Preamble
General Marking Guidance:
This mark scheme is designed to reward candidates for their achievements and Examiners should ensure that they use
it with the same intention. Examiners should be prepared to use all of the marks available on this mark scheme
judiciously; where full marks are deserved, this should always be awarded to candidates; likewise, Examiners should be
prepared to award no marks to responses that are not worthy of credit. At all times, Examiners should ensure that they
only use this mark scheme when assessing a response and not any other private schemes.
It should be remembered that a mark scheme is a working document that may be further developed or refined in the
future.
General Marking Instructions:
This question paper has 75 marks.
In this mark scheme, marks are awarded in one of three forms:
M marks - these are method marks and are awarded when candidates know a correct method and apply it correctly.
A marks - these are accuracy marks and are awarded for the correct answer. They can only be awarded in conjunction
with the relevant method mark(s).
B marks - these are accuracy marks and are independent of method marks.
Common abbreviations used in this mark scheme include:
bod - benefit of doubt
ft - follow through
cao - correct answer only
cso - correct solution only
isw - ignore subsequent working
awrt - answers which round to
sc - special case
oe - or equivalent
d... - dependent
* - answer given in question
Unless stated otherwise, it should be assumed that all accuracy marks are cao (correct answer only). Do not treat
accuracy marks as follow through unless explicitly instructed to do so by the mark scheme.
For a genuine misread that does not alter the nature or difficulty of the question, you should deduct two A or B marks
and treat any other accuracy marks as ft. Method marks are unaffected.
If a candidate makes more than one attempt at a question, you should mark the attempt that is not crossed out. If all
attempts are crossed out or no attempts are crossed out, mark all the attempts and score the highest single attempt.
Following a correct answer, any subsequent working should be ignored, including any incorrect workings or statements,
unless stated otherwise.
Unless stated in the mark scheme, workings ascribed to one part of a question by a candidate can only score marks for
that part of the question.
Use of a formula:
We require candidates to state a formula before they use it. If a candidate states a formula before they use it, you may
award method marks even if there are small errors made in the substitution of numbers, for example. If a candidate
does not state a formula before they use it, method marks can be lost through errors in their working.
Preamble
General Principles for Core Mathematics Marking:
Method Marks:
Solving 3TQs of the form ax 2+ bx + c = 0:
Way 1: Factorisation - the method mark will be awarded for factorisation of the form (mx + p)(nx + q), where |mn| = a
and |pq| = c, leading to x = ... .
Way 2: Quadratic Formula - the method mark will be awarded for an attempt to use the formula with the values of a, b
and c.
2
b
c
b
+
= 0 , leading to x = ... .
Way 3: Completing the square - the method mark will be awarded for a x +
2a
2a
a
Differentiation:
The method mark is awarded for decreasing the power of at least one term by one.
Integration:
The method mark is awarded for increasing the power of at least one term by one.
General Principles for Mechanics Marking:
Method Marks:
For method marks, we must see the correct number of terms, an equation that is dimensionally correct and all terms that
need resolving are resolved.
Omissions / Additions:
Omission or an extra g in a resolution is an accuracy error not a method error.
Omission of mass from a resolution is a method error.
Omission of a length from a moments equation is a method error.
Omission of units or incorrect units is not (usually) considered to be an accuracy error.
Use of g and signfiicant figures:
Use of g = 9.81 is a rubric error and should be penalised once per (complete) question.
Any numerical answer that comes from use of g = 9.8 should be given to 2 or 3 sf. Overaccurate or underaccurate answers
should be penalised once per (complete) question.
Premature approximation:
This should be penalised every time it occurs and penalisation is not limited to once per complete question.
Vectors:
Column vectors should be accepted in all cases.
Mechanics Abbreviations:
M(A) - Taking moments about A
N2L - Newton’s Second Law
NEL - Newton’s Experimental Law
HL - Hooke’s Law
SHM - Simple harmonic motion
COLM - Conservation of linear momentum
Variables:
Unknown values in a question are, in this mark scheme, ascribed their ‘standard’ letters. For example, R is used to denote
the normal reaction force and T for tension, but this may vary depending on the nature of the question.
Preamble
No Working Shown:
If an answer can be reasonably obtained without showing any working and it is unlikely that the correct answer may arise
from incorrect workings, then Examiners should award full marks. It is a given that candidates are aware that an incorrect
answer obtained without working will inevitably receive no marks. In the cases where working is deemed necessary,
evidence of a method is required for the candidate to be awarded any marks.
When considering whether an answer may be obtained ‘without showing any working’, Examiners should consider two
things. The first is whether or not it can be done ‘in your head’. The second is that many permitted calculators may
allow candidates to solve a question directly (this is provided the exam is a calculator exam). If Examiners are in doubt,
they should consult with us.
In many instances, a question may ask candidates to use a specific method or ‘show’ or ‘deduce’ a result. In such cases,
candidates who provide no evidence of their workings have broken the rubric of the question and should be awarded
no marks.
Question
Number
1
Way 1
Scheme
s = 0.1 m , u = 3 m s −1 , a = −g m s −2 , t = ? s
0.1 = 3t +
1
( −g ) t 2
2
4.9t 2 − 3t + 0.1 = 0 ⇒ t = ...
Way 2
Marks
Uses s = ut +
1 2
at
2
Forms a 3TQ and attempts to solve it
M1A1
dM1
t = 0.0353... or 0.576...
Correct solutions to the 3TQ
A1
0.576 – 0.0353 = 0.541 s
Correct length of time to two or three
significant figures
A1
Attempts to find velocity of P when it is
0.1m above the point of projection
M1
Correct values for v
A1
v 2 = 32 − 2g ( 0.1)
v = ± 2.653...
2.653... = −2.653 + gt ⇒ t = ...
Uses v = u + at and rearranges for t
[5]
dM1
A1
t = 0.541 s
Way 3
v 2 = 32 − 2g ( 0.1)
Correct length of time to two or three
significant figures
A1
Attempts to find velocity of P when it is
0.1m above the point of projection
M1
Correct values for v
A1
v = ± 2.653...
1
0 = ( 2.653...) t − gt 2 ⇒ t = ...
2
{t = 0} , t = 0.541 s
Uses s = ut +
[5]
1 2
dM1
at and solves for t
2
A1
Correct length of time to two or three
significant figures
A1
[5]
5
Question 1 Notes
Way 1
Way 2
Way 3
1 2
at with the magnitude of the values correct.
2
1st M1
Attempts to use s = ut +
1st A1
Substitutes all the values correctly into the formula with consistent signs.
2nd M1
Attempts to solve their 3TQ. Methods include use of the formula, completing the
square (unlikely here) or a calculator. See notes on page 4 for more detail. This is
dependent on the 1st M1.
2nd A1
Correct values of t for which P is 0.1 m above the point of projection.
3rd A1
Correct length of time given to two or three significant figures only.
st
1 M1
Attempts to find the velocity of P when it is 0.1m above the point of projection using
v2 = u2 + 2as.
1st A1
Correct values of v (must see both) – although, omission of the ± here can be
recovered through subsequent working.
2nd M1
Uses their value of v in the formula v = u + at with u = – v. This is dependent on the
1st M1.
2nd A1
For their v = u + at being correct and with consistent signs.
3rd A1
Correct length of time given to two or three significant figures only.
SC
If candidates attempt to work out the velocity of P when it is 0.1 m above the point of
projection by considering energy, then this should be allowed for the 1st M1 and A1.
1st M1
Attempts to find the velocity of P when it is 0.1m above the point of projection using
v2 = u2 + 2as.
1st A1
Correct values of v (must see both) – although, omission of the ± here can be
recovered through subsequent working.
2nd M1
Attempts to use s = ut +
2nd A1
Substitutes all the values correctly into the formula with consistent signs.
3rd A1
Correct length of time given to two or three significant figures only.
SC
1 2
at with the magnitude of the values correct.
2
If candidates attempt to work out the velocity of P when it is 0.1 m above the point of
projection by considering energy, then this should be allowed for the 1st M1 and A1.
Question
Number
2
(a)
Scheme
Marks
At t = 0 , S ( 3i + 6 j) ; at t = 4 , S ( i − j)
v=
( i − j) − ( 3i + 6 j)
4
v = −0.5i − 3.5 j
r = ( 3i + 6 j) + ( −0.5i − 3.5 j) t
∴r = ( 3 − 0.5t ) i + ( 6 − 3.5t ) j
Attempts to find the velocity
vector of S
M1
Correct velocity vector A1
Uses r = r0 + vt with their v dM1
Convincing proof A1
[4]
(b)
3 − 0.5t = 0 ⇒ t = 6
r=
{( 3 − 0.5 ( 6 )) i} + ( 6 − 3.5 ( 6 )) j
r = −15 j
(c)
Correct value for t when S is north of O
B1
Substitutes t = 6 into (a) M1
Correct position vector of S when it is A1
north of O
r = ( 3 − 0.5 ( 20 )) i + ( 6 − 3.5 ( 20 )) j
[3]
Substitutes t = 20 into (a) M1
r = −7i − 64 j
displacement of S = '− 7i − 64 j'− ( 3i + 6 j)
Attempts to work out the displacement dM1
of S after 20 seconds
displacement of S = −10i − 70 j
distance of S =
10 2 + 70 2
distance of S = 50 2
Uses Pythagoras’ on their displacement ddM1
to find the distance of S after 20 seconds
Correct distance A1 oe
[4]
11
Question 2 Notes
(a)
(b)
(c)
1st M1
Attempts to work out the velocity vector of S.
1st A1
Correct velocity vector of S.
2nd M1
Attempts to find the position vector of S. This is dependent on the 1st M1.
2nd A1
Correct and convincing proof with no errors seen.
B1
Correct value of t for when S is north of O.
M1
Substitutes their value of t into (a) to find the position vector of O.
A1
Correct position vector of O.
1st M1
nd
Attempts to find the position vector of S when t = 20.
2 M1
Subtracts their position vector at t = 20 from the initial position vector of S to find the
displacement of S. Subtraction must be the right way around. This is dependent on the
1st M1.
3rd M1
Uses Pythagoras’ on their displacement to find the distance travelled by S after 20
seconds.
A1
Correct distance.
Question
Number
Scheme
Marks
3
General
This is a general scheme. There are many different equations that candidates can use;
these are shown in the notes below. This scheme is designed to show how the marks are
awarded in the general case.
Takes moments around any valid point M1A1
Resolves vertically or takes moments to M1A1
form a second equation
Attempts to solve their moments ddM1
equation for both of the values
M = 15.714... =
RA = 55.997 =
{awrt} 16 N
{awrt} 56 N
Obtains the correct value for M or RA A1
Obtains the correct value for M and RA A1
to two or three significant figures only
7
Question 3 Notes
1st M1
Takes moments about any point to form an equation involving M or RA.
1st A1
A completely correct moments equation.
2nd M1
Forms a second equation involving M or RA by taking moments or resolving vertically.
2nd A1
A completely correct second equation.
3rd M1
Attempts to solve both of their equations either individually (or simultaneously if this is
required) to find M and RA. This is dependent on both the 1st and 2nd M1.
3rd A1
Either M or RA.
4th A1
Both M and RA to two or three significant figures only.
Notes
Table of relevant equations
Moment equations for 1st/2nd M1 A1
Pivot
Equation
A
20g (1) − 294 ( 2.5 ) + Mg ( 3.5 ) = 0
B
20g (1.5 ) − RA ( 2.5 ) − Mg (1) = 0
C
294 (1) − 20g ( 2.5 ) + RA ( 3.5 ) = 0
Centre of rod
RA (1) − 294 (1.5 ) + Mg ( 2.5 ) = 0
Other
Accept any other moments equation around another other
clearly defined point. As a reminder, it must be
dimensionally correct for the 1st M1.
Resolving equations for the 2nd M1/A1
Vertically
RA + 294 − 20g − Mg = 0
Question
Number
Scheme
Marks
4
Way 1 – resolving horizontally and vertically
(a)
R ( → + ) : TAC cos 30 − 5 cos 60 = 0
TAC =
TAC =
(b)
Resolves horizontally
5 cos 60
cos 30
5 3
= 2.8867... = 2.90
3
Correct value for the tension in AC
A1
[3]
( )
Resolves vertically
R ↑+ : TAC sin 30 + 5sin 60 − (10 + k ) g = 0
k=
M1A1
M1A1
( 2.8867...) sin 30 + 5sin 60 − 10
g
k = −9.4108... = −9.4
Correct value of k
A1
[3]
Way 2 – resolving along AB and AC
Parts (a) and (b) should be marked together for this approach
(a) / (b)
R ( AB ) : 5 − (10 + k ) g cos 30 = 0
Resolves along AB
M1A1
R ( AC ) : TAC − (10 + k ) g cos 60 = 0
Resolves along AC
M1A1
TAC =
5 3
= 2.8867... = 2.90
3
Correct value for the tension in AC
A1
Correct value for k
A1
k = −9.4108... = −9.4
[6]
Way 3 – use of a triangle of forces
(a)
T
5
= AC
sin 60 sin 30
TAC =
(b)
5 3
= 2.8867... = 2.90
3
Correct use of the sine rule (or any other M1A1
method) to form a correct expression
Correct value for the tension in AC
A1
Any valid method. See notes
6
Question 4 Notes
Resolves horizontally. Must be dimensionally correct, but condone sin/cos confusion.
Way 1
M1
(a)
1st A1
Correct equation.
2nd A1
Correct value for the tension in AC. Do not penalise accuracy here (as g not used).
M1
Resolves vertically. Must be dimensionally correct, but condone sin/cos confusion.
(b)
1st A1
Correct equation.
2nd A1
Correct value of k to two or three significant figures only.
Way 2
1st M1
Resolves along AB. Must be dimensionally correct, but condone sin/cos confusion.
(a) & (b)
1st A1
Correct equation.
nd
Resolves along AC. Must be dimensionally correct, but condone sin/cos confusion.
nd
Correct equation.
rd
Correct value for the tension in AC to two or three significant figures only.
th
4 A1
Correct value of k two or three significant figures only.
Note
Do not penalise over accuracy or under accuracy more than once.
M1
Use of a triangle of forces along with ‘sohcahtoa’ or the sine rule to form an equation
involving the tension in AC.
2 M1
2 A1
3 A1
Way 3
(a)
1st A1
nd
(b)
Correct equation (or equivalent, i.e. 5tan60).
2 A1
Correct value for the tension in AC. Do not penalise accuracy here (as g not used).
Notes
Part (b) can be done by either resolving along AB or AC or vertically, so mark using the
appropriate scheme in Way 1 or Way 2 with the 1st M1 and A1 for a correct resolution
and the 2nd A1 for a correct answer.
Question
Number
Scheme
Marks
5
R − 20g cos α = 0 {⇒ R = 20g cos α }
100 cos α −
Resolves perpendicular to the plane
Resolves parallel to the plane
1
( 20g cos α ) − 20gsin α = 0
5
M1A1
B1
M1A1
100 cos α − 4g cos α = 20gsin α
cos α (100 − 4g ) = 20gsin α ⇒ tan α = ...
tan α =
Attempts to form an expression in tan α
Correct expression
A1
Correct value of α to two or three
significant figures
A1
100 − 4g
20g
⎛ 100 − 4g ⎞
α = tan −1 ⎜
= 17.23... = 17 °
⎝ 20g ⎟⎠
ddM1
[8]
Question 5 Notes
st
1 M1
Resolves perpendicular to the plane. Must be dimensionally correct. Condone sin/cos
confusion.
1st A1
Correct equation.
B1
Use of F = 0.2 R at any stage. Can be implied or seen on a diagram perhaps.
2nd M1
Resolves parallel to the plane. Must be dimensionally correct. Condone sin/cos
confusion.
2nd A1
Correct equation.
3rd M1
Attempts to form an expression in terms of tanx using the relevant trig identity. This is
dependent on both previous method marks.
4th A1
Correct expression in terms of tanx.
5th A1
Correct value of α to two or three significant figures only.
Question
Number
Scheme
Marks
6
(a)
Initial speed of A = 3 ms-1
Correct speed of A due to impulse
3m = 2m + xm
Uses COLM
{ x =} 1 ( ms−1 )
Correct speed of A after its collision
with B
B1
M1A1
A1
(speed of A after the collision is 1 ms-1)
(b)
[4]
2m = 2my − my
Uses COLM
2 ( ms −1 )
Correct speed of B after its collision
with C
{ y =}
M1A1
A1
(speed of B after the collision is 2 ms-1)
(c)
[3]
Yes, {there will be another collision,}
because A and B are moving towards each
other.
Convincing explanation
B1
[1]
8
Question 6 Notes
(a)
B1
States or implies that the speed of A as it approaches B is 3 ms-1.
(b)
M1
Uses COLM with the correct number of terms, which must be dimensionally correct.
(c)
1st A1
Forms a correct equation. Accept any (or even no) variable in place of x.
2nd A1
Correct speed of A after its collision with B.
M1
Uses COLM with the correct number of terms, which must be dimensionally correct.
1st A1
Forms a correct equation. Accept any (or even no) variable in place of y.
2nd A1
Correct speed of B after its collision with C.
B1
Provides a convincing explanation that says there will be another collision and gives the
correct explanation for this: they are moving towards each other. The explanation may
be written but you may also give credit to candidates that explain this using a diagram.
Question
Number
Scheme
Marks
7
(a)
Considering A: T – 3g = 3a
Considers A and B and uses N2L
M1A1
Considering B: 7g – T = 7a
a = 3.92 (ms-2)
Correct value for the acceleration of
the two particles
T = 41.2 (N)
(b)
A1
Correct value for the tension in the A1
strings
Resultant force on pulley – 2T = 0
Considers the entire system and uses M1
N2L
Resultant force on pulley = 82.4 N
Correct resultant force on the pulley
[4]
A1ft
[2]
(c)
speed of A when B hits the ground =
Way 1
2 ( 3.92 ) ( 0.1)
= 0.8854...
Height reached by A when string becomes slack
2
=
2
0 − ( 0.8854 )
= 0.0399...
2 ( −g )
∴ x = 0.5 + 0.1 + ‘0.0399’
∴ x = 0.639 or 0.64
(c)
Way 2
Uses the correct ‘suvat’ equation
M1
Correct speed
A1
Uses the correct ‘suvat’ equation to
work out the height A reaches
dM1
Correct height
A1ft
Correct method to find x
Correct value for x to two or three
significant figures only
ddM1
A1
[6]
For this method, the 1st M1 and A1 and 3rd M1 and A1 are the same as ‘Way 1’. The
marks for the 2nd M1 and A1 can be alternatively also as below.
Considering energy:
1
2
m ( 0.8854 ) = m ( 9.8 ) h
2
Height reached by A when string becomes slack
= 0.0399…
Considers energy to find the height A
reaches above the point the string
becomes slack
Correct height
dM1
A1
12
Question 7 Notes
(a)
(b)
M1
Considers A and B and uses N2L to form two dimensionally correct equations.
1st A1
Forms two equations that are completely correct.
2nd A1
Obtains the correct value for the tension in the string.
3rd A1
Obtains the correct value for the acceleration of the system.
M1
Considers the whole system and applies N2L and deduces the relevant equation.
A1ft
Obtains the correct resultant force ft their value of T in (a).
(c)
1st M1
Uses v2 = u2 + 2as to find the speed of A when B hits the ground. Condone a sign error.
Way 1
1st A1
Correct value for the speed of A when B hits the ground.
2 M1
Uses v2 = u2 + 2as to find the height gained by A when the string becomes slack.
Condone a sign error. This is dependent on the 1st M1.
2nd A1
Correct value for the height gained by A when the string becomes slack ft their speed.
3rd M1
Attempts to combine their height with 0.1 and 0.5. This is dependent on both previous
method marks.
3rd A1
Correct value of x.
(c)
2nd M1
Considers energy to work out the height gained by A when the string becomes slack.
Way 2
2nd A1
Correct value for the height gained by A when the string becomes slack ft their speed.
nd
Question
Number
Scheme
Marks
8
(a)
Considering the system:
5400 − 750 − 500 = ( 2400 + 1000 ) a
a = 1.2205... = 1.22 (ms-2)
Considers the system and applies
N2L
Correct value for the acceleration of
the system
M1A1
A1
[3]
(b)
v = 0 + 10 (1.2205...)
v = 12.205... = 12 ( ms −1 )
(c)
Way 1
Considering the caravan:
5400 − 750 − T = 2400 (1.220...)
T = {awrt} 1720
Uses v = u + at
M1
Correct value for the speed of the
system when t = 10
A1
Considers the caravan and applies
N2L
M1A1
Correct tension
[2]
A1
[3]
(c)
Way 2
Considering the trailer:
Considers the trailer and applies
N2L
T − 500 = 1000 (1.220...)
T = {awrt} 1720 (N)
Correct tension
M1A1
A1
[3]
(d)
Deceleration of system =
−750 − 500
= – 0.3676
3400
0 2 − (1.22 × 30 )
Distance travelled =
−2 ( 0.3676 )
2
Distance travelled = {awrt} 1820 (m)
Correct deceleration
Uses v2 = u2 + 2as to find the
distance travelled by the system
Correct distance
B1
M1A1
A1
[4]
(e)
Way 1
Considering the caravan:
FR − 750 = 2400 ( −0.3676...)
Magnitude of force in rod = {awrt} 130 N
Since FR is negative, {or positive if they consider
FR to be in the other direction}, the rod is in
tension
Correct equation but allow FR to be
in the same direction as the 750 N
force
M1
Correct force
A1
Convincing explanation
A1
[3]
(e)
Way 2
Considering the trailer:
FR − 500 = 1000 ( −0.3676...)
Correct equation but allow FR to be
in the same direction as the 500 N
force
M1
Correct equation
A1
Convincing explanation
A1
Magnitude of force in rod = {awrt} 130 N
Since FR is positive, {or negative if they consider
FR to be in the other direction}, the rod is in
tension
[3]
Correct shape
B1
Relative steepness of the lines is correct
B1
Times indicated on the graph correctly
B1
(f)
[3]
18
Question 8 Notes
(a)
(b)
(c)
Both
ways
M1
1st A1
Correct equation.
2nd A1
Correct acceleration of the system.
M1
Attempts to work out the speed of the system at t = 10.
A1
Correct speed.
M1
Considers either the caravan or the trailer and applies N2L.
st
Correct equation.
nd
Correct value for the tension in the rod.
1 A1
2 A1
(d)
Considers the whole system and applies N2L to form a dimensionally correct equation.
B1
Obtains the correct value for the deceleration in the system.
M1
Attempts to use v2 = u2 + 2as to find the distance travelled by the system.
1st M1
Correct equation.
2nd A1
Correct distance travelled by the system as it decelerates.
Considers the caravan or the trailer and forms an equation involving the force in the
rod, which may act in any direction.
(e)
M1
Both
ways
1st A1
Correct value for the magnitude of the force acting in the rod.
2nd A1
Provides a convincing explanation that the rod is in tension.
1st B1
For a correct shape.
2nd B1
A steeper line for acceleration than deceleration.
3rd B1
t = 0 and t = 30 correctly marked onto the graph (values of v not necessary).
(f)