4/18/25, 5:05 AM
QuestionBank Test
Questions
20N.2.HL.TZ0.A
Explain, by reference to Faraday’s law of induction, how an electromotive force (emf) is induced in the coil.
[3]
EXE.2.SL.TZ0.16B
(b)
The magnetic field strength of Earth’s field at the location of the wires is 45 μT.
Discuss the assumption made in this question.
[3]
23M.2.HL.TZ1.6C
(c)
The charges Q are replaced by neutral masses M and the charge q by a neutral mass m . The mass m is displaced away from C by a
small distance x and released. Discuss whether the motion of m will be the same as that of q .
[2]
21N.1A.HL.TZ0.33
A small magnet is released from rest to drop through a stationary horizontal conducting ring.
[1]
What is the variation with time of the emf induced in the ring?
22N.2.HL.TZ0.5
(a)
State what is meant by an ideal voltmeter.
[1]
SPM.1A.HL.TZ0.25
A planet orbits the Sun in an elliptical orbit moving in the direction shown.
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[1]
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QuestionBank Test
At the position shown, which quantity is decreasing for the planet?
A. Acceleration
B. Angular momentum
C. Kinetic energy
D. Gravitational potential energy
19M.1A.SL.TZ2.23
Which graph shows the relationship between gravitational force F between two point masses and their separation r ?
[1]
21N.2.HL.TZ0.5A
(a)
Show that the speed of the loop is 20 cm s −1 .
[1]
21M.1A.HL.TZ1.31
Which is a correct unit for gravitational potential?
A. m 2 s −2
B. J kg
C. m s −2
D. N m −1 kg −1
[1]
EXE.2.SL.TZ0.12
(a)
1
Show that k = GM .
[2]
SPM.1A.HL.TZ0.32
A positive point charge of magnitude 1.0 μC and a point charge q are separated by a distance d .
[1]
An electron is placed at a distance d from the +1.0 μC charge. The electric force on the electron is zero.
What is q ?
A. −4.0 μC
B. −2.0 μC
C. 2.0 μC
D. 4.0 μC
23M.2.HL.TZ2.6BIV
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(iv)
QuestionBank Test
The gravitational potential due to the mass of M at the surface of P can be assumed to be negligible.
Estimate, using the graph, the gravitational potential at the surface of M due to the mass of M.
[2]
19M.2.SL.TZ1.5A.I
(a.i)
Outline the origin of the force that acts on Phobos.
[1]
20N.2.HL.TZ0.D
The concept of potential is also used in the context of gravitational fields. Suggest why scientists developed a common
terminology to describe different types of fields.
[1]
EXE.2.SL.TZ0.11A
(a)
A comet orbits the Sun in an elliptical orbit. A and B are two positions of the comet.
Explain, with reference to Kepler’s second law of planetary motion, the change in the kinetic energy of the comet as it moves from
A to B.
[3]
22M.1A.SL.TZ2.29
1
A simple pendulum has a time period T on the Earth. The pendulum is taken to the Moon where the gravitational field strength is 6
that of the Earth.
What is the time period of the pendulum on the Moon?
A. T√ 6
B. T
6
C. √ T
6
D.
T
6
[1]
22N.2.SL.TZ0.5C.I
(c.i)
Explain, by reference to charge carriers in the wire, how the magnetic force on the wire arises.
[2]
22N.1A.HL.TZ0.33
A resistor connects two parallel conducting rails a distance d apart. A conducting bar rolls along the rails at a constant velocity v
through a uniform magnetic field of 2.0 T perpendicular to the rails as shown.
The voltage V across the resistor is measured.
The graph shows the variation of V with d .
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QuestionBank Test
What is v ?
A. 0.33 m s −1
B. 3.0 m s −1
C. 6.0 m s −1
D. 12.0 m s −1
[1]
EXE.2.SL.TZ0.16A
Determine the magnetic force acting on the 15 Ω wire due to the current in the 30 Ω wire.
(a)
[4]
22M.1A.HL.TZ1.33
An object of mass m is launched from the surface of the Earth. The Earth has a mass M and radius r . The acceleration due to gravity
at the surface of the Earth is g . What is the escape speed of the object from the surface of the Earth?
A. √ gr
B. √ 2gr
C. √ 2Mgr
D. √ 2mgr
[1]
23M.1A.SL.TZ1.19
P and R are parallel wires carrying the same current into the plane of the paper. P and R are equidistant from a point Q. The line PQ
is perpendicular to the line RQ.
The magnetic field due to P at Q is X . What is the magnitude of the resultant magnetic field at Q due to both wires?
X
A. 2
B. X
C. X√ 2
D. 2X
[1]
22N.1A.HL.TZ0.34
Two coils of wire are wound around an iron cylinder. One coil is connected in a circuit with a cell and a switch that is initially
closed. The other coil is connected to an ammeter. The switch is opened at time t 0 .
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QuestionBank Test
What is the ammeter reading before t 0 and what is the ammeter reading after t 0 ?
[1]
19M.1A.SL.TZ2.31
A proton of velocity v enters a region of electric and magnetic fields. The proton is not deflected. An electron and an alpha particle
enter the same region with velocity v . Which is correct about the paths of the electron and the alpha particle?
[1]
19N.2.SL.TZ0.5A
Show that the electric field strength due to the point charge at the position of the electron is 3.4 × 10 8 N C –1 .
(a)
[2]
23M.1A.SL.TZ2.22
An electron is accelerated from rest through a potential difference V .
What is the maximum speed of the electron?
A. √
[1]
2eV
me
eV
B. m
e
2eV
C. m
e
2V
D. √ m
e
23M.2.SL.TZ2.4CII
(ii)
state and explain the net magnetic force acting on it due to the currents in PQ and TU.
[2]
22M.1A.SL.TZ1.20
Three point charges of equal magnitude are placed at the vertices of an equilateral triangle. The signs of the charges are shown.
Point P is equidistant from the vertices of the triangle. What is the direction of the resultant electric field at P?
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QuestionBank Test
[1]
SPM.1A.SL.TZ0.19
A charged rod is brought near an initially neutral metal sphere without touching it.
When the sphere is grounded (earthed), there is an electric current for a short time from the sphere to the ground.
[1]
The ground connection is then removed.
What are the charge on the rod and the charge induced on the sphere when the connection is removed?
23M.1A.SL.TZ2.15
Three point charges, +Q, +Q and −Q, are fixed at the three corners of a square.
What is the direction of the electric field at the fourth corner?
[1]
23M.2.HL.TZ1.6AII
(ii)
Calculate the work done to bring a small charge q from infinity to point C.
Data given:
Q = 2.0 × 10 −3 C,
q = 4.0 × 10 −9 C
D = 1.2 m
[2]
22M.1A.HL.TZ2.19
The coil of a direct current electric motor is turning with a period T . At t = 0 the coil is in the position shown in the diagram.
Assume the magnetic field is uniform across the coil.
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QuestionBank Test
Which graph shows the variation with time of the force exerted on section XY of the coil during one complete turn?
[1]
23M.2.SL.TZ2.6A
(a)
Determine
gM
gP
.
[2]
SPM.1A.SL.TZ0.18
Planets X and Y orbit the same star.
The average distance between planet X and the star is five times greater than the average distance between planet Y and the star.
orbital period of planet X
What is orbital period of planet Y ?
A. 3√ 5
B. √ 5
3
C. √52
D. √53
[1]
21M.1A.SL.TZ2.30
An object of mass m released from rest near the surface of a planet has an initial acceleration z . What is the gravitational field
strength near the surface of the planet?
A. z
z
B. m
C. mz
m
D.
z
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QuestionBank Test
[1]
19M.1A.SL.TZ1.18
Two currents of 3 A and 1 A are established in the same direction through two parallel straight wires R and S.
[1]
What is correct about the magnetic forces acting on each wire?
A. Both wires exert equal magnitude attractive forces on each other.
B. Both wires exert equal magnitude repulsive forces on each other.
C. Wire R exerts a larger magnitude attractive force on wire S.
D. Wire R exerts a larger magnitude repulsive force on wire S.
19M.1A.HL.TZ1.32
A negative charge Q is to be moved within an electric field E, to equidistant points from its position, as shown.
Which path requires the most work done?
[1]
SPM.2.HL.TZ0.4A
(a)
Draw an arrow on the diagram to represent the direction of the acceleration of the satellite.
[1]
21N.1A.SL.TZ0.18
Two parallel wires carry equal currents in the same direction out of the paper. Which diagram shows the magnetic field
surrounding the wires?
[1]
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QuestionBank Test
20N.2.HL.TZ0.8D
(d)
The concept of potential is also used in the context of gravitational fields. Suggest why scientists developed a common
terminology to describe different types of fields.
[1]
21M.1A.SL.TZ1.18
Two charges Q 1 and Q 2 , each equal to 2 nC, are separated by a distance 3 m in a vacuum. What is the electric force on Q 2 and the
electric field due to Q 1 at the position of Q 2 ?
[1]
21M.1A.SL.TZ2.19
An ion moves in a circle in a uniform magnetic field. Which single change would increase the radius of the circular path?
[1]
A. Decreasing the speed of the ion
B. Increasing the charge of the ion
C. Increasing the mass of the ion
D. Increasing the strength of the magnetic field
SPM.1A.HL.TZ0.27
P is a point in a uniform electric field.
What is the direction in which the electric potential increases at P?
[1]
EXE.2.HL.TZ0.16A
(a)
The radius of the dwarf planet Pluto is 1.19 x 10 6 m. The acceleration due to gravity at its surface is 0.617 m s −2 .
Determine the escape speed for an object at the surface of Pluto.
[4]
21N.2.HL.TZ0.5B.II
(b.ii)
about:blank
Sketch, on the axes, a graph to show the variation with time of the magnitude of the emf induced in the loop.
[1]
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QuestionBank Test
SPM.1A.SL.TZ0.20
A positive point charge of magnitude 1.0 μC and a point charge q are separated by a distance d .
[1]
An electron is placed at a distance d from the +1.0 μC charge. The electric force on the electron is zero.
What is q ?
A. −4.0 μC
B. −2.0 μC
C. 2.0 μC
D. 4.0 μC
21N.2.HL.TZ0.B.II
Sketch, on the axes, a graph to show the variation with time of the magnitude of the emf induced in the loop.
[1]
19M.1A.SL.TZ1.25
Satellite X orbits a planet with orbital radius R . Satellite Y orbits the same planet with orbital radius 2 R . Satellites X and Y have the
same mass.
centripetal acceleration of X
What is the ratio centripetal acceleration of Y ?
A.
1
4
1
B. 2
C. 2
D. 4
[1]
EXE.2.SL.TZ0.II
State and explain, using your diagram, why a force acts on B due to A in the plane of the paper.
[3]
21N.2.HL.TZ0.A
Show that the speed of the loop is 20 cm s −1 .
[1]
EXE.1A.HL.TZ0.26
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QuestionBank Test
What is the escape speed from the surface of a planet of radius r that has an acceleration of gravity g at its surface?
A. √
[1]
g
r
B. √ gr
2g
C. √ r
D. √ 2gr
19M.1A.HL.TZ2.18
A particle with a charge ne is accelerated through a potential difference V .
What is the magnitude of the work done on the particle?
A. eV
B. neV
nV
C. e
D.
[1]
eV
n
22M.1A.SL.TZ1.32
A charged sphere in a gravitational field is initially stationary between two parallel metal plates. There is a potential difference V
between the plates.
Three changes can be made:
I. Increase the separation of the metal plates
II. Increase V
III. Apply a magnetic field into the plane of the paper
What changes made separately will cause the charged sphere to accelerate?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
[1]
23M.2.HL.TZ1.A
Explain, by reference to Faraday’s law of electromagnetic induction, why there is an electromotive force (emf) induced in the loop
as it leaves the region of magnetic field.
[2]
SPM.2.HL.TZ0.5A
(a)
Outline why the magnetic flux in ring B increases.
[1]
19M.2.HL.TZ2.9AI
(ai)
Show that the total energy of the planet is given by the equation shown.
1
E = 2 mV
[2]
21M.2.HL.TZ1.2C.I
(c.i)
Show that the gravitational potential due to the planet and the star at the surface of the planet is about −5 × 10 9 J kg −1 .
[3]
20N.1A.SL.TZ0.20
A current in a wire lies between the poles of a magnet. What is the direction of the electromagnetic force on the wire?
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[1]
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QuestionBank Test
23M.1A.HL.TZ1.30
Two isolated point masses, P of mass m and Q of mass 2 m , are separated by a distance 3 d . X is a point a distance d from P and 2
d from Q.
What is the net gravitational field strength at X and the net gravitational potential at X?
Net gravitational Net gravitational
field strength at X potential at X
A.
B.
C.
D.
Gm
2
0
2
- d
d
Gm
d
Gm
Gm
0
2
2d
Gm
2d
2
-
2Gm
d
[1]
EXE.2.SL.TZ0.13
(a)
3
Show that T ∝ r 2 for the planets in a solar system where T is the orbital period of a planet and r is the radius of circular orbit of
planet about its sun.
[2]
EXE.1A.SL.TZ0.16
The force per unit length between two long parallel current-carrying wires is F .
The distance between the wires is halved and the current in each wire is doubled. What is the force per unit length of the wires
after the change?
A. F
B. 2 F
C. 4 F
D. 8 F
[1]
19M.1A.HL.TZ2.29
A circular coil of wire moves through a region of uniform magnetic field directed out of the page.
[1]
What is the direction of the induced conventional current in the coil for the marked positions?
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QuestionBank Test
23M.1A.HL.TZ1.31
A negatively charged particle is stationary halfway between two horizontal charged plates. The plates are separated by a distance
d with potential difference V between them.
What is the magnitude of the electric field and direction of the electric field at the position of the particle?
Magnitude of
electric field
A.
2V
d
B.
V
C.
2V
d
d
V
d
D.
Direction of
electric field
up
up
down
down
[1]
22N.2.HL.TZ0.8B.I
Show that the kinetic energy of the satellite in orbit is about 2 × 10 10 J.
(b.i)
[2]
21N.1A.HL.TZ0.32
A satellite of mass m orbits a planet of mass M in a circular orbit of radius r . What is the work that must be done on the satellite to
increase its orbital radius to 2r ?
A.
B.
GMm
r
GMm
C.
D.
2r
GMm
4r
GMm
8r
[1]
20N.2.HL.TZ0.8C(I)
Calculate the electric potential difference between points A and B.
(c(i))
[1]
EXE.2.SL.TZ0.14B
(b)
Calculate the current in wire Q.
[2]
23M.2.HL.TZ1.7
(a)
Explain, by reference to Faraday’s law of electromagnetic induction, why there is an electromotive force (emf) induced in the loop
as it leaves the region of magnetic field.
[2]
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QuestionBank Test
22N.1A.SL.TZ0.22
The centre of the Earth and the Moon are a distance D apart. There is a point X between them where their gravitational fields
cancel out. The distance from the centre of the Earth to X is d . The mass of the Earth is ME and the mass of the Moon is MM .
What is correct at X?
M
M
ME
= dM
A. dE = D -Md
B.
C.
M
D-d
ME
D.
2
d
ME
d
=
2 =
MM
2
D-d
MM
D2 - d
2
[1]
22N.2.SL.TZ0.5
(a)
State what is meant by an ideal voltmeter.
[1]
23M.1A.SL.TZ2.21
A negatively charged sphere is falling through a magnetic field.
[1]
What is the direction of the magnetic force acting on the sphere?
A. To the left of the page
B. To the right of the page
C. Out of the page
D. Into the page
23M.2.HL.TZ1.7B
(b)
Just before the loop is about to completely exit the region of magnetic field, the loop moves with constant terminal speed v .
The following data is available:
Mass of loop
m = 4.0 g
Resistance of loop
R = 25 mΩ
Width of loop
L = 15 cm
Magnetic flux density B = 0.80 T
Determine, in m s −1 the terminal speed v .
[4]
EXE.1A.SL.TZ0.15
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QuestionBank Test
Two long parallel wires X and Y carry equal currents I . The magnetic force exerted per unit length of each wire is F .
The current in X is halved and the current in Y is doubled. What is the force per unit length of each wire after the change?
Force per unit length of X
Force per unit length of Y
A.
F
F
B.
F
2
2F
C.
2F
F
2
D.
2F
2F
[1]
EXE.1A.SL.TZ0.14
Kepler’s Third law relates the orbital period T of a planet about its sun to its orbital radius r . The mass of the Sun is M .
What is a correct algebraic form of the law?
A. T =
B. T =
C. T =
[1]
2πr1 . 5
GM0 . 5
2πr1 . 5
GM
4πr0 . 67
2
GM
4πr0 . 67
D. T = GM
23M.1A.SL.TZ2.18
A negatively charged sphere is falling through a magnetic field.
[1]
What is the direction of the magnetic force acting on the sphere?
A. To the left of the page
B. To the right of the page
C. Out of the page
D. Into the page
21M.1A.HL.TZ1.30
A particle with charge −2.5 × 10 −6 C moves from point X to point Y due to a uniform electrostatic field. The diagram shows some
equipotential lines of the field.
What is correct about the motion of the particle from X to Y and the magnitude of the work done by the field on the particle?
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QuestionBank Test
[1]
EXE.2.HL.TZ0.IV
In one particular event, a maximum emf of 65 mV is generated in the geophone. The geophone coil has 150 turns.
Calculate the rate of flux change that leads to this emf.
[2]
EXE.2.SL.TZ0.15A
(a)
State the fundamental SI units for permeability of free space, μ0 .
[1]
21N.2.HL.TZ0.7A
(a)
Show that the charge on the surface of the sphere is +18 μC.
[1]
EXE.2.SL.TZ0.17AI
(i)
Draw the magnetic field lines due to A.
[2]
EXE.2.SL.TZ0.15BI
(i)
magnetic field at A;
[1]
20N.2.HL.TZ0.C(II)
Determine the charge Q of the sphere.
[2]
23M.2.HL.TZ2.4
(a)
The designers state that the energy transferred by the resistor every second is 15 J.
Calculate the current in the resistor.
[1]
23M.1A.SL.TZ1.18
An electron enters a region of uniform magnetic field at a speed v . The direction of the electron is perpendicular to the magnetic
field. The path of the electron inside the magnetic field is circular with radius r .
The speed of the electron is varied to obtain different values of r .
Which graph represents the variation of speed v with r ?
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[1]
22M.1A.SL.TZ1.24
P and Q are two moons of equal densities orbiting a planet. The orbital radius of P is twice the orbital radius of Q. The volume of P is
half that of Q. The force exerted by the planet on P is F . What is the force exerted by the planet on Q?
A. F
B. 2 F
C. 4 F
D. 8 F
[1]
EXE.2.HL.TZ0.15CII
(ii)
in its final orbit.
[1]
20N.2.HL.TZ0.8B
(b)
Draw, on the axes, the variation of electric potential V with distance r from the centre of the sphere.
[2]
19N.2.SL.TZ0.4A
(a)
Explain why the path of the proton is a circle.
[2]
21N.2.HL.TZ0.5
(a)
Show that the speed of the loop is 20 cm s −1 .
[1]
SPM.1A.HL.TZ0.31
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QuestionBank Test
A charged rod is brought near an initially neutral metal sphere without touching it.
When the sphere is grounded (earthed), there is an electric current for a short time from the sphere to the ground.
[1]
The ground connection is then removed.
What are the charge on the rod and the charge induced on the sphere when the connection is removed?
21N.2.SL.TZ0.4C.II
(c.ii)
Determine the magnitude and direction of the resultant magnetic field at Q.
[2]
SPM.1A.HL.TZ0.28
Planets X and Y orbit the same star.
The average distance between planet X and the star is five times greater than the average distance between planet Y and the star.
orbital period of planet X
What is orbital period of planet Y ?
A. 3√ 5
B. √ 5
3
C. √52
D. √53
[1]
22N.2.SL.TZ0.C.I
Explain, by reference to charge carriers in the wire, how the magnetic force on the wire arises.
[2]
23M.1A.HL.TZ1.34
Wire XY moves perpendicular to a magnetic field in the direction shown.
[1]
The graph shows the variation with time of the displacement of XY.
What is the graph of the electromotive force (emf) ε induced across XY?
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EXE.2.SL.TZ0.14CII
(ii)
Deduce the current in R.
[2]
23M.2.HL.TZ1.B
Just before the loop is about to completely exit the region of magnetic field, the loop moves with constant terminal speed v .
The following data is available:
Mass of loop
m = 4.0 g
Resistance of loop
R = 25 mΩ
Width of loop
L = 15 cm
[4]
Magnetic flux density B = 0.80 T
Determine, in m s −1 the terminal speed v .
21N.2.SL.TZ0.4C.I
(c.i)
On the diagram draw an arrow to show the direction of the magnetic field at Q due to wire X alone .
[1]
SPM.1A.HL.TZ0.30
A spherical planet has a radius R 0 .
The graph shows the variation of the gravitational potential due to the planet with distance r from the centre of the planet.
[1]
What is the escape speed from the surface of the planet?
A. 1.6 × 10 3 m s −1
B. 2.2 × 10 3 m s −1
C. 3.2 × 10 3 m s −1
D. 4.5 × 10 3 m s −1
SPM.2.HL.TZ0.III
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QuestionBank Test
Show that the electric charge on the oil drop is given by
ρ gV
q= o
E
where ρ o is the density of oil and V is the volume of the oil drop.
[2]
EXE.2.HL.TZ0.2AIII
(iii)
Explain why the magnitude of the emf is related to the amplitude of the ground movement.
[3]
21N.2.SL.TZ0.C.I
On the diagram draw an arrow to show the direction of the magnetic field at Q due to wire X alone .
[1]
EXE.2.HL.TZ0.15CI
(i)
in its initial circular orbit;
[1]
22N.2.HL.TZ0.8A
(a)
The diagram shows field lines for an electrostatic field. X and Y are two points on the same field line.
[2]
Outline which of the two points has the larger electric potential.
21M.1A.HL.TZ1.33
A conducting ring encloses an area of 2.0 cm 2 and is perpendicular to a magnetic field of strength 5.0 mT. The direction of the
magnetic field is reversed in a time 4.0 s. What is the average emf induced in the ring?
A. 0
B. 0.25 μV
C. 0.40 μV
D. 0.50 μV
[1]
23M.2.SL.TZ2.4CI
(i)
outline the magnetic force acting on it due to the current in PQ.
[1]
19M.2.HL.TZ1.5C
(c)
The graph shows the variation of the gravitational potential between the Earth and Moon with distance from the centre of the
Earth. The distance from the Earth is expressed as a fraction of the total distance between the centre of the Earth and the centre of
the Moon.
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Determine, using the graph, the mass of the Moon.
[3]
EXE.2.SL.TZ0.12A
1
(a)
Show that k = GM .
[2]
EXE.2.HL.TZ0.15AI
(i)
orbital speed;
[1]
21M.1A.HL.TZ1.32
A planet has radius R . The escape speed from the surface of the planet is v . At what distance from the surface of the planet is the
orbital speed 0.5 v ?
A. 0.5 R
B. R
C. 2 R
D. 4 R
[1]
22N.2.HL.TZ0.8B.II
(b.ii)
Determine the minimum energy required to launch the satellite. Ignore the original kinetic energy of the satellite due to Earth’s
rotation.
[2]
19M.2.SL.TZ1.6
(a.i)
A black body is on the Moon’s surface at point A. Show that the maximum temperature that this body can reach is 400 K. Assume
that the Earth and the Moon are the same distance from the Sun.
[2]
EXE.2.SL.TZ0.15CI
(i)
magnitude of the net force acting on the loop;
[2]
21N.2.SL.TZ0.6B.II
(b.ii)
The orbital radius of Titan around Saturn is 1.2 × 10 9 m and the orbital period is 15.9 days. Estimate the mass of Saturn. [2]
SPM.2.SL.TZ0.7AI
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(i)
QuestionBank Test
Draw the electric field lines due to the charged plates.
[2]
21M.1A.HL.TZ2.31
The points X and Y are in a uniform electric field of strength E . The distance OX is x and the distance OY is y .
[1]
What is the magnitude of the change in electric potential between X and Y?
A. Ex
B. Ey
C. E ( x + y )
D. E√x2 + y2
19N.2.SL.TZ0.4
(a)
Explain why the path of the proton is a circle.
[2]
EXE.2.SL.TZ0.17
Draw the magnetic field lines due to A.
(a.i)
[2]
EXE.2.SL.TZ0.14CI
(i)
State the direction of the current in R, relative to the current in P.
[1]
19M.2.SL.TZ1.B.II
The following data for the Mars–Phobos system and the Earth–Moon system are available:
Mass of Earth = 5.97 × 10 24 kg
The Earth–Moon distance is 41 times the Mars–Phobos distance.
The orbital period of the Moon is 86 times the orbital period of Phobos.
Calculate, in kg, the mass of Mars.
[2]
EXE.2.SL.TZ0.11B
(b)
An asteroid (minor planet) orbits the Sun in a circular orbit of radius 4.5 × 10 8 km. The radius of Earth’s orbit is 1.5 × 10 8 km.
Calculate, in years, the orbital period of the asteroid.
[2]
22N.1A.SL.TZ0.19
A loop of wire lies in a magnetic field directed into the plane of the page. The loop carries a current in a clockwise direction.
[1]
The magnetic force acting on the wire tends to
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A. rotate the loop about the X axis.
B. rotate the loop about the Y axis.
C. reduce the radius of the loop.
D. increase the radius of the loop.
21N.2.HL.TZ0.7B.II
(b.ii)
Predict the charge on each sphere.
[3]
21M.1A.HL.TZ2.35
A magnet connected to a spring oscillates above a solenoid with a 240 turn coil as shown.
The graph below shows the variation with time t of the emf across the solenoid with the period, T , of the system shown.
The spring is replaced with one that allows the magnet to oscillate with a higher frequency. Which graph shows the new variation
with time t of the current I in the resistor for this new set-up?
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[1]
19M.1A.HL.TZ1.16
Two parallel plates are a distance apart with a potential difference between them. A point charge moves from the negatively
charged plate to the positively charged plate. The charge gains kinetic energy W . The distance between the plates is doubled and
the potential difference between them is halved. What is the kinetic energy gained by an identical charge moving between these
plates?
W
A. 2
B. W
C. 2 W
D. 4 W
[1]
21M.2.HL.TZ2.10B.II
(b.ii)
Outline, using (b)(i), why it is not correct to use the equation √
2G × mass of Io
to calculate the speed required for the spacecraft to
radius of Io
reach infinity from the surface of Io .
[1]
19M.2.SL.TZ1.6C.I
(c.i)
Outline why a force acts on the Moon.
[1]
SPM.1A.HL.TZ0.26
Two long parallel wires P and Q are a distance d apart. They each carry a current.
A magnetic force per unit length F acts on P due to Q.
I
The distance between the wires is increased to 2 d and the current in Q is decreased to .
[1]
2
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What is the magnetic force per unit length that acts on P due to Q after the changes?
F
A. 8
B.
F
4
F
C. 2
D. F
20N.1A.HL.TZ0.31
P and S are two points on a gravitational equipotential surface around a planet. Q and R are two points on a different gravitational
equipotential surface at a greater distance from the planet.
The greatest work done by the gravitational force is when moving a mass from
A. P to S.
B. Q to R.
C. R to P.
D. S to R.
[1]
EXE.1A.HL.TZ0.25
A space probe moves in a circular orbit around Earth. The kinetic energy of the probe is E .
The probe will reach the escape speed when its kinetic energy is increased at least to:
[1]
A. √ 2E
B. 2E
C. 2√ 2E
D. 4E
20N.2.HL.TZ0.8C(II)
(c(ii))
Determine the charge Q of the sphere.
[2]
22N.1A.HL.TZ0.32
Two satellites are in circular orbits around the Earth. Both satellites have the same mass and satellite X is closer to Earth than
satellite Y.
What is correct for the orbital periods of X and Y and the total energies of X and Y?
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[1]
19M.1A.SL.TZ1.23
A beam of negative ions flows in the plane of the page through the magnetic field due to two bar magnets.
[1]
What is the direction in which the negative ions will be deflected?
A. Out of the page ⊙
B. Into the page X
C. Up the page ↑
D. Down the page ↓
21N.1A.HL.TZ0.30
The diagram shows equipotential lines for an electric field. Which arrow represents the acceleration of an electron at point P?
[1]
EXE.2.SL.TZ0.13BII
(ii)
Pluto is a dwarf planet of the Sun that orbits at a distance of 5.9 × 10 9 km from the Sun. Determine, in years, the orbital period of
Pluto.
[3]
21M.1A.HL.TZ2.32
A satellite orbits planet X with a speed vX at a distance r from the centre of planet X . Another satellite orbits planet Y at a speed of
v
vy at a distance r from the centre of planet Y . The mass of planet X is M and the mass of planet Y is 4M . What is the ratio of X ?
vy
A. 0.25
B. 0.5
C. 2.0
D. 4.0
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[1]
23M.2.HL.TZ1.C
The charges Q are replaced by neutral masses M and the charge q by a neutral mass m . The mass m is displaced away from C by a
small distance x and released. Discuss whether the motion of m will be the same as that of q .
[2]
19M.2.SL.TZ1.5B.I
The orbital period T of a moon orbiting a planet of mass M is given by
R3
= kM
T2
where R is the average distance between the centre of the planet and the centre of the moon.
G
Show that k = 2
(b.i)
[3]
4π
EXE.2.HL.TZ0.2
State the movement direction for which the geophone has its greatest sensitivity.
(a.i)
[1]
22M.1A.HL.TZ2.32
Two positive and two negative charges are located at the corners of a square as shown. Point X is the centre of the square. What is
the value of the electric field E and the electric potential V at X due to the four charges?
[1]
19M.1A.HL.TZ2.10
Satellite X is in orbit around the Earth. An identical satellite Y is in a higher orbit. What is correct for the total energy and the kinetic
energy of the satellite Y compared with satellite X?
[1]
SPM.2.HL.TZ0.9CIII
(iii)
Show that the electric charge on the oil drop is given by
ρ gV
q= o
E
where ρ o is the density of oil and V is the volume of the oil drop.
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[2]
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21N.2.HL.TZ0.C.I
There are 85 turns of wire in the loop. Calculate the maximum induced emf in the loop.
[2]
EXE.2.HL.TZ0.2AI
State the movement direction for which the geophone has its greatest sensitivity.
(i)
[1]
EXE.2.SL.TZ0.13A
(a)
3
Show that T ∝ r 2 for the planets in a solar system where T is the orbital period of a planet and r is the radius of circular orbit of
planet about its sun.
[2]
SPM.1A.SL.TZ0.17
Two long parallel wires P and Q are a distance d apart. They each carry a current.
A magnetic force per unit length F acts on P due to Q.
I
The distance between the wires is increased to 2 d and the current in Q is decreased to 2 .
[1]
What is the magnetic force per unit length that acts on P due to Q after the changes?
F
A. 8
B.
F
4
F
C. 2
D. F
21M.2.HL.TZ2.10B.I
(b.i)
Show that the
gravitational potential due to Jupiter at the orbit of Io
gravitational potential due to Io at the surface of Io
is about 80.
[2]
19N.1A.HL.TZ0.34
A coil is rotated in a uniform magnetic field. An alternating emf is induced in the coil. What is a possible phase relationship
between the magnetic flux through the coil and the induced emf in the coil when the variations of both quantities are plotted with
time?
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[1]
20N.1A.HL.TZ0.30
Two satellites W and X have the same mass. They have circular orbits around the same planet. W is closer to the surface than X.
What quantity is smaller for W than for X?
A. Gravitational force from the planet
B. Angular velocity
C. Orbital speed
D. Orbital period
[1]
SPM.1A.HL.TZ0.24
A solid metallic sphere is positively charged and isolated from all other charges.
The electric potential due to the sphere
[1]
A. is constant inside the sphere.
B. is constant outside the sphere.
C. is smallest at the surface of the sphere.
D. increases with distance from the sphere.
EXE.2.SL.TZ0.14A
(a)
State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P.
[2]
SPM.2.HL.TZ0.9AI
(i)
Draw the electric field lines due to the charged plates.
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20N.2.HL.TZ0.9
(a)
Explain, by reference to Faraday’s law of induction, how an electromotive force (emf) is induced in the coil.
[3]
19M.1A.HL.TZ1.34
A satellite in a circular orbit around the Earth needs to reduce its orbital radius.
What is the work done by the satellite rocket engine and the change in kinetic energy resulting from this shift in orbital height?
[1]
21M.1A.SL.TZ1.23
Which is the definition of gravitational field strength at a point?
A. The sum of the gravitational fields created by all masses around the point
B. The gravitational force per unit mass experienced by a small point mass at that point
M
C. G r2 , where M is the mass of a planet and r is the distance from the planet to the point
D. The resultant force of gravitational attraction on a mass at that point
[1]
EXE.2.SL.TZ0.15BII
(ii)
magnetic force on section AB of the loop.
[1]
23M.1A.HL.TZ2.31
The mass of Mars is about ten times that of the Moon. The radius of Mars is about twice that of the Moon.
escape speed from Mars
What is the
?
[1]
Moon
A. √ 5
B. 2 √ 5
C. 5
D. 25
SPM.2.SL.TZ0.3A
(a)
Draw an arrow on the diagram to represent the direction of the acceleration of the satellite.
[1]
22N.1A.HL.TZ0.31
An electric field is established between two electrodes separated by distance d , held at a potential difference of V . A charged
particle in this field experiences a force F .
What is the charge on the particle?
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d
A. FV
B.
FV
d
V
C. Fd
D.
Fd
V
[1]
EXE.2.SL.TZ0.12B
(b)
The table gives data relating to the two moons of Mars.
Moon
T / hour
r / Mm
Phobos
7.66
9.38
30.4
Deimos
Determine r for Deimos.
[2]
21N.1A.SL.TZ0.23
The gravitational field strength at the surface of a planet of radius R is g . A satellite is moving in a circular orbit a distance R above
the surface of the planet. What is the magnitude of the acceleration of the satellite?
A. 0
g
B. 4
g
C. 2
D. g
[1]
23M.2.HL.TZ2.II
Outline why the graph between P and O is negative.
[2]
19N.1A.SL.TZ0.17
A negatively charged particle in a uniform gravitational field is positioned mid-way between two charged conducting plates.
[1]
The potential difference between the plates is adjusted until the particle is held at rest relative to the plates.
What change will cause the particle to accelerate downwards relative to the plates?
A. Decreasing the charge on the particle
B. Decreasing the separation of the plates
C. Increasing the length of the plates
D. Increasing the potential difference between the plates
22N.2.HL.TZ0.8
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(a)
QuestionBank Test
The diagram shows field lines for an electrostatic field. X and Y are two points on the same field line.
[2]
Outline which of the two points has the larger electric potential.
21M.1A.HL.TZ1.34
The conservation of which quantity explains Lenz’s law?
A. Charge
B. Energy
C. Magnetic field
D. Mass
[1]
21M.2.HL.TZ2.B.II
Outline, using (b)(i), why it is not correct to use the equation √
2G × mass of Io
radius of Io
to calculate the speed required for the spacecraft to
reach infinity from the surface of Io .
[1]
SPM.2.SL.TZ0.3
(a)
Draw an arrow on the diagram to represent the direction of the acceleration of the satellite.
[1]
SPM.2.HL.TZ0.9
(a.i)
Draw the electric field lines due to the charged plates.
[2]
SPM.1A.HL.TZ0.29
A rectangular conducting coil rotates at a constant angular velocity in a uniform magnetic field. The rotation axis of the coil is
perpendicular to the field.
At one instant the plane of the coil is at an angle θ to the direction of the field.
The magnitude of the emf induced in the coil is
A. never zero.
B. at a maximum when θ = 0° or 180°.
C. at a maximum when θ = 45° or 225°.
D. at a maximum when θ = 90° or 270°.
[1]
19N.2.SL.TZ0.4B(I)
(b(i))
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Show that the radius of the path is about 6 cm.
[2]
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19M.2.SL.TZ1.5B.II
(b.ii)
The following data for the Mars–Phobos system and the Earth–Moon system are available:
Mass of Earth = 5.97 × 10 24 kg
The Earth–Moon distance is 41 times the Mars–Phobos distance.
The orbital period of the Moon is 86 times the orbital period of Phobos.
Calculate, in kg, the mass of Mars.
[2]
22M.1A.SL.TZ1.19
P and Q are two opposite point charges. The force F acting on P due to Q and the electric field strength E at P are shown.
[1]
Which diagram shows the force on Q due to P and the electric field strength at Q?
20N.1A.HL.TZ0.32
The graph shows the variation of electric field strength E with distance r from a point charge.
[1]
The shaded area X is the area under the graph between two separations r1 and r2 from the charge.
What is X?
A. The electric field average between r1 and r2
B. The electric potential difference between r1 and r2
C. The work done in moving a charge from r1 to r2
D. The work done in moving a charge from r2 to r1
22N.2.SL.TZ0.5C.II
(c.ii)
Identify the direction of the magnetic force on the wire.
[1]
22M.1A.SL.TZ2.19
A charge Q is at a point between two electric charges Q 1 and Q 2 . The net electric force on Q is zero. Charge Q 1 is further from Q
than charge Q 2 .
What is true about the signs of the charges Q 1 and Q 2 and their magnitudes?
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[1]
19M.1A.HL.TZ2.11
The escape speed from a planet of radius R is v esc . A satellite orbits the planet at a distance R from the surface of the planet. What
is the orbital speed of the satellite?
1
A. vesc
2
2
B. √2 vesc
C. √ 2vesc
D. 2vesc
[1]
19M.2.SL.TZ1.B.I
The orbital period T of a moon orbiting a planet of mass M is given by
R3
= kM
T2
where R is the average distance between the centre of the planet and the centre of the moon.
G
Show that k = 2
4π
[3]
22M.1A.SL.TZ1.22
A conductor is placed in a uniform magnetic field perpendicular to the plane of the paper. A force F acts on the conductor when
there is a current in the conductor as shown.
The conductor is rotated 30° about the axis of the magnetic field.
What is the direction of the magnetic field and what is the magnitude of the force on the conductor after the rotation?
[1]
23M.2.HL.TZ2.6BIII
(iii)
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Show that the gravitational potential V P at the surface of P due to the mass of P is given by V P = − g P R P where R P is the radius of
the planet.
[2]
SPM.2.SL.TZ0.I
Draw the electric field lines due to the charged plates.
[2]
19M.2.SL.TZ2.9B
(b)
The diagram shows some of the electric field lines for two fixed, charged particles X and Y.
The magnitude of the charge on X is Q and that on Y is q . The distance between X and Y is 0.600 m. The distance between P and Y
is 0.820 m.
Q
At P the electric field is zero. Determine, to one significant figure, the ratio q .
[2]
EXE.2.HL.TZ0.2AIV
(iv)
In one particular event, a maximum emf of 65 mV is generated in the geophone. The geophone coil has 150 turns.
Calculate the rate of flux change that leads to this emf.
[2]
21N.2.SL.TZ0.5C.II
(c.ii)
The resistance of the loop is 2.4 Ω. Calculate the magnitude of the magnetic force on the loop as it enters the region of magnetic
field.
[2]
22M.1A.HL.TZ2.33
The graph shows the variation with distance r of the electric potential V from a charge Q .
[1]
What is the electric field strength at distance s?
A. The area under the graph between s and infinity
B. The area under the graph between 0 and s
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C. The gradient of the tangent at s
D. The negative of the gradient of the tangent at s
20N.1A.HL.TZ0.35
A rectangular coil rotates at a constant angular velocity. At the instant shown, the plane of the coil is at right angles to the line ZZ '
. A uniform magnetic field acts in the direction ZZ ' .
What rotation of the coil about a specified axis will produce the graph of electromotive force (emf) E against time t ?
π
A. Through 2 about ZZ '
B. Through π about YY '
π
C. Through 2 about XX '
D. Through π about XX '
[1]
EXE.2.SL.TZ0.12C
(c)
Determine the mass of Mars.
[3]
23M.2.HL.TZ2.6BII
(ii)
Outline why the graph between P and O is negative.
[2]
20N.2.HL.TZ0.9A
(a)
Explain, by reference to Faraday’s law of induction, how an electromotive force (emf) is induced in the coil.
[3]
EXE.1A.SL.TZ0.13
Planets X and Y move in circular orbits around the same star.
The orbital period of planet Y is twice the orbital period of planet X. The orbital radius of planet X is R .
What is the orbital radius of planet Y?
[1]
A. 3√ 2 R
B. 3√ 4 R
C. 2R
D. 4R
EXE.2.HL.TZ0.2AII
(ii)
Outline how an emf is generated in the coil.
[2]
19M.1A.HL.TZ1.35
The graph below shows the variation with time of the magnetic flux through a coil.
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[1]
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Which of the following gives three times for which the magnitude of the induced emf is a maximum?
T T
A. 0, 4 , 2
T
B. 0,
,T
2
T
C. 0, 4 , T
T T 3T
D.
4
,
2
,
4
21N.2.SL.TZ0.4
(a)
The work done to move a particle of charge 0.25 μC from one point in an electric field to another is 4.5 μJ. Calculate the magnitude
of the potential difference between the two points.
[1]
21M.2.SL.TZ2.10A
(a)
Calculate, for the surface of Io , the gravitational field strength g Io due to the mass of Io . State an appropriate unit for your answer.
[2]
22M.1A.SL.TZ2.16
Four particles, two of charge +Q and two of charge −Q, are positioned on the x -axis as shown. A particle P with a positive charge
is placed on the y -axis. What is the direction of the net electrostatic force on this particle?
[1]
SPM.2.HL.TZ0.4
(a)
Draw an arrow on the diagram to represent the direction of the acceleration of the satellite.
[1]
EXE.2.SL.TZ0.17AII
(ii)
State and explain, using your diagram, why a force acts on B due to A in the plane of the paper.
[3]
21N.2.SL.TZ0.3A.II
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State the direction of the resultant electric field at P.
(a.ii)
[1]
EXE.2.SL.TZ0.13BI
(i)
Outline what is meant by one astronomical unit (1 AU)
[1]
23M.1A.HL.TZ1.32
R
The escape speed from the surface of earth is v esc . The radius of earth is R . A satellite of mass m is in orbit at a height 4 above the
surface of the Earth. What is the energy required to move the satellite to infinity?
A.
B.
mv2 esc
5
2mv2 esc
5
C. mv2 esc
D. 2mv2 esc
[1]
23M.1A.HL.TZ2.30
Two spheres have the same positive charge. A point M is midway between the two spheres.
[1]
Along the line joining the spheres, what is true about the electrical field and the electric potential at M?
Electric field
Electric potential
A.
zero
minimum positive value
B.
maximum
minimum positive value
C.
zero
maximum positive value
D.
maximum
maximum positive value
19M.1A.SL.TZ1.20
Two charges, +Q and −Q, are placed as shown.
What is the magnitude of the electric field strength, in descending order, at points X, Y and Z.
[1]
A. YXZ
B. ZXY
C. ZYX
D. YZX
23M.2.HL.TZ2.6
(a)
Determine
gM
gP
.
[2]
EXE.2.SL.TZ0.14
(a)
State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P.
[2]
21N.2.HL.TZ0.5C.I
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(c.i)
QuestionBank Test
There are 85 turns of wire in the loop. Calculate the maximum induced emf in the loop.
[2]
21M.1A.SL.TZ1.21
A long straight vertical conductor carries a current I upwards. An electron moves with horizontal speed v to the right.
[1]
What is the direction of the magnetic force on the electron?
A. Downwards
B. Upwards
C. Into the page
D. Out of the page
22M.1A.SL.TZ2.21
An astronaut is orbiting Earth in a spaceship. Why does the astronaut experience weightlessness?
A. The astronaut is outside the gravitational field of Earth.
B. The acceleration of the astronaut is the same as the acceleration of the spaceship.
C. The spaceship is travelling at a high speed tangentially to the orbit.
D. The gravitational field is zero at that point.
[1]
21N.2.HL.TZ0.5B.I
(b.i)
Sketch, on the axes, a graph to show the variation with time of the magnetic flux linkage Φ in the loop.
[1]
23M.1A.HL.TZ2.32
A single loop of wire of resistance 10 Ω has its plane perpendicular to a changing magnetic field.
The graph shows the variation with time of the magnetic flux linked through the loop of wire.
[1]
What is the maximum current in the loop of wire?
A. 1.0 A
B. 2.0 A
C. 4.0 A
D. 20 A
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23M.2.HL.TZ2.6BV
(v)
Draw on the axes the variation of gravitational potential between O and M.
[1]
21N.2.HL.TZ0.B.I
Sketch, on the axes, a graph to show the variation with time of the magnetic flux linkage Φ in the loop.
[1]
21N.2.SL.TZ0.6B.I
(b.i)
The orbital radius of Titan around Saturn is R and the period of revolution is T .
Show that T =
2
4π2 R 3
GM
[2]
where M is the mass of Saturn.
19M.2.HL.TZ2.9AII
(aii)
Suppose the star could contract to half its original radius without any loss of mass. Discuss the effect, if any, this has on the total
energy of the planet.
[2]
21N.2.HL.TZ0.3
(a.i)
Show that the magnitude of the resultant electric field at P is 3 MN C −1
[2]
23M.2.HL.TZ1.7A
(a)
Explain, by reference to Faraday’s law of electromagnetic induction, why there is an electromotive force (emf) induced in the loop
as it leaves the region of magnetic field.
[2]
21M.2.HL.TZ1.C.II
Estimate the escape speed of the spacecraft from the planet–star system.
[2]
21N.1A.HL.TZ0.31
Two charged parallel plates have electric potentials of 10 V and 20 V.
A particle with charge +2.0 μC is moved from the 10 V plate to the 20 V plate. What is the change in the electric potential energy
of the particle?
A. −20 μJ
B. −10 μJ
C. 10 μJ
D. 20 μJ
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QuestionBank Test
[1]
21M.2.HL.TZ2.10C
(c)
An engineer needs to move a space probe of mass 3600 kg from Ganymede to Callisto. Calculate the energy required to move the
probe from the orbital radius of Ganymede to the orbital radius of Callisto. Ignore the mass of the moons in your calculation.
[2]
EXE.1A.SL.TZ0.18
A 4.0 cm length of a conducting wire carries a current of 2.5 A. The length is parallel to another long straight wire that carries a
current of 10 A. The distance between the wires is 1.0 cm. The currents are in opposite directions.
What is the magnitude of the force and the direction of the force acting on the 4.0 cm length?
Magnitude of force / μN
A.
B.
C.
D.
1.3
20
1.3
20
Direction of force
Towards other wire
Towards other wire
Away from other wire
Away from other wire
[1]
21M.2.HL.TZ1.C.I
Show that the gravitational potential due to the planet and the star at the surface of the planet is about −5 × 10 9 J kg −1 .
[3]
19M.2.HL.TZ1.5
Outline the origin of the force that acts on Phobos.
(a.i)
[1]
19N.1A.SL.TZ0.20
When a wire with an electric current I is placed in a magnetic field of strength B it experiences a magnetic force F . What is the
direction of F ?
A. In a direction determined by I only
B. In a direction determined by B only
C. In the plane containing I and B
D. At 90° to the plane containing I and B
[1]
EXE.2.SL.TZ0.15
(a)
State the fundamental SI units for permeability of free space, μ0 .
[1]
21N.2.SL.TZ0.C.II
Determine the magnitude and direction of the resultant magnetic field at Q.
[2]
20N.2.HL.TZ0.9B(III)
(b(iii))
The frequency of the generator is doubled with no other changes being made. Draw, on the axes, the variation with time of the
voltage output of the generator.
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QuestionBank Test
[2]
21M.2.HL.TZ2.10
(a)
Calculate, for the surface of Io , the gravitational field strength g Io due to the mass of Io . State an appropriate unit for your answer.
[2]
22M.1A.HL.TZ2.24
Three statements about Newton’s law of gravitation are:
I. It can be used to predict the motion of a satellite.
II. It explains why gravity exists.
III. It is used to derive the expression for gravitational potential energy.
Which combination of statements is correct?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
[1]
19M.1A.HL.TZ2.30
An electron is fixed in position in a uniform electric field. What is the position for which the electrical potential energy of the
electron is greatest?
[1]
21N.2.SL.TZ0.6
Show that the intensity of the solar radiation at the location of Titan is 16 W m −2
(a.i)
[1]
21M.2.HL.TZ1.2
(a)
Explain why a centripetal force is needed for the planet to be in a circular orbit.
[2]
EXE.2.SL.TZ0.16
(a)
Determine the magnetic force acting on the 15 Ω wire due to the current in the 30 Ω wire.
[4]
22M.1A.SL.TZ2.23
A satellite is orbiting Earth in a circular path at constant speed. Three statements about the resultant force on the satellite are:
I. It is equal to the gravitational force of attraction on the satellite.
II. It is equal to the mass of the satellite multiplied by its acceleration.
III. It is equal to the centripetal force on the satellite.
Which combination of statements is correct?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
[1]
EXE.1A.SL.TZ0.17
Three current-carrying wires lie in the same plane and carry currents of 6 A, 2 A and 4 A. The currents are all in the same direction.
The 2 A wire is 4 cm from the 4 A wire and 6 cm from the 6 A wire.
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QuestionBank Test
What magnetic force per unit length acts on the 2 A wire?
A. 0
B. 40 μN m −1
C. 80 μN m −1
D. 160 μN m −1
[1]
19N.1A.HL.TZ0.30
The gravitational potential is V at a distance R above the surface of a spherical planet of radius R and uniform density. What is the
gravitational potential a distance 2R above the surface of the planet?
V
A. 4
B.
4V
9
V
C. 2
2V
D. 3
[1]
19M.2.HL.TZ2.9
(ai)
Show that the total energy of the planet is given by the equation shown.
1
E = 2 mV
[2]
EXE.2.SL.TZ0.17B
(b)
Both wires are 7.5 m long and are 0.25 m apart. The current in both wires is 12 A. Determine the force that acts on one wire due to
the other.
[2]
SPM.2.HL.TZ0.5
(a)
Outline why the magnetic flux in ring B increases.
[1]
19N.1A.HL.TZ0.33
X and Y are two plane coils parallel to each other that have a common axis. There is a constant direct current in Y.
X is first moved towards Y and later is moved away from Y. What, as X moves, is the direction of the current in X relative to that in
Y?
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QuestionBank Test
[1]
19N.2.SL.TZ0.5
Show that the electric field strength due to the point charge at the position of the electron is 3.4 × 10 8 N C –1 .
(a)
[2]
19N.2.SL.TZ0.5B(I)
(b(i))
Calculate the magnitude of the initial acceleration of the electron.
[2]
EXE.2.HL.TZ0.15AII
(ii)
escape speed from its orbit.
[1]
SPM.2.HL.TZ0.I
Draw the electric field lines due to the charged plates.
[2]
EXE.2.HL.TZ0.16B
Pluto rotates about an axis through its centre. Its rotation is in the opposite sense to that of the Earth, i.e. from east to west.
Explain the advantage of an object launching from the equator of Pluto and travelling to the west.
(b)
[3]
20N.2.HL.TZ0.8A
Explain why the electric potential decreases from A to B.
(a)
[2]
19N.2.SL.TZ0.5B(II)
(b(ii))
Describe the subsequent motion of the electron.
[3]
EXE.2.HL.TZ0.2B
Suggest two changes to the system that will make the geophone more sensitive.
(b)
[4]
23M.2.HL.TZ2.6BI
(i)
State and explain the magnitude of the resultant gravitational field strength at O.
[2]
21N.2.SL.TZ0.3A.I
(a.i)
Show that the magnitude of the resultant electric field at P is 3 MN C −1
[2]
21M.2.HL.TZ1.2C.II
(c.ii)
Estimate the escape speed of the spacecraft from the planet–star system.
[2]
21M.2.HL.TZ2.C
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QuestionBank Test
An engineer needs to move a space probe of mass 3600 kg from Ganymede to Callisto. Calculate the energy required to move the
probe from the orbital radius of Ganymede to the orbital radius of Callisto. Ignore the mass of the moons in your calculation.
[2]
19N.1A.SL.TZ0.23
A satellite travels around the Earth in a circular orbit. What is true about the forces acting in this situation?
A. The resultant force is the same direction as the satellite’s acceleration.
B. The gravitational force acting on the satellite is negligible.
C. There is no resultant force on the satellite relative to the Earth.
D. The satellite does not exert any force on the Earth.
[1]
23M.2.HL.TZ2.V
Draw on the axes the variation of gravitational potential between O and M.
[1]
23M.2.SL.TZ2.4
The designers state that the energy transferred by the resistor every second is 15 J.
Calculate the current in the resistor.
(a)
[1]
SPM.2.SL.TZ0.7
(a.i)
Draw the electric field lines due to the charged plates.
[2]
19N.1A.SL.TZ0.32
An electron enters a uniform electric field of strength E with a velocity v . The direction of v is not parallel to E . What is the path of
the electron after entering the field?
A. Circular
B. Parabolic
C. Parallel to E
D. Parallel to v
[1]
21M.2.HL.TZ2.B.I
Show that the
gravitational potential due to Jupiter at the orbit of Io
is about 80.
gravitational potential due to Io at the surface of Io
[2]
21N.2.SL.TZ0.4B.I
(b.i)
Determine the force on Q at the instant it is released.
[2]
23M.2.HL.TZ2.IV
The gravitational potential due to the mass of M at the surface of P can be assumed to be negligible.
Estimate, using the graph, the gravitational potential at the surface of M due to the mass of M.
[2]
19N.1A.SL.TZ0.31
The force acting between two point charges is F when the separation of the charges is x . What is the force between the charges
when the separation is increased to 3x ?
F
A. 3
F
B. 3x2
F
C. 9
F
D. 9x2
[1]
EXE.2.SL.TZ0.15CII
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(ii)
QuestionBank Test
direction of the net force acting on the loop.
[1]
22M.1A.SL.TZ1.34
The graph shows the variation of magnetic flux Φ in a coil with time t .
[1]
What represents the variation with time of the induced emf ε across the coil?
21N.2.SL.TZ0.A.I
Show that the magnitude of the resultant electric field at P is 3 MN C −1
[2]
21N.2.HL.TZ0.7
Show that the charge on the surface of the sphere is +18 μC.
(a)
[1]
EXE.2.HL.TZ0.16
The radius of the dwarf planet Pluto is 1.19 x 10 6 m. The acceleration due to gravity at its surface is 0.617 m s −2 .
Determine the escape speed for an object at the surface of Pluto.
(a)
[4]
23M.1A.SL.TZ1.23
R
The radius of the Earth is R. A satellite is launched to a height h = 4 above the Earth’s surface.
What is
gravitational force on satellite at the surface
gravitational force on satellite at height h
[1]
?
4
A. 5
B.
16
25
25
C. 16
D.
5
4
19M.2.SL.TZ1.C.I
Outline why a force acts on the Moon.
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QuestionBank Test
23M.2.HL.TZ1.6
(a)
The centres of two identical fixed conducting spheres each of charge + Q are separated by a distance D . C is the midpoint of the
line joining the centres of the spheres.
23M.2.HL.TZ1.6AI
(i)
Sketch, on the axes, how the electric potential V due to the two charges varies with the distance r from the centre of the left
charge. No numbers are required. Your graph should extend from r = 0 to r = D.
[3]
19M.2.HL.TZ2.AII
Suppose the star could contract to half its original radius without any loss of mass. Discuss the effect, if any, this has on the total
energy of the planet.
[2]
EXE.2.SL.TZ0.11
(a)
A comet orbits the Sun in an elliptical orbit. A and B are two positions of the comet.
Explain, with reference to Kepler’s second law of planetary motion, the change in the kinetic energy of the comet as it moves from
A to B.
[3]
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QuestionBank Test
Markschemes
20N.2.HL.TZ0.A
there is a magnetic flux « linkage » in the coil / coil cuts magnetic field ✓
this flux « linkage » changes as the angle varies/coil rotates ✓
« Faraday’s law » connects induced emf with rate of change of flux « linkage » with time ✓
Do not award MP2 or 3 for answers that don’t discuss flux.
EXE.2.SL.TZ0.16B
Use of F = BIl ✓
45 × 10 - 6 × 10 × 0 . 25 = 1 . 15 × 10 - 4 N ✓
Concludes that the assumption is not valid ✓
23M.2.HL.TZ1.6C
the net force will no longer be a restoring force/directed towards equilibrium
OR
the gravitational force is attractive/neutral mass would be pulled towards larger masses/ OWTTE ✓
«and so» no, motion will not be the same/no longer be SHM / OWTTE ✓
21N.1A.HL.TZ0.33
C
22N.2.HL.TZ0.5
infinite resistance
OR no current is flowing through it ✓
SPM.1A.HL.TZ0.25
D
19M.1A.SL.TZ2.23
D
21N.2.HL.TZ0.5A
70
✓
3.5
21M.1A.HL.TZ1.31
A
EXE.2.SL.TZ0.12
Equates centripetal force (with Newton’s law of gravitation mr ω2 =
GMm
r2
)
OR T = 2πω ✓
Uses both equation correctly with clear re-arrangement ✓
SPM.1A.HL.TZ0.32
A
23M.2.HL.TZ2.6BIV
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QuestionBank Test
g R
V
« M = M M = 0.75 × 0.27» = 0.20 ✓
VP
gP RP
VM
= «−6.4 × 10 7 × 0.2 =» «−»1.3 × 10 7 «J kg −1 »✓
19M.2.SL.TZ1.5A.I
gravitational attraction/force/field «of the planet/Mars» ✔
Do not accept “gravity” .
20N.2.HL.TZ0.D
to highlight similarities between « different » fields ✓
EXE.2.SL.TZ0.11A
The areas swept out in unit time by the Sun-comet line are the same at A and B ✓
At B, the distance is greater hence the orbital speed/distance moved in unit time is lower «so that the area remains the same» ✓
A decrease in speed means that the kinetic energy also decreases ✓
22M.1A.SL.TZ2.29
A
22N.2.SL.TZ0.5C.I
charge/carriers are moving in a magnetic field ✓
there is a magnetic force on them / quote F = qvB
this creates a magnetic field that interacts with the external magnetic field ✓
OR
Accept electrons.
For MP2 , the force must be identified as acting on charge / carriers.
22N.1A.HL.TZ0.33
B
EXE.2.SL.TZ0.16A
Use of combination of resistors OR V = IR ✓
To show that current in 30 Ω wire is 5.0 A ✓
F
I
=
2 × 10 - 7 I15 I30
r
=
2 × 10 - 7 × 10 × 5
0 . 20
✓
F = 5 × 10 - 5 × 0 . 25 = 1 . 3 × 10 - 5 N ✓
22M.1A.HL.TZ1.33
B
23M.1A.SL.TZ1.19
C
22N.1A.HL.TZ0.34
A
19M.1A.SL.TZ2.31
D
19N.2.SL.TZ0.5A
E=
k×q
✔
r2
E=
8 . 99 × 109 × 6 . 0 × 10 - 3
0.4
2
OR E = 3 . 37 × 108 « N C - 1 » ✔ NOTE: Ignore any negative sign.
23M.1A.SL.TZ2.22
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QuestionBank Test
A
23M.2.SL.TZ2.4CII
The magnetic fields «due to currents in PQ and TU» are in opposite directions
OR
Net force is zero ✓
There are two «repulsive» forces in opposite directions ✓
22M.1A.SL.TZ1.20
C
SPM.1A.SL.TZ0.19
C
23M.1A.SL.TZ2.15
B
23M.2.HL.TZ1.6AII
V «= 2 ×
8 . 99 × 109 × 2 . 0 × 10 - 3
0 . 60
» = 6.0 × 10 7 «V» ✓
W = «qV = 6.0 × 10 7 × 4.0 × 10 −9 =» 0.24 «J» ✓ Allow ECF from MP1
22M.1A.HL.TZ2.19
A
23M.2.SL.TZ2.6A
m
Work using g ∝ r2 ✓
gM
gP
m r 2
= mM r P = 0.75 ✓
P M
SPM.1A.SL.TZ0.18
D
21M.1A.SL.TZ2.30
A
19M.1A.SL.TZ1.18
A
19M.1A.HL.TZ1.32
D
SPM.2.HL.TZ0.4A
arrow normal to the orbit towards the Earth ✓
21N.1A.SL.TZ0.18
A
20N.2.HL.TZ0.8D
to highlight similarities between « different » fields ✓
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QuestionBank Test
21M.1A.SL.TZ1.18
A
21M.1A.SL.TZ2.19
C
SPM.1A.HL.TZ0.27
B
EXE.2.HL.TZ0.16A
vesc = √
2GM
r
AND g =
GM
r2
seen ✓
2gr2
vesc = √ r ✓
Leading to vesc = √ 2gr ✓
1.2 km s −1 ✓
21N.2.HL.TZ0.5B.II
shape as above ✓
Vertical lines not necessary to score.
shape as above ✓
Vertical lines not necessary to score.
Allow ECF from (b)(i) .
SPM.1A.SL.TZ0.20
A
21N.2.HL.TZ0.B.II
Allow ECF from (b)(i) .
19M.1A.SL.TZ1.25
D
EXE.2.SL.TZ0.II
B lies in magnetic field of A OWTTE ✓
Explained use of appropriate rule together with drawn indication of rule operating in this case ✓
To show that force on B is to left and in plane of paper ✓
OR
Magnetic field lines of B merge with those of A to give combined field line pattern ✓
Sketch of combined pattern to show null point somewhere on line between wires. ✓
Wires will move to reduce stored energy and this is achieved by moving together so force on B is to left ✓
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QuestionBank Test
21N.2.HL.TZ0.A
70
✓
3.5
EXE.1A.HL.TZ0.26
D
19M.1A.HL.TZ2.18
B
22M.1A.SL.TZ1.32
A
23M.2.HL.TZ1.A
The induced emf is equal/proportional/related to the «rate of» change of «magnetic» flux/flux linkage ✓
Flux is changing because the area pierced/enclosed by magnetic field lines changes «decreases»
OR
Flux is changing because the loop is leaving/moving out of the «magnetic» field. ✓
Need to see a connection between the EMF and change in flux for MP1 .
Need to see a connection between the area changing or leaving the field and the change in flux for MP2
SPM.2.HL.TZ0.5A
ring B cuts an increasing number of magnetic field lines ✓
OR
magnetic field from current in A increases at the position of B ✓
19M.2.HL.TZ2.9AI
1
GM
E = 2m r −
GMm
r
1 GMm
= −2
r
✔
comparison with V = −
GM
r
✔
«to give answer»
21M.2.HL.TZ1.2C.I
V planet = «−»
V star = «−»
( 6 . 67 × 10 - 11 ) ( 8 × 1024 )
9 . 1 × 10
6
( 6 . 67 × 10 - 11 ) ( 3 . 2 × 1030 )
4 . 4 × 1010
= «−» 5.9 × 10 7 «J kg −1 » ✓
9
−1
= «−» 4.9 × 10 9 «J kg −1 » ✓ V planet + V star = «−» 4.9 «09» × 10 «J kg » ✓
Must see substitutions and not just equations.
20N.1A.SL.TZ0.20
A
23M.1A.HL.TZ1.30
D
EXE.2.SL.TZ0.13
Equates centripetal force (with Newton’s law gravitation mr ω2 =
GMm
)
r2
AND
2π
T= ω ✓
leads to T2 = r3 ×
4π2
GM
hence result ✓
EXE.1A.SL.TZ0.16
D
19M.1A.HL.TZ2.29
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QuestionBank Test
C
23M.1A.HL.TZ1.31
D
22N.2.HL.TZ0.8B.I
orbital radius = 6 . 4 × 106 + 5 . 0 × 105 « = 6 . 9 × 106 m» ✓
1
KE = 2 × 8 . 0 × 102 ×
6 . 67 × 10 - 11 × 6 . 0 × 1024
6 . 9 × 106
OR 2 . 3 × 1010 «J» ✓
Award [1] max for answers ignoring orbital height (KE = 2.5 × 10 10 J ).
21N.1A.HL.TZ0.32
C
20N.2.HL.TZ0.8C(I)
W
« q =
1 . 7 × 10
- 16
1 . 60 × 10 - 19
= » 1 . 1 × 103 « V » ✓
EXE.2.SL.TZ0.14B
2 . 0 × 10
0 . 50
-5
=
4π × 10 - 7 × 5 . 0 × IQ
2π × 0 . 10
✓
IQ = 4 . 0 «A» ✓
23M.2.HL.TZ1.7
The induced emf is equal/proportional/related to the «rate of» change of «magnetic» flux/flux linkage ✓
Flux is changing because the area pierced/enclosed by magnetic field lines changes «decreases»
OR
Flux is changing because the loop is leaving/moving out of the «magnetic» field. ✓
Need to see a connection between the EMF and change in flux for MP1 .
Need to see a connection between the area changing or leaving the field and the change in flux for MP2
22N.1A.SL.TZ0.22
C
22N.2.SL.TZ0.5
infinite resistance
OR no current is flowing through it ✓
23M.1A.SL.TZ2.21
D
23M.2.HL.TZ1.7B
mg = BIL OR I = 0.33 «A» ✓ BvL = IR OR ℰ = 8.25 × 10 −3 «V» OR ℰ = 0.12 v ✓ Combining results to get v = mgR
2 2 ✓
B L
v=«
0 . 040 × 9 . 81 × 0 . 025
2
0 . 80 × 0 . 15
2
=» 0.068 «m s
−1
» Allow ECF between steps if clear work is shown.
EXE.1A.SL.TZ0.15
A
EXE.1A.SL.TZ0.14
A
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QuestionBank Test
23M.1A.SL.TZ2.18
D
21M.1A.HL.TZ1.30
D
EXE.2.HL.TZ0.IV
Use of ε = -
ΔNΦ
✓
Δt
ΔΦ
Δt
= -
= 0 . 43 mWb s −1 ✓
65 × 10 - 3
150
EXE.2.SL.TZ0.15A
kg m s −2 A −2 ✓
21N.2.HL.TZ0.7A
VR
Q= « k = »
OR Q = 18 . 2 «μC» ✓
3 . 4 × 105 × 0 . 48
8 . 99 × 109
EXE.2.SL.TZ0.17AI
At least one circle centred on centre of wire A
AND
indication of clockwise direction ✓
More than 2 circles with increasing separation between circles from centre outwards (by eye) ✓
EXE.2.SL.TZ0.15BI
Into the page ✓
20N.2.HL.TZ0.C(II)
8 . 99 × 109 × Q ×
1
5 . 0 × 10
-2
-
1
1 . 0 × 10
-1
= 1 . 1 × 103 ✓
Q = 1 . 2 × 10 - 8 « C » ✓
23M.2.HL.TZ2.4
P
I = « √ =» 1.9 «A» ✓
R
23M.1A.SL.TZ1.18
A
22M.1A.SL.TZ1.24
D
EXE.2.HL.TZ0.15CII
Positive ✓
20N.2.HL.TZ0.8B
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QuestionBank Test
curve decreasing asymptotically for r > R ✓
non - zero constant between 0 and R ✓
19N.2.SL.TZ0.4A
magnetic force is to the left «at the instant shown»
force is perpendicular to velocity/«direction of» motion
explains a rule to determine the direction of the magnetic force ✔
force is constant in magnitude ✔
OR
force is centripetal/towards the centre ✔
OR
NOTE: Accept reference to acceleration instead of force
21N.2.HL.TZ0.5
70
✓
3.5
SPM.1A.HL.TZ0.31
C
21N.2.SL.TZ0.4C.II
«using components or Pythagoras to get» B = 21 «mT» ✓
directed «horizontally» to the right ✓
If no unit seen, assume mT.
SPM.1A.HL.TZ0.28
D
22N.2.SL.TZ0.C.I
charge/carriers are moving in a magnetic field ✓
there is a magnetic force on them / quote F = qvB
this creates a magnetic field that interacts with the external magnetic field ✓
OR
Accept electrons.
For MP2 , the force must be identified as acting on charge / carriers.
23M.1A.HL.TZ1.34
C
EXE.2.SL.TZ0.14CII
The force on Q due to R must have the same magnitude «but opposite direction» as the force on Q due to P ✓
The distance is halved therefore one half of the current is needed to produce the same force, so 2.5 A ✓
23M.2.HL.TZ1.B
mg = BIL OR I = 0.33 «A» ✓ BvL = IR OR ℰ = 8.25 × 10 −3 «V» OR ℰ = 0.12 v ✓ Combining results to get v = mgR
2 2 ✓
B L
v=«
0 . 040 × 9 . 81 × 0 . 025
0 . 802 × 0 . 152
=» 0.068 «m s −1 » Allow ECF between steps if clear work is shown.
21N.2.SL.TZ0.4C.I
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QuestionBank Test
arrow of any length as shown ✓
SPM.1A.HL.TZ0.30
C
SPM.2.HL.TZ0.III
Mass of drop is ρ o V ✓
qE = ρ o Vg ✓
«hence answer»
MP1 must be shown implicitly for credit.
EXE.2.HL.TZ0.2AIII
The springs have a natural time period for the oscillation ✓
A greater amplitude of movement leads to higher magnet speed (with constant time period) ✓
So field lines cut coil more quickly leading to greater emf ✓
21N.2.SL.TZ0.C.I
arrow of any length as shown ✓
EXE.2.HL.TZ0.15CI
Negative ✓
22N.2.HL.TZ0.8A
potential greater at Y ✓
ΔV
«from E = - Δre » the potential increases in the direction opposite to field strength «so from X to Y»
OR opposite to the direction of the field lines, «so from X to Y» OR
«from W = q ∆ Ve » work done to move a positive charge from X to Y is positive «so the potential increases from X to Y» ✓
21M.1A.HL.TZ1.33
D
23M.2.SL.TZ2.4CI
«The force is» away from PQ/repulsive/to the right ✓
19M.2.HL.TZ1.5C
read off separation at maximum potential 0.9 ✔
equating of gravitational field strength of earth and moon at that location OR
✔
7.4 × 10 22 «kg» ✔
Allow ECF from MP1
EXE.2.SL.TZ0.12A
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QuestionBank Test
Equates centripetal force (with Newton’s law of gravitation mr ω2 =
GMm
)
r2
OR T = 2πω ✓
Uses both equation correctly with clear re-arrangement ✓
EXE.2.HL.TZ0.15AI
«√
6 . 67 × 10 - 11 × 5 . 97 × 1024
6 . 70 × 106
= » 7 . 71 × 103 «m s −1 » ✓
21M.1A.HL.TZ1.32
B
22N.2.HL.TZ0.8B.II
change in PE = 6 . 67 × 10 - 11 × 6 . 0 × 1024 × 8 . 0 × 102
energy needed = KE + ΔPE = 2 . 7 × 1010 «J» ✓
1
6 . 4 × 106
-
1
6 . 9 × 106
= « 3 . 6 × 109 J» ✓
Allow ECF from 8(b)(i).
19M.2.SL.TZ1.6
T= (
390 «K» ✔
1360 0.25
✔
)
σ
Must see 1360 (from data booklet) used for MP1.
Must see at least 2 s.f .
EXE.2.SL.TZ0.15CI
4π × 10 - 7 × 2 . 0 × 1 . 0
2π × 0 . 30
-
4π × 10 - 7 × 2 . 0 × 1 . 0
2π × 0 . 50
× 0 . 20 ✓
1 . 1 × 10 - 7 «N» ✓
21N.2.SL.TZ0.6B.II
T = 15 . 9 × 24 × 3600 «s» ✓
M=
4π2 1 . 2 × 10
6 . 67 × 10
- 11
93
× 15 . 9 × 24 × 36002
= 5 . 4 × 1026 «kg» ✓
Award [2] marks for a bald correct answer.
Allow ECF from MP1 .
SPM.2.SL.TZ0.7AI
equally spaced arrows «by eye» all pointing down ✓
edge effects also shown with arrows ✓
21M.1A.HL.TZ2.31
A
19N.2.SL.TZ0.4
magnetic force is to the left «at the instant shown»
force is perpendicular to velocity/«direction of» motion
explains a rule to determine the direction of the magnetic force ✔
force is constant in magnitude ✔
OR
force is centripetal/towards the centre ✔
OR
NOTE: Accept reference to acceleration instead of force
EXE.2.SL.TZ0.17
At least one circle centred on centre of wire A
AND
indication of clockwise direction ✓
More than 2 circles with increasing separation between circles from centre outwards (by eye) ✓
EXE.2.SL.TZ0.14CI
Opposite ✓
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QuestionBank Test
19M.2.SL.TZ1.B.II
R
mMars = ( R Mars )
3
Earth
2
T
( TEarth ) mEarth or other consistent re-arrangement ✔
6.4 × 10 23 «kg» ✔
Mars
EXE.2.SL.TZ0.11B
T2
4 . 53
An attempt to use Kepler’s 3 rd law, e.g., 1 = 1 . 5 ✓
4 . 51 . 5
T= « 1.5
= » 5.2 «years» ✓
22N.1A.SL.TZ0.19
D
21N.2.HL.TZ0.7B.II
q
q
k 481 = k 242 ⇒ q1 = 2q2 ✓
q1 + q2 = 18 ✓
so q1 = 12 «μC», q2 = 6 . 0 «μC» ✓
Award [3] marks for a bald correct answer.
21M.1A.HL.TZ2.35
A
19M.1A.HL.TZ1.16
A
21M.2.HL.TZ2.10B.II
«this is the escape speed for Io alone but» gravitational potential / field of Jupiter must be taken into account ✓
OWTTE
19M.2.SL.TZ1.6C.I
gravitational attraction/force/field «of the planet/Moon» ✔
Do not accept “gravity” .
SPM.1A.HL.TZ0.26
B
20N.1A.HL.TZ0.31
C
EXE.1A.HL.TZ0.25
B
20N.2.HL.TZ0.8C(II)
8 . 99 × 109 × Q ×
1
5 . 0 × 10 - 2
-
1
1 . 0 × 10 - 1
= 1 . 1 × 103 ✓
Q = 1 . 2 × 10 - 8 « C » ✓
22N.1A.HL.TZ0.32
D
19M.1A.SL.TZ1.23
A
21N.1A.HL.TZ0.30
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QuestionBank Test
D
EXE.2.SL.TZ0.13BII
r3
T2Pluto = T2Earth r3Pluto used ✓
Earth orbital radius = 1.5 × 10 11 m (from AU) AND uses 1 earth year (in any units) ✓
247 years ✓
Earth
21M.1A.HL.TZ2.32
B
23M.2.HL.TZ1.C
OR
the net force will no longer be a restoring force/directed towards equilibrium
the gravitational force is attractive/neutral mass would be pulled towards larger masses/ OWTTE ✓
«and so» no, motion will not be the same/no longer be SHM / OWTTE ✓
19M.2.SL.TZ1.5B.I
ALTERNATE 1 «using fundamental equations» use of Universal gravitational force/acceleration/orbital velocity equations ✔
equating to centripetal force or acceleration. ✔
rearranges to get k = 4πG2 ✔
ALTERNATE 2 «starting with RT = kM »
3
2
substitution of proper equation for T from orbital motion equations ✔
substitution of proper equation for M OR R from orbital motion equations ✔
rearranges to get k = 4πG2 ✔
EXE.2.HL.TZ0.2
Vertical direction / parallel to springs ✓
22M.1A.HL.TZ2.32
A
19M.1A.HL.TZ2.10
B
SPM.2.HL.TZ0.9CIII
Mass of drop is ρ o V ✓
qE = ρ o Vg ✓
«hence answer»
MP1 must be shown implicitly for credit.
21N.2.HL.TZ0.C.I
ALTERNATIVE 1 maximum flux at « 5 . 0 × 5 . 0 × 10 - 4 × 85 × 0 . 94 » = 0 . 19975 ≈ 0 . 20 «Wb» ✓
0 . 20
emf = « 0 . 25 = » 0 . 80 «V» ✓
ALTERNATIVE 2
emf = 85 × 0 . 0094 = 0 . 80 «V» ✓
emf induced in one turn = BvL = 0 . 94 × 0 . 20 × 0 . 05 = 0 . 0094 «V» ✓
Award [2] marks for a bald correct answer. Allow ECF from MP1 .
EXE.2.HL.TZ0.2AI
Vertical direction / parallel to springs ✓
EXE.2.SL.TZ0.13A
Equates centripetal force (with Newton’s law gravitation mr ω2 =
GMm
)
r2
AND
2π
T= ω ✓
leads to T2 = r3 ×
4π2
GM
hence result ✓
SPM.1A.SL.TZ0.17
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QuestionBank Test
B
21M.2.HL.TZ2.10B.I
1 . 9 × 1027
4 . 9 × 108
AND 81..98××10106 seen ✓
22
«
1 . 9 × 1027 × 1 . 8 × 106
4 . 9 × 108 × 8 . 9 × 1022
= » 78 ✓
For MP1 , potentials can be seen individually or as a ratio.
19N.1A.HL.TZ0.34
B
20N.1A.HL.TZ0.30
D
SPM.1A.HL.TZ0.24
A
EXE.2.SL.TZ0.14A
From Newton’s third law, the force on a length of Q is equal but opposite to the force on the same length of P ✓
2 . 0 × 10 - 5 N ✓
SPM.2.HL.TZ0.9AI
equally spaced arrows «by eye» all pointing down ✓
edge effects also shown with arrows ✓
20N.2.HL.TZ0.9
there is a magnetic flux « linkage » in the coil / coil cuts magnetic field ✓
this flux « linkage » changes as the angle varies/coil rotates ✓
« Faraday’s law » connects induced emf with rate of change of flux « linkage » with time ✓
Do not award MP2 or 3 for answers that don’t discuss flux.
19M.1A.HL.TZ1.34
C
21M.1A.SL.TZ1.23
B
EXE.2.SL.TZ0.15BII
Repulsive / to the right ✓
23M.1A.HL.TZ2.31
A
SPM.2.SL.TZ0.3A
arrow normal to the orbit towards the Earth ✓
22N.1A.HL.TZ0.31
D
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QuestionBank Test
EXE.2.SL.TZ0.12B
r3De =
T2De r3Ph
T2Ph
seen or correct substitution ✓
23.5 Mm ✓
21N.1A.SL.TZ0.23
B
23M.2.HL.TZ2.II
The gravitational field is attractive so that energy is required «to move away from P» ✓
the gravitational potential is defined as 0 at ∞ , (the potential must be negative) ✓
19N.1A.SL.TZ0.17
A
22N.2.HL.TZ0.8
potential greater at Y ✓
ΔV
«from E = - Δre » the potential increases in the direction opposite to field strength «so from X to Y»
OR opposite to the direction of the field lines, «so from X to Y» OR
«from W = q ∆ Ve » work done to move a positive charge from X to Y is positive «so the potential increases from X to Y» ✓
21M.1A.HL.TZ1.34
B
21M.2.HL.TZ2.B.II
«this is the escape speed for Io alone but» gravitational potential / field of Jupiter must be taken into account ✓
OWTTE
SPM.2.SL.TZ0.3
arrow normal to the orbit towards the Earth ✓
SPM.2.HL.TZ0.9
equally spaced arrows «by eye» all pointing down ✓
edge effects also shown with arrows ✓
SPM.1A.HL.TZ0.29
B
19N.2.SL.TZ0.4B(I)
qvB =
mv2
R
✔
R=
1 . 67 × 10 - 27 × 2 . 0 × 106
1 . 6 × 10 - 19 × 0 . 35
OR 0.060 « m »
NOTE: Award MP2 for full replacement or correct answer to at least 2 significant figures
19M.2.SL.TZ1.5B.II
R
mMars = ( R Mars )
Earth
3
2
T
( TEarth ) mEarth or other consistent re-arrangement ✔
6.4 × 10 23 «kg» ✔
Mars
22M.1A.SL.TZ1.19
B
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QuestionBank Test
20N.1A.HL.TZ0.32
B
22N.2.SL.TZ0.5C.II
into the plane «of the paper» ✓
22M.1A.SL.TZ2.19
A
19M.1A.HL.TZ2.11
A
19M.2.SL.TZ1.B.I
ALTERNATE 1 «using fundamental equations» use of Universal gravitational force/acceleration/orbital velocity equations ✔
equating to centripetal force or acceleration. ✔
rearranges to get k = 4πG2 ✔
ALTERNATE 2 «starting with RT = kM »
3
2
substitution of proper equation for T from orbital motion equations ✔
substitution of proper equation for M OR R from orbital motion equations ✔
rearranges to get k = 4πG2 ✔
22M.1A.SL.TZ1.22
C
23M.2.HL.TZ2.6BIII
GM
V P= - R
P
AND g P = RGM2 (at surface) ✓ Suitable working and cancellation of G and M seen ✓ V P = − g P R P
P
Must see negative sign
SPM.2.SL.TZ0.I
equally spaced arrows «by eye» all pointing down ✓
edge effects also shown with arrows ✓
19M.2.SL.TZ2.9B
kQ
( 0.600 + 0.820 ) 2
=
kq
0.8202
✔
Q
=
q
«
( 0.600 + 0.820 ) 2
0.8202
= 2.9988 ≈ » 3 ✔
EXE.2.HL.TZ0.2AIV
Use of ε = -
ΔNΦ
Δt
✓
ΔΦ
=
Δt
-
65 × 10
150
-3
= 0 . 43 mWb s −1 ✓
21N.2.SL.TZ0.5C.II
I= «
V
R
= »
0.8
2.4
OR 0 . 33 «A» ✓ F = « NBIL = 85 × 0 . 94 × 0 . 33 × 0 . 05 = » = 1 . 3 «N» ✓
Allow ECF from (c)(i) .
Award [2] marks for a bald correct answer.
22M.1A.HL.TZ2.33
D
20N.1A.HL.TZ0.35
C
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QuestionBank Test
EXE.2.SL.TZ0.12C
Converts T to 27.6 ks and converts to m from Mm ✓
k = 7 . 33 × 10 - 14 «s 2 m −3 » ✓
1
« M = kG » = 2 . 04 × 1023 «kg» ✓
MP1 can be implicit
23M.2.HL.TZ2.6BII
The gravitational field is attractive so that energy is required «to move away from P» ✓
the gravitational potential is defined as 0 at ∞ , (the potential must be negative) ✓
20N.2.HL.TZ0.9A
there is a magnetic flux « linkage » in the coil / coil cuts magnetic field ✓
this flux « linkage » changes as the angle varies/coil rotates ✓
« Faraday’s law » connects induced emf with rate of change of flux « linkage » with time ✓
Do not award MP2 or 3 for answers that don’t discuss flux.
EXE.1A.SL.TZ0.13
B
EXE.2.HL.TZ0.2AII
The magnetic field moves relative to the coil ✓
As field lines cut the coil, forces act on (initially stationary) electrons in the wire (and these move producing an emf) ✓
19M.1A.HL.TZ1.35
B
21N.2.SL.TZ0.4
«V=
4.5
0 . 25
= » 18 «V» ✓
21M.2.SL.TZ2.10A
GM
« r2 =
6 . 67 × 10 - 11 × 8 . 9 × 1022
1 . 8 × 106
2
= »1.8✓
N kg −1 OR m s −2 ✓
22M.1A.SL.TZ2.16
D
SPM.2.HL.TZ0.4
arrow normal to the orbit towards the Earth ✓
EXE.2.SL.TZ0.17AII
B lies in magnetic field of A OWTTE ✓
Explained use of appropriate rule together with drawn indication of rule operating in this case ✓
To show that force on B is to left and in plane of paper ✓
OR
Magnetic field lines of B merge with those of A to give combined field line pattern ✓
Sketch of combined pattern to show null point somewhere on line between wires. ✓
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Wires will move to reduce stored energy and this is achieved by moving together so force on B is to left ✓
21N.2.SL.TZ0.3A.II
directed vertically up «on plane of the page» ✓
Allow an arrow pointing up on the diagram.
EXE.2.SL.TZ0.13BI
«mean» Distance from centre of Sun to centre of Earth ✓
OR Suitable ratio in terms of parsec and arcsecond ✓
23M.1A.HL.TZ1.32
A
23M.1A.HL.TZ2.30
A
19M.1A.SL.TZ1.20
C
23M.2.HL.TZ2.6
m
Work using g ∝ r2 ✓
gM
gP
m r 2
= mM r P = 0.75 ✓
P M
EXE.2.SL.TZ0.14
From Newton’s third law, the force on a length of Q is equal but opposite to the force on the same length of P ✓
2 . 0 × 10 - 5 N ✓
21N.2.HL.TZ0.5C.I
ALTERNATIVE 1 maximum flux at « 5 . 0 × 5 . 0 × 10 - 4 × 85 × 0 . 94 » = 0 . 19975 ≈ 0 . 20 «Wb» ✓
0 . 20
emf = « 0 . 25 = » 0 . 80 «V» ✓
ALTERNATIVE 2
emf = 85 × 0 . 0094 = 0 . 80 «V» ✓
emf induced in one turn = BvL = 0 . 94 × 0 . 20 × 0 . 05 = 0 . 0094 «V» ✓
Award [2] marks for a bald correct answer. Allow ECF from MP1 .
21M.1A.SL.TZ1.21
A
22M.1A.SL.TZ2.21
B
21N.2.HL.TZ0.5B.I
shape as above ✓
23M.1A.HL.TZ2.32
B
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QuestionBank Test
23M.2.HL.TZ2.6BV
Line always negative, of suitable shape and end point below −8 and above −20 unless awarding ECF from b(iv) ✓
21N.2.HL.TZ0.B.I
shape as above ✓
21N.2.SL.TZ0.6B.I
correct equating of gravitational force / acceleration to centripetal force / acceleration ✓
correct rearrangement to reach the expression given ✓
GM
Allow use of √ R =
2πR
T
for MP1 .
19M.2.HL.TZ2.9AII
ALTERNATIVE 1
«at the position of the planet» the potential depends only on the mass of the star /does not depend on the radius of the star ✔
the potential will not change and so the energy will not change ✔
ALTERNATIVE 2
r / distance between the centres of the objects / orbital radius remains unchanged ✔
1 GMm
, energy will not change ✔
r
since ETotal = − 2
21N.2.HL.TZ0.3
«electric field at P from one charge is
kQ
r2
= »
8 . 99 × 109 × 44 × 10 - 6
0 . 482
OR 1 . 7168 × 106 «NC −1 » ✓
« net field is » 2 × 1 . 7168 × 106 × cos 30 ° = 2 . 97 × 106 «NC −1 » ✓
23M.2.HL.TZ1.7A
The induced emf is equal/proportional/related to the «rate of» change of «magnetic» flux/flux linkage ✓
Flux is changing because the area pierced/enclosed by magnetic field lines changes «decreases»
OR
Flux is changing because the loop is leaving/moving out of the «magnetic» field. ✓
Need to see a connection between the EMF and change in flux for MP1 .
Need to see a connection between the area changing or leaving the field and the change in flux for MP2
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QuestionBank Test
21M.2.HL.TZ1.C.II
use of v esc = √ 2V ✓
v = 9.91 × 10 4 «m s −1 » ✓
21N.1A.HL.TZ0.31
D
21M.2.HL.TZ2.10C
- GMJupiter
1
1 . 88 × 10
9
1
-
1 . 06 × 10
9
= « 5 . 21 × 107 J kg - 1 » ✓ « multiplies by 3600 kg to get » 1.9 × 10 11 «J» ✓
Award [2] marks if factor of ½ used, taking into account orbital kinetic energies, leading to a final answer of 9.4 x 10 10 «J».
Allow ECF from MP1 Award [2] marks for a bald correct answer.
EXE.1A.SL.TZ0.18
D
21M.2.HL.TZ1.C.I
V planet = «−»
V star = «−»
( 6 . 67 × 10 - 11 ) ( 8 × 1024 )
9 . 1 × 106
( 6 . 67 × 10 - 11 ) ( 3 . 2 × 1030 )
4 . 4 × 1010
= «−» 5.9 × 10 7 «J kg −1 » ✓
9
−1
= «−» 4.9 × 10 9 «J kg −1 » ✓ V planet + V star = «−» 4.9 «09» × 10 «J kg » ✓
Must see substitutions and not just equations.
19M.2.HL.TZ1.5
gravitational attraction/force/field «of the planet/Mars» ✔
Do not accept “gravity” .
19N.1A.SL.TZ0.20
D
EXE.2.SL.TZ0.15
kg m s −2 A −2 ✓
21N.2.SL.TZ0.C.II
«using components or Pythagoras to get» B = 21 «mT» ✓
directed «horizontally» to the right ✓
If no unit seen, assume mT.
20N.2.HL.TZ0.9B(III)
peak emf doubles ✓
T halves ✓
Must show at least 1 cycle.
21M.2.HL.TZ2.10
GM
« r2 =
6 . 67 × 10 - 11 × 8 . 9 × 1022
1 . 8 × 10
62
= »1.8✓
N kg −1 OR m s −2 ✓
22M.1A.HL.TZ2.24
B
19M.1A.HL.TZ2.30
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QuestionBank Test
D
21N.2.SL.TZ0.6
incident intensity
1360
9.3
2
OR 15 . 7 ≈ 16 «W m −2 » ✓
Allow the use of 1400 for the solar constant.
21M.2.HL.TZ1.2
«circular motion» involves a changing velocity ✓
«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓
there must be a force/acceleration towards centre/star ✓ without a centripetal force the planet will move in a straight line ✓
EXE.2.SL.TZ0.16
Use of combination of resistors OR V = IR ✓
To show that current in 30 Ω wire is 5.0 A ✓
F
I
=
2 × 10 - 7 I15 I30
r
=
2 × 10 - 7 × 10 × 5
0 . 20
✓
F = 5 × 10 - 5 × 0 . 25 = 1 . 3 × 10 - 5 N ✓
22M.1A.SL.TZ2.23
D
EXE.1A.SL.TZ0.17
A
19N.1A.HL.TZ0.30
D
19M.2.HL.TZ2.9
1
GM
E = 2m r −
GMm
=
r
1 GMm
✔
r
comparison with V = −
−2
GM
✔
r
«to give answer»
EXE.2.SL.TZ0.17B
F
l
=
2 × 10 - 7 IA IB
r
=
2 × 10 - 7 × 122
0 . 25
✓
8 . 6 × 10 - 4 N ✓
SPM.2.HL.TZ0.5
ring B cuts an increasing number of magnetic field lines ✓
OR
magnetic field from current in A increases at the position of B ✓
19N.1A.HL.TZ0.33
C
19N.2.SL.TZ0.5
E=
k×q
✔
r2
E=
8 . 99 × 109 × 6 . 0 × 10 - 3
0.4
2
OR E = 3 . 37 × 108 « N C - 1 » ✔ NOTE: Ignore any negative sign.
19N.2.SL.TZ0.5B(I)
F=q×E
OR F = 1 . 6 × 10 - 19 × 3 . 4 × 108 = 5 . 4 × 10 - 11 « N » ✔ a = « 5 . 4 × 10 -- 1131 = » 5 . 9 × 1019 « m s - 2 » ✔
9 . 1 × 10
NOTE: Ignore any negative sign.
Award [1] for a calculation leading to a = « m s - 2 »
Award [2] for bald correct answer
EXE.2.HL.TZ0.15AII
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«√
QuestionBank Test
2 × 6 . 67 × 10
- 11
× 5 . 97 × 10
6 . 70 × 10
24
6
= » 1 . 09 × 104 « m s - 1 » ✓
SPM.2.HL.TZ0.I
equally spaced arrows «by eye» all pointing down ✓
edge effects also shown with arrows ✓
EXE.2.HL.TZ0.16B
Object at equator has the maximum linear/tangential speed possible ✓
It therefore has maximum kinetic energy before takeoff (and this is not required from the fuel) ✓
Idea that the object is already moving in direction of planet before takeoff ✓
20N.2.HL.TZ0.8A
ALTERNATIVE 1
work done on moving a positive test charge in any outward direction is negative ✓
potential difference is proportional to this work « so V decreases from A to B » ✓
ALTERNATIVE 2
potential gradient is directed opposite to the field so inwards ✓
the gradient indicates the direction of increase of V « hence V increases towards the centre/decreases from A to B » ✓
ALTERNATIVE 3
kQ
V = R so as r increases V decreases ✓
V is positive as Q is positive ✓
ALTERNATIVE 4
the work done per unit charge in bringing a positive charge from infinity ✓
to point B is less than point A ✓
19N.2.SL.TZ0.5B(II)
the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔
NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force as attractive so concludes that the acceleration will
increase
EXE.2.HL.TZ0.2B
Any two suggestions from:
Increase number of turns in coil ✓
Because more flux cutting per cycle ✓
Increase field strength of magnet ✓
So that there are more field lines ✓
Change mass-spring system so that time period decreases ✓
So magnet will be moving faster for given amplitude of movement ✓
23M.2.HL.TZ2.6BI
g=0✓
As g «= -
ΔVg
Δr
which» is the gradient of the graph
OR
As the force of attraction/field strength of P and M are equal ✓
21N.2.SL.TZ0.3A.I
«electric field at P from one charge is
kQ
=
r2
»
9
8 . 99 × 10 × 44 × 10
0 . 48
2
-6
OR 1 . 7168 × 106 «NC −1 » ✓
« net field is » 2 × 1 . 7168 × 106 × cos 30 ° = 2 . 97 × 106 «NC −1 » ✓
21M.2.HL.TZ1.2C.II
use of v esc = √ 2V ✓
v = 9.91 × 10 4 «m s −1 » ✓
21M.2.HL.TZ2.C
- GMJupiter
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1
1 . 88 × 109
-
1
1 . 06 × 109
= « 5 . 21 × 107 J kg - 1 » ✓ « multiplies by 3600 kg to get » 1.9 × 10 11 «J» ✓
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QuestionBank Test
Award [2] marks if factor of ½ used, taking into account orbital kinetic energies, leading to a final answer of 9.4 x 10 10 «J».
Allow ECF from MP1 Award [2] marks for a bald correct answer.
19N.1A.SL.TZ0.23
A
23M.2.HL.TZ2.V
Line always negative, of suitable shape and end point below −8 and above −20 unless awarding ECF from b(iv) ✓
23M.2.SL.TZ2.4
P
I = « √ =» 1.9 «A» ✓
R
SPM.2.SL.TZ0.7
equally spaced arrows «by eye» all pointing down ✓
edge effects also shown with arrows ✓
19N.1A.SL.TZ0.32
B
21M.2.HL.TZ2.B.I
1 . 9 × 1027
4 . 9 × 108
AND 81..98××10106 seen ✓
22
«
1 . 9 × 1027 × 1 . 8 × 106
4 . 9 × 108 × 8 . 9 × 1022
= » 78 ✓
For MP1 , potentials can be seen individually or as a ratio.
21N.2.SL.TZ0.4B.I
F=
8 . 99 × 109 × 68 × 10 - 6 × 0 . 25 × 10 - 6
0 . 482
✓
F = 0 . 66 «N» ✓
Award [2] marks for a bald correct answer.
Allow symbolic k in substitutions for MP1 . Do not allow ECF from incorrect or not squared distance.
23M.2.HL.TZ2.IV
«
g R
VM
= M M = 0.75 × 0.27» = 0.20 ✓
VP
gP RP
V M = «−6.4 × 10 7 × 0.2 =» «−»1.3 × 10 7 «J kg −1 »✓
19N.1A.SL.TZ0.31
C
EXE.2.SL.TZ0.15CII
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QuestionBank Test
Repulsive / to the right ✓
22M.1A.SL.TZ1.34
A
21N.2.SL.TZ0.A.I
«electric field at P from one charge is
kQ
r2
= »
8 . 99 × 109 × 44 × 10 - 6
0 . 482
OR 1 . 7168 × 106 «NC −1 » ✓
« net field is » 2 × 1 . 7168 × 106 × cos 30 ° = 2 . 97 × 106 «NC −1 » ✓
21N.2.HL.TZ0.7
VR
Q= « k = »
3 . 4 × 105 × 0 . 48
8 . 99 × 109
OR Q = 18 . 2 «μC» ✓
EXE.2.HL.TZ0.16
2GM
GM
vesc = √ r AND g = r2 seen ✓
vesc = √
2gr2
r
✓
Leading to vesc = √ 2gr ✓
1.2 km s −1 ✓
23M.1A.SL.TZ1.23
C
19M.2.SL.TZ1.C.I
gravitational attraction/force/field «of the planet/Moon» ✔
Do not accept “gravity” .
23M.2.HL.TZ1.6
Constant, non-zero within spheres ✓
A clear, non-zero positive minimum at C ✓
Symmetric bowl shaped up curved shape in between ✓
Do not allow a bowl shaped down curve for MP3 .
23M.2.HL.TZ1.6AI
Constant, non-zero within spheres ✓
A clear, non-zero positive minimum at C ✓
Symmetric bowl shaped up curved shape in between ✓
Do not allow a bowl shaped down curve for MP3 .
19M.2.HL.TZ2.AII
ALTERNATIVE 1
«at the position of the planet» the potential depends only on the mass of the star /does not depend on the radius of the star ✔
the potential will not change and so the energy will not change ✔
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ALTERNATIVE 2
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QuestionBank Test
r / distance between the centres of the objects / orbital radius remains unchanged ✔
1 GMm
, energy will not change ✔
r
since ETotal = − 2
EXE.2.SL.TZ0.11
The areas swept out in unit time by the Sun-comet line are the same at A and B ✓
At B, the distance is greater hence the orbital speed/distance moved in unit time is lower «so that the area remains the same» ✓
A decrease in speed means that the kinetic energy also decreases ✓
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