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The object is undergoing uniform circular motion hence the net force (centripetal) must be directed towards the centre of the circle.
Diffraction and interference were the direct points of evidence for Huygen’s longitudinal wave model of light (Young’s Double Slit and Poisson’s spot). Photoelectric effect was evidence for Einstein’s photon model of light. Emission spectra were explained by electron transitions in Bohr’s model of atom. Black body radiation relates to UV catastrophe and Planck’s innovations. Hence, diffraction is the best answer.
Hadrons, protons and neutrons all consist of a substructure of quarks, hence they are not fundamental. Photons mediate the electromagnetic force as a boson and are a fundamental particle.
A split ring commutator should be used in a DC motor where the direction of the current needs to be reversed when the coil is in the vertical position. Option D best matches this whereby the gap in the split ring commutator is aligned with the brushes when the coil is at the vertical position.
A cluster of stars consists of stars that are formed at approximately the same time. Along the main sequence branch on a HR diagram, the more luminous, high temperature stars have a shorter lifetime compared to less luminous, low temperature main sequence stars as they burn faster. Therefore, the more luminous main sequence stars will progress to the next evolutionary stage before the less luminous main sequence stars.
Therefore, Cluster Y is the youngest as all stars are still in the main sequence stage followed by Cluster X where the most luminous main sequence stars have evolved, followed by Cluster Z where less luminous main sequence stars have evolved into red giant/supergiants and white dwarfs.
Symbol Φ represents the work function which is the minimum energy required to liberate an electron from a metal material. Option C best matches this definition.
The half life of 138 days represents the time required for half the sample to decay. Hence, over 276 days only ¼ of the original sample of polonium-210 remains and ¾ have decayed into lead-206. Therefore, the ratio is 1:3.
The ratio of input to output voltage is directly proportional to the ratio of turns from primary to secondary coil as described by Vp / Vs = Np / Ns. Therefore, to double the voltage from 6 V to 12 V using the same input voltage, the ratio of turns from the secondary to primary coil must increase. B matches this requirement by decreasing the number of turns in the primary coil.
Both objects P and Q have the same initial vertical velocity (uy) however Q has some initial horizontal velocity (ux) while P does not. There are two methods to analyse this scenario:
Option A is incorrect because they both experience the same acceleration of g = 9.8 ms-2
Option C is incorrect because time of flight is dependent only on launch height and initial vertical velocity (uy) which is the same for both.
Option D is incorrect because they both have zero initial vertical velocity (uy = 0).
The force experienced by a current carrying conductor in a magnetic field is given by F = lIBsinθ (motor effect). In the initial position, θ = 90º, therefore, to halve the force experienced, θ should equal 30º as sin(30) = ½. Since θ represents the angle between the current carrying conductor and the magnetic field, it should be rotated by 90 - 30 = 60º.
By the orbital speed equation for satellites, vorb = √(GM / r), vorb is proportional to 1 / √r. Hence, vorb is inversely proportional to √r.
Hence, option D is correct.
The contracted length of the rod is given by l = l0√(1 - v2 / c2). Hence, the greater the speed of the rod, the smaller the length becomes, eliminating W.
It is clear from the equation that the relationship is nonlinear, eliminating Y.
At low velocities, v2 / c2 is almost zero and thus l is very close to l0. At larger velocities, the effect of length contraction becomes significant, where when v approaches c, v2 / c2 approaches 1 and thus l approaches zero. Z seems to asymptote towards zero, however, the graph needs to reach zero at c. Thus, X is the best answer. This can also be seen with a plot of the equation for length contraction:
Potential energy can be given by U = -GMm / r. Thus, the greater the orbital radius, the greater the potential energy. UA > UB
Kinetic energy can be given by K = ½mv2 where v is the orbital velocity v = √(GM / r). Thus, the greater the orbital radius, the lesser the orbital velocity and the lesser the kinetic energy. KA < KB.
These relationships between the different energies and orbital radius of a satellite is also depicted in the graph below.
The wavelength of the proton can be found by:
λH = h / mHvH
An alpha particle contains 2 protons and 2 neutrons and is thus four times the mass of a proton: mHe = mH / 4. The question tells us the velocity is half vHe = vH / 2
Hence, its wavelength can be found by:
λHe = h / mHevHe
= h / (4mH)(vH / 2)
= h / 2mHvH
= ½h / mHvH
= ½λH
There are two ways to solve this question:
Photoelectrons are only emitted when the incident photon provides enough energy to overcome the work function , i.e. the energy required to eject a single photoelectron. The photon’s remaining energy becomes the photoelectron’s kinetic energy K after.
When photons strike the material: Ephoton = Φ + K, and Ephoton = hf
At the critical point K = 0, a photon with frequency f has just enough energy to eject a photoelectron from a material with a work function Φ. This can be seen as the X intercept of each line on the given graph, representing the Ephoton where y = K = 0.
hf = Φ + 0
⇒ Φ = hf = (6.626 x 10-34)(7 x 1014)
= 4.638… x 10-19 J
= (4.638… x 10-19 J) / (1.602 x 10-19 JeV-1)
= 2.895 eV
Thus, the photons can eject photoelectrons from a material with a work function of 2.895 eV at most. Materials with work functions beyond that, such as Mg, Ag, will not be affected, as seen on the graph with Ephoton = Φ + 0 > 3 eV. Conversely, K and Li have work functions below 2.895 eV, and thus can have photoelectrons ejected.
Hence, the answer is A.
Inside the dees, a uniform magnetic field is applied which results in the charged particle undergoing uniform circular motion. Thus, whilst the speed of the charge remains the same, the velocity is changing as the direction is changing. Hence, it is undergoing acceleration in the dees (centripetal acceleration).
Between the dees, an electric field acts on the particles which results in the charged particle increasing its velocity. Thus, it is undergoing acceleration between the dees.
Hence, the charged particle accelerates both inside and between the dees.
When the magnet moves to the right, there is an increase in flux through the solenoid X. This induces an emf by Faraday’s Law which by Lenz’s Law will create a field to oppose the change. Thus, a field must be created to the left. Using the right hand grip rule, it can be determined that the induced current must be anti-clockwise as viewed from the left hand side.
Option A increases the flux through X, however, the field is in the opposite direction compared to the magnet which can be found using the right hand grip rule. Thus, the induced current will be in the opposite direction compared to the given case.
Option B has the current going the opposite way, but also the coils wrapped in the opposite way such that the field created is exactly the same as option A. Thus, the induced current will also be in the opposite direction compared to the given case.
Option C decreases the flux through X, however, the field is in the same direction compared to the magnet which can be found using the right hand grip rule. Thus, the induced current will be in the opposite direction to oppose the change compared to the given case.
Thus, the answer is option D, where it decreases the flux through X in the opposite direction compared to the magnet. Thus, the induced current will be in the same direction to oppose the change compared to the given case.
The first proton has a velocity v such that FE = qE is balanced with FB = qvB in the opposite direction, so that Fnet = 0, allowing it to travel through the chamber at constant velocity.
The second proton has velocity 2v, indicating that FB > FE, making Fnet > 0 and initially point perpendicularly to velocity by right hand palm rule. FE is constant in direction and magnitude, whereas FB is constant in magnitude but differs in direction to stay perpendicular to velocity. This means that the strong perpendicular force of FB will gradually change the direction of velocity, and hence the direction of FB in the next instant, such that it cancels less with FE. Hence Fnet will vary in both magnitude and direction, if it stays within the chamber.
By Fnet = ma, where the proton’s mass is constant, its acceleration must also vary in magnitude and direction which is option C.
*Note: There was a lot of debate for the answer, due to the specific wording of the question. However, the consensus was that while option C is a more scientifically correct answer, due to the specific (and scientifically incorrect) assumptions the question makes - the more likely answer to be marked correct is option B.
New reasoning for option B:
The question specifies to "assume that the satellites are inertial frames of reference", this means that we need to discount the rotational motion of the Earth and satellites, effectively treating it as a flat Earth.
As Z is a geostationary satellite, TZ is 24 hours. Thus, from an observer at X, they would see satellite Z to be stationary and thus in the same frame of reference. Again, this is assuming inertial frames of reference with essentially a flat Earth. This means that Z will have no time dilation as it is in the same frame of reference, so an observer at X would observe the clock at X and the clock at Z to tick at the same rate.
We know that Y has a shorter orbital period than Z using Kepler's 3rd Law:
rZ3 / TZ2 = rY3 / TY2
As rY > rZ, then TY > TZ.
Thus, Y is moving faster than Z which is in the same frame of reference at X. Hence, the observer at X would also see Y to be moving faster and hence experiencing time dilation. As “moving clocks run slow”, the clock at Y will tick slower than X which is the same as Z. Thus, Y would tick slower than both X and Z which matches option B.
Original reasoning for option C:
An object’s linear velocity relative to the centre of Earth determines the extent of relativistic effects. Faster clocks will experience more time dilation by t = t0 / √(1 - v2 / c2), meaning the fastest-moving clock will tick the slowest.
Comparing v = s / t for X and Y:
Let X have sx = 2πr, tx = 24 hours
Y will have sy = 2π(r + altitudey), ty = 12 hours
Thus, vy = sy / 0.5tx > sx / 0.5tx = 2vx.
Therefore vy > 2vx.
Z will have sz = 2π(r + altitudez), tz = 24 hours (geostationary)
Thus, vz = sz / tx > sx / tx = vx.
Therefore vz > vx.
Therefore, vx < vy and vx < vz, but it’s still unclear which is the fastest.
Now comparing vorb = √(GM / r) for orbiting satellites Y and Z:
Y orbits closer than Z, and hence Y has a greater orbital velocity. This can be affirmed using the equation from Kepler’s 3rd law, and linking r to given T. (All this does not apply to X because it is on Earth’s surface where normal force acts, i.e. not in orbit).
Thus, vy > vz > vx. Thus Y ticks slowest, X ticks fastest.
This makes options A and D incorrect. Option B is also wrong because Y would tick slower than both of them, not ‘either’ of them. Option C is the best answer.
Exemplar Answer
τ = rFsinθ = (0.18)(75)sin(40) = 8.7 Nm (2 s.f.)
Exemplar Answer
Torque produced by a coil in a simple DC motor can be given by τ = NIABsinθ. Thus, two ways of increasing torque are:
Marker’s Notes
Other ways that may be accepted are:
Exemplar Answer
The graph shows how the recessional velocity of galaxies increases proportionally to their distance from the Earth. This relationship is described by Hubble’s Law v = H0d. This suggests that galaxies further away from the earth are moving away faster.
The significance of this relationship is that it supports an expanding model of the universe as predicted by the Big Bang. In an expanding universe where space is stretching, galaxies that are further have more space between them and the Earth. Hence more space is expanding and the galaxy would be measured to be moving away faster. The proportional relationship can only be accounted for by space itself expanding, and not due to the movement of galaxies through space.
Thus, the graph is significant as it acts as evidence supporting the Big Bang model of the universe.
Marker’s Notes
Although Hubble’s Law is not provided on the data sheet, it is useful to remember and include. Alternatively, students could have described the relationship as recessional velocity being proportional to distance (v ∝ d).
Exemplar Answer
The recessional velocities of each galaxy can be determined by analysing the amount of redshift in the spectra they emit.
As a galaxy moves away from Earth, the wavelengths of its light are stretched, shifting it towards the red end of the spectrum which is an effect known as the Doppler shift.By comparing spectral lines of known elements to their spectra produced on Earth where there is no redshift, the extent of the redshift can be measured and quantified by a change in the measured frequency.
The extent of red shift can be used to calculate the recessional velocity where the greater the amount of redshift, the greater its recessional velocity. In the 1920s, Hubble followed the principles of this method when observing stars in distant galaxies, allowing him to determine their recessional velocities.
Marker’s Notes
The inclusion of a sample diagram and reference to the Doppler shift equation (from the data sheet) may not be necessary for full marks. However, it is useful to illustrate the method being explained.
The question’s past tense of “how were” may indicate that an explicit link to Hubble’s historical procedure is important for a complete answer.
Exemplar Answer
The paraffin was used as a target to detect and analyse the properties of the unknown radiation. Paraffin wax is a hydrocarbon which is proton-rich.
The unknown radiation was difficult to detect due to its neutral properties. However, when the unknown radiation struck the paraffin, it caused protons to be ejected. The protons could be detected and their energies measured, using the elastic collision to determine the properties of the unknown radiation.
Thus the paraffin wax enabled Chadwick to discover the unknown radiation to be neutral particles with a similar mass to the proton, the neutron.
Exemplar Answer
Chadwick applied the Laws of Conservation of Energy and Momentum to the collision between the unknown radiation and a proton, discovering the unknown radiation to be neutral particles with a similar mass to the proton, the neutron.
Prior to the discovery of the neutron in 1932, the model of the atom consisted only of protons in the nucleus and electrons on the outside. One issue with this model of the nucleus was that while the protons accounted for all the charge of the nucleus, they couldn’t account for all the nucleus’ mass.
However, the discovery of the neutron through Chadwick’s experiment changed the model of the atom to include neutrons in the nucleus. The updated model of the atom, with both protons and neutrons, could now account for the full mass of the nucleus.
Marker’s Notes
Specific date details are not essential, however do help to contextualise the role of Chadwick’s experiment in changing the model of the atom.
Exemplar Answer
The limitations of the Bohr-Rutherford atom was that Rutherford predicted electrons to be in circular orbits around the nucleus, due to a centripetal force supplied by electrostatic attraction. However, this suggests the electrons’ velocity is constantly changing and thus they must be accelerating.
By Maxwell’s predictions, accelerating charges emit electromagnetic radiation due to the mutual induction of changing electric and magnetic fields (Ampere-Maxwell Law and Faraday’s Law). Thus the electrons were predicted to emit EMR, losing energy and spiralling into the nucleus. This suggests that no atom can be stable, which is not the case. Bohr attempted to address this by postulating that electrons exist in stationary states with discrete energy levels, given by a quantised angular momentum (L = mvr = nh/2𝜋). However, Bohr was unable to provide a theoretical explanation behind why and how these stationary states and discrete energy levels exist.
De Broglie’s matter-wave hypothesis addressed this limitation, providing a theoretical explanation. The stationary states where the electrons can exist around the atom without losing energy, can be explained by modelling the electron as a standing wave. Further, the discrete energy levels correspond to the specific standing wave patterns that can form, where the circumference of the electron orbit must correspond to an integer multiple of the wavelength (2πr = nλ) of the electron by λ = h / mv for constructive interference to occur and a standing wave to be produced. This can be used to derive Bohr’s quantisation condition (L = mvr = nh/2𝜋). Hence De Broglie provided a theoretical explanation to address the limitation of electron orbital stability in the Bohr-Rutherford model.
Marker’s Notes
Specific detail about Bohr’s postulates and the link to quantised angular momentum have been included for completeness, but are likely not necessary for a full response.
Exemplar Answer
λmax = b / T
T = b / λmax
= (2.898 x 10-3) / (490 x 10-9)
= 5914
= 5900 K (2 s.f.)
Exemplar Answer
1 / λ = R(1 / nf2 - 1 / ni2) = (1.097 x 107)(1 / 22 - 1 / 62) = 2437778 m-1
λ = 4.102 x 10-7 m
c = fλ
λ = c / f
= (3 x 108) / (4.102 x 10-7)
= 7.3135 x 1014 Hz
= 7.3 x 1014 Hz (2 s.f.)
Marker’s Notes
During the exam, students could have verified their answer to this question by reading off the wavelength of W from the graph and calculating the frequency using c = fλ.
Exemplar Answer
An absorption spectrum is produced when continuous light (white light containing all wavelengths) shines through an unexcited gas. This is produced by stars as the hotter layers of the star which emit light due to their temperature, emitting a continuous spectrum of light approximating a black body. The cooler outer photosphere layers of a star extend outwards, decreasing in density and temperature.
Hence, the star’s continuous black body spectrum shines through the outer layers of the star, where electrons in the cooler gas can absorb specific wavelengths of light, corresponding to energy differences between discrete energy levels. Only specific wavelengths are absorbed and rescattered, as they must correspond to the energy levels difference by Ei - Ef = Ephoton = hf = hc / λ.
Thus, most of the wavelengths of light pass through, except for the specific wavelengths that are absorbed, resulting in discrete dips in the graph (or darker lines on the spectrum). Hence an absorption spectrum can be produced.
Exemplar Answer
The gravitational force acts as the centripetal force keeping a satellite in a circular orbit:
Fc = Fg
⇒ mv2 / r = GMm / r2
⇒ v2 = GM / r
The orbital velocity can be related to the orbital period by v = 2πr / T:
(2πr / T)2 = GM / r
⇒ 4π2r2 / T2 = GM / r
⇒ r3 / T2 = GM / 4π2
Hence the model is derived.
Exemplar Answer
gradient = (y2 - y1) / (x2 - x1) = (0.32 - 0.08) / (400 x 1012 - 100 x 1012) = 8 x 10-16 days2km-3
⇒ gradient = 8 x 10-16 x (24 x 60 x 60)2 = 5.97196 x 10-6 s2km-3
⇒ gradient = 5.97196 x 10-6 / (10003) = 5.97196 x 10-15 s2m3
gradient = T2 / r3 = 4π2 / GM
⇒ M = 4π2 / (G x gradient)
= 4π2 / (6.67 x 10-11 x 5.97196 x 10-15)
= 9.911 x 1025
= 9.9 x 1025 kg (2 s.f.)
Exemplar Answer
Einstein’s Theory of Special Relativity can be applied to explain why the muons can reach the Earth’s surface. As the muons travel “at almost the speed of light” significant relativistic effects apply.
From the muon’s frame of reference:
From the muon’s frame of reference, the muon is stationary while the Earth is moving past. Hence, the muons’ half-life will remain the same as compared to when they are stationary. However, as the Earth is moving relative to the muon, the length contraction will occur to the distance between the muon and the Earth. The distance to the Earth will be contracted by l = l0√(1 - v2 / c2). Although the half-life remains very short, by v = d / t, the muons can successfully travel the shortened distance due to the contracted distance.
From the Earth’s frame of reference:
From the Earth’s frame of reference, the muons are moving while the Earth is stationary. Instead of the muon’s journey being length contracted, since the muons are moving relative to the Earth’s frame of reference, time dilation will occur to the muon. Hence the muon’s half life will be dilated by t = t0 / √(1 - v2 / c2) and hence the muons can travel the same long distance to the Earth’s surface as their half-life is now longer by v = d / t.
As the Lorentz factor √(1 - v2 / c2) is the same for both frames of reference, these relativistic effects apply equally for both frames of reference. Thus, regardless of whether the motion is considered from the muon’s or the Earth’s frame of reference, the muons can reach the surface.
Exemplar Answer
π-: One up quark and one down antiquark
π+: One down quark and one up antiquark
π0: One up quark and one up antiquark
Marker’s Notes
Whilst π0 can also be composed of one down quark and one anti-down quark, this is not acceptable as an answer as the question states that only rearrangement of quarks occur in this process. The reactants are a proton and an antiproton which is composed of 2 up quarks, 2 up antiquarks, 1 down quark and 1 down antiquark. Taking away the quarks that compose π- and π+ the only left over particles is one up quark and one up antiquark which forms the π0.
Exemplar Answer
E = ∆mc2
∆m = mproducts - mreactants
= (140 + 140 + 140) - (940 + 940)
= -1460 MeV/c2
Etotal = -1460 MeV = 1460 MeV released
Epion = Etotal / 3
= 1460 / 3
= 490 MeV (2 s.f.)
Exemplar Answer
The prediction of the speed exceeding 3 x 108ms-1 made by classical physics arises when the energy released is directly equated to the classical formula for kinetic energy (KE = ½mv2). This is a problem as that is faster than the speed of light which violates Einstein’s Theory of Special Relativity which predicts that no objects with mass can travel at or faster than the speed of light.
This prediction can be resolved by applying the more accurate model of special relativity to determine the velocity of the pions. This can be achieved by equating the energy released to the difference between Etotal = m0c2 / √(1 - v2 / c2) and Erest = m0c2, KE = Etotal- Erest. Solving for v, will arrive at a more accurate prediction for the velocity of the pions which will be less than 3 x 108 ms-1.
Exemplar Answer
Fnet = ma = qE
a = qE / m
= (1.602 x 10-19)(1.5 x 104) / (9.109 x 10-31)
= 2.638 x 1015
= 2.6 x 1015 ms-2 (2 s.f.)
Exemplar Answer
sx = uxt
t = sx / ux = 0.005 / (2.0 x 106) = 2.5 x 10-9 s
sy = uyt + ½ayt2
= (0)(2.5 x 10-9) + ½(2.6 x 1025)(2.5 x 10-9)2
= 0.008125 m
= 8.1 mm (2 s.f.)
Exemplar Answer
Finding the direction of the electron beam as it exits the plates:
vy = uy + ayt = 0 + (2.6 x 1015)(2.5 x 10-9) = 6.5 x 106 ms-1
vx = 2.0 x 106 ms-1
tanθ = vy / vx
= (6.5 x 106) / (2.0 x 106)
= 3.25
Finding the vertical displacement after the plates:
tanθ = sy / sx = 3.25
sy = sxtanθ
= 30 x 3.25
= 97.5 mm
Finding total vertical displacement:
stotal = 8.1 (from part (b)) + 97.5 (from above)
= 105.6
= 110 mm (2 s.f.)
Exemplar Answer
Marker’s Notes
There are multiple ways this can be drawn, as long as it achieves the following goals:
Exemplar Answer
Rod A has a greater current than Rod B as a result of its lower resistance by V = IR (they have the same voltage across them due to being connected in parallel). However, this statement is inaccurate as the larger current in rod A does not cause a larger force on rod B, in fact the forces on both rods are equal in magnitude.
Each rod creates a field that the other rod is inside which results in a repulsive force between them by F / ℓ = (µ0 / 2π) * (I1I2 / r). However, as seen in the equation, and in accordance with Newton’s Third Law, the force is the same on either rod. The reason rod B displaces more than rod A, must be due to its lower mass, due to being different materials. By Newton’s Second Law, Fnet = ma ⇒ a = Fnet / m, a lower mass will result in a larger acceleration and subsequent displacement despite having the same force acting on it.
Thus, in contrast to the statement, the differing displacements of the rods in Position 2 is not a result of differing forces but rather of differing masses. Thus, this statement is inaccurate.
Exemplar Answer
Ignoring friction, there are three forces acting on the object: the weight force, the normal force, and a force from the walls of the cylinder. If the period of the cylinder’s rotation is halved, below is the effect on the forces:
Weight force:
This comes from the gravitational field of the Earth, and can be given by Fg = mg. However, the period of rotation does not affect m, the mass of the object, nor does it affect g, the gravitational field strength. Thus, there will be no change to the weight force.
Normal force:
The normal force acts upwards perpendicularly away from the ground to prevent the object from sinking into the surface. The rotation of the object is only in the horizontal plane, and thus changing the rotational speed will not affect the vertical motion. Since there is no vertical motion, the net vertical force is zero. As the weight force does not change, there will also be no change to the normal force which balances the weight force.
Force from the walls:
The force from the walls acts perpendicularly away from the walls towards the centre of the circle. This is the force that supplies the centripetal force Fc = mv2 / r which keeps the object moving in uniform circular motion. If the period of rotation is halved, the speed is doubled by v = 2πr / T. Thus, Fc will increase by a factor of 4 since the v term is squared. Assuming the walls can sustain such a force, the force from the walls will thus increase by a factor of 4.
Exemplar Answer
Model A:
Model A assumes a uniform field strength and hence the application of U = mgh. The initial mechanical energy and the final mechanical energy must be the same by the Law of Conservation of Energy:
Ei = Ef
⇒ KEi + Ui = KEf + Uf
⇒ ½mu2 + 0 = ½mv2 + mgh (defining ground as U = 0)
Initially, the projectile has its highest kinetic energy. At the maximum height h = hmax, all of the kinetic energy has been converted into gravitational potential energy (due to the negative work done by gravity) and thus the final vertical velocity is zero v = 0. Thus:
½mu2 = mghmax
⇒ hmax = u2 / 2g
Model B:
Similarly to Model A, the initial mechanical energy and the final mechanical energy must be the same by the Law of Conservation of Energy:
Ei = Ef
⇒ KEi + Ui = KEf + Uf
Whilst KE = ½mv2 is still true for model B, unlike model A, U = mgh cannot be used to express gravitational potential energy. This is because mgh assumes a constant gravitational field strength. Instead, U = -GMm / r must be used:
½mu2 - GMm / rE = ½mv2 - GMm / (rE + h)
At the maximum height h = hmax, similarly to model A, all of the kinetic energy would have converted into gravitational potential energy and thus the final vertical velocity is zero v = 0. Thus:
½mu2 - GMm / rE = -GMm / (rE + hmax)
⇒ ½u2 = GM / rE = -GM / (rE + hmax)
Solving for hmax will arrive at the maximum height of the projectile. This will be a greater value compared to model A for the same initial velocity. This is because the gravitational field strength decreases by the inverse square law g = GM / r2 and thus, a greater height would correspond to the same amount of initial kinetic energy converted into gravitational potential energy..
Marker’s Notes
Projectile motion analysis can also be used for model A:
vy2 = uy2 + 2aysy
At the maximum height sy = hmax, the final vertical velocity is zero vy = 0. Thus:
0 = uy2 + 2ayhmax
⇒ hmax = -uy2 / 2ay
Defining up as positive, ay = -g where g = 9.8 ms-2. Thus:
hmax = uy2 / 2g
However, students must also describe the energy changes of the projectile. Projectile motion analysis cannot be used for model B as a key assumption in projectile motion is that the acceleration due to gravity is constant.
Exemplar Answer
Throughout history, many experiments have profoundly advanced our understanding of the interaction between light and matter, uncovering the complex, dual nature of light and matter and laying the foundations of quantum mechanics.
Thomas Young’s Double Slit Experiment
One significant experiment is Thomas Young’s double-slit experiment, conducted in 1801. In this experiment, Young passed light through two narrow, closely spaced slits, projecting it onto a screen. Instead of producing two distinct bright spots as would be expected if light interacted with the slits as a particle as Isaac Newton had proposed, the result was an interference pattern of alternating bright and dark bands, characteristic of wave behaviour. This observation was significant as it demonstrated that light could undergo diffraction around the slits and exhibit constructive and destructive interference, which were wave behaviours. Young’s experiment provided strong evidence for the wave theory of light, challenging Newton’s earlier particle theory (based on matter) and supporting Huygens’ wave model. This discovery laid the groundwork for the classical wave model of light.
Black Body Radiation Experiments
As the 20th century approached, scientists were studying black body radiation, the emission of light from matter as a result of its temperature. They encountered discrepancies, where classical physics would predict that the intensity of radiation would approach infinity as wavelengths approached zero which violated the Law of Conservation of Energy and did not agree with the experimental results. This discrepancy was coined the “Ultraviolet Catastrophe”. In 1900, Max Planck proposed a radical solution by mathematically treating the energy to be quantised by E = hf. This notion of quantised energy exchange between light and matter marked a shift from classical ideas, as it implied that energy is discrete rather than continuous. Planck’s work was the first step towards quantum theory, introducing the concept of quantisation that would later prove essential in explaining atomic interactions with light.
Photoelectric Effect Experiments
The next major contribution came in 1905, when Albert Einstein built on Planck’s work to explain the photoelectric effect. In this phenomenon, light shining on a metal surface causes electrons to be ejected, a process that classical wave theory could not adequately explain. This highlighted a key interaction between light and matter, where light was not merely produced by matter (Maxwell’s predictions), but light could release matter particles from a metal surface. According to classical predictions, the energy of the emitted electrons should depend on the intensity (brightness) of the light. However, experiments by Philipp von Lenard showed that if the light’s frequency was above a certain threshold, electrons were emitted regardless of intensity, while below this threshold, no electrons were emitted, even at high intensities. Furthermore, the kinetic energy of the ejected electrons were found to be related to the frequency of light rather than its intensity as classical physics had predicted. Einstein proposed that light consists of discrete packets, or photons, each carrying an energy E = hf. This experiment analysing the interaction of light with matter transformed our understanding of physics as it provided direct evidence of light’s particle-like behaviour, challenging the classical wave model (Previously, Maxwell had modelled light as a continuous electromagnetic wave, produced by accelerating charged matter). This ultimately shifted our understanding of light as a wave-particle duality, where light can be modelled as a wave or a particle depending on the context.
Atomic Emission Spectra Experiments
Further evidence of light’s interaction with matter emerged from the study of atomic emission spectra. When elements are excited, they emit light at specific, discrete frequencies, producing a unique line spectrum for each element. One such spectrum was the Balmer series which is the emission spectrum of hydrogen. Classical physics could not explain why these emissions appeared as distinct lines rather than a continuous spectrum. Rydberg derived an empirical equation to relate the wavelengths of these lines, but could not explain the theory behind it. In 1913, Niels Bohr addressed this by proposing a model of the atom in which electrons exist in stationary states with quantised energy levels. According to Bohr’s model, when electrons transition between these levels, they emit or absorb photons with energy precisely equal to the difference between the initial and final energy states, resulting in the emission of light of specific frequencies. This model not only explained the emission spectra but also linked the behaviour of light to the internal structure of the atom. Bohr’s theory of quantised energy levels provided an understanding of atomic structure that integrated light-matter interactions, reinforcing the notion that energy exchange occurs in discrete units and further validating the quantum framework.
These experiments collectively advanced our understanding of the interaction between light and matter. Together, they reveal that light exhibits both wave and particle properties, depending on the conditions, and interacts with matter in quantised ways.
Marker’s Notes
Experiments that have explored the interaction between light and matter may include:
An answer could include some of the above or even experiments that are not mentioned here. However, as the prompt says “many scientists” it would make sense to name experiments where we may have learnt the specific scientist who performed them. Likewise, “contributed to our understanding of physics” is a broad application where we can extend the impact of these experiments beyond the immediate understanding of what light and matter are and how they interact.
Exemplar Answer
1. Initial position of the magnet
Consider the initial position of the magnet on the left hand side, with a North pole facing towards the centre. As a permanent magnet produces a magnetic field directed out of North and into South, a magnetic field directed towards the bottom right will thread through the portion of the can’s top left area.
At its highest point on the left hand side, the magnet’s motion is momentarily stationary. At this highest point, the gravitational potential energy (GPE) of the magnet is maximum (U = mgh), while the kinetic energy is 0 (KE = ½mv2). Tension force and gravitational force act on the magnet, producing a net force perpendicular to the string. Hence the magnet begins to accelerate to the right hand side in a circular path, with a component of tension force supplying a centripetal force (Fc = mv2 / r) and a component of weight force producing a net torque. Meanwhile, the can is initially stationary, and is not rotating and thus has zero kinetic energy.
2. As the magnet approaches the can
As the magnet gains linear and angular velocity, following the circular path traced by the string, the work done by gravity is positive (W = F||s, as it acts in the same direction as displacement). Hence gravitational potential energy is converted into kinetic energy. As gravity is a conservative force, the total mechanical energy of the magnet initially remains constant by the Law of Conservation of Energy.
However, as the magnet moves to the right, its distance to the aluminium can decreases. As the magnetic field strength around a permanent magnet is stronger closer to its poles, the decreasing distance between the magnet and can will lead to the area of the can experiencing an increase in flux (Φ = BA). By Faraday’s Law, this change in flux will induce an emf (ε = -NΔ𝜙 / Δt), which will be directed to oppose the initial change in flux by Lenz’s Law. As the aluminium can is a conductive area, the induced emf leads to the formation of eddy currents. By the Right Hand Grip Rule, the eddy currents flow anticlockwise on the surface of the can (as viewed from the top).
As the eddy currents flow, they produce their own magnetic field by Ampere’s Law (B = μ0I / 2𝜋r). Hence a North pole will be induced above the top left corner of the aluminium can. As like poles repel, the can will thus experience a force towards the right hand side, producing a net torque (𝝉 = r⟂F) that causes the can to begin rotating clockwise about its fixed axis of rotation (as viewed from the front). By Newton’s 3rd Law, the magnet will also experience an equal and opposite force towards the left, opposing a component of its original net force. Thus the magnet will lose speed as it approaches the can.
Overall, the electromagnetic induction between the can and magnet converts kinetic energy from the magnet’s motion into electrical potential energy of the induced emf in the can then into heat produced by the eddy currents. The resultant force leads to a net increase in the overall mechanical energy of the can as it begins moving and gains kinetic energy, while it leads to a net decrease in the overall mechanical energy of the magnet. This interaction continues as the magnet approaches the central vertical position, with its swinging motion opposed by the induced force from the can, and the can gaining clockwise rotational velocity.
3. As the magnet swings past the can
As the magnet swings past the vertical central position, the change in flux through the can at that moment is zero. Hence no emf nor eddy currents are induced. As the magnet continues past the central position its motion begins to swing upwards to the right-hand side. However, as weight force continues to act vertically down, the component of gravity now does negative work on the magnet. Hence the magnet’s kinetic energy and velocity decreases as it rises up, gaining height. By a similar logic to the aforementioned explanation of electromagnetic induction, a force towards the right continues to act on the can (anticlockwise eddy currents opposing the change in flux experienced by the top right area of the can, as the magnet moves away and magnetic field strength decreases). Again, a North pole is induced above the top right of the can, attracting the South pole of the magnet as it moves away. Hence the net torque on the can continues to act in the same direction as before, and the can’s clockwise rotational velocity further increases. By Newton’s 3rd Law, an equal and opposite force acts on the magnet, hence towards the left. In addition to gravity, this induced magnetic force increases the deceleration of the magnet. As the can reaches its fastest motion, rotating clockwise, the magnet reaches its highest height on the right hand side.
4. As the magnet is left to swing
However, due to the conversion of the magnet’s initial mechanical energy into the can’s rotational kinetic energy (via electromagnetic induction) and then heat, the magnet’s total mechanical energy continually decreases and it reaches a lower height than when it started on the left hand side (Etotal = U + KE). As the magnet begins swinging back to the left hand side, the net torque on the can is anticlockwise, opposing its clockwise rotation. Hence as the magnet swings back to the left, the rotational velocity of the can decreases, although it does not stop completely. Hence the magnet continues to swing back and forth, however it reaches a lower and lower maximum height until it eventually comes to a stop. Likewise, the can rotates clockwise with its velocity alternating between slow and decreasing maximum values. Eventually, the magnet and the can's motion both approach stationary. This can be explained as the overall mechanical energy of the system decreases to zero, as the induced eddy currents dissipate energy into the surrounding environment in the form of heat and power loss Ploss = I2R.
Marker’s Notes
Note that whether an object in motion merely slows down or completely stops is dependent on the strength of the force produced by the eddy currents.
A link to the work done by the forces is likely unnecessary, but has been included for complete detail.
Some students may have interpreted the magnet’s swinging motion, and automatically concluded that the can must oscillate the same. This is understandable, however a closer analysis of the provided scenario shows that this is inaccurate.
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