OCR A Physics OCR Paper 1 2019

  • SECTION A

  • You should spend a maximum of [math]30\ \text{minutes}[/math] on this section.

  • Write your answer to each question in the box provided.

  • Answer all the questions

  • 1. Which set of prefixes A, B, C or D are in order of increasing magnitude?
  • A. micro, milli, centi, kilo
  • B. milli, centi, micro, kilo
  • C. kilo, centi, milli, micro
    D. centi, micro, milli, kilo
  • Answer: A
  • Reason:
  • SI prefixes values are:
  • [math]\text{micro}\ (\mu)=10^{-6}=0.000001[/math]
  • [math]\text{milli}\ (m)=10^{-3}=0.001[/math]
  • [math]\text{centi}\ (c)=10^{-2}=0.01[/math]
  • [math]\text{kilo}\ (k)=10^{3}=1000[/math]
  • Increasing magnitude means smallest → largest:
  • [math]\text{micro}<\text{milli}<\text{centi}<\text{kilo}[/math]
  • So correct order is:
  • [math]10^{-6}<10^{-3}<10^{-2}<10^{3}[/math]
  • Therefore, Option A is correct.
  • 2. A paper cone is held above the ground and dropped. It falls vertically and reaches terminal velocity before it hits the ground.
  • Which statement correctly describes the resultant force on the falling cone before it reaches terminal velocity?
  • A. decreasing and upwards
  • B. decreasing and downwards
  • C. increasing and downwards
  • D. increasing and upwards
  • Answer: B
  • Reason:
  • Forces on cone:
  • – Weight acts downward (constant).
  • – Air resistance acts upward.
  • As speed increases:
  • – Air resistance increases.
  • – Resultant force decreases.
  • Before terminal velocity:
  • [math]\text{Weight}>\text{Air Resistance}[/math]
  • So resultant force is:
  • – downward
  • – decreasing
  • 3. A solid cylindrical glass rod has length [math]20.0\pm0.1\ \text{cm}[/math] and diameter [math]5.00\pm0.01\ \text{mm}[/math].
  • What is the percentage uncertainty in the calculated volume of this rod?
  • A. [math]0.1\text{%}[/math]
  • B. [math]0.2\text{%}[/math]
  • C. [math]0.7\text{%}[/math]
  • D. [math]0.9\text{%}[/math]
  • Answer: D
  • Reason:
  • Length [math]=l=20.0\pm0.1\ \text{cm}[/math]
  • Diameter [math]=d=5.00\pm0.01\ \text{mm}[/math]
  • Volume of cylinder:
  • [math]V=\pi r^{2}l[/math]
  • Percentage uncertainty in length:
  • [math]\frac{0.1}{20.0}\times100[/math]
  • [math]=0.5\text{%}[/math]
  • Percentage uncertainty in diameter:
  • [math]\frac{0.01}{5.00}\times100[/math]
  • [math]=0.2\text{%}[/math]
  • Since diameter is squared:
  • [math]2(0.2\text{%})=0.4\text{%}[/math]
  • Add uncertainties:
  • [math]0.5\text{%}+0.4\text{%}=0.9\text{%}[/math]
  • 4. A simple harmonic oscillator has maximum speed [math]24\ \text{m}\ \text{s}^{-1}[/math] and amplitude [math]5.6\ \text{cm}[/math].
  • What is its angular frequency?
  • A. [math]0.23\ \text{rad}\ \text{s}^{-1}[/math]
  • B. [math]21\ \text{rad}\ \text{s}^{-1}[/math]
  • C. [math]68\ \text{rad}\ \text{s}^{-1}[/math]
  • D. [math]430\ \text{rad}\ \text{s}^{-1}[/math]
  • Answer: D
  • Reason:
  • A simple harmonic oscillator has:
  • Maximum speed [math]= 24\ \text{m}\ \text{s}^{-1}[/math]
  • Amplitude [math]= 5.6\ \text{cm} = 0.056\ \text{m}[/math]
  • Find angular frequency.
  • For SHM:
  • [math]v_{\text{max}} = \omega A[/math]
  • So:
  • [math]\omega = \frac{v_{\text{max}}}{A}[/math]
  • Substitute values:
  • [math]\omega = \frac{24}{0.056}[/math]
  • [math]\omega = 428.57 \approx 430\ \text{rad}\ \text{s}^{-1}[/math]
  • 5. A car is dripping oil at a steady rate on a straight road. The road is divided into four sections A, B, C, and D.
  • Which section of the road shows the car travelling at a constant speed?
  • Answer: C
  • Reason:
  • Since oil drops fall at equal time intervals, the distance between drops shows how far the car travels in equal times.
  • – Large spacing = high speed
  • – Small spacing = low speed
  • – Equal spacing = constant speed
  • In section C, the oil drops are equally spaced.
  • This means:
  • – In every equal time interval, the car covers the same distance.
  • – So, speed is constant.
  • Hence, constant speed = Section C.
  • 6. The Earth is surrounded by a gravitational field.
  • Which of the following statements is/are correct about the gravitational field lines near the surface of the Earth.
  • 1. They are parallel.
  • 2. They show the direction of the force on a small mass.
  • 3. They are equally spaced.
  • A. Only 1
  • B. Only 1 and 2
  • C. Only 2 and 3
  • D. 1, 2 and 3
  • Answer: D
  • Reason:
  • “They are parallel.”
  • True
  • Near Earth’s surface field lines are nearly parallel.
  • “They show direction of force on a small mass.”
  • True
  • Field lines indicate force direction.
  • “They are equally spaced.”
  • True
  • Uniform field → equal spacing.
  • D (1, 2 and 3)
  • 7. A pendulum bob is oscillating in a vacuum.
  • The maximum height of the bob from the ground is [math]1.3\ \text{m}[/math] and its minimum height is [math]1.1\ \text{m}[/math].
  • What is the maximum speed of the pendulum bob?
  • A. [math]2.0\ \text{m}\ \text{s}^{-1}[/math]
  • B. [math]3.9\ \text{m}\ \text{s}^{-1}[/math]
  • C. [math]5.1\ \text{m}\ \text{s}^{-1}[/math]
  • D. [math]26\ \text{m}\ \text{s}^{-1}[/math]
  • Answer: A
  • Reason:
  • Highest Point: [math]h_{1} = 1.3\ \text{m}[/math]
  • Lowest Point: [math]h_{2} = 1.1\ \text{m}[/math]
  • Height Fallen: [math]\Delta h = 0.2\ \text{m}[/math]
  • Use Conservation of energy:
  • Loss in GPE:
  • [math]GPE = mg\Delta h[/math]
  • Gain in KE:
  • [math]KE = \frac{1}{2}mv^{2}[/math]
  • [math]mg\Delta h = \frac{1}{2}mv^{2}[/math]
  • [math]2g\Delta h = v^{2}[/math]
  • [math]v = \sqrt{2g\Delta h}[/math]
  • [math]v = \sqrt{2(9.8)(0.2)}[/math]
  • [math]v = 1.98\ \text{m}.\ \text{s}^{-1}[/math]
  • 8. An object is falling.
  • The weight of the object is [math]4.5\ \text{N}[/math].
  • The wind provides a horizontal force of magnitude [math]F[/math] on the object.
  • The resultant force on the object is [math]5.8\ \text{N}[/math].
  • Air resistance and upthrust on the object are negligible.
  • What is the value of [math]F[/math]?
  • A. [math]1.3\ \text{N}[/math]
  • B. [math]3.7\ \text{N}[/math]
  • C. [math]7.3\ \text{N}[/math]
  • D. [math]13\ \text{N}[/math]
  • Answer: B
  • Reason:
  • Weight: [math]W = 4.5\ \text{N}[/math]
  • Horizontal force: [math]F = ?[/math]
  • Resultant: [math]R = 5.3\ \text{N}[/math]
  • According to Pythagoras theorem:
  • [math]R^{2} = W^{2} + F^{2}[/math]
  • [math](5.8)^{2} = (4.5)^{2} + F^{2}[/math]
  • [math]F^{2} = 33.64 – 20.25[/math]
  • [math]F^{2} = 13.39[/math]
  • [math]F = 3.66\ \text{N}[/math]
  • [math]F = 3.7\ \text{N}[/math]
  • 9. A solid molecular substance is supplied with energy and it starts to melt.
  • Which of the following pairs of quantities remains the same as the substance melts?
  • A. Kinetic energy of molecules and internal energy of molecules.
  • B. Potential energy of molecules and internal energy of molecules.
  • C. Kinetic energy of molecules and temperature of substance.
  • D. Potential energy of molecules and temperature of substance.
  • Answer: C
  • Reason:
  • During melting:
  • – Temperature remains constant.
  • – Average kinetic energy remains constant.
  • Energy supplied increases intermolecular separation.
  • Therefore:
  • – Potential energy increases.
  • – Internal energy increases.
  • 10. Which of the following shows the correct base units for pressure?
  • A. [math]\text{kg}\ \text{m}^{-2}[/math]
  • B. [math]\text{kg}\ \text{m}^{-2}\text{s}^{-2}[/math]
  • C. [math]\text{kg}\ \text{m}^{-1}\text{s}^{-2}[/math]
  • D. [math]\text{kg}\ \text{m}^{2}\text{s}^{-3}[/math]
  • Answer: C
  • Reason:
  • Pressure:
  • [math]P=\frac{F}{A}[/math]
  • Force:
  • [math]F=ma[/math]
  • [math]F=(\text{kg})(\text{m}.\ \text{s}^{-2})[/math]
  • [math]F=\text{kg}\text{m}.\ \text{s}^{-2}[/math]
  • Therefore:
  • [math]P=\frac{F}{A}[/math]
  • [math]P=\frac{\text{kg}\text{m}.\ \text{s}^{-2}}{\text{m}^{2}}[/math]
  • [math]P=\text{kg}\text{m}^{-1}.\ \text{s}^{-2}[/math]
  • 11. A student has collected some data on the Solar System.
  • The student plots a graph, but only two data points are shown below.
  • The distance from the centre of the Sun is [math]r[/math].
  • Which quantity [math]y[/math] is represented on the vertical axis?
  • A. Speed of a planet.
  • B. Period of a planet.
  • C. Gravitational potential of the Sun.
  • D. Gravitational field strength of the Sun.
  • Answer: B
  • Reason:
  • The graph shows:
  • – [math]y[/math] increases as [math]r[/math] increases.
  • Check each quantity:
  • A Speed:
  • [math]v\propto\frac{1}{\sqrt{r}}[/math]
  • Decreases with [math]r[/math].
  • 12. A graph showing the variation of the stress [math]\sigma[/math] with strain [math]\varepsilon[/math] for a material is shown below.
  • What is the Young modulus of the material?
  • A. [math]6.0\times10^{4}\ \text{Pa}[/math]
  • B. [math]1.2\times10^{9}\ \text{Pa}[/math]
  • C. [math]8.0\times10^{10}\ \text{Pa}[/math]
  • D. [math]1.2\times10^{11}\ \text{Pa}[/math]
  • Answer: D
  • Reason:
  • Young modulus is:
  • [math]E=\frac{\text{stress}}{\text{strain}}[/math]
  • Use the straight-line (elastic) region of the graph.
  • From the graph:
  • At
  • [math]\varepsilon=0.10\text{%}[/math]
  • And
  • [math]\sigma=120\ \text{MPa}[/math]
  • Convert units:
  • [math]120\ \text{MPa}=120\times10^{6}\ \text{Pa}[/math]
  • [math]0.10\text{%}=\frac{0.10}{100}=0.001[/math]
  • Calculation:
  • [math]E=\frac{\text{stress}}{\text{strain}}[/math]
  • [math]E=\frac{120\times10^{6}}{0.001}[/math]
  • [math]E=1.2\times10^{11}\ \text{Pa}[/math]
  • 13. Which column A, B, C or D, shows the correct sequence for the evolution of the Universe between the Big Bang and the formation of stars?
  •             A              B             C             D
       Universe starts to           expand

                    ↓

     quarks and leptons

                form

                    ↓

         hadrons form

                   ↓

         nuclei form

                  ↓

         atoms form

    		 Universe starts to            expand

                    ↓

        hadrons form

                    ↓

     quarks and leptons               form

                     ↓

              nuclei form

                     ↓

              atoms form

    		quarks and leptons

                  form

                     ↓

             nuclei form

                     ↓

    Universe starts to            expand

                      ↓

             atoms form

                      ↓

           hadrons form

    		quarks and leptons

                  form

                     ↓

           hadrons form

                     ↓

    Universe starts to             expand

                     ↓

             nuclei form

                     ↓

              atoms form

  • Answer: A
  • Reason:
  • After the Big Bang:
  • 1. Universe starts to expand
  • 2. Quarks and leptons form
  • 3. Hadrons form (protons and neutrons)
  • 4. Nuclei form
  • 5. Atoms form
  • 6. Stars form later
  • Only A follows the correct order:
  • [math]\text{Expansion}\rightarrow\text{Quarks and leptons}\rightarrow\text{Hadrons}\rightarrow\text{Nuclei}\rightarrow\text{Atoms}[/math]
  • Immediately after the Big Bang the Universe expanded rapidly.
  • As it cooled:
  • – Quarks and leptons were produced.
  • – Quarks combined to form hadrons.
  • – Protons and neutrons combined to form nuclei.
  • – Electrons joined nuclei to form atoms.
  • – Gas clouds later collapsed to form stars.
  • 14. Some stars will evolve into white dwarfs.
  • The mass of the Sun is [math]2.0\times10^{30}\ \text{kg}[/math].
  • Which of the following cannot be the mass of a white dwarf?
  • A. [math]1.2\times10^{30}\ \text{kg}[/math]
  • B. [math]2.0\times10^{30}\ \text{kg}[/math]
  • C. [math]2.7\times10^{30}\ \text{kg}[/math]
  • D. [math]3.2\times10^{30}\ \text{kg}[/math]
  • Answer: D
  • Reason:
  • Mass of Sun:
  • [math]M_{\odot}=2.0\times10^{30}\ \text{kg}[/math]
  • Important Physics:
  • A white dwarf cannot exceed the Chandrasekhar Limit:
  • [math]1.4M_{\odot}=1.4(2.0\times10^{30})=2.8\times10^{30}\ \text{kg}[/math]
  • White dwarfs are supported by electron degeneracy pressure.
  • If the mass exceeds approximately
  • [math]1.4M_{\odot}[/math]
  • The white dwarf becomes unstable and may collapse into a neutron star.
  • Since
  • [math]3.2\times10^{30}\ \text{kg}[/math]
  • Is greater than the Chandrasekhar limit, it cannot be the mass of a white dwarf.
  • 15. An astronomer analyses the light from a distant galaxy.
  • One of the spectral lines in the spectrum observed from the galaxy has wavelength [math]610\ \text{nm}[/math].
  • The same spectral line has a wavelength of [math]590\ \text{nm}[/math] when measured in the laboratory.
  • What is the speed of this galaxy?
  • A. [math]9.8\times10^{6}\ \text{m}\ \text{s}^{-1}[/math]
  • B. [math]1.0\times10^{7}\ \text{m}\ \text{s}^{-1}[/math]
  • C. [math]2.9\times10^{8}\ \text{m}\ \text{s}^{-1}[/math]
  • D. [math]3.0\times10^{8}\ \text{m}\ \text{s}^{-1}[/math]
  • Answer: B
  • Reason:
  • Speed of the Galaxy:
  • Given:
  • Observed wavelength:
  • [math]\lambda_{0}=610\ \text{nm}[/math]
  • Laboratory wavelength:
  • [math]\lambda=590\ \text{nm}[/math]
  • Speed of light:
  • [math]c=3.0\times10^{8}\ \text{m}.\ \text{s}^{-1}[/math]
  • Calculate Redshift:
  • [math]z=\frac{\Delta\lambda}{\lambda}[/math]
  • [math]z=\frac{610-590}{590}[/math]
  • [math]z=\frac{20}{590}[/math]
  • [math]z=0.0339[/math]
  • Use Doppler Approximation:
  • For small redshifts:
  • [math]z=\frac{v}{c}[/math]
  • Therefore:
  • [math]v=zc[/math]
  • [math]v=(0.0339)(3.0\times10^{8})[/math]
  • [math]v=1.02\times10^{7}\ \text{m}.\ \text{s}^{-1}[/math]
  • [math]v=1.0\times10^{7}\ \text{m}.\ \text{s}^{-1}[/math]
  • Q No.16
  • (a) Explain what is meant by the ultimate tensile strength of a material.
  • Answer:
  • Ultimate Tensile Strength
  • Ultimate tensile strength (UTS) is the maximum tensile stress that a material can withstand when it is stretched before it begins to fail (neck) or fracture.
  • When a material is subjected to a tensile force:
  • 1. It first deforms elastically.
  • 2. It may then deform plastically.
  • 3. The stress continues to increase until it reaches a maximum value.
  • 4. This maximum value of stress is called the ultimate tensile strength.
  • 5. After this point, the material starts to neck and eventually breaks.
  • [math]\text{Ultimate tensile strength}=\frac{\text{Maximum Tensile Force}}{\text{Original cross sectional Area}}[/math]
  • (b) A footbridge is supported by a number of metal cables of the same length.
  • Each cable has uniform cross-section and diameter [math]4.20\ \text{mm}[/math] as shown in Fig. 16.1.
  • A group of engineers investigate how the extension [math]x[/math] varies with applied force [math]F[/math] for one of the cables.
  • The results of the investigation are shown in Fig. 16.2.
  • The cable breaks when the force is [math]2.2\ \text{kN}[/math].
  • (i) Describe how a suitable measuring device may have been used by the engineers to demonstrate that the cable had uniform cross-section.
  • Answer:
  • Use a micrometer screw gauge to measure the diameter of the cable at several positions along its length and in different directions.
  • A cable has a uniform cross-section if its diameter is the same throughout.
  • The engineers could:
  • 1. Measure the diameter at many points along the cable using a micrometer.
  • 2. Rotate the micrometer through [math]90^\circ[/math] at each position to check for circularity.
  • 3. Compare all measurements.
  • If all measured diameters are the same (within experimental uncertainty), the cable has a uniform cross-section.
  • (ii) State any value of [math]F[/math] when the cable behaves
  • 1. Elastically
  • 2. Plastically.
  • Answer:
  • Elastically:
  • The graph is a straight line from [math]0[/math] to approximately [math]4\ \text{mm}[/math] extension and [math]2.0\ \text{kN}[/math] force.
  • In this region:
  • – Force is proportional to extension.
  • – Hooke’s Law is obeyed.
  • – The cable returns to its original length when the force is removed.
  • This is the elastic region.
  • Therefore, a force such as:
  • [math]1.5\ \text{kN}[/math]
  • is acceptable.
  • Plastically:
  • After about [math]2.0\ \text{kN}[/math], the graph begins to curve.
  • This means:
  • – Force is no longer proportional to extension.
  • – Hooke’s Law is no longer obeyed.
  • – Permanent deformation occurs.
  • This is the plastic region.
  • Hence:
  • [math]2.1\ \text{kN}[/math]
  • is a suitable answer.
  • (iii) Use Fig. 16.2 to determine the force constant [math]k[/math] in [math]\text{N}.\ \text{m}^{-1}[/math] of the cable.
  • Answer:
  • Given from graph:
  • Choose a point in the straight-line region:
  • [math]F=2.0\ \text{kN}[/math]
  • [math]x=4.0\ \text{mm}[/math]
  • Convert units:
  • [math]F=2000\ \text{N}[/math]
  • [math]x=4.0\times10^{-3}\ \text{m}[/math]
  • Formula:
  • [math]F=kx[/math]
  • [math]k=\frac{F}{x}[/math]
  • [math]k=\frac{2000}{4.0\times10^{-3}}[/math]
  • [math]k=5.0\times10^{5}\ \text{N}.\ \text{m}^{-1}[/math]
  • The force constant is the gradient of the straight-line section of the force-extension graph.
  • [math]k=\frac{\Delta F}{\Delta x}[/math]
  • Using any point from the elastic region gives:
  • [math]5.0\times10^{5}\ \text{N}.\ \text{m}^{-1}[/math]
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