A2 Force and Momentum
323 minutes
246 marks
Q1.
A girl kicks a ball along the ground at a wall 2.0 m away. The ball strikes the wall normally
at a velocity of 8.0 m s–1 and rebounds in the opposite direction with an initial velocity of 6.0 m
s–1. The girl, who has not moved, stops the ball a short time later.
(a)
Explain why the final displacement of the ball is not 4.0 m.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(1)
(b)
Explain why the average velocity of the ball is different from its average speed.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(c)
The ball has a mass of 0.45 kg and is in contact with the wall for 0.10 s. For the period of
time the ball is in contact with the wall,
(i)
calculate the average acceleration of the ball.
.............................................................................................................
.............................................................................................................
(ii)
calculate the average force acting on the ball.
.............................................................................................................
(iii)
state the direction of the average force acting on the ball.
.............................................................................................................
(5)
(Total 8 marks)
Q2.
The diagram represents part of an experiment that is being used to estimate the speed of
an air gun pellet.
The pellet which is moving parallel to the track, strikes the block, embedding itself. The trolley
and the block then move along the track, rising a vertical height, h.
(a)
Using energy considerations explain how the speed of the trolley and block immediately
after it has been struck by the pellet, may be determined from measurements of h.Assume
frictional forces are negligible.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(b)
The following data is collected from the experiment
mass of trolley and block
pellet
immediately after impact
Calculate
0.50 kgmass of
0.0020 kgspeed of trolley and block
0.40 m s–1
(i)
the momentum of the trolley and block immediately after impact,
.............................................................................................................
.............................................................................................................
(ii)
the speed of the pellet just before impact.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(c)
(i)
State what is meant by an inelastic collision.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Use the data from part (b) to show that the collision between the pellet and block is
inelastic.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 11 marks)
Q3.
A golf club undergoes an inelastic collision with a golf ball and gives it an initial velocity
of 60 m s–1. The ball is in contact with the club for 15 ms and the mass of the ball is
4.5 × 10–2 kg.
(a)
Explain what is meant by an inelastic collision.
......................................................................................................................
......................................................................................................................
(1)
(b)
Calculate
(i)
the change in momentum of the ball,
.............................................................................................................
(ii)
the average force the club exerts on the ball.
.............................................................................................................
.............................................................................................................
(4)
(c)
(i)
State the value of the force exerted by the ball on the club and give its direction.
.............................................................................................................
.............................................................................................................
(ii)
Explain how your answer to part (i) follows from an appropriate law of motion.
You may be awarded marks for the quality of written communication in your answer.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 9 marks)
Q4.
In a football match, a player kicks a stationary football of mass 0.44 kg and gives it a speed
of 32 m s–1.
(a)
(i)
Calculate the change of momentum of the football.
.............................................................................................................
(ii)
The contact time between the football and the footballer’s boot was 9.2 m s.
Calculate the average force of impact on the football.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(b)
A video recording showed that the toe of the boot was moving on a circular arc of radius
0.62 m centred on the knee joint when the football was struck. The force of the impact
slowed the boot down from a speed of 24 m s–1 to a speed of 15 m s–1.
Figure 1
(i)
Calculate the deceleration of the boot along the line of the impact force when it
struck the football.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Calculate the centripetal acceleration of the boot just before impact.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
Discuss briefly the radial force on the knee joint before impact and during the
impact.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 7 marks)
Q5.
The graph shows how the momentum of two colliding railway trucks varies with time.
TruckA has a mass of 2.0 × 104 kg and truck B has a mass of 3.0 × 104 kg. The trucks are
travelling in the same direction.
(a)
Calculate the change in momentum of
(i)
truck A,
.............................................................................................................
(ii)
truck B.
.............................................................................................................
(4)
(b)
Complete the following table.
Initialvelocity/m Finalvelocity/m
Initial
Final
s–1
s–1
kineticenergy/J kineticenergy/J
truck A
truck B
(4)
(c)
State and explain whether the collision of the two trucks is an example of an elastic
collision.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 11 marks)
Q6.
(a)
State two quantities that are conserved in an elastic collision.
quantity 1: ....................................................................................................
quantity 2: ....................................................................................................
(2)
(b)
A gas molecule makes an elastic collision with the walls of a gas cylinder. The molecule is
travelling at 450 m s–1 at right angles towards the wall before the collision.
(i)
What is the magnitude and direction of its velocity after the collision?
.............................................................................................................
.............................................................................................................
(ii)
Calculate the change in momentum of the molecule during the collision if it has a
mass of 8.0 × 10–26 kg.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(c)
Use Newton’s laws of motion to explain how the molecules of a gas exert a force on the
wall of a container.
You may be awarded additional marks to those shown in brackets for the quality of written
communication in your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(4)
(Total 10 marks)
Q7.
Communications satellites are usually placed in a geo-synchronous orbit.
(a)
State two features of a geo-synchronous orbit.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(b)
The mass of the Earth 6.00 × 1024 kg and its mean radius is 6.40 × 106 m.
(i)
Show that the radius of a geo-synchronous orbit must be 4.23 × 107 m,
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Calculate the increase in potential energy of a satellite of 750 kg when it is raised
from the Earth’s surface into a geo-synchronous orbit.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(6)
(c)
Satellites in orbits nearer the Earth than geo-synchronous satellites may be used in the
future to track road vehicles.
(i)
State and explain one reason why geo-synchronous satellites would not be suitable
for such a purpose.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Give two points you would make in arguing for or against tracking road vehicles.
Explain your answers.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 12 marks)
Q8.
(a) A spring, which hangs from a fixed support, extends by 40 mm when a mass of
0.25 kg is suspended from it.
(i)
Calculate the spring constant of the spring.
.............................................................................................................
.............................................................................................................
(ii)
An additional mass of 0.44 kg is then placed on the spring and the system is set into
vertical oscillation. Show that the oscillation frequency is 1.5 Hz.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(b)
With both masses still in place, the spring is now suspended from a horizontal support rod
that can be made to oscillate vertically, as shown in the diagram below, with amplitude
30 mm at several different frequencies.
The response of the masses suspended from the spring to the vertical oscillations of the
support rod varies with frequency. Describe and explain, as fully as you can, the motion of
the masses when the support rod oscillates at a frequency of (i) 0.2 Hz, (ii) 1.5 Hz and (iii)
10 Hz.
The quality of your written answer will be assessed in this question.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(6)
(Total 10 marks)
Q9.
A golf club undergoes an inelastic collision with a stationary golf ball and gives it an initial
velocity of 60 m s–1. The ball is in contact with the club for 15 ms and the mass of the ball is4.5 ×
10–2 kg.
(a)
Explain what is meant by an inelastic collision.
......................................................................................................................
......................................................................................................................
(1)
(b)
Calculate
(i)
the change in momentum of the ball,
.............................................................................................................
.............................................................................................................
(ii)
the average force the club exerts on the ball.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 5 marks)
Q10.
Near the surface of a planet the gravitational field strength is uniform and for two points, 10
m apart vertically, the gravitational potential difference is 3 J kg–1. How much work must be done
in raising a mass of 4 kg vertically through 5 m?
A
3J
B
6J
C
12 J
D
15 J
(Total 1 mark)
Q11.
Which one of the following graphs correctly shows the relationship between the
gravitational force, F, between two masses and their separation, r?
(Total 1 mark)
Q12.
The Earth has density ρ and radius R. The gravitational field strength at the surface is g.
What is the gravitational field strength at the surface of a planet of density 2ρ and radius 2R?
A
g
B
2g
C
4g
D
16 g
(Total 1 mark)
Q13.
Which one of the following statements is not true for a body vibrating in simple harmonic
motion when damping is present?
A
The damping force is always in the opposite direction to the velocity.
B
The damping force is always in the opposite direction to the displacement.
C
The presence of damping gradually reduces the maximum potential energy of the system.
D
The presence of damping gradually reduces the maximum kinetic energy of the system.
(Total 1 mark)
Q14.
The time period of a simple pendulum is doubled when the length of the pendulum is
increased by 3.0 m. What is the original length of the pendulum?
A
1.0 m
B
1.5 m
C
3.0 m
D
6.0 m
(Total 1 mark)
Q15.
A body moves with simple harmonic motion of amplitude 0.50 m and period 4π seconds.
What is the speed of the body when the displacement of the body from the equilibrium position
is 0.30 m?
A
0.10 m s–1
B
0.15 m s–1
C
0.20 m s–1
D
0.40 m s–1
(Total 1 mark)
Q16.
A particle of mass m moves horizontally at constant speed v along the arc of a circle from
P1 to P2 under the action of a force. What is the work done on the particle by the force during this
displacement?
A
zero
B
C
D
2 mv2
(Total 1 mark)
Q17.
A model car moves in a circular path of radius 0.8 m at an angular speed of
its displacement from point P, 6 s after passing P?
A
zero
B
1.6 m
C
0.4 πm
D
1.6 πm
rad s–1. What is
(Total 1 mark)
Q18.
What is the value of the angular velocity of a point on the surface of the Earth?
A
1.2 × 10–5 rad s–1
B
7.3 × 10–5 rad s–1
C
2.6 × 10–1 rad s–1
D
4.6 × 102 rad s–1
(Total 1 mark)
Q19.
The rate of change of momentum of a body falling freely under gravity is equal to its
A
weight.
B
power.
C
kinetic energy.
D
potential energy.
(Total 1 mark)
Q20.
A force, F, varies with time, t, as shown by the graph and is applied to a body initially at rest on
a smooth surface. What is the momentum of the body after 5.0 s?
A
zero.
B
12.5 N s.
C
25 N s.
D
50 N s.
(Total 1 mark)
Q21.
A particle of mass m strikes a rigid wall perpendicularly from the left with velocity v.
If the collision is perfectly elastic, the change in momentum of the particle which occurs as a
result of the collision is
A
2mv to the right.
B
2mv to the left.
C
mv to the left.
D
zero.
(Total 1 mark)
Q22.
For the two physical quantities, impulse and force, which one of the following is correct?
A
Impulse is a scalar and force is a scalar.
B
Impulse is a scalar and force is a vector.
C
Impulse is a vector and force is a scalar.
D
impulse is a vector and force is a vector.
(Total 1 mark)
Q23.
Which row, A to D, in the table correctly shows the quantities conserved in an inelastic
collision?
mass
momentum
kinetic energy
total energy
A
conserved
not conserved
conserved
conserved
B
not conserved
conserved
conserved
not conserved
C
conserved
conserved
conserved
conserved
D
conserved
conserved
not conserved
conserved
(Total 1 mark)
Q24.
Water of density 1000 kg m–3 flows out of a garden hose of cross-sectional area 7.2 × 10–
m at a rate of 2.0 × 10–4 m3 per second. How much momentum is carried by the water leaving
the hose per second?
4
2
A
5.6 × 10–5 N s
B
5.6 × 10–2 N s
C
0.20 N s
D
0.72 N s
(Total 1 mark)
Q25.
The graph shows the variation with time, t, of the force, F, acting on a body.
What physical quantity does the area X represent?
A
the displacement of the body
B
the acceleration of the body
C
the change in momentum of the body
D
the change in kinetic energy of the body
(Total 1 mark)
Q26.
Deep space probes often carry modules which may be ejected from them by an explosion.
A space probe of total mass 500 kg is travelling in a straight line through free space at 160 m s–
1
when it ejects a capsule of mass 150 kg explosively, releasing energy. Immediately after the
explosion the probe, now of mass 350 kg, continues to travel in the original straight line but
travels at 240 m s–1, as shown in the figure below.
(a)
Discuss how the principles of conservation of momentum and conservation of energy
apply in this instance.
The quality of your written communication will be assessed in this question.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(6)
(b)
(i)
Calculate the magnitude of the velocity of the capsule immediately after the
explosion and state its direction of movement.
magnitude of velocity = ....................................... m s–1
direction of movement ............................................................
(3)
(ii)
Determine the total amount of energy given to the probe and capsule by the
explosion.
answer = ....................................... J
(4)
(Total 13 marks)
Q27.
A projectile moves in a gravitational field. Which one of the following is a correct statement
about the gravitational force acting on the projectile?
A
The force is in the direction of the field.
B
The force is in the opposite direction to that of the field.
C
The force is at right angles to the field.
D
The force is at an angle between 0° and 90° to the field.
(Total 1 mark)
Q28.
The time period of oscillation of a simple pendulum of length l is the same as the time
period of oscillation of a mass M attached to a vertical spring. The length and mass are then
changed.Which row, A to D, in the table would give a simple pendulum with a time period twice
that of the spring oscillations?
new pendulum length
new mass on spring
A
2l
2M
B
2l
C
2M
D
(Total 1 mark)
Q29.
Which graph, A to D, shows the variation of the kinetic energy, Ek, with displacement x for
a particle performing simple harmonic motion?
(Total 1 mark)
Q30.
A body moves with simple harmonic motion of amplitude 0.90 m and period 8.9 s. What is
the speed of the body when its displacement is 0.70 m?
A
0.11 m s–1
B
0.22 m s–1
C
0.40 m s–1
D
0.80 m s–1
(Total 1 mark)
Q31.
A particle of mass m moves in a circle of radius r at uniform speed, taking time T for each
revolution. What is the kinetic energy of the particle?
A
B
C
D
(Total 1 mark)
Q32.
A mass on the end of a string is whirled round in a horizontal circle at increasing speed
until the string breaks. The subsequent path taken by the mass is
A
a straight line along a radius of the circle.
B
a horizontal circle.
C
a parabola in a horizontal plane.
D
a parabola in a vertical plane.
(Total 1 mark)
Q33.
A gas molecule of mass m in a container moves with velocity v. If it makes an elastic
collision at right angles to the walls of the container, what is the change in momentum of the
molecule?
A
B
zero
mv
C
mv
D
2 mv
(Total 1 mark)
Q34.
The graph shows how the force on a glider of mass 2000 kg changes with time as it is
launched from a level track using a catapult.
Assuming the glider starts at rest what is its velocity after 40 s?
A
2.5 m s–1
B
10 m s–1
C
50 m s–1
D
100 m s–1
(Total 1 mark)
Q35.
The graph shows how the force on a glider of mass 2000 kg changes with time as it is
launched from a level track using a catapult.
Assuming the glider starts at rest what is its velocity after 40 s?
A
2.5 m s–1
B
10 m s–1
C
50 m s–1
D
100 m s–1
(Total 1 mark)
M1.
(a) displacement is a vector (1)
ball travels in opposite directions (1)
max 1
(b)
velocity is rate of change of displacement
average speed is rate of change of distance
velocity is a vector [or speed is a scalar)
velocity changes direction
any two (1) (1)
2
(c)
(i)
a=
(1)
= (–)140.m s–1 (1)
(allow C.E. for incorrect values of Δv)
(ii)
F = 0.45 × (–) 140 = (–) 63N (1)(allow C.E for value of a)
(iii)
away from wall (1)at right angles to wall (1)[or back to girl (1) (1)][or opposite to
direction of velocity at impact (1) (1)]
5
[8]
M2.
(a) kinetic energy changes to potential energy (1)
potential energy calculated by measuring h (1)
equate kinetic energy to potential energy to find speed (1)
[or use h to find s (1)
use g sinθ for a (1)
use v2 = u2 + 2as (1)]
[or use h to find s (1)
time to travel s and calculate vav (1)
v = 2vav (1)]
3
(b)
(i)
p(= mv) = 0.5(0) × 0.4(0) = 0.2(0) (1) N s(or kg m s–1) (1)
(ii)
(use of mpvp = mtvt gives) 0.002(0) v = 0.2(0) (1)
v = 100 m s–1 (1)
4
(c)
(i)
kinetic energy is not conserved (1)
(ii)
initial kinetic energy =
final kinetic energy =
× 0.002 × 1002 = 10 (J) (1)
× 0.5 × 0.42 = 0.040 (J) (1)
hence change in kinetic energy (1)
(allow C.E. for value of v from (b))
4
[11]
M3.
(a) kinetic energy not conserved (1)
[or velocity of approach is equal to velocity of separation]
1
(b)
(i)
(use of p = mv gives) p = 4.5 × 10–2 × 60 (1)
= 2.7 kg m s–1 (1)
(ii)
(use of F =
gives)
F=
(1)
= 180 N (1)
[or a =
=
= 4000 (ms–1)
F = (ma) = 4.5 × 10–2 × 4000 = 180 N]
4
(c)
(i)
180 N (1)
(allow C.E. for value of F from (b) (ii)
in opposite direction (to motion of the club) (1)
(ii)
body A (or club) exerts a force on body B (or ball) (1)
(hence) body B (or ball) exerts an equal force on body
A (or club) (1)
correct statement of Newton’s third law (1)
max 4
QWC 1
[9]
M4.
(a)
(ii)
(i)
change of momentum (= 0.44 × 32) = 14(.1) kg m s1 (1)
(use of F =
gives)
F=
(1)
= 1.5(3) × 103N (1)
(allow C.E. for value of Δ(mv) from (i)
3
(b)
(i)
deceleration =
= 9.8 × 102m s–2 (1)
(9.78 × 102m s–2)
(ii)
(use of a =
gives)
centripetal acceleration =
(9.29 × 102 m s–2)
(iii)
= 9.3 × 102m s–2 (1)
before impact: radial pull on knee joint due to centripetalacceleration of
boot (1)during impact: radial pull reduced (1)
4
[7]
M5.
(a)
(ii)
(i)
(change in momentum of A) = – (1) 25 × 103 (1)
kg m s–1 (or N s) (1)
(change in momentum of B) = 25 × 103 kg m s–1 (1)
4
(b)
initial vel/m s–1
final vel/m s–1
initial k.e./J
final k.e./J
truck A
2.5
1.25
62500
15600
truck B
0.67
1.5
6730
33750
(1)
(1)
(1)
(1)
4
(c)
not elastic (1)
because kinetic energy not conserved (1)
kinetic energy is greater before the collision (or less after) (1)
[or justified by correct calculation]
3
[11]
M6.
(a)
momentum (1)
kinetic energy (1)
2
(b)
(i)
450ms–1 (1)
in the opposite direction (1)
(ii)
Δp = 8.0 × 10–26 × 900 (1)
= 7.2 × 10–23Ns (1)
4
(c)
force is exerted on molecule by wall (1)to change its momentum (1)molecule must exert
an equal but opposite force on wall (1)in accordance with Newton's second or third law (1)
4
[10]
M7.
(a)
period is 24 hours (or equal to period of Earth’s rotation) (1)
remains in fixed position relative to surface of Earth (1)
equatorial orbit (1)
same angular speed as Earth (or equatorial surface) (1)
max 2
(b)
(i)
= mω2 r (1)
T=
(1)
(1)
(gives r = 42.3 × 103 km)
(ii)
ΔV =
(1)
= 6.67 × 10–11 × 6 × 1024 ×
= 5.31 × 107 (J kg–1) (1)
ΔEP = mΔV (= 750 × 5.31 × 107) = 3.98 ×1010 J (1)
(allow ecf for value of ΔV)
6
(c)
(i)
signal would be too weak at large distance (1)
(or large aerial needed to detect/transmit signal, or any other
acceptable reason)
the signal spreads out more the further it travels (1)
(ii)
for
road pricing would reduce congestion
stolen vehicles can be tracked and recovered
uninsured/unlicensed vehicles can be apprehended
against
road pricing would increase cost of motoring
possibility of state surveillance/invasion of privacy
(1)(1) any 2 valid points (must be for both for or against)
4
[12]
M8.
(a)
(i)
mg = ke (1)
k=
(ii)
= 61(.3) N m–1 (1)
T=
(1)
(1)
(= 0.667 s)
(= 1.5(0) Hz)
4
(b)
The marking scheme for this part of the question includes an overallassessment for the
Quality of Written Communication (QWC). Thereare no discrete marks for the assessment
of QWC but the candidates’QWC in this answer will be one of the criteria used to assign a
leveland award the marks for this part of the question.
Level
Descriptor
an answer will be expected to meet most of the criteria in the
level descriptor
Good 3
Mark
range
– answer supported by appropriate range of relevant points
– good use of information or ideas about physics, going
beyond those given in the question
– argument well structured with minimal repetition or
irrelevant points
– accurate and clear expression of ideas with only minor
errors of spelling, punctuation and grammar
5-6
Modest 2
– answer partially supported by relevant points
– good use of information or ideas about physics given in
the question but limited beyond this
– the argument shows some attempt at structure
3-4
– the ideas are expressed with reasonable clarity but with a
few errors of spelling, punctuation and grammar
Limited 1
– valid points but not clearly linked to an argument structure
– limited use of information or ideas about physics
1-2
– unstructured
– errors in spelling, punctuation and grammar or lack of
fluency
0
– incorrect, inappropriate or no response
0
examples of the sort of information or idea that might be used tosupport an argument
•
forced vibrations (at 0.2 Hz) (1)
•
amplitude fairly large (≈ 30 mm) (1)
•
in phase with driver (1)
•
resonance (at 1.5 Hz) (1)
•
amplitude very large (> 30 mm) (1)
•
oscillations may appear violent (1)
•
phase difference at 90º (1)
•
forced vibrations (at 10 Hz) (1)
•
small amplitude (1)
•
out of phase with driver or phase lag of π on driver (1)
[10]
M9.
(a)
kinetic energy is not conserved (1)
(or velocity of approach equals velocity of separation)
1
(b)
(i)
(use of p = mv gives) p = 4.5 × 10–2 × 60 (1)
= 2.7kg m s–1 (1)
(ii)
(use of F =
gives) F =
(1)
= 180 N (1)
[or a =
= 400 (m s–1) (1)
F = ma = 4.5 × 10–2 × 4000 = 180N (1)
4
[5]
M10.
B
[1]
M11.
D
[1]
M12.
C
[1]
M13.
B
[1]
M14.
A
[1]
M15.
C
[1]
M16.
B
[1]
M17.
B
[1]
M18.
A
[1]
M19.
A
[1]
M20.
C
[1]
M21.
B
[1]
M22.
D
[1]
M23.
D
[1]
M24.
B
[1]
M25.
C
[1]
M26.
(a) The candidate’s writing should be legible and the spelling,punctuation and
grammar should be sufficiently accurate for themeaning to be clear.
The candidate’s answer will be assessed holistically. The answer will beassigned to one
of three levels according to the following criteria.
High Level (Good to excellent): 5 or 6 marks
The information conveyed by the answer is clearly organised, logical andcoherent, using
appropriate specialist vocabulary correctly. The form andstyle of writing is appropriate to
answer the question.
The candidate states that momentum is conserved, supported by reasoningto explain why
the conditions required for momentum conservation aresatisfied in this case.
The candidate also gives a statement that total energy is conserved, givingdetailed
consideration of the energy conversions which take place,described in the correct
sequence, when there is an explosion on a bodythat is already moving.
Intermediate Level (Modest to adequate): 3 or 4 marks
The information conveyed by the answer may be less well organised and
not fully coherent. There is less use of specialist vocabulary, or specialist
vocabulary may be used incorrectly. The form and style of writing is less
appropriate.
The candidate states that momentum is conserved, but the reasoning is
much more limited.
and/or
There is a statement that (total) energy is conserved, with basic
understanding that some energy is released by the explosion.
Low Level (Poor to limited): 1 or 2 marks
The information conveyed by the answer is poorly organised and may not
be relevant or coherent. There is little correct use of specialist vocabulary.
The form and style of writing may be only partly appropriate.
The candidate indicates that either momentum or energy is conserved, or
that both are conserved. There are very limited attempts to explain either
of them.
The explanation expected in a competent answer should include acoherent
selection of the following points concerning the physicalprinciples involved and
their consequences in this case.
Momentum
•
momentum is conserved because there are no external forcesacting on the overall
system (probe plus capsule) – or because it’sfree space
•
they are moving in free space and are therefore so far from largemasses that
gravitational forces are negligible
•
during the explosion, there are equal and opposite forces actingbetween the probe
and the capsule
•
these are internal forces that act within the overall system
•
because momentum has to be conserved, and it is a vector, thecapsule must move
along the original line of movement after theexplosion
Energy
•
total energy is always conserved in any physical process because
energy can be neither created nor destroyed
•
however, energy may be converted from one form to another
•
the probe is already moving and has kinetic energy
•
in the explosion, some chemical energy is converted into kinetic
energy (and some energy is lost in heating the surroundings)
•
the system of probe and capsule has more kinetic energy than the
probe had originally, because some kinetic energy is released by
the explosion
max 6
(b)
(i)
conservation of momentum gives (500 × 160)
= 150 v + (350 × 240) (1)
from which v = (−)26(.7) (m s−1) (1)
direction: opposite horizontal direction to larger fragment
[or to the left, or backwards] (1)
3
(ii)
initial Ek = ½ × 500 × 1602 (1) (= 6.40 × 106 J)
final Ek = (½ × 350 × 2402) + (½ × 150 × 26.72) (1) (= 1.01 × 107 J)
energy released by explosion = final Ek − initial Ek (1)
= 3.7 × 106 (J) (1)
4
[13]
M27.
A
[1]
M28.
B
[1]
M29.
A
[1]
M30.
C
[1]
M31.
D
[1]
M32.
D
[1]
M33.
D
[1]
M34.
C
[1]
M35.
C
[1]
E1.
This question was answered well and the only common problem was the calculation of the
acceleration of the ball. Only the best candidates appreciated that in part (c) the change in
velocity was (–)14 m s –1. This was not a major handicap however as allowance for
consequential errors enabled most of the remaining marks to be scored.
E2.
This question was also done well although a significant minority of candidates could not
explain in part (a) how the speed of the trolley and block might be determined from energy
considerations. Part (b) was done well apart from the usual confusion over the unit for
momentum.
The final part of the question produced more variable responses. Many candidates were able to
explain correctly what is meant by an inelastic collision but were unable to carry out the
necessary calculation to show that the collision of the pellet and the block was inelastic.
E3.
Candidates seemed quite confident in dealing with this sort of problem and there were many
excellent answers. However, a significant proportion of candidates still have problems with the
units for momentum and this was a commonly applied penalty. Newton’s Third Law was, for the
most part, well understood, and candidates seem more confident applying this law than they are
applying the First and Second Laws.
E4.
In part (a) most candidates calculated the momentum correctly, although N s–1 was
commonly given as the unit of momentum. Many correct answers were seen in part (ii),
although a significant number of candidates misread the impact time as 9.2 s. The final answer
was often presented with too many significant figures.
Many correct calculations were seen in part (b) (i) although some candidates attempted to
usev2 = u2 + 2as with s = 0.62 m. In part (ii), most candidates calculated the centripetal
acceleration correctly. In both parts, incorrect units or provision of answers with too many
significant figures were not uncommon. In part (iii), few candidates realized that the radial force
pulled on the knee joint although a significant number of candidates knew that the force after
impact was less because the speed was less. Many candidates failed to confine their answer to
the limits set by the question and discussed features not relevant to the question.
E5.
This was an unusual question and a considerable amount of work was required in parts (a)
and (b). There were many opportunities to make errors. Part (a) was answered quite well but
common errors were omitting the 103 factor, quoting both changes as positive and the usual unit
problem that appears in questions involving momentum.
Part (b) caused real problems for a significant proportion of candidates and calculation errors
were common. A significant proportion of candidates confused momentum with velocity and
although they were then were able to score marks for a correct calculation of kinetic energy,
arithmetic errors were common.
Part (c) provided evidence that there is a common misunderstanding of what is meant by an
elastic collision. A relatively frequent response was that this was an example of an elastic
collision because momentum was conserved.
E6.
This question proved to be quite discriminating. The calculation in part (b) caused some
confusion and a significant proportion of candidates calculated the actual momentum of the gas
molecule and not the change in momentum. Those who did calculate the change often had
problems with the sign and consequently calculated the change in momentum to be zero. This
was in part due to thinking that as this was an elastic collision, momentum was conserved and
hence there should be no change. An incorrect unit for momentum was a common mistake.
The more able candidates had few problems with part (c) and were obviously very familiar with
the derivation of the gas laws from kinetic theory. They made use of this to produce very
sophisticated answers. Less able candidates were less confident in their answers and found it
difficult to apply Newton’s laws appropriately. There was the usual confusion with Newton’s third
law, with a significant proportion of candidates thinking it referred to the forces acting on the
molecule and not the action of the molecule on the wall producing a reaction from the wall.
E23.
This was a straightforward test of candidates’ knowledge. It required candidates to decide
whether or not mass, momentum, kinetic energy and total energy would be conserved in an
inelastic collision. 85% of the candidates appreciated that everything except kinetic energy
would be conserved. Incorrect responses were fairly evenly spread around the other three
distractors.
E24.
This question required candidates to determine the momentum of the water flowing out of
a garden hose in one second. This called for mathematical application as well as knowledge
and it was therefore much more demanding. 41% of the candidates selected the correct
answer, and the question was not a strong discriminator. The most popular incorrect distractor,
chosen by 28%, was C (0.20). This numerical value could be found by multiplying the density of
water by the flow rate, ignoring the cross-sectional area value given in the question.
E25.
This question, on factual knowledge of the impulse – momentum relationship, was an easy
starter with a facility of 85%.
E26.
It was evident from their attempts at part (a) that during their courses many candidates had
considered the application of conservation of momentum to events involving an explosion. It
was less clear that they had ever considered an explosion that takes place in a moving object,
or considered how conservation of energy applies in an explosion. Consequently, part (a) of the
question proved to be difficult, not least because it was unfamiliar territory for so many. Part (b),
which was formulaic and involved much less original thinking, brought much more success for
the majority.
In part (a) only a very small proportion of the candidates were able to produce answers that
were well organised, coherent, detailed and contained correct physics to merit a ‘high level’
mark of five or six. More answers fell into the ‘intermediate level’ (three or four marks) and even
more into the ‘low level’ (one or two marks). A major failing in most answers was to overlook the
question’s requirement to address the two conservation laws ‘in this instance’. For a high level
answer, it was necessary to consider an explosion on a moving space vehicle travelling in a
straight line in deep space. All of the italicised section is significant. The system has momentum
before exploding (unlike a straightforward recoil example); this momentum has to be conserved
because there are no external forces in deep space. Hence the probe speeds up and the
capsule must be ejected along the original line of movement (although it may not be possible to
tell that this is ‘backwards’ until the calculation has been done). Forces between probe and
capsule during the explosion are equal and opposite, but they are internal forces for the system.
When considering momentum, it was common for candidates to conclude that ‘momentum must
be conserved because momentum is always conserved’.
In the explosion, chemical energy is converted into kinetic energy; this increases the total kinetic
energy of the system, which is shared between probe and capsule. Examiners saw many very
weak answers that showed total confusion – such as momentum being converted into energy,
mass being converted into energy, or energy not being conserved. A serious omission in many
answers was that of the word ‘kinetic’ before ‘energy’, whilst many answers referred to the event
as an ‘inelastic collision’. There was seldom any reference to conservation of the total energy of
the system taken as a whole.
Most candidates recovered from their poor attempts at part (a) to gain all three marks for the
calculation in part (b) (i). There were also many awards of full marks in part (b) (ii), where the
main mistake was to calculate only the kinetic energy of the system (probe + capsule) after the
explosion, and to regard this as the answer to the question. Apparently, the candidates who did
this had not realised that the system had an initial kinetic energy.
E27.
The candidates in 2010 found this question to be slightly easier than their predecessors,
with the facility advancing from 55% to 59%. One in four candidates demonstrated their
confusion with magnetic fields by opting for distractor C, where the force was perpendicular to
the field.
E28.
Knowledge of both the simple pendulum and the mass spring system was required in this
question. Application of the time period formulae should have led students to conclude that,
since under the new conditions Tpendulum = 2Tspring, the new pendulum length should be four times
greater than the new mass on the spring. 62% of the candidates made this correct choice, but
(like the preceding question) this question did not discriminate well between the candidates.
E29.
In this question, 64% of the responses were for the correct shape of Ek against x curve. By
selecting distractor B, 25% of the candidates showed that they understood the trend of the Ek
cbehaviour but not the exact form of it.
E30.
This question does not appear to be easy – albeit it that a correct answer only requires
care when working out the result from three values substituted into v = ± 2πf (A2 – x2). Yet, as on
the previous occasion when this question was used, it turned out to have the highest facility on
the paper – it was 88% this time.
E31.
Candidates found the quantitative content of this question on circular motion more to their
liking, because 63% of them chose the correct answer. Both of these questions gave statistics
which were very similar to those obtained when last used.
E32.
This question was the first of a pair of questions on circular motion, both of which had
appeared in previous examinations. The main failing exhibited in the responses was the fact
that the ball, once it had broken away from the string, would fall under gravity. Only answer D
offered the possibility of some vertical motion and it was chosen by 40% of the candidates.
Distractors A and C each attracted 28% of the answers.
E33.
This question tested the change of momentum of a gas molecule making an elastic
collision with the walls of its container. Misunderstanding the vector nature of momentum, and
therefore of the change of momentum, was responsible for the 21% of candidates who chose
distractor A.
Their reasoning is likely to have been that mv – mv = 0, rather than the correct mv – (–mv) =
2mv, which 69% of the candidates selected.
E34.
The surprising outcome of this question was that just as many candidates chose the wrong
answer as the right one. This seems to have been caused by careless working. Forgetting that
the mass of the glider was 2000 kg could have led candidates to the conclusion that its velocity
would be 100 m s–1 (distractor D) instead of 50 m s–1, or perhaps they forgot the factor of ½ when
finding the area of the triangle under the graph line. The facility of this question was 40%.
E35.
The surprising outcome of this question was that just as many candidates chose the wrong
answer as the right one. This seems to have been caused by careless working. Forgetting that
the mass of the glider was 2000 kg could have led candidates to the conclusion that its velocity
would be 100 m s–1 (distractor D) instead of 50 m s–1, or perhaps they forgot the factor of ½ when
finding the area of the triangle under the graph line. The facility of this question was 40%.