UNIVERSITY OF MALTA G.F. ABELA JUNIOR COLLEGE FIRST YEAR END-OF-YEAR EXAMINATION SUBJECT: PHYSICS DATE: JUNE 2010 LEVEL: INTERMEDIATE TIME: 09.00h to 12.00h Show ALL working Write units where appropriate Directions to Candidates Answer ALL questions in Section A Answer the single question in Section B Answer any TWO questions from Section C You have been provided with two booklets. Use one booklet for Section A, the other for Sections B and C. Take g = 10 m s -2 or 10 N kg -1
Section A Attempt ALL questions in this section. Please do not forget to write the question number in the margin. Each question carries 7 marks. This section carries 50% of the total marks of the exam. 1. The speed v of surface ocean waves is given by the expression: v = k g λ g is the acceleration due to gravity and λ is the wavelength of the waves. (a) State the base units of v, g and λ. (3 marks) Base units of v = m s -1 Base units of g = m s -2 Base units of λ = m (b) Show that the constant k has no units. (4 marks) Since equation is homogeneous w.r.t. units, then: m s -1 = (k)(m s -2. m) 1/2 m s -1 = (k)(m 2 s -2 ) 1/2 m s -1 = (k) m s -1 k has no units 4 marks Note: Just award marks for the correct working. Correct units for v, g and λ have already been tested for (and awarded marks) in part (a). 2. A body is undergoing free vibrations about a fixed position O. A O B A and B are the extreme points of this vibratory motion. It is found that as the body vibrates between A and B, it travels with constant velocity for most of the time.
(a) Would you describe the motion of this vibrating body as simple harmonic motion? Explain. (3 marks) No, since a body performing s.h.m. does not travel with constant velocity for most of the time. It accelerates throughout its motion, except instantaneously at its equilibrium position. 3 marks Or A body performing s.h.m. has its velocity varying sinusoidally with time, so it is not constant for most of the time. Or SIMILAR Note: Please do NOT award marks if the student writes NO but does not provide a valid reason. (b) The body travels from A to B in 0.5s. What is the frequency of vibrations in Hz? (2 marks) Period, T = 1s f = 1/T = 1/1 = 1 Hz State the effect on the amplitude of the body if the vibrations were to be subjected to an ever-increasing level of damping. (2 marks) The amplitude would progressively decrease to zero. 2 marks 3. Water emerges horizontally from a hose-pipe at a constant rate of 2m 3 per minute. The pipe is 4m away from a wall and the water hits the wall as shown. 4m T Hose-pipe Water P The water hits the wall at point P as shown.
(a) Determine the rate at which water emerges from the hose-pipe in metres cubed per second (m 3 s -1 ). () Rate of flow = 2 m 3 per minute = 2 m 3 / 60 s = 0.033 m 3 s -1 (b) The pipe has a cross-sectional area of 1.8 10-3 m 2. Determine the speed (in m s -1 ) with which the water emerges. (2 marks) v = 0.033 m 3 s -1 / 1.8 10-3 m 2 = v = 18.5 m s -1 Determine the time the water takes to travel to P once it emerges from the hose-pipe. (2 marks) Using s = ut + ½ a t 2 along the HORIZONTAL 4 = 18.5 t t = 0.216 s (d) How far is P below T (a point horizontally aligned with the initial direction of the water)? (2 marks) Using s = ut + ½ a t 2 along the VERTICAL s = ½ 10 0.216 2 s = 0.23m 4. State clearly what you understand by the following statements: (a) A car is accelerating at 5 m s -2. (2 marks) The velocity of the car is changing by 5 m s -1 every second. 2 marks Accept also: The body is being subjected to a net force of 5N kg -1.
(b) To every action there is an equal and opposite reaction. (3 marks) If a body A exerts a force on a body B, then B exerts an equal but opposite force on body A. or similar 3 marks All bodies have inertia. (2 marks) All bodies have a natural tendency to resist a change in their state of motion. or All bodies tend to resist being accelerated. or Bodies do not accelerate out of their own accord; they have to be forced to do so. or similar 2 marks 5. A falling ball strikes the ground at a velocity of 6 m s -1 and rebounds vertically upwards at 4 m s -1. The ball has a mass of 50g. (a) Determine the change in velocity of the ball after it hits the ground. (2 marks) Taking upward vectors as +ve and downward vectors as ve v = v - u v = 4 (-6) v = 10 m s -1 Note: Accept also -10m s -1 (b) Hence determine its change in momentum. (2 marks) momentum = mv = m v = 0.050 10 = 0.50 kg m s -1 Estimate the time it stays in contact with the ground, if it is known that it exerts a force of about 10N on impact. (3 marks) Newton s 2 nd Law: F = mv / t t = mv / F t = 0.50/10 t = 0.05 s
6. The diagram below shows a cross-section through a rectangular lightemitting diode (LED). When current passes through the LED, light is emitted from its transparent surface, of refractive index 1.5, into the surrounding air. Transparent material 40 o Q air P R (a) Determine the critical angle for light passing from the transparent material to air. (3 marks) η 1 sin θ 1 = η 2 sin θ 2 1.5 sin c = 1 sin 90 o sin c = 1/1.5 = 0.667 c = 41.8 o (b) Light ray PQ is incident at point Q, so that the angle of refraction in air is 40 o. Calculate the angle of incidence at point Q. (2 marks) η 1 sin θ 1 = η 2 sin θ 2 1.5 sin i = 1 sin 40 o sin i = sin 40 o / 1.5 = 0.429 i = 25.4 o Ray PR is incident at an angle of 45 o at point R. What happens to this light ray? Explain. (2 marks) It is reflected back inside the transparent medium - - since the angle of incidence is greater than the critical angle.
7. A converging lens has a focal length of 20.0cm. (a) Explain with the aid of a diagram the statement in italics. (2 marks) Note: Award only if focal length is not clearly labelled. 2 marks (b) When the lens is placed at a certain distance from an illuminated object, an image, having the same size as the object, is formed on a screen placed on the other side of the lens. What is the distance between the object and its image? (3 marks) The object is placed on 2F. Hence distance between object and image = 4f = 4 20.0cm = 80.0cm For what object distances from the lens, would it be impossible for the lens to form images on a screen? (2 marks) For all object distances less than 20.0cm from the lens. 2 marks 8. An alpha particle having a speed of 1.0 10 6 m s -1 collides with a stationary proton. The proton is pushed forward with an initial speed of 1.6 10 6 m s -1. (a) What is the speed of the alpha particle immediately after collision? (3 marks) Conservation of momentum Principle: (mu) alpha + (mu) proton = (mv) alpha + (mv) proton (6.64 10-27 )(1.0 10 6 ) = (6.64 10-27 v alpha ) + (1.66 10-27 )(1.6 10 6 ) 2 marks v alpha = 6 10 5 m s -1
(b) How much kinetic energy is gained by the proton? (2 marks) KE gained = ½ m v 2 = ½ 1.66 10-27 (1.6 10 6 ) 2 = 2.12 10-15 J It is known that the collision is perfectly elastic. Explain what this means. (2 marks) Total KE is conserved after collision 2 marks (Mass of alpha particle = 6.64 10-27 kg; mass of proton = 1.66 10-27 kg) 9. The following graphs refer to simple harmonic motion. For all graphs, the x-axis represents displacement from the equilibrium position, x. Copy the graphs and indicate clearly the quantity that is being plotted on the y-axis. You may ignore the effects of damping. Potential Energy Total Energy x x 2 marks Acceleration Kinetic Energy x x 2 marks 2 marks
10. The energy E stored in a compressed spring is given by E = ½ k x 2 where k is the spring constant and x is the amount of compression. A spring is mounted at an angle θ = 30 o on a frictionless incline as illustrated in the figure below. The spring is compressed by 15 cm and then released to propel a mass of 4.5 kg up the incline. (a) If the spring constant is 575 N m -1, how fast is the mass moving when it leaves the spring? (3 marks) Energy stored in spring is converted to kinetic energy of the propelled mass Hence ½ k x 2 = ½ mu 2 575 0.15 2 = 4.5 u 2 2 marks u 2 = 2.875 u = 1.7 m s -1 (b) To what maximum distance from the starting point will the mass rise up the incline? (4 marks) Using v 2 = u 2 + 2 as 0 = 2.875 + 2 (-10 sin 30 o ) s 2 marks s = 0.29 m
Section B. Attempt the question in this section. PLEASE START A NEW BOOKLET. This question carries 20 marks which is equivalent to 14% of the total mark of the exam. 11. A diffraction grating may be considered to consist of a very high number of close, narrow, parallel slits. The separation between two adjacent slits is denoted by d. When a fine parallel beam of light is incident normally on the grating, the beam is diffracted. It can be shown that for first-order diffraction (see diagram below), the angle θ by which light is diffracted depends on the wavelength of light used (λ) and the grating spacing (d) according to the equation: d sin θ = λ grating beam of light θ zero-order first-order Using a particular grating, the angle of diffraction θ (in degrees) was measured for five different wavelengths (in nanometres) of light. The results obtained are shown below. λ/nm Angle of Diffraction, θ/ o sin θ (a) (b) 400 14.5 0.25 550 20.0 0.34 600 22.0 0.37 650 23.9 0.41 700 25.9 0.44 2 marks Copy the table and fill in the missing values. (2 marks) Plot a graph of sin θ against λ. (6 marks) Reduce if line plotted is not line of BEST fit Reduce for each unlabelled axis Reduce for each awkward scale Reduce if graph is λ against sin θ
Use the graph to determine a value for d, the grating spacing. (10 marks) 5 marks for finding gradient (about 6.43 10 5 m -1 ) Note: If nm were not converted to m and value obtained is 6.43 10-4 award 3 marks not 5. 3 marks for any proof that candidate has realised that gradient = 1/d 2 marks for working d (should be about 1.6 10-6 m) (d) Why, do you think, it is so important that the light striking the grating be in the form of a fine parallel beam? (2 marks) It would not be possible to identify the exact amount by which the grating has diffracted the beam if the beam did not have one specific direction before hitting the grating. 2 marks Note: Answers like To be accurate - award only
Section C Continue this section on the booklet used for Section B. You are to attempt only TWO questions from this section. Each question carries 25 marks. This section carries 36% of the total mark of the exam. 12.(a) Explain what is meant by gravitational field strength. In what units is it measured? (3 marks + ) The gravitational field strength at a point in a gravitational field is equivalent to the force exerted on a unit mass placed at that point. 3 marks N kg -1 (b) (i) A person weighing 800N on Earth would have a weight mass of 256N on Mars. What is the gravitational field strength on the surface of Mars? (3 marks) Using W = mg On Earth: 800 = m X 10 m = 80 kg On Mars: 256 = 80 g g = 3.2 N kg -1 (ii) What is the mean radius of Mars given that its mass is 6.4 10 23 kg? (5 marks) g = G M / R 2 3 marks R 2 = G M / g R 2 = (6.7 10-11 X 6.4 10 23 ) / 3.2 R = 3.66 10 6 m (i) Mars orbits the sun with an average orbital radius of 228 million km. Given that the mass of the sun is 1.9 10 30 kg, determine the size of the gravitational force which the sun and Mars exert on each other. (4 marks) F = G Mm/r 2 F = (6.7 10-11 X 6.4 10 23 X 1.9 10 30 )/(228 10 9 ) 2 F = 1.57 10 21 N 3 marks
(ii) Work out a value for the speed of Mars. (5 marks) F = mv 2 / r 2 marks 1.57 10 21 = 6.4 10 23 v 2 / 228 10 9 2 marks v = 2.36 10 4 m s -1 OR v = GM/r 2 marks v = 6.7 10-11 X 1.9 10 30 /228 10 9 v = 2.36 10 4 m s -1 2 marks (iii) Hence determine the time it takes Mars to complete one full revolution around the sun. (4 marks) (G = 6.7 10-11 N m 2 kg -2 ) T = 2πr / v 2 marks T = 2π 228 10 9 / 2.36 10 4 T = 6.06 10 7 s (= 1.92 years) 13.(a) With the help of appropriate diagrams, distinguish between constructive and destructive interference. (4 marks) Constructive Interference Destructive Interference 2 marks 2 marks
(b) 12m loudspeaker 10m P The diagram shows two loudspeakers that operate in phase to emit sound of frequency 660Hz and velocity 330m s -1. The wave-trains from the two speakers spread out in space and interfere with one another. Describe the type of interference occurring at a point P, 12m away from one speaker and 10m away from the other. Show clearly how you arrived at your conclusion. (6 marks) Using c = f λ 330 = 660 λ λ = 0.5 m The actual path difference = 12m 10m = 2m This is equivalent to 4 λ i.e. a whole number of wavelengths 2 marks Thus waves interfere CONSTRUCTIVELY at P. (i) (ii) Another form of interference (or superposition) occurs when two similar waves, travelling with equal and opposite velocities, cross paths to form a stationary wave. Why is the resultant wave said to be stationary? (2 marks) The wave profile does not propagate forward but is trapped in stationary mode between two boundaries. 2 marks State two differences between stationary and progressive waves. (4 marks) In stationary waves, different points along the wave vibrate with different amplitudes while in progressive all points may have the same amplitude. The points between two successive nodes in a stationary wave vibrate in phase with one another while in progressive waves, only two points that are exactly a whole number of wavelengths apart are in phase. 4 marks Note: Accept any other valid differences.
(iii) If the antinode of the stationary wave formed has an amplitude of 4cm, what was the amplitude of the two superposing waves? (2 marks) 2cm 2 marks (d) A progressive wave has a frequency of 250Hz and velocity of 30 m s -1. (i) Comment on the phase difference between two vibrating points along the wave that are exactly 6cm apart. (4 marks) Using c = f λ 30 = 250 λ λ = 0.12 m 2 marks The wavelength is 12cm. Hence two points that are 6cm apart are half a wavelength apart. Thus they are in ANTI-PHASE or OUT OF PHASE BY HALF AN OSCILLATION or by T/2. (ii) How many oscillations does any point along the wave perform in 2s? (3 marks) Frequency = 250 Hz i.e. 250 oscillations per second Therefore, 500 oscillations in 2s. 2 marks 14.(a) State Newton s first law of motion. What does the law imply about bodies travelling along a circular path? (2 marks + 2 marks) A body will remain at rest or continue to move with constant speed in a straight line unless acted upon by a resultant force. 2 marks A body performing circular motion is doing neither of the two; hence, it MUST be acted by a resultant force (or, it must be accelerating). 2 marks (b) An aircraft of mass 60 000 kg, cruising at 250 m s -1, banks at angle of 40 o to the vertical, so as to move in a circular path of radius r. At this point, the lift force, L, on the plane acts as shown.
(i) What is the weight of the plane? State its direction. (2 marks + ) W = mg W = 60000 kg 10 N kg -1 W = 600000 N Vertically downwards OR Towards centre of the earth (ii) Hence, determine the size of L during banking. (You may assume that the plane is not changing altitude). (3 marks) L cos 40 o = 600000 L = 600000 / cos 40 o 2 marks L = 7.8 10 5 N (iii) Use the value of L to work out a value for r, the radius of the circular path. (6 marks) L sin 40 o is the centripetal force L sin 40 o = m v 2 / r 4 marks 7.8 10 5 sin 40 o = (60 000 250 2 ) / r r = 7448m r = 7450 m
(iv) What should the pilot do to stop the plane from turning and start cruising forward in a straight line? Why? (2 marks) Gradually decrease the angle 40 o to 0 o. No component of L towards centre; no centripetal force. (i) (ii) When cruising in a straight line, at constant altitude, the aeroplane encounters a drag force of 24kN. What is the forward thrust produced by the engines, if the aeroplane is not accelerating? Explain your reasoning. (2 marks + 2 marks) Forward thrust = 24 kn 2 marks This makes resultant force on plane equal to 0N. Hence, the plane will not accelerate in accordance with first law. If the cruising speed is still 250 m s -1, determine the power developed by the aeroplane. (3 marks) P = F v P = 24000 250 P = 6 10 6 W 15. (a) In an experiment to determine the Young s modulus of steel, a long, thin wire of steel is loaded and its extension determined for many different loads. (i) What information about a material does the Young Modulus give? (2 marks) It gives a measure of the stiffness of the material. 2 marks Note: Answer Young modulus = stress/strain (ii) What is the advantage of using a long and thin wire for this particular experiment? (2 marks) The extension would be larger than if a short, thick wire were used - - and hence can be measured to a higher degree of accuracy. (iii) Apart from force and extension what other quantities would need to be known for the Young Modulus to be determined? (4 marks) Cross-sectional area 2 marks Original Length 2 marks
(b)(i) Sketch a force-extension graph for a sample of rubber as it is taken through a loading-unloading cycle. (4 marks) 3 marks for correct shape for correct labelling (ii) State two major differences between the elastic properties of rubber and those of a metal. (4 marks) Rubber does not obey Hooke s law; a metal does. Rubber retains a degree of energy inside it for each loading unloading cycle; a metal does not. Rubbers are much more flexible than metals of the same dimensions 2 marks for each correct comparison. (i) A gymnast of mass 70kg hangs by one arm from a high bar. If the gymnast s whole weight is assumed to be taken by the humerus bone, in the upper arm, calculate the stress in the humerus if it has a radius of 1.5cm. (4 marks) Stress = force/area = 700/(π 0.015 2 ) 2 marks = 9.9 10 5 Pa (ii) extension of the humerus. You may consider the bone to be 30.0cm long. The Young Modulus for bone is 1.8 10 10 Pa. (5 marks) E = stress/strain Strain = stress/e l = (stress l)/e l = (9.9 10 5 0.30)/1.8 10 10 l = 1.65 10-5 m 4 marks