ZIMSEC A Level Physics Paper 3 November 2004

Question 101

[1 marks]work and energy

Define work done by a force.

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Question 102

[2 marks]work and energy

Starting from W = Fx, with F = ma and v^2 = u^2 + 2ax for a body accelerated from rest (u = 0), what equation for kinetic energy Ek does this derivation give, in terms of mass m and speed v?

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Question 103

[2 marks]work and energy

A cyclist and bicycle of total mass 69.0 kg move at 5.5 m/s at the top of a rise. Calculate the cyclist's kinetic energy at the top of the rise.

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Question 104

[2 marks]work and energy

The same cyclist and bicycle (total mass 69.0 kg, kinetic energy 1.04x10^3 J at the top) roll down without pedalling to level ground 4.6 m below the top. Taking g = 9.81 m/s^2, calculate the kinetic energy at the bottom of the rise.

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Question 105

[3 marks]work and energy

Using a kinetic energy of 4.15x10^3 J at the bottom of the rise and a total mass of 69.0 kg, calculate the cyclist's speed at the bottom.

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Question 106

[2 marks]work and energy

The cyclist's speed is 5.5 m/s at the top of the rise and about 11.0 m/s at the bottom. Which statement best compares these two speeds?

AThe final speed is roughly double the initial speed, since lost gravitational PE becomes kinetic energy.
BThe two speeds are approximately equal, since the rise is too short to matter.
CThe final speed is about four times the initial speed, matching the change in kinetic energy gained from the fall down the slope.
DThe final speed is about half the initial speed, since friction removes most of the energy.

Question 107

[2 marks]work and energy

While cycling at a constant 5.5 m/s along level ground, the cyclist's kinetic energy is not increasing. Where does the energy from pedalling go?

AIt is stored as elastic strain energy in the tyres and released later as the bicycle slows.
BIt builds up as extra kinetic energy too small to be measured with ordinary instruments.
CIt converts into heat from muscles and friction, and heat and sound lost to air resistance.
DIt converts entirely into gravitational potential energy stored in the bicycle frame.

Question 201

[1 marks]gravitational fields and orbits

Define gravitational field strength at a point.

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Question 202

[1 marks]gravitational fields and orbits

The newton (N) expands to the SI base units kg m s^-2. Which set of base units does N kg^-1 (gravitational field strength) reduce to?

Akg m^-1 s^-2 kg^-1 = m^-1 s^-2
Bkg^2 m s^-2 kg^-1 = kg m s^-2
Ckg m s^-2 kg^-1 = m s^-2
Dkg m^2 s^-2 kg^-1 = m^2 s^-2

Question 203

[1 marks]gravitational fields and orbits

State the value of the gravitational field strength at the Earth's surface.

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Question 204

[2 marks]gravitational fields and orbits

The mass of the Earth is 6.0x10^24 kg and the gravitational field strength at its surface is 9.81 N/kg. Using g = GM/r^2, with G = 6.67x10^-11 N m^2 kg^-2, calculate the radius of the Earth.

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Question 205

[2 marks]gravitational fields and orbits

Using the Earth's radius (6.39x10^6 m) and mass (6.0x10^24 kg), with G = 6.67x10^-11 N m^2 kg^-2, calculate the gravitational field strength at a point 2.4x10^5 m above the Earth's surface.

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Question 206

[1 marks]gravitational fields and orbits

In calculating g above the Earth's surface from g = GM/r^2, what assumption about the Earth must be made for this formula to apply?

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Question 207

[2 marks]gravitational fields and orbits

Why can objects inside an orbiting spaceship be considered weightless, even though gravity still acts on them?

AGravity supplies exactly the centripetal force for the orbit, so there is no support force.
BAir resistance in orbit balances the weight of objects inside the cabin.
CGravity is essentially zero at typical orbital altitudes above the Earth, which is why astronauts and objects inside the cabin appear to float freely.
DThe spaceship's engines cancel out the pull of gravity during orbit.

Question 208

[2 marks]gravitational fields and orbits

A satellite orbits the Earth with a period of 4.0 hours. Using GM(Earth) = 6.67x10^-11 x 6.0x10^24 N m^2 kg^-1 and w = 2*pi/T, calculate the radius of its orbit.

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Question 209

[2 marks]gravitational fields and orbits

Using the orbital radius found above (1.28x10^7 m) and w = 4.36x10^-4 rad/s, calculate the satellite's orbital speed.

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Question 210

[2 marks]gravitational fields and orbits

A satellite needs a large orbital speed to stay in orbit. Why is this speed built up in multiple launch stages rather than given to the satellite all at once?

AA single rocket stage cannot physically generate enough force to move the satellite at all, no matter how much fuel it is given to burn.
BMulti-stage launches are simply cheaper, with no effect on the satellite's safety.
CInternational treaties on space launches require staged acceleration by law.
DReaching full speed in one burst low in the atmosphere risks overheating and burning up.

Question 301

[1 marks]wave motion and sound

For a wave travelling through a medium, how does the phase of different particles compare between a progressive wave and a stationary wave?

APhase differences only exist in stationary waves, never in progressive waves.
BIn a progressive wave nearby particles are generally out of phase; in a stationary wave, particles between two nodes are in phase.
CIn both wave types, all particles in the medium vibrate exactly in phase with each other at every instant, with no relative phase difference ever appearing between any two particles.
DIn a progressive wave all particles are in phase; in a stationary wave none of them are.

Question 302

[1 marks]wave motion and sound

How does the amplitude of vibration of different particles compare between a progressive wave and a stationary wave?

AIn a stationary wave every particle has the same, constant amplitude of vibration, identical to every other particle regardless of its position between two adjacent nodes.
BAmplitude is undefined for stationary waves, since their particles do not move.
CIn a progressive wave amplitude is constant for all particles; in a stationary wave it varies from zero at nodes to a maximum at antinodes.
DIn a progressive wave the amplitude varies from zero to a maximum along the wave.

Question 303

[1 marks]wave motion and sound

How does the motion of the wave profile itself differ between a progressive wave and a stationary wave?

ABoth wave profiles travel at the same constant speed through the medium.
BIn a progressive wave the profile stays fixed in place while the medium moves through it; in a stationary wave the whole profile instead travels bodily along the medium.
CNeither wave type has a profile that can be said to move or stay still.
DIn a progressive wave the profile travels through the medium; in a stationary wave it appears to stay in place.

Question 304

[2 marks]wave motion and sound

Fig. 3.1 shows a snapshot of a sound wave's displacement against distance. Read off the graph to find the wavelength of the wave.

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Question 305

[1 marks]wave motion and sound

Fig. 3.1 shows the wave's displacement axis labelled in micrometres (um), reaching a peak displacement of 10 on that axis. State the amplitude of the wave.

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Question 306

[2 marks]wave motion and sound

Fig. 3.1 shows two particles, A (near x = 0.7 m) and B (near x = 1.5-1.8 m), on a wave of wavelength 2.0 m. Using the path difference between A and B as a fraction of the wavelength, calculate the phase difference between the vibrations of A and B, in radians.

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Question 307

[1 marks]wave motion and sound

Fig. 3.2 shows the same wave's displacement plotted against time, with a repeating pattern every 6.0 ms. State the period of oscillation.

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Question 308

[2 marks]wave motion and sound

Using a wavelength of 2.0 m and a period of 6.0 ms, calculate the speed of this sound wave.

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Question 309

[2 marks]wave motion and sound

In a resonance-tube experiment to find the speed of sound, a tuning fork of frequency f is held over a tube whose air-column length L is adjusted for resonance. A graph of L against 1/f is plotted. What does the gradient of this graph represent?

AFour times the speed of sound, 4V
BOne quarter of the speed of sound, V/4
CThe speed of sound itself, V
DThe end correction of the tube, c, which is added to the tube's physical length rather than being the resonant length itself

Question 310

[1 marks]wave motion and sound

In the resonance-tube experiment, how is the correct air-column length for resonance at a given frequency identified?

AIt is calculated first, then the tube is cut to that exact length.
BIt is found by timing an echo travelling down and back up the tube.
CIt is read directly off a ruler marked on the side of the tube.
DThe length is adjusted until the loudest note (resonance) is heard.

Question 311

[2 marks]wave motion and sound

When a 340 Hz piano key is struck, the 680 Hz and 1020 Hz keys are also found to vibrate. What term describes 680 Hz and 1020 Hz in relation to the 340 Hz fundamental?

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Question 401

[1 marks]electric circuits

Define potential difference between two points in a circuit.

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Question 402

[1 marks]electric circuits

A battery has an emf of 12 V. What does this value mean in terms of energy per unit charge?

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Question 403

[2 marks]electric circuits

A 12 V supply is connected across a 10 kilohm resistor in series with a 5 kilohm resistor. An ideal voltmeter is connected across the 5 kilohm resistor. Calculate the voltmeter reading.

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Question 404

[2 marks]electric circuits

In the circuit above, the calculation predicts 4.0 V across the 5 kilohm resistor, but the voltmeter actually reads only 2.5 V. What explains this difference?

AThe 10 kilohm resistor has a printed value that is lower than its true resistance, causing the voltmeter to read a smaller share of the supply voltage.
BThe voltmeter draws current because its own resistance is comparable to 5 kilohms, so it is not ideal.
CThe voltmeter is faulty and is reading incorrectly regardless of the circuit.
DThe 12 V supply must actually be delivering less than 12 V to the circuit.

Question 405

[3 marks]electric circuits

The voltmeter above, connected across the 5 kilohm resistor in the 12 V, 10 kilohm + 5 kilohm series circuit, reads 2.5 V. Calculate the resistance of the voltmeter.

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Question 406

[2 marks]electric circuits

The same voltmeter (resistance 5.6 kilohms) is now connected across the 10 kilohm resistor instead, in the same 12 V, 10 kilohm + 5 kilohm series circuit. Calculate the new voltmeter reading.

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Question 407

[1 marks]electric circuits

What consideration about a voltmeter's resistance is important when using it to measure potential difference accurately?

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Question 408

[1 marks]electric circuits

Name one instrument, other than an ordinary voltmeter, that can be used to obtain an accurate reading of potential difference without drawing significant current.

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Question 409

[2 marks]electric circuits

A superconductor carries current with zero electrical resistance. State one advantage this gives over an ordinary copper cable in terms of power loss.

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Question 410

[1 marks]electric circuits

State one advantage a superconducting cable can have over a copper cable in terms of its physical size.

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Question 501

[1 marks]operational amplifiers and logic gates

State one advantage of negative feedback over positive feedback in an amplifier.

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Question 502

[1 marks]operational amplifiers and logic gates

State a second advantage of negative feedback over positive feedback in an amplifier.

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Question 503

[1 marks]operational amplifiers and logic gates

State a third advantage of negative feedback over positive feedback in an amplifier.

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Question 504

[2 marks]operational amplifiers and logic gates

Fig. 5.1 shows a light-dependent resistor (LDR) and resistors forming a potential divider that feeds an op-amp; the op-amp output can energise a coil through a diode. What happens at the op-amp output when the light intensity falling on the LDR is low?

AThe input voltage from the LDR branch rises above the reference, so the output saturates positively, forward-biasing the diode and energising the coil.
BThe output stays close to 0 V, so the coil is never energised.
CThe output saturates negatively, reverse-biasing the diode so the coil stays off, exactly as it does in the high-light-intensity case described elsewhere.
DThe output oscillates rapidly between positive and negative saturation.

Question 505

[1 marks]operational amplifiers and logic gates

Using the same circuit (Fig. 5.1), what happens at the op-amp output when the light intensity on the LDR is high?

AThe output saturates negatively, so the diode is reverse-biased and the coil is not energised.
BThe output saturates positively, energising the coil exactly as in low light.
CThe output stays at exactly 0 V, holding the coil in a partly-energised state that draws half the normal current indefinitely.
DThe output amplitude simply increases in proportion to the light intensity.

Question 506

[2 marks]operational amplifiers and logic gates

In a light-operated switch circuit, a diode is placed in series with the coil driven by the op-amp output. Large currents flowing when the coil switches off could damage the op-amp if this diode were absent. What does the diode protect the op-amp from?

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Question 507

[1 marks]operational amplifiers and logic gates

Suggest one practical use for a circuit where an op-amp compares an LDR's potential divider output to a fixed reference and switches a coil accordingly.

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Question 508

[1 marks]operational amplifiers and logic gates

A logic circuit takes two inputs A and B and produces a single output Q. For every combination of A and B, Q = 1 only when A = 1 and B = 1; for every other combination, Q = 0. What single logic gate has this truth table?

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Question 509

[1 marks]operational amplifiers and logic gates

What is the minimum number of two-input NAND gates needed to build a circuit that behaves as an AND gate?

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Question 601

[1 marks]kinetic theory of gases

State one assumption made about the molecules of an ideal gas.

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Question 602

[1 marks]kinetic theory of gases

State a second assumption made about the molecules of an ideal gas.

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Question 603

[1 marks]kinetic theory of gases

State a third assumption made about the molecules of an ideal gas.

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Question 604

[1 marks]kinetic theory of gases

A molecule of mass m moves with speed u directly towards face ABCD of a cubic container (Fig. 6.1) and rebounds elastically with the same speed u, reversed in direction. What is the change in the molecule's momentum, taking the initial direction as positive?

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Question 605

[2 marks]kinetic theory of gases

The molecule above (mass m, speed u) travels a round trip of distance 2L between successive collisions with face ABCD of a cube of side L, taking time 2L/u. Using force = rate of change of momentum, find the force this one molecule exerts on face ABCD, in terms of m, u and L.

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Question 606

[1 marks]kinetic theory of gases

Using the force mu^2/L found above, and the area L^2 of face ABCD, find the pressure this one molecule exerts on face ABCD.

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Question 607

[1 marks]kinetic theory of gases

If N molecules, each of mass m and speed u, strike face ABCD in the same way, what total pressure do they exert on that face, in terms of N, m, u and L?

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Question 608

[2 marks]kinetic theory of gases

A gas molecule collides elastically with a container wall and rebounds with reversed velocity. Is the momentum of the molecule itself conserved in this collision?

AYes, because kinetic energy is conserved in an elastic collision.
BNo, the molecule's own momentum reverses; only the molecule-plus-wall system conserves momentum overall.
CYes, because the molecule's speed is unchanged by the collision.
DNo, momentum is simply destroyed when the molecule strikes the wall, since the wall itself is assumed to be infinitely rigid and immovable.

Question 609

[2 marks]kinetic theory of gases

The equation p = (1/3)*rho*<c^2> gives the pressure exerted by an ideal gas. Why does this equation include the factor 1/3?

ABecause molecular motion is shared equally between three perpendicular directions, so only a third of the mean square speed acts along any one axis.
BBecause the container is assumed to always have exactly three pairs of opposite faces, one perpendicular to each of the three coordinate directions used in the derivation.
CBecause pressure has three different possible directions it can act in in a gas.
DBecause only one third of the gas molecules are moving at any given instant.

Question 610

[1 marks]kinetic theory of gases

In the equation p = (1/3)*rho*<c^2>, what does the term <c^2> represent?

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Question 611

[2 marks]kinetic theory of gases

A container holds 1.6 kg of oxygen gas (molar mass 32 g/mol) at 30 degrees C, exerting a pressure of 5.54x10^5 Pa. Using PV = nRT with R = 8.31 J/(K mol), calculate the volume occupied by the gas.

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Question 612

[2 marks]kinetic theory of gases

Using a pressure of 5.54x10^5 Pa, a volume of 0.227 m^3 and a gas mass of 1.6 kg, calculate the root mean square speed of the oxygen molecules from p = (1/3)*rho*<c^2>.

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Question 701

[2 marks]photoelectric effect

State the photoelectric equation, using phi (or W) for the work function, f for the light frequency, h for the Planck constant and v(max) for the maximum speed of the emitted electrons.

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Question 702

[2 marks]photoelectric effect

In an experiment verifying the photoelectric effect, a stopping potential V0 is measured for several light frequencies f above the threshold, and eV0 is plotted against f. What does the gradient of this graph give?

AThe threshold frequency, f0
BThe maximum photoelectron speed, v max
CThe Planck constant, h
DThe work function of the metal, phi

Question 703

[1 marks]photoelectric effect

In the same experiment, what is observed about the photocurrent when light of frequency below the threshold frequency f0 shines on the metal, however intense the light is made?

AA current flows only after a short time delay, then stays constant.
BA current flows only when the metal surface is heated as well.
CA small current flows, growing steadily with the light's intensity.
DNo current flows at all, no matter how intense the light is.

Question 704

[2 marks]photoelectric effect

Fig. 7.1 shows current I against potential difference V for a photocell, where I becomes constant once V is positive. Why does the current stay constant (saturate) for positive V?

AA positive V pushes some of the photoelectrons back towards the metal surface before they can reach the collecting electrode and contribute to the current.
BEvery photoelectron emitted from the surface is already being collected at that point.
CThe circuit's resistance limits the current to a fixed maximum value.
DThe metal surface runs out of free electrons permanently once V is positive.

Question 705

[2 marks]photoelectric effect

Fig. 7.1 shows two current-voltage curves, A and B, for the same metal and light frequency but different intensities, and both curves meet the voltage axis at the same stopping potential V0. Why do they meet at the same V0?

AThe stopping potential depends only on the light's frequency, which is the same for A and B.
BThe stopping potential depends only on the light's intensity, which happens to match here even though the two light sources have different frequencies.
CBoth curves were measured using the exact same photocell current.
DThe stopping potential is fixed by the circuit's resistance, not by the light at all.

Question 706

[2 marks]photoelectric effect

Fig. 7.1 shows curve B lying above curve A everywhere except at the shared stopping potential V0. Away from V0, why does the current differ between the two curves at the same value of V?

ACurve B corresponds to light of a higher frequency than curve A.
BCurve B was measured with a larger stopping potential than curve A.
CCurve B corresponds to a more intense beam, so more photoelectrons are emitted per second.
DCurve B was measured with a lower work-function metal than curve A, which would also shift the stopping potential rather than just the saturation current.

Question 707

[2 marks]photoelectric effect

Ultraviolet radiation of wavelength 1.2x10^-7 m, intensity 12 W/m^2, shines on a silicon memory chip of area 9.5 cm^2, which absorbs 20% of the radiation. Assuming one electron is released per absorbed photon, calculate the number of electrons released per second.

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Question 708

[2 marks]photoelectric effect

Using the electron emission rate found above (about 1.4x10^15 electrons per second), calculate the electric current this generates.

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Question 709

[2 marks]photoelectric effect

The silicon chip's memory is not erased when exposed to red light, only to the ultraviolet radiation described above. Why does red light fail to release photoelectrons from the chip?

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Physics Paper 3 November 2004

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The questions

Question 101

Question 102

Question 103

Question 104

Question 105

Question 106

Question 107

Question 201

Question 202

Question 203

Question 204

Question 205

Question 206

Question 207

Question 208

Question 209

Question 210

Question 301

Question 302

Question 303

Question 304

Question 305

Question 306

Question 307

Question 308

Question 309

Question 310

Question 311

Question 401

Question 402

Question 403

Question 404

Question 405

Question 406

Question 407

Question 408

Question 409

Question 410

Question 501

Question 502

Question 503

Question 504

Question 505

Question 506

Question 507

Question 508

Question 509

Question 601

Question 602

Question 603

Question 604

Question 605

Question 606

Question 607

Question 608

Question 609

Question 610

Question 611

Question 612

Question 701

Question 702

Question 703

Question 704

Question 705

Question 706

Question 707

Question 708

Question 709

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