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| Question | Answer | Explanation |
| 1 | C | In uniform circular motion, the net force always acts towards the centre of the circle. |
| 2 | B | Huygens proposed that light was a wave. Diffraction is a wave behaviour. |
| 3 | C | Photons are one of the gauge bosons responsible for mediating forces. Hadrons, neutrons and protons are all made up of quarks. |
| 4 | D | For a DC motor, a split-ring commutator reverses the direction of current in the coil every half-turn and ensures torque is always in the same direction. The split should disconnect the rotor coil when the coil is oriented vertically. |
| 5 | A | All the stars in Cluster Y are still on the main sequence. In Cluster X, the largest main sequence stars have evolved into the red giant phase. In Cluster Z, some have further evolved into supergiants and others have ceased nuclear fusion to become dwarf stars. |
| 6 | C | The symbol \(\phi\) represents the work function of the metal. It is the smallest amount of energy required to liberate an electron from the metal. |
| 7 | B | After two half-lives, three-quarters of the polonium atoms will have undergone decay into lead-206, while one-quarter will remain in their original form. This gives a 1:3 ratio. |
| 8 | B | According to the transformer equation, the ratio of input voltage to output voltage equals the ratio of primary coils to secondary coils: \(\frac{V_P}{V_S} = \frac{N_P}{N_S}\) We must either increase the number of turns in the secondary coil (not one of the options) or decrease the number of turns in the primary coil. |
| 9 | B | Both projectiles will land with the same vertical velocity. The final velocity of Q has both a vertical component and a horizontal component, giving it a greater final velocity than P. |
| 10 | C | The force on the conductor is given by \(F=LIB \sin \theta\). |
| 11 | D | The formula for orbital velocity is: \(v = \sqrt{\frac{GM}{r}}\) Hence the velocity is inversely proportional to the square root of r. |
| 12 | B | The length of the rod will appear to contract. The factor by which it contracts is initially small, with the contracted length varying only slightly from rest length \(L_0\), but increases as the speed approaches c. |
| 13 | A | Satellites have a lower speed and hence a lower kinetic energy at higher orbital radii. Potential energy is greater at larger radii. |
| 14 | B | De Broglie’s equation for the wavelength of a particle is: \(\lambda = \frac{h}{mv}\) The alpha particle has (approximately) four times as much mass but half the velocity of the proton, causing the denominator to be twice as large overall. |
| 15 | D | As the conductor rotates, all the charges in it undergo circular motion with their instantaneous velocity being tangential to the circle. Applying the Right Hand Palm Rule, we find that positive charges always experience a force towards P and negative charges always experience a force towards Q. This leads to a consistent emf along the length of the conductor. |
| 16 | A | The photon energy when the frequency is \(7 \times 10^{14} \text{Hz}\) can be calculated from Planck’s equation: \(E = hf = (6.626 \times 10^{-34}) \times (7 \times 10^{14} \text{Hz})\) \(E = 4.6382 \times 10^{-19} \text{J}\) This is equivalent to 2.9 eV. At this photon energy, only potassium and lithium will emit electrons. |
| 17 | C | The electric field between the dees accelerates the charge by speeding it up. The magnetic field in the dees accelerates the charge by changing its direction of motion. |
| 18 | D | The original action of the magnet moving towards the coil induces a current such that coil X has a magnetic field that points to the left. The only situation that will induce this is option D, where the white coil will produce a decreasing amount of flux to the left and hence coil X will act to replace the missing flux. |
| 19 | C | The first proton is undeflected because the magnetic force equals the electric force. The second proton will experience twice as much magnetic force as electric force, and hence it will accelerate. The electric force depends only on the charge of the proton, so it will remain constant. The magnetic force will not change the speed of the particle, so its magnitude will remain constant. However, its direction will change as the particle’s path changes direction. The net force on the second proton will vary in magnitude and direction, and so will its acceleration. |
| 20 | B | According to special relativity, the faster a clock is travelling relative to a stationary observer, the more time dilation it will experience, i.e. it will tick more slowly. Our observer is at X on the surface of the Earth and hence considers clock X to be stationary. Clock Z is in a geostationary orbit and hence also does not move relative to the earths’ surface. The satellite at Y has a period of 12 hours, so it must have a smaller radius orbit than Z and hence travel with a higher linear velocity. It is the only clock that appears to be in motion relative to the observer and hence it is seen to tick more slowly than the others. |
a) Using the torque equation:
\(\tau = rF \sin \theta\)
\(\tau = 0.18 \times 75 \times \sin 40^{\circ} = 8.7 \text{Nm clockwise}\)
b) The torque in a DC motor is given by \(\tau = nIAB \sin \theta\).
Increasing the current will increase the force exerted by the magnetic field on the sides of the loop and hence increase the maximum torque.
Increasing the number of turns in the motor coil will also increase the total force and torque exerted on each side.
(Note: other answers such as increasing the magnetic field strength are possible)
a) The graph showed that the entire universe was expanding, causing other galaxies to move away at higher speeds the further away they were from Earth. This contradicted the previous theory that the universe was static over long time scales
b) The spectra of galaxies were measured and the absorption lines of elements identified. The absorption lines were shifted to lower frequencies due to the Doppler effect on light waves, which allowed the recession velocities of the galaxies to be calculated.
a) The unknown radiation was difficult to detect as it had zero charge. When it collided with paraffin, positive protons were ejected from the paraffin and their velocity could be easily measured. Chadwick could then calculate the mass of the neutrons using conservation of energy and momentum.
b) The Rutherford model of the atom showed that there was a dense, positively charged nucleus at the centre of the atom, orbited by electrons. However, the only nuclear particle previously known was the proton, which could account for the charge but not the mass of the nucleus. Chadwick’s discovery of the neutron allowed scientists to completely determine the composition of the nucleus.
c) The Bohr-Rutherford model described electrons as being in fixed, stable orbits around the nucleus in which they would not radiate any energy despite being accelerating charges. The angular momentum of each electron in the hydrogen atom as: \(L_n = \frac{nh}{2 \pi}\)
However, the model could not explain why these particular states were stable. De Broglie’s hypothesis said that particles could have wave properties with \(\lambda = \frac{h}{mv}\). He described electrons in the atom as circular standing waves, where an integer number of wavelengths fit around the circumference of the orbit: \(n \lambda = 2 \pi r\).
Substituting the formula for electron wavelength gave Bohr’s result for angular momentum, explaining why these were the allowed electron orbits.
a) The spectrum intensity peaks at a wavelength of 500 nm.
Using Wien’s displacement law and rearranging for temperature:
\(\lambda = \frac{b}{T}\)
\(T = \frac{b}{\lambda} = \frac{2.898 \times 10^{-3}}{500 \times 10^{-9}} = 5800 \text{K}\)
b) Using the Rydberg equation for \(n_i = 6, n_f = 2\):
\(\frac{1}{\lambda} = R(\frac{1}{n^{2}_{f}} – \frac{1}{n^{2}_{i}})\)
This gives \(lambda = 410.2 nm\)
Corresponding to a frequency \(f = \frac{c}{\lambda} = 7.3 \times 10^{14} \text{Hz}\)
c) To produce the spectrum, there must be a continuum emission source such as a black body. As photons of all wavelengths pass through the sample, only some are absorbed because their energy matches a possible electron transition in the sample. These wavelengths are then removed from the beam of light. The light is then passed through a diffraction grating or prism to separate it by wavelength and the intensity at each wavelength is recorded.
a) When a satellite orbits around a central mass, the centripetal force is provided by gravitational force.
\(F_C = F_G\)
\(\frac{m v^2}{r}=\frac{GMm}{r^2}\)
This simplifies to
\(v^2 = \frac{GM}{r}\)
We then substitute in the fact that speed in circular motion is equal to the circumference of the circle divided by period \(v = \frac{2 \pi r}{T}\):
\(\frac{4 \pi^2 r^2}{T^2} = \frac{GM}{r}\)
This can be rearranged to give Kepler’s Third Law as required.
b) Kepler’s Third Law can be expressed as:
\(T^2 = \frac{4 \pi^2}{GM} \times r^3\)
Therefore the gradient of this graph, when converted to SI units, will be \(\frac{4 \pi^2}{GM}\).
Taking two points on the graph, the gradient is equal to:
\(\frac{0.2 days^2}{250 \times 10^{12} km^3} = \frac{1.493 \times 10^9 seconds^2}{250 \times 10^{21} m^3}\)
Hence
\(\frac{4 \pi^2}{GM} = 5.972 \times 10^{-15} s^2 m^{-3}\)
This gives the mass of the planet \(M = 9.91 \times 10^{25} \text{kg}\).
Consider two events: the formation of the muon, and the decay of the muon. In the muon’s frame of reference, the time interval is the short proper time, but the distance that they travel appears to be contracted as the space moves relative to the muon.
In the Earth’s frame of reference, the muons travel the much longer proper distance but they undergo time dilation, allowing them to exist long enough to reach the Earth’s surface.
Both frames of reference agree that the muons are able to reach the surface. They disagree on how much time it takes and how much distance is travelled.
a) \(\pi^+: u \bar{d}\)
\(\pi^-: d \bar{u}\)
\(\pi^0: u \bar{u}\)
b) Energy released depends on the mass defect of the reaction.
Initial mass = \(940 + 940 = 1880 \text{MeV}/c^2\)
Final mass = \(140 + 140 + 140 = 420 \text{MeV}/c^2\)
The mass defect is \(1460 \text{MeV}/c^2\) corresponding to 1460 MeV being released. Each pion receives a third of this = 487 MeV.
c) The classical relationship between kinetic energy, velocity and mass is
\(K = \frac{1}{2} mv^2\)
It assumes that any work done on the particle results in an increase in velocity. This leads to the stated result that it exceeds the speed of light. The Special Theory of Relativity predicts that no object with mass can ever reach or exceed this speed, so the conclusion cannot be correct.
As objects approach relativistic speeds, an increasing proportion of the work done goes towards increasing the mass of the particle instead of its velocity. The velocity of the pions should be calculated using the formula for relativistic kinetic energy.
a) Charges in an electric field experience a force \(F=qE\) and hence an acceleration \(a = \frac{qE}{m}\)
Substituting in the given field strength as well as the mass and charge magnitude of an electron:
\(a = \frac{1.602 \times 10^{-19} \times 1.5 \times 10^4}{9.109 \times 10^{-31}} = 2.6 \times 10^{15} ms^{-2}\)
As required.
b) The time taken to reach the end of the parallel plates is found from \(t = \frac{s_x}{u_x}\)
\(t = \frac{5 \times 10^{-3}}{2.0 \times 10^6} = 2.5 \times 10^{-9} s\)
Let the direction of vertical acceleration be positive. The vertical displacement at this time is given by \(s_y = u_y t + \frac{1}{2} a t^2\)
\(s_y = 0 + \frac{1}{2} \times 2.6 \times 10^{15} \times (2.5 \times 10^{-9})^2 = 0.081 m\)
As required.
c) After the electron leaves the field, it will continue travelling with a constant velocity. Its final vertical velocity when leaving the plates is:
\(v_y = u_y + at\)
\(v_y = 0 + 2.6 \times 10^{15} \times 2.5 \times 10^{-9} = 6.5 \times 10^6 ms^{-1}\).
The time it will take to reach the plate is given from the remaining horizontal motion: \(t = \frac{s_x}{u_x}\)
\(t = \frac{30 \times 10^{-3}}{2.0 \times 10^6} = 1.5 \times 10^{-8} s\)
Therefore the vertical displacement travelled in this time is:
\(s_y = v_y t = 6.5 \times 10^6 \times 1.5 \times 10^{-8} = 0.098 m\)
This adds to the existing 0.081 m to give a final vertical displacement from X of 0.18 m.
a) The currents must be in opposite directions to give a repulsive force. However the rods cannot be connected in series if they are to have different magnitudes of current.
b) Ampere’s Law is used to calculate the force between two parallel current-carrying conductors.
\(\frac{F}{l} = \frac{\mu_0}{2 \pi} \frac{I_1 I_2}{r}\)
According to this law, the conductors will exert equal and opposite forces on each other even if they carry different magnitudes of current. However, their accelerations do not have to be equal if their masses are not the same. Rod B will accelerate more if it has a lower mass than Rod A.
The object always experiences a force downward due to weight and a normal force upwards due to contact with the floor. Because the object does not accelerate vertically, these two forces always cancel each other out. Weight (mg) is constant and so the vertical normal force (mg) is constant.
The horizontal normal force from the cylinder wall is responsible for the object’s circular motion and is given by \(F_c = \frac{mv^2}{r}\).
Halving the period means that the object must travel at twice the speed in the same circle. This horizontal normal force must then increase by a factor of 4 to provide the appropriate centripetal force
In both models, as the projectile travels upwards from the Earth’s surface, its kinetic energy will be converted into gravitational potential energy (GPE). When the projectile’s kinetic energy reaches zero, it will stop and begin to fall downwards at increasing speed.
Model A assumes that the strength of the gravitational field is uniform at any height. The GPE increases linearly with height: \(U = mgh\).
In Model B, the gravitational field gets weaker as distance from the Earth increases, and the GPE is given by \(U = – \frac{GMm}{r}\). It is no longer linear, and the rate at which it increases becomes smaller at greater heights. Model B will therefore predict a greater maximum height for the projectile than Model A, as it is able to travel further before all kinetic energy is converted to GPE.
The study of blackbody radiation was an attempt to understand the light emitted by hot matter. When scientists observed the spectrum, they discovered certain distinctive features including a zero-intensity cutoff at short wavelengths and a peak wavelength that decreased with temperature.
When they attempted to explain the radiation using the classical electromagnetic wave model, they found the theory could not replicate the experimental results and predicted infinite emission at short wavelengths. Planck proposed that the light was emitted in quantised energy packets which corresponded to frequency, governed by the equation \(E = hf\). This represented the first step in the development of quantum theory.
Another significant experiment was the photoelectric effect, in which light incident on a metal would cause electrons to be emitted but only if the light frequency was above a certain threshold. Einstein applied Planck’s energy relationship to the problem and concluded that light travelled in the form of discrete, localised massless particles that he called photons. This represented a completely different model of light from the classical wave theory and cemented quantum mechanics as an important field of study.
Scientists discovered that the absorption and emission spectra of atoms consisted of discrete spectral lines instead of a continuum. Given what was already known about the photon model of light, the presence of specific frequencies suggested that the energy level structure of the atom was also made up of fixed, discrete states. Bohr’s model of the atom in which he postulated electrons orbited in fixed, stable states was the first model to apply quantum ideas to the structure of matter. This eventually led to the fully quantum Schrodinger model of the atom.
When the magnet is first held up and released, it converts gravitational potential energy to kinetic energy.
As the magnet gets closer to the can, the increasing flux in the can walls leads to the formation of eddy currents; some of the magnet’s kinetic energy has been converted to electrical energy.
The magnet slows down because Lenz’s law predicts an opposing force will result from the induced currents. This repulsion also leads to rotation of the can in the same sense that the magnet is swinging, indicating that the can’s mechanical energy has increased.
When the magnet begins to move away from the can, eddy currents are induced again and cause a force of attraction between the can and the magnet. This causes the magnet to lose more of its mechanical energy. The magnet will not reach the same height at the other end of its swing.
This process repeats when the magnet begins to swing the other way. The kinetic energy of the magnet is always converted to electrical energy of the eddy currents and kinetic energy of the can’s rotation.
After some time both the magnet and the can come to a complete stop, as all the electrical energy is dissipated as heat according to \(P_{loss} = I^2 R\).
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