Conservation of Momentum - Physics - Solved Paper, Exams of Physics

Download Conservation of Momentum - Physics - Solved Paper and more Exams Physics in PDF only on Docsity! 2005 Leaving Cert Physics Solutions (Higher Level) 1 In an experiment to verify the principle of conservation of momentum, a body A was set in motion with a constant velocity. It was then allowed to collide with a second body B, which was initially at rest and the bodies moved off together at constant velocity. The following data was recorded. Mass of body A = 520.1 g Mass of body B = 490.0 g Distance travelled by A for 0.2 s before the collision = 10.1 cm Distance travelled by A and B together for 0.2 s after the collision = 5.1 cm (i) Draw a diagram of the apparatus used in the experiment. See diagram (ii) Describe how the time interval of 0.2 s was measured. It corresponded to 10 intervals on the ticker-tape. (iii) Using the data calculate the velocity of the body A before and after the collision. Velocity before: v = s/t = 0.101/0.2 v = 0.505 m s-1 ≈ 0.51 m s-1 Velocity after: v = 0.051/0.2 v = 0.255 m s-1 ≈ 0.26 m s-1 (iv) Show how the experiment verifies the principle of conservation of momentum. Momentum before p = mv = (0.5201)(0.505) = 0.263 ≈ 0.26 kg m s-1 Momentum after p = mv = (0.5201 + 0.4900)(0.255) p = 0.258 ≈ 0.26 kg m s-1 Momentum before ≈ momentum after (v) How were the effects of friction and gravity minimised in the experiment? Friction: sloped runway // oil wheels or clean track Gravity: horizontal track // frictional force equal and // tilt track so that trolley moves with constant velocity 2 In an experiment to measure the specific latent heat of vaporisation of water, cool water was placed in an insulated copper calorimeter. Dry steam was added to the calorimeter. The following data was recorded. Mass of calorimeter = 50.5 g Mass of calorimeter + water = 91.2 g Initial temperature of water = 10 oC Temperature of steam = 100 oC Mass of calorimeter + water + steam = 92.3 g Final temperature of water = 25 oC (i) Calculate a value for the specific latent heat of vaporisation of water. mslw + mscwΔθs = mwcwΔθw+ mcccΔθc Δθs = 75 0C and Δθw (= Δθc) = 15 0C (0.0011) lw + (0.0011)(4200)(75) = (0.0407)(4200)(15) + (0.0505)(390)(15) [(0.0011) lw + 346.5 = 2564.1 + 295.425] lw = 2.28 × 10 6 J kg-1 (ii) Why was dry steam used? Calculations assume that only steam is added, not water. (iii) How was the steam dried? Use a steam trap / insulated delivery tube / sloped delivery tube / allow steam to issue freely initially (iv) A thermometer with a low heat capacity was used to ensure accuracy. Explain why. It absorbs little heat from system in calorimeter and calculations assume that no energy is transferred to the thermometer. 3 In an experiment to verify Snell’s law, a student measured the angle of incidence i and the angle of refraction r for a ray of light entering a substance. This was repeated for different values of the angle of incidence. The following data was recorded. i/degrees 20 30 40 50 60 70 r/degrees 14 19 26 30 36 40 (i) Describe, with the aid of a diagram, how the student obtained the angle of refraction. See diagram, plus ray-box. Mark the position of the incident and exit rays and also the outline of the block. Remove the block then measure the angle between the refracted ray and the normal using a protractor. (ii) Draw a suitable graph on graph paper and explain how your graph verifies Snell’s law. sin i 0.34 0.50 0.64 0.77 0.87 0.94 sin r 0.24 0.33 0.44 0.50 0.59 0.64 (iii) From your graph, calculate the refractive index of the substance. Refractive index = slope = y2 – y1 / x2 – x1 ⇒ n = 1.49 (iv) The smallest angle of incidence chosen was 200. Why would smaller values lead to a less accurate result? There would be a greater percentage error associated with measuring smaller angles. 4 A student investigated the variation of the current I flowing through a filament bulb for a range of different values of potential difference V. (i) Draw a suitable circuit diagram used by the student. See diagram (ii) Describe how the student varied the potential difference. By adjusting the voltage on the power supply. (iii) The student drew a graph, as shown, using data recorded in the experiment. With reference to the graph, explain why the current is not proportional to the potential difference. Because the graph is not a straight line. (iv) With reference to the graph, calculate the change in resistance of the filament bulb as the potential difference increases from 1 V to 5 V. At 1 V: R = V/I = 1/0.028 = 35.7 Ω At 5 V: R = (5/0.091) = 54.9 Ω (vii) Calculate the decay constant of cobalt−60. Formula: T1/2 = ln 2/λ ⇒ λ = T1/2/ln 2 T1/2 = 5.26 y = 1.66 × 10 8 s and ln 2 = 0.693 λ = 1.66 × 108 / 0.693 ⇒ λ = 4.18 × 10-9 s-1 (viii) Calculate the rate of decay of a sample of cobalt−60 when it has 2.5 × 1021 atoms. dN/dt = (-) λN = (4.18 × 10-9)( 2.5 × 1021) = 1.04 × 1013 Bq 9 (i) Define potential difference. Potential difference is the work done in bringing unit charge from one point to another. (ii) Define resistance. Resistance of a conductor is the ratio of the potential difference across it to the current passing through it. (iii) Two resistors, of resistance R1 and R2 respectively, are connected in parallel. Derive an expression for the effective resistance of the two resistors in terms of R1 and R2. IT = I1 + I2 (apply Ohm’s law ) V = IR V/RT= V/R1+ V/R2 1/RT= 1/R1+ 1/R2 (iv) In the ciruit diagram, the resistance of the thermistor at room temperature is 500 Ω. At room temperature calculate the total resistance of the circuit. 1/Rp = 1/500 + 1/750 Rp = 300 Ω RTot = 600 Ω (v) At room temperature calculate the current flowing through the 750 Ω resistor. ITotal = (VTot / RTot)/ = 6 ÷ 600 / = 0.01 A V300 = (0.01)(300) = 3 V ⇒ Vp = 6 – 3 = 3 V. I750 = 3 ÷ 750 = 4 × 10 -3 A = 4 mA (vi) As the temperature of the room increases, explain why the resistance of the thermistor decreases. More energy is added to the thermistor therefore more electrons are released and are available for conduction. (vii) As the temperature of the room increases, explain why the potential at A increases. The resistance of thermistor (and 750 Ω combination) decreases Therefore potential difference across thermistor and 750 Ω combination decreases Therefore potential at A increases 10 (i) Define electric field strength. Electric field strength is defined as force per unit charge. (ii) State Coulomb’s law of force between electric charges. The force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. (iii) Why is Coulomb’s law an example of an inverse square law? Force is inversely proportional to distance squared. (iv) Give two differences between the gravitational force and the electrostatic force between two electrons. Gravitational force is much smaller than the electrostatic force. Gravitational force is attractive, electrostatic force (between two electrons) is repulsive. (v) Describe an experiment to show an electric field pattern. High voltage and two metal plates /electrodes Semolina and oil in container Connect a (high) voltage to the plates in container Semolina lines up in the field (vi) Calculate the electric field strength at the point B, which is 10 mm from an electron. E = Q/4πεd2 = (1.6 × 10-19)/4π(8.9 × 10-12)(0.01)2 E = 1.4 × 10-5 N C-1 (vii) What is the direction of the electric field strength at B? Towards the electron / to the right (viii) A charge of 5 μC is placed at B. Calculate the electrostatic force exerted on this charge. F = Eq or F = (1.4 × 10-5)(5 × 10-6) = 7.2 × 10-11 N Towards the electron 11 (a) Read the following passage and answer the accompanying questions. Ernest Rutherford made the following point: If the particles that come out naturally from radium are no longer adequate for my purposes in the laboratory, then maybe the time had come to look at ways of producing streams of fast particles artificially. High voltages should be employed for the task. A machine producing millions of alpha particles or protons would be required. These projectiles would be released close to a high voltage and would reel away at high speed. It would be an artificial particle accelerator. Potentially such apparatus might allow physicists to break up all atomic nuclei at will. (Adapted from “The Fly in the Cathedral” Brian Cathcart; 2004) (i) What is the structure of an alpha particle? An alpha particle is identical to a helium nucleus (2 protons and 2 neutrons). (ii) Rutherford had bombarded gold foil with alpha particles. What conclusion did he form about the structure of the atom? The atom was mostly empty space with a dense positively-charged core and with negatively-charged electrons in orbit at discrete levels around it. (iii) High voltages can be used to accelerate alpha particles and protons but not neutrons. Explain why. Alpha particles and protons are charged, neutrons are not. (iv) Cockcroft and Walton, under the guidance of Rutherford, used a linear particle accelerator to artificially split a lithium nucleus by bombarding it with high-speed protons. Copy and complete the following nuclear equation for this reaction. 11H + 7 3Li → 4 2 4 2 HeHe + + K.E. (v) Circular particle accelerators were later developed. Give an advantage of circular accelerators over linear accelerators. Circular accelerators result in progressively increasing levels of energy and occupy much less space than an equivalent linear accelerator. (vi) In an accelerator, two high-speed protons collide and a series of new particles are produced, in addition to the two original protons. Explain why new particles are produced. The kinetic energy of the two protons gets converted into mass. (vii) A huge collection of new particles was produced using circular accelerators. The quark model was proposed to put order on the new particles. List the six flavours of quark. Up, down, strange, charm, top and bottom. (viii) Give the quark composition of the proton. Up, up, down. 12 (a) (i) State the principle of conservation of energy. Energy cannot be created or destroyed but it can only be changed from one form to another . (ii) A basketball of mass 600 g which was resting on a hoop falls to the ground 3.05 m below. What is the maximum kinetic energy of the ball as it falls? KE = PE (at height of 3.05 m) v2 = u2 +2as ⇒ v2 = 0 + 2(9.8)(3.05) ⇒ v2 = 59.78 {you could also have use P.E. = mgh} Ek = ½mv 2 = ½ (0.60)(59.78) ⇒ Ek = 17.9 J (iii) On bouncing from the ground the ball loses 6 joules of energy. What happens to the energy lost by the ball? It changes into sound and heat. (iv) Calculate the height of the first bounce of the ball. [retained energy = 17.9 – 6] ⇒ E = 11.9 J E = mgh ⇒ h = E / mg ⇒ h = 11.9 /(0.600)(9.8) ⇒ h = 2.02 m 12 (b) (i) Define magnetic flux. Magnetic flux is defined as the product of magnetic flux density multiplied by area. (ii) State Faraday’s law of electromagnetic induction. The size of the induced emf is proportional to the rate of change of flux. (iii) A square coil of side 5 cm lies perpendicular to a magnetic field of flux density 4.0 T. The coil consists of 200 turns of wire. What is the magnetic flux cutting the coil? A = (0.05)2 = 0.0025 φ (= BA ) = (4)(0.0025) φ = 0.01 Wb (iv) The coil is rotated through an angle of 90o in 0.2 seconds. Calculate the magnitude of the average e.m.f. induced in the coil while it is being rotated. E = N(Δφ/Δt) Δφ /Δt = (0.01 – 0 )/0.2 = 0.05 E= 200(0.05) ⇒ E = 10 V 12 (c) (i) The frequency of a stretched string depends on its length. Give two other factors that affect the frequency of a stretched string. Tension and mass per unit length (ii) The diagram shows a guitar string stretched between supports 0.65 m apart. The string is vibrating at its first harmonic. The speed of sound in the string is 500 m s–1. What is the frequency of vibration of the string? λ = 2 × 0.65 = 1.3 m v = fλ ⇒ f = v/ λ = 500 / 1.3 ⇒ f = 384.6 Hz (iii) Draw a diagram of the string when it vibrates at its second harmonic. (iv) What is the frequency of the second harmonic? f2nd = 2f1st = 769.2 Hz 12 (d) (i) One hundred years ago, Albert Einstein explained the photoelectric effect. What is the photoelectric effect? The photoelectric effect is the emission of electrons from the surface of a metal when light of suitable frequency / energy shines on it. (ii) Write down an expression for Einstein’s photoelectric law. hf = φ + ½mv2 (iii) Summarise Einstein’s explanation of the photoelectric effect. Light is composed of packets (or bundles) of energy which he called photons. All of energy from one photon is given to one electron. Energy must be greater than the work function of the metal for the photoelectric effect to occur. (iv) Give one application of the photoelectric effect. Sound track in film, photography, counters, photocell, burglar alarm, automatic doors, etc.