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St Josephs College Geelong 2021 physics lecture

Andrew SmithFolgen

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- Physics 2021
- Any questions smitha@ignatius.vic.edu.au To view presentation: https://www.slideshare.net/secret/cVUaf5V3uyDYsf
- The examiners aren’t set out to trick you. Make sure that you are familiar with the setting and layout of the 2019 exam. This will be more useful than 2020. Also 2021 NHT exam is a good guide and available now.
- No marks are awarded for writing a correct formula. You need to show correct substitution into the formula. NO formula or NO substitution equals NO working marks. However, is there is some attempt at working out and the answer is correct you will generally get full marks. No marks have been taken for incorrect decimal places or significant figures in the recent past. However, the practical questions will require correct uncertainty responses.
- Some general hints: Approximately 30% will be written responses. Pay attention to how many marks are awarded b. Use dot points in written response: ie. 2 clear points two marks: role and operation: The best responses are concise and address the question. Try to link your ideas together. • The role of the commutator is to keep the motor spinning in the same direction. • It does this by reversing the direction of current in the coil every 180o. • When the orientation of the coil is vertical.
- Study score 30 C+ - B 80/130 62% Study score 40 A+ 114/130 88%
- Practice exams VCAA exams back to about 2000 with solutions (The best resource) iTute Great resource for past exams detailed solutions. Their exams tend to be difficult.
- Unit 3: How do fields explain motion and electricity? Outcome 1 On completion of this unit the student should be able to analyse gravitational, electric and magnetic fields, and use these to explain the operation of motors and particle accelerators and the orbits of satellites. Key knowledge Fields and interactions • describe gravitation, magnetism and electricity using a field model. • investigate and compare theoretically and practically gravitational, magnetic and electric fields, including directions and shapes of fields, attractive and repulsive fields, and the existence of dipoles and monopoles
- • investigate and compare theoretically and practically gravitational fields and electrical fields about a point mass or charge (positive or negative) with reference to: – the direction of the field – the shape of the field – the use of the inverse square law to determine the magnitude of the field – potential energy changes (qualitative) associated with a point mass or charge moving in the field
- Field Lines never touch and never cross These are monopole fields Gravity is missing but it can only be attraction
- Attraction Repulsion Dipole negative charges will be similar with arrows going towards the charges.
- Note that Magnetism is an example of a dipole force field. There are NO monopole magnetic fields.
- Question 1 A. B. C. D. Must be showing repulsive forces (like charges) Field lines are away from the sources Correct C
- r i t e i n t h i s a r e a Question 2 X Y Z A. B. C. D. Question 3 G × 10–11 2 –2 A. –3 B. –5 C. X can be B Y can be E or B or G Z can be E or B only C only option
- You need to realise the assump1on that H has one proton and that the electron is also an elementary par1cle For some reason many students forgot to square the radius in the denominator of the frac1on.
- The radius is the orbital radius of the atom as given in part a. This connec1on needs to be made. Also circular mo1on. Think about what the electron would look like! It is travelling about 10,000 1mes the speed of a jet aircraF and is circling a H atom. It would just be a complete blur and you would never really be able to locate its posi1on. This is what the quantum world states the electron would look like.
- Question 3 G × 10–11 2 –2 A. –3 B. –5 C. D. Question 4 X A. B. F = GMm r2 F = 6.67 ×10−11 ( ) 1 ( ) 100 ( ) 0.1 ( )2 F = 6.67 ×10−7 N C do not use cm as a measurement
- Question 5 × 10–8 –9 k 9 A. B. C. D. –5 Question 6 A. B. C. D. Question 7 F = kQ1Q2 r2 F = 9 ×109 ( ) 1×10−8 ( ) 1×10−9 ( ) 0.3 ( )2 F = 1.0 ×10−6 N B do not use cm as a measurement
- Question 1 23 G –11 2 –2 a. –2 g = GM r2 g = 6.67 ×10−11 ( ) 6.4 ×1023 ( ) 3.4 ×106 ( ) 2 g = 3.69 N / kg
- b. U EK energy (J) height (m) 10 0 0 1 2 3 4 5 6 7 8 9 8 7 6 5 4 3 2 1 This question is different to others in the past: Energy v Height graphs are usually satellites, so the H is large and the graph is hyperbolic in shape. In this case H is small (10m) and the graph will be approx. linear i.e. E=mgh, m and g are constant. NOTE: E = mgh = 20 x 3.7 x 10 = 740J
- EK energy (J) height (m) 10 0 0 1 2 3 4 5 6 7 8 9 8 7 6 5 4 3 2 1 J x100 Gravitational Potential Energy Kinetic Energy Energy v Height on mars surface
- This type of question students need to: • adjust the scale of the axis given. • provide a key for the graphs. • Give a heading for the graph
- • investigate and apply theoretically and practically a vector field model to magnetic phenomena, including shapes and directions of fields produced by bar magnets, and by current-carrying wires, loops and solenoids • identify fields as static or changing, and as uniform or non-uniform. d o n o t w r i t e i G × 10–11 2 –2 A. –3 B. –5 C. D. Question 4 X A. B. C. D. Use Right Hand Grip rule. B is into the coil at X. Correct response is A
- Effects of fields • analyse the use of an electric field to accelerate a charge, including: – electric field and electric force concepts: and – potential energy changes in a uniform electric eld: W = qV, – the magnitude of the force on a charged particle due to a uniform electric eld: F = qE E = k Q r2 F = k q1 ⋅q2 r2 E = V d
- • analyse the use of a magnetic field to change the path of a charged particle, including: – the magnitude and direction of the force applied to an electron beam by a magnetic field: F = qvB, in cases where the directions of v and B are perpendicular or parallel – the radius of the path followed by a low-velocity electron in a magnetic field: qvB = m⋅v2 r • model the acceleration of particles in a particle accelerator (limited to linear acceleration by a uniform electric field and direction change by a uniform magnetic field).
- In General: Electrons fired into B Fields will have a perpendicular force towards the centre of the path. (circular motion) Electrons travelling through an E field will have constant speed in one direction and perpendicular force in another. (parabolic / projectile motion)
- Question 2 Data –31 –19 region of uniform magnetic field plate plate 10 000 V 0.10 m Figure 1
- region of uniform magnetic field plate plate 10 000 V 0.10 m Figure 1
- a. b. –1 E = V d E = 10,000 0.1 = 100,000 V / m or N /C E = 10,000 eV E = 1.6 ×10−15 J v = 2E m = 3.2 ×10−15 9.1×10−31 = 5.93×107 m / s E must be in Joules and NOT eV.
- The Magnetic force acting on a wire carrying a current F = B⋅ I ⋅l Writing I as charge per second F = B⋅ Q t ⋅l Change position of t as being under length F = B⋅Q⋅ l t Length divided by time is speed F = B⋅Q⋅v Force on a moving charge in a B field
- t h i s a r e a c. d. Bqv = mv2 r r = m⋅v B⋅q r = 9.1×10−31 ( )⋅ 5.93×107 ( ) 0.02 ( )⋅ 1.6 ×10−19 ( ) = 0.017 m The magnetic force on the electrons is equal to the centrifugal force acting on them.
- d. Figure 2a Figure 2b In the electric field the deflection will be from negative towards positive. The shape will be parabolic in nature. a ↓ v →, parabolic In B field the direction of Force is always perpendicular to the velocity. The shape will be circular. RH slap rule. Note Direction is - current.
- • For circular motion. • Force is perpendicular to the direction of motion. • Force MUST be constant magnitude.
- Application of field concepts • apply the concepts of force due to gravity, Fg, and normal reaction force, FN, including satellites in orbit where the orbits are assumed to be uniform and circular • model satellite motion (artificial, Moon, planet) as uniform circular orbital motion: a = v2 r = 4π2 r T 2
- • analyse the use of gravitational fields to accelerate mass, including: – gravitational eld and gravitational force concepts: and – potential energy changes in a uniform gravitational eld: Eg = mgΔh – the change in gravitational potential energy from area under a force-distance graph and area under a field- distance graph multiplied by mass. F = G m1 ⋅m2 r2 g = G M r2
- For Gravitational Motion write these formulae on your Cheat Sheet DO NOT try to manipulate the equations in the exam. v = GM r r = GM ⋅T 2 4π2 3 T = 4π2 ⋅r3 GM Use T is seconds not minutes, hours or days Use v in m/s not km/h Know how your calculator uses scientific notation
- This is oFen confused with Force, the ques1on is asking for Field Strength. The units are N/kg and on earth it is equivalent to the accelera1on.
- You should have a formula sheet that will give the correct formula. Trying to derive the correct formulas under the pressure of the exam is a possible disaster. Students oFen transpose these equa1ons incorrectly or make mistakes like forgeKng to cube the radius or such.
- The small m (mass of satellite) does not appear in any of the equa1ons for period, radius or velocity of the satellite around a body. As such satellites with diﬀerent mass will travel at exactly the same period and velocity if they are in the same orbit. Melissa is the only correct student. For 3 marks you should approach the ques1on as: Rick is incorrect because….. Melisa is correct because… Nam is incorrect because….. This follows the instruc1on to “Evaluate these three opinions”.
- • describe the interaction of two fields, allowing that electric charges, magnetic poles and current carrying conductors can either attract or repel, whereas masses only attract each other • investigate and analyse theoretically and practically the force on a current carrying conductor due to an external magnetic eld, F = nIlB , where the directions of I and B are either perpendicular or parallel to each other • investigate and analyse theoretically and practically the operation of simple DC motors consisting of one coil, containing a number of loops of wire, which is free to rotate about an axis in a uniform magnetic eld and including the use of a split ring commutator
- Question 3 10–2 A + – F G E H N S 6.0 V Figure 3 a. Common Question ZERO I is parallel to B.
- A Commutator: -Enables the direction of the current in the loop to be reversed. -Every half turn or 180o. -This reverses the direction of the force on the sides of the motor loop. *To keep the motor spinning in the same direction. S b.
- Area of Study 2 How are fields used to move electrical energy? The production, distribution and use of electricity has had a major impact on human lifestyles. In this area of study students use empirical evidence and models of electric, magnetic and electromagnetic effects to explain how electricity is produced and delivered to homes. They explore magnetic fields and the transformer as critical to the performance of electrical distribution systems.
- Generation of electricity • calculate magnetic flux when the magnetic field is perpendicular to the area, and describe the qualitative effect of differing angles between the area and the field: ΦB = B⊥ A • investigate and analyse theoretically and practically the generation of electromotive force (emf) including AC voltage and calculations using induced emf: , with reference to: – rate of change of magnetic flux – number of loops through which the flux passes – direction of induced emf in a coil Ε = −N ΔΦ Δt • explain the production of DC voltage in DC generators and AC voltage in alternators, including the use of split ring commutators and slip rings respectively.
- This is an easier ques1on than where the frequency is given. The loop will make 4 area changes in one complete period. In this ques1on the t will be 1/16 of a second for a quarter turn. This is the main diﬃculty with this ques1on.
- The area is not required in this ques1on as the magne1c ﬂux is given and NOT the magne1c ﬁeld strength. This is an example when reading the ques1on carefully is important. A li]le bit of a trick ques1on but this happens all of the 1me. This is a good example of how the exam ques1ons can just change a li]le and confuse students from year to year. Read the ques1ons carefully and take in all the informa1on.
- Question 4 N S oscilloscope Figure 4 2 a.
- There are tricks to this question: -The area changes x4 per rotation so while f=10 Hz the time for the area will be 1/40 sec ξ = n⋅ B⋅ΔA Δt ξ = 20⋅ 0.5⋅0.02 1/ 40 = 8.0 V
- SECTION B – conti b. time voltage output 0 The period will get larger and larger while the amplitude gets smaller and smaller.
- Transmission of electricity • compare sinusoidal AC voltages produced as a result of the uniform rotation of loop in a constant magnetic field with reference to frequency, period, amplitude, peak-to-peak voltage (Vp–p ) and peak-to-peak current (Ip–p ) • compare alternating voltage expressed as the root-mean-square (rms) to a constant DC voltage developing the same power in a resistive component • convert between rms, peak and peak-to-peak values of voltage and current
- Question 5 Figure 5 a. Vpeak = 30 V VRMS = 21.2 V
- b. f = 1/T T= 80 ms = 0.08 sec f = 12.5 Hz
- • analyse transformer action with reference to electromagnetic induction for an ideal transformer: • analyse the supply of power by considering transmission losses across transmission lines • identify the advantage of the use of AC power as a domestic power supply. NP NS = VP VS = IS IP
- This is the deriva1ve of the graph shown. There is a nega1ve sign but in general the nega1ve that represents the concepts of Len’s law and can be leF out. The only possible answer is D. In the past, when drawing EMF graphs that have a posi1ve and nega1ve constant value it has not ma]ered which is posi1ve and nega1ve, only that there is a dis1nct diﬀerence.
- Question 10 250 VRMS AC to town transformer primary coils secondary coils alternator 20 km power plant transmission line 50000 VRMS AC Figure 11 a.
- b. V Ploss = I2 R Ploss = 152 × 40 = 25,000 W VOUT = VIN − VDROP VOUT = 50,000 − 15 × 40 VOUT = 49,400 V
- c. To transmit Power (P=VI) If V ↑ , then I ↓ For the same amount of P. Power loss = I2 R, therefore at HIGH V and LOW I In order to change the voltage in the power lines transformers are needed. Transformers required a △ Flux to operate. This can only be supplied by AC and not DC. Voltage is stepped UP at transmission and stepped DOWN closer to the customer.
- Part A is fairly straight forward.
- Part B This is a ra1o problem that strong students could possibly do with no working out. -If the voltage is halved, the current is doubled to keep the power the same. -As Ploss = I2R, doubling current will increase power loss by x4. Just by wri1ng x4 with no working or jus1ﬁca1on (1 mark).
- . Area of Study 3 How fast can things go? In this area of study students use Newton’s laws of motion to analyse relative motion, circular motion and projectile motion. Newton’s laws of motion give important insights into a range of motion both on Earth and beyond. At very high speeds, however, these laws are insufficient to model motion and Einstein’s theory of special relativity provides a better model. Students compare Newton’s and Einstein’s explanations of motion and evaluate the circumstances in which they can be applied. They explore the relationships between force, energy and mass.
- • investigate and analyse theoretically and practically the uniform circular motion of an object moving in a horizontal plane: , including: – a vehicle moving around a circular road – a vehicle moving around a banked track – an object on the end of a string FNET = m⋅v2 r • investigate and apply theoretically Newton’s second law to circular motion in a vertical plane (forces at the highest and lowest positions only)
- Question 12 –1 L Figure 14a 4.00 m
- L Figure 14b a. b. The arrow must be pointing toward the centre of the circle as it is travelling in a circle and the FNET is always towards the centre.
- SECTION B – continued a. b. F = mv2 r ∑ ⇒ FN − mg = mv2 r ⇒ FN = mg + mv2 r ⇒ FN = 5⋅10 + 5⋅52 4 = 81.3 N
- • investigate and analyse theoretically and practically the motion of projectiles near Earth’s surface, including a qualitative description of the effects of air resistance No calculations about air resistance: -Path will no longer be parabolic/ symmetrical. -Air resistance will reduce the range and max. height of projection.
- Question 13 –1 15 m 20 m s–1 30° cliff sea Figure 15 vertical component only u = 20sin30 = 10ms−1 v = − a = −10 ms−2 s = −15 m t = ? s = ut + 1 2 at2 −15 = 10t − 5t2 t2 − 2t − 3 = 0 t = 3.0 sec
- • investigate and apply theoretically and practically the laws of energy and momentum conservation in isolated systems in one dimension. Key knowledge Newton’s laws of motion • investigate and apply theoretically and practically Newton’s three laws of motion in situations where two or more coplanar forces act along a straight line and in two dimensions
- Relationships between force, energy and mass • investigate and analyse theoretically and practically impulse in an isolated system for collisions between objects moving in a straight line: FΔt = mΔv • investigate and apply theoretically and practically the concept of work done by a constant force using: – work done = constant force × distance moved in direction of net force – work done = area under force-distance graph
- • analyse transformations of energy between kinetic energy, strain potential energy, gravitational potential energy and energy dissipated to the environment (considered as a combination of heat, sound and deformation of material): – kinetic energy at low speeds: ; elastic and inelastic collisions with reference to conservation of kinetic energy – strain potential energy: area under force-distance graph including ideal springs obeying Hooke’s Law: – gravitational potential energy: Eg = mgΔh or from area under a force-distance graph and area under a eld- distance graph multiplied by mass E = 1 2 m⋅v2 E = 1 2 k ⋅ x2
- Make sure that you read the ques1on carefully and you are not just looking at the picture and trying to ﬁgure out what the ques1ons states from that. The “car comes to rest when spring is compressed to 0.5 m” statement is cri1cal. Q12. k = gradient of the graph (400 N/m)
- Q13. E= ½ k x2 or the area under the graph = 50 J. It was a common error to give the area under the en1re graph which is 200 J.
- Question 11 –1 –1 A 6.0 kg Before After B 2.0 kg u = 2.0 m s–1 u = 0 m s–1 A 6.0 kg B 2.0 kg v = 1.0 m s–1 v = ? m s–1 Figure 12 a. –1 pbefore = pafter 12 = 6 + 2v v = 3 ms−1
- SECTION B – Question 11 b. KEbefore = 1 2 mv2 = 1 2 6 2 ( )2 = 12J KEafter = 1 2 mv2 + 1 2 mv = 1 2 6 1 ( )2 + 1 2 2 3 ( )2 = 3 + 9 = 12 J Therefore it is Elastic. In general if object stick together INELASTIC You must follow direction to show working.
- w r i t e i n t h i s a r e a c. t Figure 13a t Figure 13b Pat is correct. Some kinetic energy stored as elastic energy while the spring is deformed.
- Question 14 k 70 60 50 40 unstretched spring ruler (cm) stretched spring masses added 30 20 10 0 Figure 16 Number of 0.50 kg masses Length of spring 0
- Number of 0.50 kg masses Length of spring 0 1 2 3 5 The readings are approximately to the nearest cm, by the 53, 57 and 63 cm reading. Therefore ± 0.5 cm
- a. Error bars of approx 2.5 cm each way. You must follow the instructions:
- SECTION B – Question 14 TURN OVER b. part a. k –1 There would be a range of correct answers, but you will need to justify your answer. ± 0.5 cm over approx. 50 cm ±1% 0.5 kg masses ± 20g ±4% Gravity ± 2% Total error 7% k = 1 Gradient k = 81 N / m ± 7%
- c. The 40 cm mark is 30cm above the rest position. When released the masses will go 30cm below the rest position. This will be 60cm below where they are released. However the question requires some working . The lost GPE when falling is gained as EPE. mgh = 1 2 k ⋅ x2 2.5 ×10 × h = 1 2 81⋅h2 h = 0.62 m
- The mistake in this ques1on is to confuse the point where the spring comes to permanent rest and the point where it is at rest momentarily. For permanent rest If you understand springs very well, the momentarily at rest point is double this. You can also show this through the conserva1on of energy. But you could also simply state the correct answer, if you have the equa1on from above correct.
- Conserva1on of Energy
- For the full 4 marks list the three posi1ons and the total energy. At top: GPE=max EPE=0 KE=0 At mid: GPE=some EPE=some KE=some At bo]om: GPE=0 EPE=max KE=0 Total E is always constant
- Question 12 A. B. C. D. Question 13 Transverse wave Longitudinal wave A. B. C. D. Question 14 The proper length of an object needs to be measured at rest to the observer. B The same reference frame could be moving at a constant velocity compared to the object. D is the distractor.
- . Einstein’s theory of special relativity • describe Einstein’s two postulates for his theory of special relativity that: – the laws of physics are the same in all inertial (non-accelerated) frames of reference – the speed of light has a constant value for all observers regardless of their motion or the motion of the source • compare Einstein’s theory of special relativity with the principles of classical physics • describe proper time (t0) as the time interval between two events in a reference frame where the two events occur at the same point in space
- • interpret Einstein’s prediction by showing that the total ‘mass-energy’ of an object is given by: Etot = Ek + E0 = γ mc2 where E0 = mc2, and where kinetic energy can be calculated by: Ek = (γ – 1)mc2 • describe how matter is converted to energy by nuclear fusion in the Sun, which leads to its mass decreasing and the emission of electromagnetic radiation.
- • describe proper length (L0) as the length that is measured in the frame of reference in which objects are at rest • model mathematically time dilation and length contraction at speeds approaching c using the equations: t = t0γ and L = L0 γ whereγ = 1− v2 c2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 2
- • explain why muons can reach Earth even though their half-lives would suggest that they should decay in the outer atmosphere. The muons are moving at great speed compared to the earth. The height of the upper atmosphere is smaller from the muons perspective and the time they take to reach the ground is longer than for a stationary observer on earth.
- Question 15 c muons top of mountain ground level 2627 m Figure 17
- SECTION B – Question 15 Figure 17 a. t0 = 2.2 ×10−6 s v = 0.995 c t = t0 1− v2 c2 = 2.2 ×10−6 1− 0.9952 = 2.2 ×10−5 Don’t use c = 3.0 x 108 . If measured in term of c they will cancel. Note: v = 0.995c has factor of x10 v = 0.99 c has factor of x 7 v =v 0.9 c has factor of x 2.3 Time Dilation will be greater
- b. c c. Length contraction will be smaller l0 = 2627 m v = 0.995 c l = l0 1− v2 c2 = 2627 10 = 263 m Note : use previous result
- c. According to classical physics (2627 m @ 3 x 108) the time will be 8.8 x 10-6 sec. (more than a half life 2.2 x10-6 sec) From the muons perspective they are travelling 263 m at 0.995 c. This will take 8.8 x 10-7 sec. (much less than half life not so many muons disintegrate) More strike the surface of earth than predicted by classical physics. Make sure that you compare the two.
- Unit 4: How can two contradictory models explain both light and matter? Outcome 1 On completion of this unit the student should be able to apply wave concepts to analyse, interpret and explain the behaviour of light.
- Key knowledge Properties of mechanical waves • explain a wave as the transmission of energy through a medium without the net transfer of matter • distinguish between transverse and longitudinal waves • identify the amplitude, wavelength, period and frequency of waves • calculate the wavelength, frequency, period and speed of travel of waves using: v = f λ = λ/T
- Waves transfer energy without transfer of mass. Transverse Waves The particles vibrate perpendicular to the motion. Think of T for perpendicular to help remember. Easy to draw Water waves are an easy example.
- Longitudinal Waves The particles vibrate parallel to the motion. Difficult to draw Sound waves Gases cannot support shear forces. The particles must BUMP into each other.
- Question 13 Transverse wave Longitudinal wave A. B. C. D. Question 14 A. B. C. D. A is Correct
- • investigate and analyse theoretically and practically constructive and destructive interference from two sources with reference to coherent waves and path difference: nλ and (n-1/2)λ respectively
- n is the number of the antinodal line from the centre of the pattern P is the point in question S1 and S2 are the sources of the waves d(PS1) is the distance from P to S1. S1 S2 trough crest A A A N N n = 0 n = 1 n = 1 n = 2 n = 2 central antinodal line PN PA Similaranalysisshowsthat,forapointonanodalline,thedifferenceindistance
- Nodal lines are lines where destructive interference occurs on a surface, resulting in no displacement of the surface. source, and vice versa. Any point on a minimum, because of the minimum dis Between the nodal lines are regions w The centres of these regions are called antinodal line, a crest from one sourc from the other source, or a trough from as a trough from the other source, and s sometimes called a local maximum, bec occurs there. When the two sources are in phase, a page, the interference pattern produced line. Note that any point on the central a each source. Since the sources produce two sources will arrive at any point on the Similar analysis will show that, for an either side of the centre of the pattern, exactly one wavelength further from o means that crests from one source stil although they were not produced at the Point PA is on the first antinodal line fro seen that PA is 4.5 wavelengths from S1 a A way to establish whether a point is a Antinodal lines are lines where constructive interference occurs on a surface. The path difference The path difference nλ n − 1 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ λ
- Question 18 S1 S2 P X laser double slits screen central bright band Figure 22 1 2 a.
- 1 2 a. nm b. 2 1 nm a. nm b. 2 1 nm X is the second dark fringe and this represents a path difference of 1.5 waves. One wave is 500 nm P is the second bright fringe and this represents a path difference of 2 waves. i.e. 1000 nm Both average 1.2 / 2 marks
- • explain qualitatively the Doppler effect If the ambulance is stationary both observers hear the same frequency. The person moving towards the ambulance (relatively) hears a higher frequency. The person moving away from the ambulance hears lower frequency. The Doppler Effect explains red shift and is evidence of an expanding universe.
- Moving towards Moving Away f = f0 ⋅vS vS − v ( ) f = f0 ⋅vS vS + v ( ) f0 is the stationary frequency vs is the speed of sound ( 340 ms-1)
- D. Question 14 A. B. C. D. The speed of waves in a particular medium is constant. Doppler Effect will explain the change in wavelength or distance between crests. D is correct.
- • explain resonance as the superposition of a travelling wave and its reflection, and with reference to a forced oscillation matching the natural frequency of vibration • analyse the formation of standing waves in strings fixed at one or both ends Fixed end reflection reverses the orientation of the wave
- Open end reflection does not reverse the orientation of the wave. Standing waves are due to superposition of waves that have the SAME wavelength.
- Resonance occurs when a particular wavelength matches the item generating the sound. This is a musical instrument.
- Resonance occurs in many objects. You may have seen images/film of Tacoma Narrows bridge. Build in the 1930’s before the engineers knew about resonance. Particular wind speeds matched the length of the bridge and caused the bridge to fail. Electrons shells or energy levels are another example of resonance.
- Superposition of waves and other examples of wave interference is evidence for the wavelike nature of something. In particular light. Only waves can produce an interference pattern. Both Transverse and Longitudinal waves can produce interference. Longitudinal are difficult to draw. Only particular frequencies can be supported by resonance. The term is the “Natural frequency of vibration.”
- Use the following information to answer Questions 15 and 16. ends of guitar string S Question 15 A. B. C. D. Question 16 immediately after Standing waves are formed by coherent waves. Travelling in opposite directions in a medium. Due to opposing sources or reflections. In 2017 exam a simple diagram like above was enough.
- Question 15 A. B. C. D. Question 16 immediately after A. B. C. D. Question 17 A. B. C. D. In this case the wave is a Transverse wave. The wave must be travelling in opposite directions. B is Correct
- SECTION A – continu Question 16 immediately after A. B. C. D. Question 17 A. B. C. D. The particles DO NOT move left or right only UP or DOWN. At the extreme position as shown. The particle is at it’s highest point. Only possible direction is DOWN. D is correct.
- • investigate and explain theoretically and practically diffraction as the directional spread of various frequencies with reference to different gap width or obstacle size, including the qualitative effect of changing the λ/w ratio.
- heard other than from directly in front of the speaker cone. Diffraction of water waves The diffraction of sound can be modelled with water waves in a ripple tank. The next figure shows the way water waves diffract in various situations. The diagrams apply equally well to the diffraction of sound waves. aves: round velength ct, hrough ngth , round nd ound the ier (c) shadow shadow shadow (a) (e) shadow (d) (b) (f) shadow e tank tion t summary ctice ns
- Some rules about diffraction: Longer waves diffract more than short waves. (This is the DOOF DOOF effect). If the wavelength is smaller than the gap width there will be minimal diffraction. Maximum diffraction is when the wavelength is equal to or larger than the gap width The width of a diffraction pattern is proportional to λ w
- 35 c. w y laser single slit screen w y Figure 23 w A.–D. A. y B. y C. y D. 1 2 y
- If the slit width get larger the diffraction width will be smaller. OR even better Spacing is proportional to D is correct λ w
- . Light as a wave • describe light as an electromagnetic wave which is produced by the acceleration of charges, which in turn produces changing electric fields and associated changing magnetic fields • identify that all electromagnetic waves travel at the same speed, c, in a vacuum • compare the wavelength and frequencies of different regions of the electromagnetic spectrum, including radio, microwave, infrared, visible, ultraviolet, x-ray and gamma, and identify the distinct uses each has in society
- In 1865 James Clerk Maxwell proposed mathematically that light is a combination of electric and magnetic waves self propagating each other. He used maths to suggest the speed was 3.0 x108 ms-1 A number of scientists including Foucault and Michelson-Morley have since verified the speed of light.
- Question 8 Longest Shortest Radio micro Infra-red visible UV X-ray
- • explain polarisation of visible light and its relation to a transverse wave model Polarising filters have a lens structure that only allows light of one orientation to pass, either vertically or horizontally.
- Polarising filters are evidence of the transverse nature of electro-magnetic radiation.
- • investigate and analyse theoretically and practically the behaviour of waves including: – refraction using Snell’s Law: n1 sin(θ1) = n2 sin(θ2) and n1v1 = n2v2 – total internal reflection and critical angle including applications: n1 sin(θ) = n2 sin(90°)
- More generally this would be expressed as follows: n1 sin θ1 = n2 sin θ2. f he light use the he ray tion. medium 1 refractive index n1 medium 2 refractive index n2 boundary normal θ2 θ1 n1 sin θ1 = n2 sin θ2 Sample problem 10.1 A ray of light strikes a glass block of refractive index 1.45 at an angle of inci- dence of 30°. What is the angle of refraction? nair = 1.0, θair = 30°, nglass = 1.45, θglass = ? 1.0 × sin 30° = 1.45 × sin θglass (substitute values into Snell’s Law) sin sin 30 θ = ° olution: Angles of incidence and refraction are always measured from the NORMAL. Light is bent at the boundary as the light changes velocity Slowing down light bends towards the normal. Speeding up light bends away from the normal.
- The refractive index of a material is related to the speed of light in the material. n1 = c v1 If the refractive index of a material is 1.5 (water), then the speed of light in the material is: n1 = c v1 1.5 = c vw vw = 2 3 c = 2.0 ×108 m / s
- Total internal reflection ONLY occurs going from slower medium to faster medium i.e. water into air. For total internal reflection Ør = 90o. θc = sin−1 n2 n1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ If going into air (which is common) θc = sin−1 1 n1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ If ø is greater than øc then the light will have Total Internal Reflection.
- Question 7 a. 45° 25° glass slab air n = 1.00 laser Figure 8 n1 sinθi = n2 sinθr 1⋅sin45 = n2 sin25 n2 = 1.67
- r i t e i n t h i s a r e a incoming ray air n = 1.00 water n = 1.33 Figure 9 b. c. i θc = sin−1 1 n1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ θc = sin−1 1 1.33 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ θc = 48.60
- c. i incoming ray air n = 1.00 water n = 1.33 ic The incoming ray has angle of incidence greater than critical angle. Therefore, Total Internal Reflection will occur.
- The diagram should show the par1ally reﬂected and par1ally refracted rays, this is typical of what happens in the real world.
- Total internal reﬂec1on occurs. Therefore, no light leaves the block and the observed cannot see the laser. It was very common for students to make the ﬁrst point and not con1nue to explain why the observer could not see it.
- • investigate and explain theoretically and practically colour dispersion in prisms and lenses with reference to refraction of the components of white light as they pass from one medium to another
- ROYGBIV The dispersion (spectrum) is a result of the relative refractive index of each colour and ultimately the speed of each colour in the glass. While the speed of all light in a vacuum will be the same. Each colour will have slightly different speed in glass. Hence the angle of refraction will be slightly different resulting in dispersion.
- • explain the results of Young’s double slit experiment with reference to: – evidence for the wave-like nature of light – constructive and destructive interference of coherent waves in terms of path differences: nλ and (n − 1/2) λ – effect of wavelength, distance of screen and slit separation on interference patterns: ∆x = λL/d Young’s double slit experiment relies on coherent light. It is irrefutable evidence of wavelike properties of light: It is an interference pattern. ONLY light is capable of an interference pattern. If light was a particle there would be only two maxima in the double slit experiment.
- Question 18 a. b.
- There is a number of components to Young’s experiment: -A colour filter to allow one wavelength/ frequency. -A small single slit for the waves to diffract through. This is to ensure the waves are coherent. -The double slits for the coherent waves to pass through and produce the interference pattern.
- Key knowledge Behaviour of light • investigate and describe theoretically and practically the effects of varying the width of a gap or diameter of an obstacle on the diffraction pattern produced by light and apply this to limitations of imaging using light Very similar dot point covered earlier. However, if using optical instrument such as microscope. It is difficult to see objects smaller than one wavelength as the light will diffract around the object. This is a limitation. Hence, higher resolution in electron microscopes as the DeBroglie wavelength is much smaller.
- Question 6 a. –1 b. P v = λ ⋅ f v = 0.1× 5 v = 0.5 ms−1 Don’t use cm as a measurement
- b. Y X P Figure 6 As P is the second min. the path difference is 1.5 λ. As XP is 16 cm, YP must be 31cm.
- c. Figure 7 wavefront meets a small gap, e which means the next wavefron Now let us investigate what h ling at an angle θ such that: θ λ = w sin . Point sources in a diffraction gap Diffraction patterns change with gap width. As the gap width gets smaller, coming down the figure, the pattern spreads out more. Diffraction spreading If gap larger than wavelength not a lot of diffraction If gap equal or smaller than wavelength then a lot of diffraction Diffraction spreading is at a maximum when the gap is equal or smaller than the wavelength. Increasing the frequency will reduce the wavelength and reduce the angle of diffraction.
- Elli is assuming that with two speakers will produce sound that is twice as loud as one speaker, this is incorrect. Sam is correct in that an interference pa]ern will be produced between the speakers and this will have areas of loudness (construc1ve interference) and low volume (destruc1ve interference).
- The rule for the spacing between fringes is: Student 2 will hear a high volume being at a point of construc1ve interference. Student 5 will hear low volume being at a point of destruc1ve interference. It is also possible to use Pythagora’s Theorem with this ques1on and arrive at correct answers. It is not enough to say Student 2 will hear high volume and Student 5 low without any jus1ﬁca1on.
- You could draw a diagram to show the ﬁrst point or state: λ 2 = 4 m λ = 8m
- This is a li]le bit of a trick ques1on need to ﬁnd fo: fo = 30 Hz this is the fundamental freq. For a string the second-lowest resonance is 60 Hz. All mul1ples of the fundamental are possible in strings and open pipes.
- The waves are travelling in opposite direc1ons or have been reﬂected from the ends of the string. The waves need be coherent. (same wavelength and amplitude) The waves interfere and cause the observed pa]ern.
- Area of Study 2 How are light and matter similar? In this area of study students explore the design of major experiments that have led to the development of theories to describe the most fundamental aspects of the physical world – light and matter. Outcome 2 On completion of this unit the student should be able to provide evidence for the nature of light and matter, and analyse the data from experiments that supports this evidence.
- • analyse the photoelectric effect with reference to: – evidence for the particle-like nature of light – experimental data in the form of graphs of photocurrent versus electrode potential, and of kinetic energy of electrons versus frequency – kinetic energy of emitted photoelectrons: Ek max = hf – φ, using energy units of joule and electron-volt – effects of intensity of incident irradiation on the emission of photoelectrons There are two values for Planck’s constant: 6.63 x 10-34 J.s (S.I. units) 4.14 x 10-15 eV.s (non S.I. units) Know when to use each.
- • describe the limitation of the wave model of light in explaining experimental results related to the photoelectric effect. The major points are: Classical Theory means waves Quantum Theory means particles -Classical theory would predict that waves of any frequency would eventually eject electrons from the surface of the metal as the energy builds up. -Classical theory would predict that the stopping voltage is also related to the intensity of the light as the wave energy is related to the amplitude. These observations are not made in PE.
- b. The photo-electric effect shows how a single photon will behave like a particle. However, when a single photon is used in Young’s experiment the photon splits into a wave of potential, passes the slits and interferes with itself causing a interference pattern. Only waves can cause interference patterns. The photon is both wave and particle. Q 18
- c. If light is a wave then: 1 The more intense the light, the more energy the electrons will have when the fly off the plate. 2 If the light is very feeble, one may have to expose the source plate for several seconds or minutes until enough waves strike it to knock electrons loose. 3 Waves of any frequency ought to knock electrons free.
- 1 The energy of the electrons does NOT depend on the intensity of the light. ◦ Each electron absorbs only one photon at a time. If the absorbed energy is large enough to expel the electron from the metal, it leaves. If not, the electron dissipates its energy in collisions with nearby electrons and atoms before it can absorb another photon. And yes, this implies that the time it takes for an electron to lose the energy gained in one absorption is much smaller than the interval between absorptions. Under ordinary circumstances, it is. 2 The electrons always appear AS SOON AS the light reaches the plate. ◦ As soon as a single photon containing sufficient energy strikes the source plate, it will knock an electron free. There is no need to wait for multiple waves to build up enough energy. 3 NO electrons are produced if the frequency of the light waves is below a critical value. ◦ Since the energy of each photon is E=h f below some critical frequency, no photon has enough energy to knock an electron free.
- ... We shall assume that in leaving the body, each electron must perform an amount of work W characteristic of the substance. The ejected electrons leaving the body with the largest normal velocity will be those that were [located exactly on] the surface. The kinetic energy of such electrons is given by h*f - W. Albet Einstein
- Quantum mechanics is all about par1cles as waves so the possible answers are C or D. Electron diﬀrac1on explains interference pa]erns seem by electrons. Electron shells (energy levels) are explained by the standing wavelength pa]erns. Therefore D.
- Question 17 A V metal plate collector electrode light source filter voltmeter variable DC voltage source photocell + – ammeter Figure 19
- a. I ( A) V (volts) X O Figure 20 The point X is the stopping voltage. The voltage X stops the most energetic photoelectron that is ejected from the plate and the current falls to zero.
- The difficulty with the previous question is that the diagram usually used is like the one below. Students did not recognise the difference and were unable to apply their knowledge of the situation.
- Frequency (Hz) Stopping voltage (Vs) × 10 b. 2.0 2.5
- b. 0 0.5 1.0 1.5 2.0 2.5 4.0 5.0 6.0 frequency (Hz × 1014) 7.0 8.0 EK max of photoelectrons (eV) c. part b. h + + + + +
- SECTION B – Question 17 c. part b. h h = rise run h = 2 3.75 ×1014 = 5.3×10−15 eVs You can not just quote a known value for h. The instructions are “from the graph plotted…”.
- d. e. one f. 2 3 EK max of photoelectrons (eV) The Frequency of light used Stopping voltage OR max E of photoelect The target metal would be controlled. Different metals have a different work function.
- f. 0 1 2 3 frequency (Hz × 1014) EK max of photoelectrons (eV)
- SECTION B – continued TURN OVER Draw in the error bars that have the same length as the one already shown. Draw a straight line that can pass through the errors bars. If you can then a straight line is appropriate
- f. 0 1 2 3 frequency (Hz × 1014) EK max of photoelectrons (eV)
- To get the answer in eV you will need to use the correct value of Planck’s constant.
- . Matter as particles or waves • interpret electron diffraction patterns as evidence for the wave-like nature of matter • distinguish between the diffraction patterns produced by photons and electrons • calculate the de Broglie wavelength of matter: λ = h/p
- w r i t e i n t h i s a r e a Question 16 –31 –1 crystal e– fluorescent screen electron gun Figure 18a Figure 18b a.
- SECTION B – Question 16 a. b. λ = h p λ = 6.63×10−34 9.1×10−31 ×1.5 ×107 λ = 4.86 ×10−11 m The momentum will increase. Therefore the wavelength will be smaller. The observed pattern will become smaller as well.
- c. –1 If it has the same pattern it must have the same wavelength. However X-rays and electrons will have different energies. E = h⋅c λ E = 6.63×10−34 ⋅3.0 ×108 4.86 ×10−11 E = 4.09 ×10−15 J
- . Similarities between light and matter • compare the momentum of photons and of matter of the same wavelength including calculations using: p = h/λ • explain the production of atomic absorption and emission line spectra, including those from metal vapour lamps • interpret spectra and calculate the energy of absorbed or emitted photons: ΔE = hf
- d o n o t w r i t e i n t h i s Question 20 Data h –15 c 8 –1 A. –19 B. C. D. E = h⋅c λ E = 4.14 ×10−15 ⋅3.0 ×108 4.14 ×10−7 E = 3.0 eV
- w r i t e i n t h i s a r e a Question 9 ionisation energy n = 6 n = 5 n = 4 n = 3 n = 2 n = 1 (ground state) 0 10.2 12.1 12.8 13.1 13.2 13.6 energy (eV) Figure 10 n 12.8, 2.6, 0.7 12.1, 1.9 10.2 Arrows MUST go DOWN
- • analyse the absorption of photons by atoms, with reference to: – the change in energy levels of the atom due to electrons changing state – the frequency and wavelength of emitted photons: E = hf = hc/λ
- • describe the quantised states of the atom with reference to electrons forming standing waves, and explain this as evidence for the dual nature of matter
- • interpret the single photon/electron double slit experiment as evidence for the dual nature of light/matter
- • explain how diffraction from a single slit experiment can be used to illustrate Heisenberg’s uncertainty principle • explain why classical laws of physics are not appropriate to model motion at very small scales.
- d o n o t w r i t e i n t h i s a Question 10 y O x y electron’s path barrier with a slit A. B. C. y D. x D from the previous video
- . Production of light from matter • compare the production of light in lasers, synchrotrons, LEDs and incandescent lights. Light Amplification by Stimulated Emmission of Radiation Laser light is COHERENT All photons are the same wavelength and are all in phase with each other.
- Incandescent light globes work by heating the electrons in the wire. Electrons move into higher energy levels and release energy as photons. Incandescent light globes produces INCOHERENT light of many colors (frequencies) and are not in phase.
- LED’s are semi conductor materials with a P-N junction. Electrons move from the conductive band to the valence band and give extra energy as photons. The light is INCOHERENT although there will be discrete colors the light will NOT be in phase.
- Fluorescent lights absorb energy into a gas vapour and then release the energy in the form of photons. There will be many frequencies of light due to the discrete energy levels. Think of the Sodium lamp or the energy rungs of a Hydrogen atom.
- A synchrotron is a particle accelerator. Beams of particles (usually electrons) are accelerated and bent in Magnetic fields. They are capable of making a very wide range of photon wavelengths. However, usually make very very high intensity X- rays. The intensity can be x100,000’s the brightness of the sun.
- e i n t h i s a r e a Question 18 A. B. C. D. Question 19 A. B. C. D. Question 20 Data h –15 c 8 –1 t h i s a r e a Question 18 A. B. C. D. Question 19 A. B. C. D. Question 20 Data D is the only correct statement C LED’s move electrons from the conduction band to the valence band.
- Area of Study 3 Practical investigation A student-designed practical investigation related to waves, fields or motion is undertaken either in Unit 3 or Unit 4, or across both Units 3 and 4. The investigation relates to knowledge and skills developed across Units 3 and 4 and is undertaken by the student through practical work.
- • the characteristics of scientific research methodologies and techniques of primary qualitative and quantitative data collection relevant to the selected investigation, including experiments and/or the construction and evaluation of a device; precision, accuracy, reliability and validity of data; and the identification of, and distinction between, uncertainty and error
- • methods of organising, analysing and evaluating primary data to identify patterns and relationships including sources of uncertainty and error, and limitations of data and methodologies • models and theories, and their use in organising and understanding observed phenomena and physics concepts including their limitations • the nature of evidence that supports or refutes a hypothesis, model or theory • the key findings of the selected investigation and their relationship to concepts associated with waves, fields and/or motion
- • the conventions of scientific report writing and scientific poster presentation, including physics terminology and representations, symbols, equations and formulas, units of measurement, significant figures, standard abbreviations and acknowledgment of references.
- Key knowledge • independent, dependent and controlled variables • the physics concepts specific to the investigation and their significance, including definitions of key terms, and physics representations A. B. C. D. –5 Question 6 A. B. C. D. Question 7 The investigator changes the value of the independent variable. A
- Question 7 A. B. C. D. Question 8 A. B. C. D. D. Question 7 A. B. C. D. Question 8 A. B. C. D. The dependent variable changes with the independent D The control variables are fixed by the investigator D
- i t e i n t h i s a r e a Question 9 –2 –2 –2 –2 –2 –2 A. –2 B. –2 C. –2 D. –2 Question 10 y O y Highest value = 9.84 Lowest value = 9.78 Difference 0.06 Uncertainty = ± 0.03 A
- Q18. Accurate: is close the expected value. Precise: the measured values are close to each other. A

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