UNIVERSITY OF MALTA G.F. ABELA JUNIOR COLLEGE FIRST YEAR MARKING END-OF-YEAR EXAMINATION SCHEME SUBJECT: PHYSICS DATE: JUNE 2010 LEVEL: ADVANCED TIME: 09.00h to 12.00h Directions to Candidates Show ALL working. Write units where appropriate. Answer ALL questions in Section A. Answer any FOUR questions from Section B You have been provided with two booklets. Use one booklet for Section A, the other for Section B. Take g = 10 m s -2 or 10 N kg -1
Section A Answer ALL questions in one of the booklets provided. Please make sure that you write the question number in the margin. Each question carries 10 marks 1. (a) Explain why the homogeneity of a physical equation does not necessarily guarantee the correctness of the equation. () Homogeneity does not take into account any dimensionless constants that could be present in the equation. (b) The root mean square speed for gas molecules is given the symbol c rms and has the same units as speed. It is found to be related to mass per mole, M of the gas through the equation c rms = 3RT M where R is the so-called molar gas constant and T is the absolute temperature. Work out the units of R. (8 marks) Marks should be awarded for all correct steps and correct units. Not more than 5 marks should be awarded to students who do NOT arrive at the correct units for R. The correct units are J mol -1 K -1. Accept also kg m 2 s -2 mol -1 K -1. 2
2. A horizontal force of 2000N is applied to a vehicle of mass 400kg which is initially at rest on a horizontal surface. If the total force opposing motion is constant at 800N, calculate (a) the acceleration of the vehicle, F = m a 2000 800 = 400 a a = 3 m s -2 (b) (i) At an instant 5 seconds after the force is first applied, what is the momentum of the vehicle, Using v = u + at to find velocity at the 5 th second: v = 0 + (3)(5) v = 15 m s -1 for this part p = m v p = 400 15 p = 6000 kg m s -1 for this part (ii) the kinetic energy of the vehicle, () K.E. = ½ m v 2 K.E. = ½ 400 15 2 K.E. = 45000 J (iii) the power developed? () P = F v P = 2000 15 P = 30000 W 3
3. For a body undergoing simple harmonic motion, the displacement y and the velocity v at an arbitrary time t are given by the equations y = A sin ω t v = A ω cos ω t where ω is the angular frequency in rad s -1. (a) (b) Write down an expression for the maximum velocity of the oscillator in terms of the frequency f of the oscillations, and the amplitude, A. v max = A ω v max = A 2π f v max = 2π f A Note: Students writing v max = 2π f A straightaway should be awarded the FULL 3 marks. (b) Sketch a graph to show how the acceleration, a, of the oscillator varies with time. (4 marks) a 4 marks NOTE: Any other form of sinusoidal variation should be awarded a maximum of. (c) Comment on the phase relationship that exists between y and a. y and a are in anti-phase with one another. Accept also π or 180 o out of phase. 3 marks Note: An answer such as: y and a are out of phase with one another should be awarded. 4
4. A person attached to a parachute is towed over the sea by a tow-rope attached to a motor boat as shown in Figure 1. Figure 2 shows the directions of the forces acting on a person of weight 650N when being towed at a constant horizontal velocity. The 1.5kN force is the pull exerted by the tow-rope on the person, while D is the pull of the parachute on the person. (a) State why the horizontal resultant force on the person must be zero. () A horizontal resultant force would have caused acceleration in accordance with the laws of motion. Question says that boat is being towed at a constant horizontal velocity. (b) Hence, determine the value of D. D cos 55 o = 1500 sin 50 o D = 2003N D = 2.0 kn (c) By considering the forces along the vertical acting on the person, determine whether he is in a state of equilibrium. (5 marks) Resultant Upward force = D sin 55 o = 2003 sin 55 o = 1641 N Resultant Downward force = 1500 cos 50 o + 650 5
= 1614 N No. Forces are not balanced. He is accelerating upwards. 5. The elastic limit of copper is to be estimated by stretching a specimen of thin copper wire of diameter 0.28mm. The specimen, of length 2.0m, is to be suspended vertically and stressed by hanging weights to its lower end. It is known that the Young modulus of copper is 1.3 10 11 Pa and its ultimate tensile stress is 2.0 10 8 Pa. (a) What information about the elastic properties of a material can be deduced from the value of its Young Modulus? The Young Modulus gives a measure of stiffness. 3 marks Note: Young modulus gives the ratio stress/strain for the material should be awarded. (b) What is the maximum load, in N, which can be applied to the lower end of the wire without breaking it? (4 marks) U.T.S. = Maximum load / cross-sectional area Therefore, Max. Load = U.T. S. cross-sectional area = 2.0 10 8 π (0.14 10-3 ) 2 = 12.3N (c) It is thought that the copper will remain elastic with stresses up to 90% of the ultimate tensile stress. Work out the maximum extension which the wire would then be expected to exhibit before it becomes permanently deformed. Maximum stress for wire to remain elastic = 0.90 2.0 10 8 = 1.8 10 8 Pa Young Modulus, E = stress / strain Strain = stress / E L / L = stress / E L = (stress L) / E L = 1.8 10 8 2.0 / 1.3 10 11 6
L = 2.8 10-3 m or 2.8 mm 3 marks Use your discretion to award marks for correct steps. Full marks only to be awarded if correct answer is obtained. 6. The rings of the planet Saturn consist of vast numbers of particles, each in a circular orbit about the planet. Two of the rings are shown in the diagram. The inner edge of the inner ring is 70 000 km from the centre of the planet while the outermost edge of the outer ring is 140 000 km from the centre. The speed of the outermost particles is 17 km s -1. (a) Show that the speed v of any particle that is at a distance R from the centre of the planet, of mass M, is given by v = GM R where the universal gravitational constant G has a value of, 6.7 10-11 N m 2 kg -2. (4 marks) The gravitational force on a particle of mass m is given by GMm/R 2. This force is the centripetal force acting on the particle. Hence, GMm/R 2 = mv 2 / R v 2 = GM / R Therefore v = G M / R (b) Determine the mass of Saturn. Using v = G M / R M = v 2 R / G M = (17 000) 2 (140 000 000) / 6.7 10-11 M = 6 10 26 kg (c) Calculate the orbital speed of the particles nearest to Saturn. 7
Using v = G M / R v = 6.7 10-11 6 10 26 /70000000 v = 2.4 10 4 m s -1 or 24 km s -1 7. (a) Draw a graph that shows how the receding velocity of distant galaxies varies with their distance from an observer. Receding velocity distance (b) Explain briefly how the concept of receding galaxies leads to a Big Bang Theory about the origins of the universe? (4marks) The graph shows that galaxies move away from any observer at a speed that is proportional to their distance from the observer. This leads to the picture of an expanding universe. By extrapolating back in time, the entire Universe can be regarded to have started from a single point of phenomenal density via a Big Bang. 4 marks Note: Use your discretion in awarding marks. (c) Describe one other piece of evidence that supports such a theory. As the universe expanded, following the Big Bang, the early radiation red-shifted to longer wavelengths that should today be in the microwave region of the electromagnetic spectrum. Such radiation has indeed been discovered and thus further supports the Big Bang theory. 8
8. The diagram shows a grinding wheel used to sharpen a chisel. 3.0N 0.15m grinding wheel chisel The grinding wheel has a radius of 0.15m and its moment of inertia is 0.70 kg m 2. The wheel is first switched on and accelerated to an angular speed of 50 rad s -1. The chisel is then pressed against the rim of the wheel, producing a constant tangential force of 3.0 N. (a) Calculate the rotational energy of the wheel at 50 rad s -1. () K.E. rot. = ½ I ω 2 = ½ 0.70 50 2 = 875 J (b) Determine the size of the torque on the wheel due to the chisel. () τ = F r = 3 0.15 = 0.45 N m (c) How much power would have to be supplied to maintain the angular speed at 50 rad s -1? P = τ ω = 0.45 50 = 22.5 W 9
(d) If the chisel is still pressed against the rim when the grinding wheel is switched off, determine how long it takes for the wheel to be brought to rest. τ = I α -0.45 = 0.70 α α = - 0.643 rad s -2 ω = ω o + α t 0 = 50 + (-0.643) t t = 78 s 10
Section B. Answer any FOUR questions from this section. USE THE OTHER BOOKLET FOR THIS SECTION. DO NOT forget to write the question number in the margin. Each question carries 25 marks 9. (a) The graph below shows how the refractive index of water varies with the wavelength of light. (i) Explain the term refractive index of water. The refractive index of water gives the ratio of the speed of light in free space (air) to the speed of light in water. 3 marks (ii) How can the graph help explain the phenomenon of rainbows, caused when sunlight passes through droplets of moisture in the Earth s atmosphere? Refractive Index of Water at Different Wavelengths Refractive Index 1.354 1.349 1.344 1.339 1.334 1.329 350 400 450 500 550 600 650 700 750 Wavelength / nm The graph shows that the refractive index of water is wavelength-dependent. This means that the different wavelengths that constitute white light will be refracted differently when they pass through the droplets of moisture in the air. The shorter wavelengths (violet) are the most refracted (highest refractive index); the longer wavelengths (red), the least. This leads to dispersion of the different wavelengths, as they pass through the water droplets, which eventually creates a rainbow of colours. 3 marks (iii) Sketch a labelled diagram to show the behaviour of white light as it passes through a glass prism. () Only award full marks if the dispersion shown reflects the correct sequence of colours. Also deduct if the refraction inside the prism is not shown to be towards the normal. 11
(b) (i) An illuminated object is placed 35cm in front of a convex lens of focal length 20cm and an image is formed on a screen, placed on the other side of the lens. Sketch a ray diagram to illustrate the meaning of the term focal length. () Award only if focal length is not clearly labelled on diagram (ii) Calculate the image distance. 1/f = 1/v + 1/u 1/20 = 1/v + 1/35 v = 47cm (iii) What is the magnification m of the image? () m = v / u m = 47 / 35 m = 1.3 (iv) (c) The distance between object and screen are kept fixed and the lens is moved between them until a second image appears on the screen. State the respective object and image distances at which this will happen. () Values of u and v are interchanged i.e. u = 47cm and v = 35cm A converging lens, in a slide projector, is used to magnify a square-shaped slide, of area 4cm 2, to an image covering a total area of 6400 cm 2 on the projector screen. Calculate the focal length of this lens if the screen is 82cm away from the slide. (8 marks) 12
Since object is square-shaped, height of object = 2cm and height of image = 80cm magnification = 80 / 2 = 40 Therefore if u = x then v = 40x But u + v = 82 cm (given) Hence x + 40x = 82cm 41x = 82cm x = 2cm Hence u = 2cm and v = 80cm 3 marks 1/f = 1/v + 1/u 1/f = 1/80 + 1/2 f = 1.95cm 13
10. The diagram shows an arrangement used to determine the wavelength λ of monochromatic light emitted by a laser. S 1 S 2 screen Light from laser NOT TO SCALE S 1 and S 2 are slits at right angles to the plane of the diagram. When illuminated by light from the laser, they behave as two coherent sources of light. An interference pattern is formed on the screen from which measurements can be taken to determine the fringe separation which may then be used to calculate the value of λ. (a)(i) What property do wave-trains have in common when they are monochromatic? () Wavelength (Accept also colour ) (ii) What are coherent sources? () These are sources that emit wave-trains with the same frequency a zero (or a constant) phase difference (iii) Name two wave phenomena that occur as the light passes through the two slits and into the region beyond. () Diffraction Interference (b)(i) Describe briefly, with the aid of a graph, the intensity variation of the pattern observed on the screen. (5 marks) The pattern consists of alternate bright and dark fringes. Their intensity varies as shown by the graph: 14
4 marks Note: Award only if bright fringes are depicted with equal intensities. (ii) State one major difference in the pattern obtained, if the laser light had fallen on just one slit only. Again use a graph to illustrate your answer. (4 marks) The fringes around the brightest central fringe make way to a very wide central maximum. (The intensity pattern becomes that shown by the dotted envelope in the graph above). 4 marks (c) The distance y between the centres of neighbouring bright images in the interference pattern is given by: y = Dλ / d (i) Name the instruments you would use to measure D and d. () D: metre ruler or measuring tape d: travelling microscope with vernier scale 15
(ii) The wavelength of the laser light used is known to be 620 nm. Given that 2mm is the minimum fringe spacing which can be observed and that D is set to 3.000m, determine the value of d which will provide this minimum fringe spacing. y = Dλ / d d = Dλ / y d = (3.000 620 10-9 ) / 0.002 d = 9.3 10-4 m (ii) When the laser is replaced by a white light source, having wavelengths ranging from 400nm to 780nm, the first order bright fringe is seen to have violet and red edges respectively. Estimate the width of this first order bright fringe. (5 marks) Central first-order y violet y red For violet y = Dλ / d y = (3.000 X 400 10-9 ) / 9.3 10-4 = 1.29 10-3 m For red y = Dλ / d y = (3.000 X 780 10-9 ) / 9.3 10-4 = 2.52 10-3 m Width of first order bright fringe = 2.52 10-3 m - 1.29 10-3 m = 1.23 10-3 m 16
11. (a) A shell of mass 0.2kg is fired vertically upwards with a speed of 100 m s -1 and 15s later arrives back at ground level with a speed of 80 m s -1. (i) Draw a sketch, showing all the forces (all duly named) that act on the shell on its way up. () Air resistance Weight (ii) Draw a similar sketch, this time for the shell on its way down. () Air resistance Weight (iii) Find the total work done against air resistance during flight. Initial KE = ½ 0.2 100 2 = 1000 J Final KE = ½ 0.2 80 2 = 640 J Work done against air resistance = 1000 J 640 J = 360 J (iv) Find the total change in momentum of the shell during the flight. Initial momentum = 0.2 kg (+ 100 m s -1 ) = +20 kg m s -1 Final momentum = 0.2 kg (-80 m s -1 ) = - 16 kg m s -1 Change in momentum = -16 20 = - 36 kg m s -1 Note: Accept also + 36 kg m s -1 Calculations for initial momentum and final momentum should be awarded the full ONLY if they carry opposite signs. If both are shown with the same sign, award just. 17
(v) Hence, determine the size of the average force acting on it during the entire 15s. By Newton 2 nd Law, the force is equal to the rate of change of momentum. Hence, F = 36 / 15 = 2.4 N 3 marks Note: Award if working is correct but value for change in momentum is incorrect. (b) (i) The same shell is now fired at an angle of 50 o to the horizontal with the same speed of 100 m s -1. It strikes the same level ground a time t later. Determine t, ignoring the effects of air resistance in your calculation. Using s = u t + ½ a t 2 along the VERTICAL 0 = (100 sin 50 o ) t + ½ (-10) t 2 0 = 76.6 t - 5 t 2 t = 15.3 s (ii) What is the horizontal range of the shell? s = u t + ½ a t 2 along the HORIZONTAL Range = (100 cos 50 o )(15.3) + 0 Range = 983 m (iii) Sketch a graph to show how the vertical acceleration of the shell varies with time during the entire trajectory. Vertical acceleration/ m s -2 Time /s -10 3 marks Award only if sign does not tally with the sign convention used by candidate in the earlier working. 18
(iv) Sketch a graph to show how the vertical velocity of the shell varies with time. Vertical velocity / m s -1 Time/s 3 marks Note: Only to be awarded if candidate draws a speed-time graph, i.e. ignores change in direction. 19
12. (a) Define the capacitance of a capacitor. () The amount of charge a capacitor can store per unit p.d. across it. Answers such as: the ability to store charge should NOT be awarded any marks. (b) Two identical capacitors differ only in the type of dielectric used. One uses mica with a dielectric constant (relative permittivity) of 3, the other glass, with a dielectric constant of 9. In terms of charge storage, how does the glass capacitor differ from the mica capacitor when both are charged by a 6V-battery? The glass capacitor stores three times more charge than the mica capacitor. 3 marks Note: Answers such as the glass capacitor stores more charge than the mica capacitor should be awarded only. (c) The diagram shows a circuit where a 0.47µF capacitor may be connected by a two-way switch S either to an 11.0 V d.c. supply or to a 2200Ω resistor. A B S 11.0 V 0.47µF 2200Ω (i) The capacitor is charged with switch S in position A. Calculate the maximum charge stored in the capacitor. Q = C V Q = 0.47 10-6 11.0 Q = 5.2 10-6 C (ii) What is the energy stored by the charged capacitor? Energy stored = ½ C V 2 = ½ 0.47 10-6 11.0 2 = 2.8 10-5 J The switch is moved to position B to discharge the capacitor. 20
(iii) Calculate the initial current passing through the resistor. I o = V/R = 11.0/2200 = 5 10-3 A (iv) How long will this current take to fall to about 37% of the initial value determined in (iii)? The time is equivalent to 1 time-constant T = CR = 0.47 10-6 2200 = 1 X 10-3 s (v) Does the capacitor take the same time to charge and discharge? Explain. No. The time-constants for the charging and discharging paths respectively are different. The 2200 ohm resistor is present in the discharging path only. The capacitor thus takes longer time to discharge than to charge. 3 marks If the answer is no, but the explanation is incorrect, then NO marks should be awarded. (vi) Sketch a current-time graph that covers both the charging as well as the discharging of the capacitor. (5 marks) current during charging time During discharging 5 marks Note: Award a maximum of 3 marks if an exponential decay is shown for either the charging or discharging process. 21
13. A water molecule consists of one oxygen atom and two hydrogen atoms. A student suggests that the oxygen atom acts as a point charge of magnitude -1.1 10-19 C and that each hydrogen atom acts as a point charge of magnitude +0.55 10-19 C. The oxygen atom is separated from each hydrogen atom by a distance of 1.0 10-10 m. The arrangement is a shown in the diagram. Oxygen atom 108 o Hydrogen atom Hydrogen atom Assuming the student s model is correct, (a) calculate the magnitude of the force exerted on the oxygen atom by one of the hydrogen atoms. F = 1/4πε Qq/r 2 F = 1/(4π 8.85 10-12 ) (1.1 10-19 0.55 10-19 ) / (1 10-10 ) 2 F = 5.4 10-9 N (b) Draw a diagram showing each of the forces exerted on the oxygen atom by the two hydrogen atoms. () Oxygen atom 54 o 54 o 5.4 10-9 N 5.4 10-9 N 22
(c) Determine the magnitude and direction of the resultant of these forces. (4 marks) The resultant force points towards the midpoint between the two hydrogen atoms. The resultant force = 2 5.4 10-9 cos 54 o = 6.3 10-9 N Note: Award mark if direction is shown by means of a diagram. (The permittivity of free space, ε o = 8.85 10-12 F m -1 ) An electron having a negative charge of magnitude 1.6 10-19 C is accelerated from rest by a potential difference of 200V. (c) What is the kinetic energy gained by the electron in (i) ev () KE = 200 ev (ii) J? () KE = 1.6 10-19 200 KE = 3.2 10-17 J (d) If the mass of the electron is 9.1 10-31 kg, calculate its final velocity. KE = ½ m v 2 3.2 10-17 = ½ 9.1 10-31 v 2 v = 8.4 10 6 m s -1 (e) The electron then enters the region between two charged parallel plates, 1cm apart, across which a potential difference of 1000V is maintained. The electron enters at right angles to the field between the two plates. Determine the (i) electric field strength between the plates, E = V/d E = 1000 / 0.01 E = 100000 V m -1 (ii) the force experienced by the electron. (3marks) F = E e F = 100000 1.6 10-19 F = 1.6 10-14 N 23
(iii) Explain why the force in (ii) remains constant anywhere in the region between the two charged plates. In between the plates the electric field is uniform. Hence, the force on any charged particle in this region will remain constant. 3 marks 24
14. (a) A body travelling with constant speed v along a circular path r must necessarily be acted upon by a force. Explain how this may be inferred from Newton s first law of motion. Newton s first law makes it clear that in the absence of a resultant force a body can only be at rest OR moving uniformly in a straight line. Hence a body undergoing circular motion MUST be acted upon by a force. 3 marks (b) With the aid of a suitable diagram show that the acceleration a caused by this force is given by a = v 2 / r (1) There are various forms of this derivation. 1 Use your discretion when awarding marks. Reward every correct step. No more than 8 marks should be awarded if final expression is not successfully attained. (c) The cyclist in the diagram below is riding round a corner, his angle of inclination to the vertical being equal to θ. x G h The centre of gravity G is at a height h above the ground and at a horizontal distance x from the point of contact. (i) Copy the simplified diagram of the cyclist (shown below) and mark and name the three main forces acting. 25
x G θ h Normal reaction R θ x G h Weight, mg Friction, f 3marks ( for EACH correct force) (ii) Show that the angle θ is given by: tan θ = v 2 / rg (5 marks) R = mg (cyclist is not accelerating in vertical plane) By principle of moments Taking moments about G: f h = R x f h = mg x Hence f/mg = x/h f/mg = tan θ But f = mv 2 / r (it is the centripetal force) Thus (mv 2 /r)/ mg = tan θ v 2 / rg = tan θ 26
(iii) A cyclist is negotiating a bend of radius 4m at a speed of 5 m s -1. Determine the angle of inclination to the vertical. () tan θ = v 2 / rg tan θ = 5 2 / (4 10) θ = 32 o THE END The Junior College Physics Department 2010 27