5054_w22_qp_21
A paper of Physics, 5054
Questions:
9
Year:
2022
Paper:
2
Variant:
1
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1
shows a steel spring. Describe an experimental procedure for obtaining the results needed to plot an extension–load graph for this spring. In an experiment to obtain results for an extension–load graph, a spring is stretched beyond its limit of proportionality. On , sketch the extension–load graph for this spring. load extension On your graph in , mark and label with a P the limit of proportionality. The limit of proportionality for this spring occurs at a load of 8.5 N. The extension of the spring is equal to 0.014 m when the load is equal to 3.5 N. Calculate the extension for a load of 5.5 N. extension =
1. Motion, forces and energy  →  1.5.3. Elastic deformation
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2
A child’s toy consists of a flexible track and a model car. shows a diagram of the toy. D B E F C G track 0.45 m A model car The child first holds the car stationary at point A which is 0.45 m above the horizontal sections of track BC and FG. The mass of the car is 0.12 kg. The child then releases the car which travels towards point B. Both air resistance and friction between the car and the track are negligible. The gravitational field strength g is 10 N / kg. Calculate the change in gravitational potential energy (g.p.e.) of the car as it travels from A to B. change in g.p.e. = Calculate the speed of the car when it reaches B. speed = After releasing it, the child expects the car to follow the track along the route ABCDEFG. In fact, the model car does not reach F. Explain, in terms of energy, why the car does not go past D, which is also 0.45 m above the horizontal track. Immediately after being released at A, the car travels to B, to C and then to D. Describe the motion of the car after it reaches D.
1. Motion, forces and energy  →  1.7.1. Energy
1. Motion, forces and energy  →  1.2. Motion
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3
A curved, glass tube is open at one end and sealed at the other. A dense liquid is poured into the tube. The liquid traps air in the sealed end. shows the tube, the liquid and the trapped air. x h liquid trapped air curved tube The difference between the liquid levels is h. At room temperature, h is 0.57 m. The density of the liquid is 1.4 × 104 kg / m3. The gravitational field strength g is 10 N / kg and the atmospheric pressure is 1.0 × 105 Pa. Calculate the pressure of the trapped air. pressure = The trapped air in the tube is heated. The height of the trapped air in the tube is x. Explain, in terms of molecules, why x changes when the air is heated. The trapped air reaches a constant temperature that is greater than its initial temperature. Describe and explain the change in h in terms of the pressures involved.
1. Motion, forces and energy  →  1.8. Pressure
2. Thermal physics  →  2.1.2. Particle model
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5
is a diagram that shows the number of particles in a charged atom of beryllium (Be). nucleus electron The charge on an electron is –1.6 × 10–19 C. The charge on the ion is +3.2 × 10–19 C. Deduce, from the charge on the ion, the number of protons in its nucleus. Show your working. number of protons = Write down, in nuclide notation, the symbol for the nucleus of this charged atom. The isotope in is the only stable isotope of beryllium. Explain the term ‘isotope’. Explain what prevents electrons in an atom from escaping.
5. Nuclear physics  →  5.1.2. The nucleus
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6
An electromagnet is used to separate objects that are magnetic from objects that are non-magnetic. shows the electromagnet suspended from the arm of a crane. electromagnet crane electric cable State the name of the metal from which the core of the electromagnet is made. The electromagnet is powered by a 220 V d.c. (direct current) supply. The electromagnet is switched on and the current in the circuit is 39 A. Calculate the power transferred to the electromagnet. power = The d.c. supply is a set of batteries. The batteries power the magnet for 4.5 hours before they need to be replaced. Calculate the charge driven around the complete circuit in this time. charge = On a very cold morning, when the electromagnet is first switched on, the current is greater than 39 A before decreasing to the usual value. Explain why the current is initially greater than 39 A.
4. Electricity and magnetism  →  4.5.3. Magnetic effect of a current
4. Electricity and magnetism  →  4.1. Simple magnetism and magnetic fields
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7
shows a child sitting on a sledge on a snow-covered hill of constant slope. sledge brake child snow-covered hill wall (not to scale) At time t = 0, the child and the sledge begin to move down the hill in a straight line. When the child sees a wall ahead, he applies the brake. The child and sledge continue to travel in a straight line until they come to a stop before hitting the wall. is the speed-time graph for the journey. t / s speed m / s The brake is applied at t = 26 s. Fig 7.2 shows how the speed of the child and sledge varies over the whole of the journey. Explain why, between t = 0 and t = 26 s, the speed varies in the way shown by the graph. At t = 26 s, the front of the sledge is 35 m from the wall. Determine the distance between the front of the sledge and the wall when the sledge stops. distance = At t = 26 s, the child and sledge begin to decelerate. Determine the size of the deceleration. deceleration = The mass of the child is 46 kg and the mass of the sledge is 9.0 kg. Calculate the resultant force on the child and sledge as they decelerate. resultant force = State the energy transfer that is taking place as the child and sledge decelerate. At t = 26 s, when the brake is first applied, the child jerks forwards on the sledge. Explain why.
1. Motion, forces and energy  →  1.2. Motion
1. Motion, forces and energy  →  1.7.2. Work
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8
A parallel beam of light travelling in air is incident on a glass lens. The beam is perpendicular to the lens as shown in . The dashed line P indicates the position of the lens. The centre of the lens is dot C. P C rays of light The lens is a diverging lens. On : indicate the shape of the lens by drawing the outline of the lens around dashed line P draw the path taken by each ray of light after it passes through the lens. Diverging lenses are used to correct short-sight. is a simplified diagram of a short-sighted eye. Light from a distant object strikes the eye lens and enters the eye. On , continue the three rays in the eye until they reach the back of the eye. eye lens light from distant object back of eye short-sighted eye State how the image of a distant object detected by a normal eye differs from the image detected by the short-sighted eye. Explain how a diverging lens corrects the sight of a short-sighted eye viewing a distant object. The focal length of the diverging lens is 4.0 cm. An object of height 3.5 cm is placed 6.0 cm from the centre of the lens. is a full-scale diagram drawn on a grid, on which the dashed line L represents the lens and the arrow O the object. O L 1 cm 1 cm By drawing on , find the position of the image I of object O. Draw image I and label it I. Explain whether the image produced is real or virtual. On the grid in , write an E in a position from which an eye can see the image. Determine the linear magnification produced. linear magnification =
3. Waves  →  3.2.3. Thin lenses
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9
The power supply in an electric circuit is a battery of electromotive force (e.m.f.) 12 V. State two ways in which the e.m.f. of a battery differs from that of an alternating current (a.c.) power supply. The circuit includes three resistors and two open switches, S1 and S2. shows the circuit. 30 Ω 20 Ω 28 Ω S1 S2 12 V N S Z Y YZ is a straight, horizontal section of connecting wire that lies between two magnets. S1 is now closed. Calculate the current in YZ. current = Explain why YZ experiences a force. Tick the box which describes the direction of the force on YZ. towards N towards Z towards S downwards towards Y upwards Explain how the direction of the force on YZ is determined. Switch S2 in the circuit in is now also closed. Calculate the total resistance of the circuit. resistance = Explain what happens to the force on YZ as switch S2 is closed. The current in the 20 Ω resistor is I20. The current in the 30 Ω resistor is I30. State a value for the ratio I20 / I30. ratio I20 / I30 =
4. Electricity and magnetism  →  4.5.4. Forces on a current-carrying conductor
4. Electricity and magnetism  →  4.3.2. Series and parallel circuits
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