eddy currents will form within each lamination. This reduces the energy consumed by eddy currents, increasing the efficiency of the transformer.

Increasing the thickness of the wires in the coils will reduce their resistance and the energy lost as heat, again increasing the efficiency of the transformer.

Maxwell, through the equations named after him, showed that electric and magnetic fields are related and parts of the one phenomenon – electromagnetism.

He argued that just as a changing magnetic field produces an electric field (through induction and Faraday’s Law), a changing electric field would also produce a magnetic field. Hence he proposed that electromagnetic waves would form whereby changing electric and magnetic fields will produce each other resulting in a propagating wave.

Maxwell calculated the speed of electromagnetic waves and showed that they would travel at the speed of light. Hence he proposed that light is an electromagnetic wave, part of a wide spectrum of such waves.

In order to create an electromagnetic wave the electric field and magnetic field are both required to change. This can be achieved by an accelerating electric charge. An oscillating charge will produce electromagnetic waves as the charge continuously accelerates during the oscillation. The change in position of the charge produces a changing electric field, and the change in velocity of the charge produces a changing current and hence a changing magnetic field. The fields then continue to mutually generate each other.

The points appear to lie on a straight line, thus the student concludes that that the torque is proportional to the angle.

The gradient is approximately \(\frac{42-8}{25-5}=1.7\)

Hence the equation of the line of best fit would be \(τ=1.7×θ\).

We know that \(\tau = rF \sin \theta\) , and hence \(\tau \propto \sin \theta\) .

For small angles, so the student’s model will be valid as \(\sin \theta \approx \theta\).

For larger angles the model will become increasingly inaccurate as it does not have the correct dependence on angle.

Have a laser on a train moving at high speed. A pulse of light leaves the laser at one end of the carriage, reflects off a mirror at the other end of the carriage, and then returns to the laser.

Two observers measure the time it takes for the light to return to the laser. One observer is in the train carriage with the laser. A second observer is stationary on the ground outside the train.

The two observers will calculate the same speed for the light, but will measure different times for the light’s journey due to time dilation. The observer on the train will measure proper time for the light’s journey, and the observer on the ground will measure the dilated time which will be longer and will depend on the speed of the train relative to the ground.

[There are other acceptable answers.]

In the Hafele-Keating experiment three high precision atomic clocks were used. One remained stationary on the ground, one was flown around the world eastwards and one was flown around the world westwards.

Each clock, when considered in the reference frame of the Earth’s axis, is travelling at a different speed, and hence experiences a different amount of time dilation. The eastwards moving clock experiences the most, and the westwards moving clock the least.

When the clocks arrived back at the same place, their time readings were compared and they were consistent with the theories of relativity, including time dilation.

Before the rotation, the current through \(XY\) is parallel to the magnetic field, so \(XY\) experiences no force, \(F=ILB \sinθ=ILB \sin0°=0\).

After the rotation, the angle between \(XY\) and the magnetic field increases to 90°, so the force increases to \(F=ILB \sin90°=ILB\)

The current through \(WX\) is always perpendicular to the magnetic field, and the wire experiences a constant force of \(F = ILB\) both before and after the rotation.

The work done is \(W=qV.\)

The work done will increase the charge’s kinetic energy such that \(W=ΔK=\frac{1}{2} mv^2-\frac{1}{2} mu^2\).

Since \(u=0\), we get:

\(qV=\frac{1}{2} mv^2\) which gives \(v= \sqrt{\frac{2qV}{m}}\).

The potential energy at P is converted to kinetic energy at \(Q\):

\(\frac{1}{2} mv^2=mgh\),

where \( v \) is the speed as it leaves the track, and \( h \) is the initial height above point \(Q\). Hence:

\(h=\frac{v^2}{2g}=\frac{1.5^2}{2×9.8}=0.115 \ m\)

The ball’s initial vertical velocity is \(u_y= 1.5 \sin50°=1.149 \textrm{ms}^{-1}\) downwards.

Take down to be the positive direction. Using the equation \(s=ut+\frac{1}{2}at^2\) we solve for \(s\) as follows:

\(s=1.149×0.5+\frac{1}{2×9.8}×0.5^2\)

Giving a height of \(s = 1.8 \text{ m}\) above the floor.

As the fan’s rotation speed increases, it will apply an increasingly strong force on the air molecules to push them downwards.

According to Newton’s Third Law, the air will apply an equal and opposite force on the fan.

The fan will hence experience two forces: its weight downwards which is constant, and the force from the air upwards which increases with the fan’s speed.

The spring balance will initially read the fan’s weight. When the fan is switched on the reading will decrease over the first 10 seconds as the fan speeds up and the upwards force increases. It will then remain constant.

The student’s prediction is incorrect.

The current through the fan’s motor will be affected by the back EMF which depends on rotation speed. Initially the coil of the motor is stationary, and the coil experiences no change in flux. Hence there is no induced (back) EMF in the coil. The power supply to the coil will then result in maximum current, which results in maximum torque and the fan’s rotation speed increasing.

As the speed increases, the induced (back) EMF increases due to flux changing at a faster rate (Faraday’s Law). The back EMF opposes the power supply (Lenz’s Law) and reduces the current, and in turn, the torque.

Hence the current will start at maximum and will reduce over the first 10 s as the fan speeds up. Once the fan rotates at constant speed, there will be a constant current that produces just enough torque to overcome any friction and result in a net zero torque on the fan.

Thomson’s experiments with cathode rays allowed the charge-to-mass ratio of electrons to be measured. Thomson used an electric field to accelerate the electrons and used a magnetic field to deflect them. He then used an electric field to cancel the deflection caused by the magnetic field. By considering the motion of the electrons under the influence of the magnetic field and the force balance between the electric and magnetic forces, Thomson was able to calculate the charge-to-mass ratio.

Millikan’s oil drop experiments measured the charge of the electron. Millikan used charged drops of oil between two metal plates. By observing the drops using a microscope and measuring their terminal velocity when falling under the influence of gravity, Millikan calculated the size and mass of the drops.

By applying an electric field between the plates, Millikan balanced the electric and gravitational forces and calculated the charge of the oil drops. He calculated the charge of an electron from the smallest difference in charge between two oil drops. Since the charge-to-mass was already known, Millikan was able to also calculate the electron’s mass.

Firing electrons at protons at high speed, in the Stanford Linear Accelerator, in an experiment called Deep Inelastic Scattering, showed that protons were not fundamental particles.

When the scattering was inelastic, some of the kinetic energy of the collision was used to remove a quark from the proton. This led to the discovery of quarks.

[There are other acceptable answers for the second part, however they need to relate to a fundamental particle, like a quark or a lepton.]

**Similarity:** Both particles will undergo uniform circular motion. They experience a constant force \(F=qvB\) perpendicular to their velocity. As the force is perpendicular to their motion it does no work, so their speed is constant, meaning their direction changes at a constant rate and they follow a circular path at constant speed. The magnetic force provides the centripetal force: \(qvB=\frac{mv^2}{r}\) .

**Difference:** The radius of the circular path will be different. Given the mass of the alpha is approximately four times the mass of the proton, and its charge is double the charge of the proton, and we get \(r=\frac{mv}{qB}\) from above:

The radius of the proton’s path will be \(r_{\textrm{proton}}=\frac{m_{\textrm{proton}} v}{eB}\)

The radius of the alpha’s path will be \(r_{\textrm{alpha}}=\frac{4m_{\textrm{proton} }v}{2eB}=2r_{\textrm{proton}}\)

Production of energy is through nuclear fusion, specifically the fusion of hydrogen (protons) into helium in the core of the star. In the core where temperature is very high, protons travel at high speed and can overcome the electrostatic repulsion between them. They are able to get sufficiently close for the nuclear strong force to bind them into a larger nucleus.

The mass of helium is less than the mass of four protons, and the difference in mass is converted to energy through Einstein’s energy mass equivalence \(E=mc^2\).

The fusion reaction can proceed directly through the proton-proton chain to produce helium, or it may be catalysed by a large nucleus through the CNO cycle.

The energy released from fusion heats up the core of the Sun, and this energy propagates through the Sun’s surface. Approximating the Sun as a black body, the black body spectrum and Wien’s Law allows the surface temperature to be estimated:

\(λ_{\textrm{max}}=510 \ \textrm{nm}\)\(T=\frac{b}{λ_{\textrm{max}}} =\frac{2.898×10^{-3}}{510×10^{-9}}=5682 \ \text{K}\)

The Sun would radiate energy in the form of black body radiation, which will spread out over a sphere as it travels away from the Sun. The intensity of the radiated energy follows an inverse square law. From the value of the intensity at Earth, the total power emitted by the Sun can be calculated:

\(I=\frac{P}{A}=\frac{P}{4πr^2 }\) \(P=I×A=1360×4π(1.5×10^11 )^2=3.85×10^{26} \ \text{W}\)

The Sun is in thermal equilibrium and maintains a constant temperature, meaning that the power radiated is equal to the power generated by fusion in the core.

The apparatus is an accelerometer that can measure the car’s acceleration, in the direction perpendicular to the sides of the car (i.e. left and right).

If the string is hanging at an angle to the right, then there is a net force to the left due to tension. Consider the forces on the mass hanging from the string. The vertical forces on the mass will be the vertical component of tension (\(T \cos θ\)) and it’s weight. These forces cancel since there is no vertical net force. The horizontal force on the mass is the horizontal component of the tension (\(T \sin θ\)) which will be the net force on the mass.

This gives:

\(T \cosθ=mg\), based on the vertical forces, hence \(T=\frac{mg}{\cosθ}\)

Substituting into an expression for the net force:

\(T \sinθ =ma\), hence \(a=g \tan θ\).

The angle at which the mass hangs can give the magnitude of acceleration (perpendicular to the sides of the car) and the side it leans towards can give the direction (turning left or right). In the example described, the car is in uniform circular motion, turning left, as it accelerates the left whilst travelling at constant speed, hence \(T \sinθ =\frac{mv^2}{r}\).

Calculate the mass defect:

\(Δm= m_R- m_p=197.999-(193.988+4.00260)=0.0084 \text{ u}\)

Calculate the energy released in \(\textrm{MeV}\), then joules:

\(E= 0.0084×931.5=7.8246 \text{ MeV} \\=7.8246×10^6 \ \textrm{eV} \\

=(7.8246×10^6×1.6×10^{-19}) \ \textrm{J} \\

=1.25×10^{-12} \text{ J}\)

Conservation of energy means that the reaction products must have a total kinetic energy equal to this. Therefore the kinetic energy of the alpha particle is \(1.25×10^{-12}-2.55×10^{-14}=1.23×10^{-12} \text{ J}\) which is around 50 times larger than the energy of the polonium.

Since the radon was initially at rest the initial momentum of this reaction was 0. To conserve momentum, the final momentum of the polonium and alpha must add to zero. This means they have to be equal in magnitude and opposite in direction.

Since the mass of polonium is around 50 times larger than the mass of helium, the speed of the helium atom must be 50 times larger than the polonium to give the same momentum, \(p=mv\).

\(K=\frac{1}{2} mv^2=\frac{1}{2}mv×v\).

Since their momenta \(mv\) are equal, and the helium has a speed that is 50 times larger, it must also have a kinetic energy that is 50 times larger.

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