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2009 Leaving Cert Physics Solutions (Higher Level)
In an experiment to measure the acceleration due to gravity, the time t for an object to fall from rest through a distance s was measured. The procedure was repeated for a series of values of the distance s. The table shows the recorded data.
(i) Draw a labelled diagram of the apparatus used in the experiment. Timer, ball, release mechanism, trap door (ii) Indicate the distance s on your diagram. (Perpendicular) distance indicated between bottom of ball and top of trap door. (iii) Describe how the time interval t was measured. Timer starts when ball leaves release mechanism Timer stops when ball hits trap door. (iv) Calculate a value for the acceleration due to gravity by drawing a suitable graph based on the recorded data.
- Axes correctly labelled
- At least 6 points correctly plotted
- Straight line with a good distribution
- Correct slope method
- Slope = 5.02 // 0.
- g = (10.04 ± 0.20) m s– (v) Give two ways of minimising the effect of air resistance in the experiment. Small (object)/ smooth(object)/ no draughts/ in vacuum/ distances relatively short / h eavy (object) / dense / spherical/ aerodynamic
A student was asked to measure the focal length of a converging lens. The student measured the image distance v for each of three different object distances u. The student recorded the following data. u /cm 20.0 30.0 40. v /cm 65.2 33.3 25. (i) Describe how the image distance was measured. Object, (converging) lens, screen /search pin Sharp image (state/imply) // no parallax (between image and search pin) Measure (distance) from image/screen to (centre of) lens (ii) Give two precautions that should be taken when measuring the image distance. Measure from the centre of the lens (to the screen) / measure perpendicular distance /avoid parallax error (iii) Use all of the data to calculate the focal length of the converging lens. 1/u + 1/v = 1/f Correct substitution (once) f = 15.3 cm, 15.8 cm, 15.4 cm fave = (15.5 ± 0.4) cm (iv) What difficulty would arise if the student placed the object 10 cm from the lens? Object would be inside the focal point so an image cannot be formed on a screen
Alternative (graphical method): Inverse values for u and for v Plot points Read intercept(s) f = (15.87 ± 0.40) cm
s / cm 30 50 70 90 110 130 150 t /ms 247 310 377 435 473 514 540
s / cm 30 50 70 90 110 130 150 t /ms 247 310 377 435 473 514 540 t 2 / s^2 0.0610 0.0961 0.1421 0.1892 0.2237 0.2642 0.
1/u 0.050 0.033 0. 1/v 0.0153 0.0300 0.
A student investigated the variation of the fundamental frequency f of a stretched string with its tension T. The following is an extract of the student’s account of the experiment. “I fixed the length of the string at 40 cm. I set a tuning fork of frequency 256 Hz vibrating and placed it by the string. I adjusted the tension of the string until resonance occurred. I recorded the tension in the string. I repeated the experiment using different tuning forks.” (i) How was the tension measured? A newton balance // weight of pan + contents (ii) How did the student know that resonance occurred? Paper rider jumped vigorously / the string vibrated at maximum amplitude
The following data were recorded.
(iii) Draw a suitable graph to show the relationship between the fundamental frequency of a stretched string and its tension.
- Six correct values for √T
- Both axes correctly labelled
- Six points correctly plotted
- Straight line with a good fit (iv) State this relationship and explain how your graph verifies it. f is proportional to square root of T because the graph was a straight line through the origin. (v) Use your graph to estimate the fundamental frequency of the string when its tension is 11 N If tension is 11 N ⇒ √T = 3. Use the graph to get f = 542 Hz (vi) Use your graph to calculate the mass per unit length of the string.
Compare to the formula y = mx ⇒ slope = 1/(2l√μ), where l = 0.4 m Mass per unit length (μ) = 5.86 × 10–5^ kg m–
In an experiment to measure the resistivity of nichrome, the resistance, the diameter and appropriate length of a sample of nichrome wire were measured. The following data were recorded: Resistance of wire = 7.9 Ω Length of wire = 54.6 cm Average diameter of wire = 0.31 mm (i) Describe the procedure used in measuring the length of the sample of wire. Straighten the wire, clamp it to a bench and measure the distance between the points for which the resistance was measured. (ii) Describe the steps involved in finding the average diameter of the wire.
- Zero the micrometer / digital callipers
- Place wire between jaws
- Tighten jaws and take reading
- Repeat at different points on wire
- Get average diameter (iii) Use the data to calculate the resistivity of nichrome. A = πr^2 ⇒ A = π(0.155 × 10-3)^2 = 7.55 × 10-8^ m^2 ρ = RA/l ⇒ ρ = (7.9)(7.55 × 10-8)/0.546) ⇒ ρ = 1.09 × 10-6^ Ωm (iv) The experiment was repeated on a warmer day. What effect did this have on the measurements? Resistance increased / length increased (or wire expands) / diameter increased
f /Hz 256 288 320 341 384 480 512 T /N 2.4 3.3 3.9 4.3 5.7 8.5 9.
μ
T l
f 2
=^1
(iv) Calculate the component of the skateboarder’s weight that is parallel to the ramp. W = mgsinθ = mgsin20 = 234.63 N (v) Calculate the force of friction acting on the skateboarder on the ramp. Force down (due to gravity) – Resistive force (due to friction) = Net force Force down (due to gravity) = 234.63 N Net force= 70(2.98) = 208.38 N Friction force = 234.63 – 208.38 = 26.25 N (vi) The skateboarder then maintains a speed of 10.5 m s–1^ until he enters a circular ramp of radius 10 m. What is the initial centripetal force acting on him?
r
mv
Fc
2
= = 70 (10.5)^2 /10 = 771.75 N
(vii) What is the maximum height that the skateboarder can reach? (acceleration due to gravity = 9.8 m s–2) v^2 = u^2 + 2as ⇒ u^2 = 2g(s) ⇒ s = 5.63 m
(viii) Sketch a velocity-time graph to illustrate his motion. Velocity on vertical axis, time on horizontal axis, with appropriate numbers on both axes.
When light shines on a compact disc it acts as a diffraction grating causing diffraction and dispersion of the light. (i) Explain diffraction Diffraction is the spreading out of a wave when it passes through a gap or passes by an obstacle. (ii) Explain dispersion****. Dispersion is the splitting up of white light into its constituent colours. (iii) Derive the diffraction grating formula. For constructive interference path difference = nλ, where n is an integer From diagram we can see that path difference = d sin θ ⇒ nλ = d sin θ (iv) An interference pattern is formed on a screen when green light from a laser passes normally through a diffraction grating. The grating has 80 lines per mm and the distance from the grating to the screen is 90 cm. The distance between the third order images is 23.8 cm. Calculate the wavelength of the green light. d = 1/80000 = 1.25 × 10-5^ m θ = tan-1^ (0.238/0.90) n = 3 nλ = d sin θ ⇒ λ = d sin θ/n ⇒ λ = 551 (± 5) × 10-9^ m. (v) Calculate the maximum number of images that are formed on the screen. For maximum number θ = 90^0 ⇒ sin θ = 1 nλ = d sin θ ⇒ nλ = d ⇒ n = d/λ ⇒ n = 22.7 so the greatest whole number of images is 22. But this is on one side only. In total there will be 22 on either side, plus one in the middle, so total = 45 (vi) The laser is replaced with a source of white light and a series of spectra are formed on the screen. Explain how the diffraction grating produces a spectrum. Different colours have different wavelengths so constructive interference occurs at different positions for each separate wavelength. (vii) Explain why a spectrum is not formed at the central (zero order) image. At central image θ = 0 so constructive interference occurs for all separate wavelengths at the same point so no separation of colours.
An investigation was carried out to establish the relationship between the current flowing in a photocell and the frequency of the light incident on it. The graph illustrates the relationship. (i) What is a photon? A photon consists of a discrete (specific) amount of energy/electromagnetic radiation. (ii) Draw a labelled diagram of the structure of a photocell. See diagram; A = cathode, B = anode Also label glass case and vacuum inside (iii) Using the graph, calculate the work function of the metal. The graph indicates that current only flows when the frequency of the radiation reached 5.2 × 10^14 Hz, so this corresponds to the threshold frequency (f 0 ). φ = hf 0 = (6.6 × 10-34)(5.2 × 10^14 ) = 3.432 × 10-19^ J (iv) What is the maximum speed of an emitted electron when light of wavelength 550 nm is incident on the photocell? h(c/λ) = φ + ½mv^2 (6.6 × 10-34)(3 × 10^8 /550 × 10-9) = 3.432 × 10-19^ + ½ (9.1× 10-31)(v)^2 ⇒ v = 1.922 × 10^5 m s- (v) Explain why a current does not flow in the photocell when the frequency of the light is less than 5.2 × 10^14 Hz****. Because the frequency is less than the threshold frequency so does not contain enough energy to cause an electron to be released from an atom. (vi) The relationship between the current flowing in a photocell and the intensity of the light incident on the photocell was then investigated. Readings were taken and a graph was drawn to show the relationship. Draw a sketch of the graph obtained. Current is directly proportional to Intensity, so a straight line graph with the line going through the origin is required. (vii) How was the intensity of the light varied? Vary the distance from the light source to the photocell. (viii) What conclusion about the nature of light can be drawn from these investigations? Light is made up of discrete amounts of energy called photons.
(Planck constant = 6.6 × 10–34^ J s; speed of light = 3.0 × 10^8 m s–1; charge on electron = 1.6 × 10–19^ C; mass of electron = 9.1 × 10–31^ kg)
9.
(i) Define potential difference
Potential difference is the work done in moving unit charge from one place to another.
(ii) Define capacitance
The capacitance of a conductor is the ratio of the charge on the conductor to its potential.
(iii) A capacitor stores energy.
Describe an experiment to demonstrate that a capacitor stores energy.
- Set up as shown.
- Close the switch to charge the capacitor.
- Remove the battery and connect the terminals together to ‘short’ the circuit.
- The bulb will flash as the capacitor discharges, showing that it stores energy.
The ability of a capacitor to store energy is the basis of a defibrillator. During a heart attack the chambers of the heart fail to pump blood because their muscle fibres contract and relax randomly. To save the victim, the heart muscle must be shocked to re-establish its normal rhythm. A defibrillator is used to shock the heart muscle. A 64 μF capacitor in a defibrillator is charged to a potential difference of 2500 V. The capacitor is discharged through electrodes attached to the chest of a heart attack victim.
(iv) Calculate the charge stored on each plate of the capacitor.
q = CV ⇒ q = (64 × 10-6)(2500) ⇒ q = 0.16 C
(v) Calculate the energy stored in the capacitor.
E = ½ CV^2 = ½ (64 × 10-6)(2500)^2 = 200 J
(vi) Calculate the average current that flows through the victim when the capacitor discharges in a time of 10
ms.
Why are there a large number of turns in the secondary coil?
(v) Explain why the primary coil has thick wire.
(vi) Why are both coils wrapped on the same soft-iron core?
(vii) Radio waves are much less energetic than light waves. List two other types of electromagnetic waves with
energy less than that of light waves.
(viii) Give one property that is common to all types of electromagnetic waves.
(ix) The telephone used to transmit the messages to Dublin contained a moving-coil loudspeaker. Describe, with the
aid of a labelled diagram, how a loudspeaker operates.
Read the following passage and answer the accompanying questions. The sun is a major source of ‘green’ energy. In Ireland solar heating systems and geothermal systems are used to get energy from the sun. There are two main types of solar heating systems, flat-plate collectors and vacuum-tube collectors.
A flat-plate collector is usually an aluminium box with a glass cover on top and a blackened plate on the bottom. A copper pipe is laid on the bottom of the box, like a hose on the ground; water is passed through the pipe and transfers the absorbed heat to the domestic hot water system.
In a vacuum-tube collector, each tube consists of an evacuated double-walled silvered glass tube in which there is a hollow copper pipe containing a liquid. The liquid inside the copper pipe is vaporised and expands into the heat tip. There the vapour liquefies and the latent heat released is transferred, using a heat exchanger, to the domestic hot water system. The condensed liquid returns to the copper pipe and the cycle is repeated. In a geothermal heating system a heat pump is used to extract solar energy stored in the ground and transfer it to the domestic hot water system.
(a) What is the max energy that can fall on an area of 8 m^2 in one hour if the solar constant is 1350 W m–2? 1350 × 8 × 3600 Emax = 3.9 × 10^7 J (b) Why is the bottom of a flat-plate collector blackened? Dark surfaces are good absorbers of heat/energy/radiation (c) How much energy is required to raise the temperature of 500 litres of water from 20 0 C to 50 0 C? Density = mass/volume ⇒ mass = (density)(volume) ⇒ m = (1000)(500 × 10–3) = 500 kg. E = mcθ = (500)(4200)(30) = 6.3 × 10^7 J (d) The liquid in a vacuum-tube solar collector has a large specific latent heat of vaporisation. Explain why. So that a lot of energy gets absorbed (and then released) per kg in the heat exchanger during a change of state. (e) Name the three ways that heat could be lost from a vacuum-tube solar collector. Conduction, convection, radiation (f) How is the sun’s energy trapped in a vacuum-tube solar collector? Silvered walls prevent radiation and evacuated walls prevent conduction and convection (g) Describe, in terms of heat transfer, the operation of a heat pump. Energy is taken from one place (making it colder) by allowing the liquid to change state to a gas. Then in another place the gas condenses to a liquid releasing the heat to another place making it hotter. (h) Give an advantage of a geothermal heating system over a solar heating system. Geothermal system functions all the time whereas a solar heating system works only during sunshine.
(specific heat capacity of water = 4200 J kg–1^ K–1; density of water = 1000 kg m–3; 1 litre = 10–3^ m^3 )
12 (a) (i) State Hooke’s law. When a string is stretched the restoring force is proportional to the displacement. (ii) When a sphere of mass 500 g is attached to a spring of length 300 mm, the length of the spring increases to 330 mm. Calculate the spring constant. When the mass of 500 g is attached the new force down = mg = (0.5)(g). Because the spring is in equiblibrium this must be equal to the force up (which is the restoring force). Hooke’s law in symbols: F = k x ⇒ (0.5)(g) = kx ⇒ k = F/x = 0.5g/0.030 ⇒ k = 163.3 N m- (iii) The sphere is then pulled down until the spring’s length has increased to 350 mm and is then released. Describe the motion of the sphere when it is released. It executes simple harmonic motion because the displacement is proportional to t he acceleration. (iv) What is the maximum acceleration of the sphere? F = ma = kx ⇒ a = kx/m = (163.3)(0.02)/(0.5) = 6.532 m s- OR a = ω^2 x ⇒ ω^2 = k/m = 163.3/0.5 ⇒ a = 6.532 m s-
12 (b) A semiconductor diode is formed when small quantities of phosphorus and boron are added to adjacent layers of a crystal of silicon to increase its conduction. (i) Explain how the presence of phosphorus and boron makes the silicon a better conductor. When phosphorus is added more electrons become available as charge carriers. When boron is added more positive holes become available as charge carriers. (ii) What happens at the boundary of the two adjacent layers? Electrons and holes cross the junction cancelling each other out and recombine and as a result there are no free charge carriers. A depletion layer is therefore formed between the n-type and p-type regions and as a result a junction voltage is created. (iii) Describe what happens at the boundary when the semiconductor diode is forward biased. The depletion layer breaks down and the diode conducts. (iv) Describe what happens at the boundary when the semiconductor diode is reverse biased. The width of depletion layer gets increased and the region acts as an insulator. (v) Give a use of a semiconductor diode. Rectifier
12 (c) Information is transmitted over long distances using optical fibres in which a ray of light is guided along a fibre. Each fibre consists of a core of high quality glass with a refractive index of 1.55 and is coated with glass of a lower refractive index. (i) Explain, with the aid of a labelled diagram, how a ray of light is guided along a fibre.
- An optical fibre consists of a glass pipe coated with a second material of lower refractive index.
- Light enters one end of the fibre and strikes the boundary between the two materials at an angle greater than the critical angle, resulting in total internal reflection at the interface.
- This reflected light now strikes the interface on the opposite wall and gets totally reflected again.
4. This process continues all along the glass pipe until the light emerges at the far end.
(ii) Why is each fibre coated with glass of lower refractive index? Because total internal reflection can only occur for rays travelling from a denser to a rarer medium. (iii) What is the speed of the light as it passes through the fibre? n = cair/cglass ⇒ cglass = 3.0 × 10^8 /1. cglass = 1.94 × 10^8 m s- (iv) Light passing through optical fibres must travel through an enormous length of glass. Impurities in the glass reduce the power transmitted by half every 2 km. The initial power being transmitted by the light is 10 W. What is the power being transmitted by the light after it has travelled 8 km through the fibre? After 2 km power has dropped to 5 W; after 4 km power has dropped to 2.5 W; after 6 km power has dropped to 1.25 W; after 8 km power has dropped to 0.625 W.
