A2. This question is about motion in a magnetic field.
M11/TZ1/A2
An electron, that has been accelerated from rest by a potential difference of 250 V, enters a
region of magnetic field of strength 0.12 T that is directed into the plane of the page.
(a) The electron’s path while in the region of magnetic field is a quarter circle. Show that the
(i) speed of the electron after acceleration is 9.4×106 ms−1.
1
▪ ∆𝑈 = 𝑒𝑉 = 𝐾𝐸 = 𝑚𝑣 2
2
𝑣=
2𝑒𝑉
=
𝑚
2 × 1.6 × 10−19 × 250
9.1 × 10−31
▪ 𝑣 = 9.4 × 106 𝑚𝑠 −1
[2]
(ii) radius of the path is 4.5×10−4 m.
(b) The diagram below shows the momentum of the electron as it enters and leaves the
region of magnetic field. The magnitude of the initial momentum and of the final
momentum is 8.6×10−24 Ns.
(i)
On the diagram above, draw an arrow to indicate the vector representing the change
in the momentum of the electron.
[1]
[2]
(ii) Show that the magnitude of the change in the momentum of the electron is 1.2×10−23 Ns. [1]
(iii) The time the electron spends in the region of magnetic field is 7.5 ×10−11 s.
Estimate the magnitude of the average force on the electron.
[1]
A3. This question is about circular motion.
M11/TZ1/A3
A ball of mass 0.25 kg is attached to a string and is made to rotate with constant speed v
along a horizontal circle of radius r = 0.33m. The string is attached to the ceiling and makes
an angle of 300 with the vertical.
(a) (i) On the diagram above, draw and label arrows to represent the forces on the ball
in the position shown.
[1] each for correct arrow and (any reasonable) labelling;
Award [1 max] for arrows in correct direction
but not starting at the ball.
[2]
M11/TZ1/A3
(ii) State and explain whether the ball is in equilibrium.
[2]
▪ no;
▪ because the two forces on the ball can never cancel out
/ there is a net force on the ball
/ the ball moves in a circle
/ the ball has acceleration
/it is changing direction;
Award [0] for correct answer with no or wrong argument.
(b) Determine the speed of rotation of the ball.
[3]
▪ Tsin 600 = 𝑚𝑔
Circular motion: there must be a net
force in the direction of center of circle
And that force acts as centripetal force
𝑚𝑣 2
▪ Tcos 60 =
𝑟
0
𝑚𝑔
𝑚𝑣 2
0
cos 60 =
sin 600
𝑟
𝑣2
cot 60 =
𝑔𝑟
0
▪ 𝑣=
𝑔𝑟 cot 600 =
𝑣 = 1.4 𝑚 𝑠 −1
9.81 × 0.33 × cot 600
A2. This question is about kinematics.
M11/TZ2/A2
Lucy stands on the edge of a vertical cliff
and throws a stone vertically upwards.
The stone leaves her hand with a speed of 15 m s–1 at the instant her
hand is 80 m above the surface of the sea. Air resistance is negligible
and the acceleration of free fall is 10 m s–2.
(a) Calculate the maximum height reached by the stone as measured
from the point where it is thrown.
[2]
𝑢𝑝 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒: 𝑔 = −10 𝑚𝑠 −2
▪ 𝑣 2 = 𝑢2 + 2𝑔ℎ
𝑢 = 15 𝑚𝑠 −1
𝑣=0
0 = 225 − 20ℎ
▪ ℎ = 11𝑚
Award [1 max] for 91 m or 91.25 m (candidate adds cliff height incorrectly).
(b) Determine the time for the stone to reach the surface of the sea after leaving Lucy’s hand.
▪ time to reach maximum height: v = u + gt → 𝑡 = 1.5 s;
▪ time to fall 91 m: y = ut + (g/2) t2
→
or
→
→ t = 4.3 s;
Answer can be alternatively expressed as
3.0 s (to return to hand) + 2.8 s (to fall 80 m).
▪ total time = 5.8 s;
total time:
y = ut + (g/2) t2
t = 5.8 s ;
−91 = 0 − 5t2
−80 = 15𝑡 − 5t2
→ 5t2 −15𝑡 − 80 = 0 →t2 −3𝑡 − 16 = 0
[3]
A3. This question is about internal energy and thermal energy (heat).
M11/TZ2/A3
(a) Distinguish between internal energy and thermal energy.
[3]
▪ internal energy is the total kinetic and potential energy of the molecules of a body;
▪ thermal energy is a (net) amount of energy transferred between two bodies;
▪ at different temperatures;
(b) Describe, with reference to the energy of the molecules, the difference in internal energy
of a piece of iron and the internal energy of an ideal gas.
[2]
▪ the internal energy of the iron is equal to the total KE plus PE of the molecules;
▪ the molecules of an ideal gas have only KE so internal energy is the total KE of the molecules;
(c) A piece of iron is placed in a kiln until it reaches the temperature θ of the kiln.
M11/TZ2/A3
The iron is then quickly transferred to water held in a thermally insulated container.
The water is stirred until it reaches a steady temperature. The following data are available.
Thermal capacity of the piece of iron = 60 J K–1
Thermal capacity of the water = 2.0 ×103 J K–1
Initial temperature of the water = 16 °C
Final temperature of the water = 45 °C
The thermal capacity of the container and insulation is negligible.
(i) State an expression, in terms of θ and the above data, for the energy transfer of the iron in
cooling from the temperature of the kiln to the final temperature of the water.
[1]
▪ Q = Ciron ∆ Tiron = 60 (θ – 45)
(ii) Calculate the increase in internal energy of the water as the iron cools in the water.
[1]
▪ Q = Cwater ∆ Twater = 2.0 ×103 x 29 = 5.8 x 104 J
(iii) Use your answers to (c)(i) and (c)(ii) to determine θ.
▪ Ciron ∆ Tiron = Cwater ∆ Twater → 60 (θ – 45) = 5.8 x 104
60 θ – 2700 = 58000
▪ θ = 10000 C
(allow 10100 C to 3 sig. fig)
[2]
A2. This question is about the greenhouse effect.
M12/TZ1/A2
The following data are available for use in this question:
(a) Explain why the power absorbed by the Earth is
[3]
▪ intensity of the Sun’s radiation at the Earth’s orbit
▪ fraction absorbed by the Earth is (1 – 𝛼)
▪ the surface area of the disc (absorbing the radiation) π r2
Look for statements that correctly describe each term.
M12/TZ1/A2
(b) The equation in (a) leads to the following expression which can be used to predict
the Earth’s average surface temperature T.
(i) Calculate the predicted temperature of the Earth.
[2]
▪ T = 250 K
(ii) Explain why the actual average surface temperature of the Earth is in fact higher
than the answer to (b)(i).
▪ greenhouse gases in the atmosphere absorb some of the energy radiated by the Earth;
▪ and radiate some of it back to the surface of the Earth;
[2]
M12/TZ1/A3
A3. This question is about thermal energy transfer.
A hot piece of iron is placed into a container of cold water. After a time the iron and
water reach thermal equilibrium. The heat capacity of the container is negligible.
(a) Define specific heat capacity. [2]
▪ the energy required to change the temperature (of a substance)
by 1K/ C0/unit degree;
▪ of mass 1 kg / per unit mass;
M12/TZ1/A3
(b) The following data are available.
Mass of water = 0.35 kg
Mass of iron = 0.58 kg
Specific heat capacity of water = 4200 J kg–1 K–1
Initial temperature of water = 20 °C
Final temperature of water = 44 °C
Initial temperature of iron = 180 °C
(i) Determine the specific heat capacity of iron.
▪ Qfrom iron = Qto water
[3]
Q=mc∆T
▪ miron Ciron ∆ Tiron = mwater Cwater ∆ Twater
0.58 x c x (180 – 44) = 0.35 x 4200 x (44 – 20)
▪ C = 447 J kg-1 K-1 ≈ 450 J kg−1 K−1
(ii) Explain why the value calculated in (b)(i) is likely to be different from the accepted value.
[2]
▪ energy would be given off to surroundings/environment / energy would be absorbed by container
/ energy would be given off through vaporization of water;
▪ hence final temperature would be less;
hence measured value of (specific) heat capacity (of iron) would be higher;
[2 max]
A2. This question is about kinematics.
M12/TZ2/A2
(a) State the difference between average speed and instantaneous speed.
▪ average speed is the speed over a period of time
▪ instantaneous speed is the speed at a particular instant in time
(b) The graph shows how the acceleration a of a particle varies with time t.
At time t = 0 the instantaneous
speed of the particle is zero.
(i) Calculate the instantaneous speed of
the particle at t = 7.5 s.
[2]
▪ change in speed = area under graph
= ½ x7.5x3 = 11 ms-1
initial speed is zero,
so instantaneous speed at t = 7.5 s is:
▪ v = 11 ms-1
[2]
not definition,
but difference
M12/TZ2/A2
(ii) Using the axes below, sketch a graph to show how the instantaneous speed v
of the particle varies with t.
▪ suitable curve approximating to v = kt2
acceleration is linearly proportional to time
that means that speed is quadratic ally proportional to time
[1]
A3. This question is about nuclear reactions.
M12/TZ2/A3
(a) The nuclide U-235 is an isotope of uranium. A nucleus of U-235 undergoes radioactive decay
to a nucleus of thorium-231 (Th-231). The proton number of uranium is 92.
(i) State what is meant by the terms nuclide and isotope.
[2]
Nuclide
or
or
▪ a species of atom that is characterized by the specific constitution of its nucleus
▪ a species of atom that is characterized by its number of protons and neutrons in the nucleus
▪ A particular combination of protons and neutrons that form a nucleus.
It is used to distinguish isotopes among nuclei.
Isotope:
▪ nuclides with the same number of protons but different number of neutrons;
or
▪ atoms of the same element that have different numbers of neutrons/neutron number;
(ii) One of the particles produced in the decay of a nucleus of U-235 is a gamma photon.
State the name of another particle that is also produced.
▪ alpha particle / helium nucleus / 4
2𝐻𝑒
[1]
(b) The daughter nuclei of U-235 undergo radioactive decay until eventually a stable isotope of lead is reached.
Explain why the nuclei of U-235 are unstable whereas the nuclei of the lead are stable.
[3]
▪ protons repel/break nucleus apart;
▪ binding energy/strong force holds nucleus together;
▪ neutron excess / n:p ratio is greater in lead therefore overall balance of forces
is more attractive
/ (magnitude of) binding energy per nucleon is greater in lead /
/ binding energy per nucleon more negative in lead than uranium
(c) Nuclei of U-235 bombarded with low energy neutrons can undergo nuclear fission.
The nuclear reaction equation for a particular fission is shown below.
Show, using the following data, that the kinetic energy of the fission products is about 200 MeV.
Mass of nucleus of U-235 = 235.04393 u
Mass of nucleus of Ba-144 = 143.922952 u
Mass of nucleus of Kr-89 = 88.91763 u
Mass of neutron
= 1.00867 u
▪ ∆ m = 235.04393 u – [143.922952 + 88.91763 + 2 x 1.00867] u
▪ ∆ m = 0.1860u
▪ energy = 0.1860 x 931.5 =173.9 MeV
energy ≈ 200 MeV
(must see the u to award this mark)
[3]
A2. This question is about circular motion.
M13/TZ1/A2
The diagram shows a car moving at a constant speed over a curved bridge. At the position
shown, the top surface of the bridge has a radius of curvature of 50 m.
(a) Explain why the car is accelerating even though it is moving with a constant speed.
[2]
▪ direction changing;
▪ velocity changing so accelerating;
(b) On the diagram, draw and label the vertical forces acting on the car in the position shown.
▪ weight/ gravitational force/ mg/ Fg
weight between wheels (in box) from centre of mass
(do not allow “gravity” for “weight”.)
▪ reaction/normal reaction/perpendicular contact force/N/R/FN/FR
at both wheels and both labelled;
[2]
M13/TZ1/A2
(c) Calculate the maximum speed at which the car will stay in contact with the bridge. [3]
𝐹𝑛𝑒𝑡
𝑚𝑣 2
=
𝑟
𝑚𝑣 2
𝑚𝑔 − 2𝐹𝑁 =
𝑟
𝑣2
▪ 𝑔=
𝑟
▪ 𝑣 = 𝑟𝑔 = 50 × 9.8
▪ 𝑣 = 22 𝑚𝑠 −1
Circular motion: there must be a net
force in the direction of center of circle
And that force acts as centripetal force
At maximum speed the car will hardly
touch the bridge. It is on the verge of
continuing straight. In math this
condition for maximum speed is
expressed as: FN = 0
A3. This question is about the superposition of waves.
M13/TZ1/A3
(a) State what is meant by the principle of superposition of waves. [1]
▪ when two or more waves overlap, the resultant displacement at any point and at any instant is
the vector sum of the displacements of the individual waves at that point:
(b) The diagram shows two point sources of sound, X and Y. Each source emits waves of
wavelength 1.1 m and amplitude A. Over the distances shown, any decrease in amplitude
can be neglected. The two sources vibrate in phase.
Points O and P are on a line 4.0 m
from the line connecting X and
Y. O is opposite the midpoint of XY
and P is 0.75 m from O.
(i) Explain why the intensity of the sound at O is 4A2 .
▪ constructive interference gives amplitude 2A
▪ intensity is proportional to square of total amplitude [(2A)2]
[2]
M13/TZ1/A3
(ii) Deduce that no sound is detected at P.
[3]
dX
dY
▪ 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 is dX – dy
▪ 𝑑𝑥 − 𝑑𝑦 =
4.02 + 2.352 − 4.02 + 0.852 = 0.55 𝑚
▪ path difference is equal to λ /2
(λ= 1.1 m)
▪ therefore at point P there is destructive interference
M13/TZ2/A2
A2. This question is about kinematics.
(a) Fiona drops a stone from rest vertically down a water well. She hears the splash of the
stone striking the water 1.6 s after the stone leaves her hand. Estimate the
(i) distance between Fiona’s hand and the water surface.
down positive: g = 9.8 ms-2
[1]
u=0
▪ y = ut + (g/2) t2 → 𝑦 = 12.5 𝑚
(ii) speed with which the stone hits the water.
▪ 𝑣 2 = 𝑢2 + 2𝑔𝑦
→ 𝑣=
2𝑔𝑦 → 𝑣 =
▪ 𝑣 = 15.7 𝑚 𝑠 −1
or
▪ 𝑣 = 𝑢 + 𝑔𝑡 → 𝑣 = 0 + 9.8 1.6
▪ 𝑣 = 15.7 𝑚 𝑠 −1
[2]
2(9.8)(12.5)
(b) After the stone in (a) hits the water surface it rapidly reaches a terminal speed as it falls through
the water. The stone leaves Fiona’s hand at time t = 0. It hits the water surface at t1 and it comes
to rest at the bottom of the water at t2.
Using the axes below, sketch a graph to show how the speed v of the stone varies from
time t = 0 to just before t = t2. (There is no need to add any values to the axes.)
[3]
straight line to water surface;
clear decrease after hitting surface;
constant non-zero speed reached smaller than maximum
(c) Draw and label a free-body diagram representing the forces acting on the stone as it falls
through the water at its terminal speed.
[2]
▪ correctly labelled forces
▪ correct direction and equal lengths
(judge by eyes)
A3. This question is about power production.
M13/TZ2/A3
(a) The Drax coal-fired power plant has a power output of 4.0 GW. The efficiency of the plant is 40 %.
The energy density of the coal used is 24 MJ kg–1.
Estimate the minimum mass of coal that is burned each year (1 year = 3.2×107 s).
[3]
▪ energy output in a year = 4.0 x 109 x 3.2 x 107 = 1.28 x 1017 J
1.28×1017
▪ energy input =
= 3.2 x 1017 J
0.4
3.2×1017
▪ mass of coal =
= 1.3 x 1010 kg
7
2.4×10
Allow approach using power output.
or
▪ power required from coal =
4.0
= 10 GW
0.4
10×109
▪ mass of coal required every second =
= 417 kg
2.4×107
▪ mass of coal required every year = 417 x 3.2 x 107 = 1.33 x 10 10 kg
Allow alternative working leading to correct answer.
(b) Discuss one advantage and one disadvantage of using nuclear power production
compared to using coal-fired power production.
M13/TZ2/A3
[4]
Advantage:
or
▪ (nuclear power) does not produce carbon dioxide;
▪ therefore it does not add to the greenhouse effect/global warming
▪ energy density of U-235 (fuel) is very high / small mass is required;
▪ fuel is likely to last a long time/easier to transport / OWTTE
Disadvantage:
▪ waste products (of U-235 fuel) are radioactive;
▪ no safe method of disposal / long half-life; (do not accept “lasts a long time”)
or
or
or
▪ allows development of nuclear weapons;
▪ mention of plutonium/uranium enrichment/dirty bomb;
▪ accidents are potentially catastrophic;
▪ leading to widespread mutations/cancers/contamination/other named effect;
▪ power plant is more expensive;
plausible reason for the expense for example safety/complex plant/decommissioning / OWTTE;