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UVI Further Maths Vacation Work
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 1
UVI FURTHER MATHS VACATION WORK
The topics you will study next term rely on a good understanding of the material you have studied this year. You need to complete the following vacation work, and mark your work with the answers provided. Use your notes and textbook to help you. Your work is due in on the first lesson back, and you will be tested on this in your start up test.
Please also complete the UVI MATHS VACATION WORK
UVI Further Maths Vacation Work
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1. Show that can be written in the form
Find the values of the constants A and B.
(5 marks) ___________________________________________________________________________________
2. Use proof by contradiction to show that there exist no integers a and b for which 25a + 15b = 1.
(4 marks) ___________________________________________________________________________________
3. A curve has parametric equations x = cos 2t, y = sin t, t−π π .
(a) Find an expression for ddyx
in terms of t.
Leave your answer as a single trigonometric ratio. (3 marks)
(b) Find an equation of the normal to the curve at the point A where 56
t π= − .
(5 marks) ___________________________________________________________________________________
4. Showing all steps, find cot 3 dx x∫ .
(3 marks) ___________________________________________________________________________________
5. A triangle has vertices A(−2, 0, −4), B(−2, 4, −6) and C(3, 4, 4). By considering the side lengths of the triangle, show that the triangle is a right-angled triangle.
(6 marks) ___________________________________________________________________________________
6. The functions p and q are defined by 2p : x x→ and q : 5 2x x→ − .
(a) Given that pq(x) = qp(x), show that 23 10 10 0x x− + = (4 marks)
(b) Explain why 23 10 10 0x x− + = has no real solutions.
(2 marks) ___________________________________________________________________________________
7. Prove by contradiction that there are infinitely many prime numbers.
(6 marks) ___________________________________________________________________________________
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8. In a rainforest, the area covered by trees, F, has been measured every year since 1990. It was found that the
rate of loss of trees is proportional to the remaining area covered by trees. Write down a differential equation relating F to t, where t is the numbers of years since 1990.
(2 marks) ___________________________________________________________________________________
9. At the beginning of each month Kath places £100 into a bank account to save for a family holiday. Each subsequent
month she increases her payments by 5%. Assuming the bank account does not pay interest, find (a) the amount of money in the account after 9 months.
(3 marks) Month n is the first month in which there is more than £6000 in the account.
(b) Show that log 4log1.05
n >
(4 marks) Maggie begins saving at the same time as Kath. She initially places £50 into the same account and plans to increase her payments by a constant amount each month. (c) Given that she would like to reach a total of £6000 in 29 months, by how much should Maggie
increase her payments each month? (2 marks)
___________________________________________________________________________________
10. Find 2cos 6 dx x∫ .
(5 marks) ___________________________________________________________________________________
11. (a) Prove that .
(3 marks)
(b) Hence solve, in the interval 0 2x π , the equation .
(3 marks) ___________________________________________________________________________________
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12. A large arch is planned for a football stadium. The parametric equations of the arch are 8( 10)x t= + ,
2100y t= − ,–19 ≤ t ≤ 10 where x and y are distances in metres. Find (a) the cartesian equation of the arch,
(3 marks)
(b) the width of the arch, (2 marks)
(c) the greatest possible height of the arch.
(2 marks) ___________________________________________________________________________________
13.
Find the values of the constants A, B, C and D.
(5 marks) ___________________________________________________________________________________
14. The volume of a sphere V cm3 is related to its radius r cm by the formula 343
V r= π . The surface area of the
sphere is also related to the radius by the formula 24S r= π . Given that the rate of decrease in surface area,
in cm2 s–1, is d 12dSt= − ,
find the rate of decrease of volume ddVt
(4 marks) ___________________________________________________________________________________
15. Find
3sin dx x∫ .
(4 marks) ___________________________________________________________________________________
16. 21h( ) 40ln( 1) 40sin , 0 5 4tt t t t = + + −
.
The graph y = h(t) models the height of a rocket t seconds after launch. (a) Show that the rocket returns to the ground between 19.3 and 19.4 seconds after launch.
(2 marks) (b) Using t0 = 19.35 as a first approximation to α, apply the Newton–Raphson procedure once to h(t) to
find a second approximation to α, giving your answer to 3 decimal places. (5 marks)
(c) By considering the change of sign of h(t) over an appropriate interval, determine if your answer to part
(b) is correct to 3 decimal places. (3 marks)
___________________________________________________________________________________
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17. (a) Show that in KLM∆ with 3 0 6KL = + −i j k
and 2 5 4LM = + +i j k
, 66.4KLM∠ = ° to one decimal place.
(7 marks) (b) Hence find LKM∠ and LMK∠ .
(3 marks)
___________________________________________________________________________________
TOTAL FOR PAPER IS 93 MARKS
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SECTION B: MECHANICS
Answer ALL questions.
9. At time t seconds, a 2 kg particle experiences a force F N, where 28 64 12
t t= +−
F
(a) Find the acceleration of the particle at time t seconds.
(3) The particle is initially at rest at the origin. (b) Find the position of the particle at time t seconds.
(6) (c) Find the particle’s velocity when t = 1.
(3)
(Total 12 marks) ____________________________________________________________________________
10. An archer shoots an arrow at 10 m s−1 from the origin and hits a target at (10, −5) m. The initial velocity of the arrow is at an angle θ above the horizontal. The arrow is modelled as a particle moving freely under gravity.
(In this question, take g = 10 m s−2.)
(a) Show that (tan θ − 1)2 = 1.
(11) (b) Find the possible values of .
(3)
(Total 14 marks) _____________________________________________________________________________
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11. Figure 3 shows a 5000 kg bus hanging 12 m over the edge of a cliff with 1000 kg of gold at the front. The gold sits on a wheeled cart. A group of n people, each weighing 70 kg, stands at the other end. The bus is 20 m long.
Figure 3 (a) Write down the total clockwise moment about the cliff edge in terms of n.
(7) (b) Find the smallest number of people needed to stop the bus falling over the cliff.
(2) (c) One person needs to walk to the end of the bus to retrieve the gold. Find the smallest number
of people needed to stop the bus falling over the cliff in this situation, including the one retrieving the gold.
(4)
(Total 13 marks) _____________________________________________________________________________
12. A car travels along a long, straight road for one hour, starting from rest. After t hours, its acceleration is a km h−2, where a = 180 − 360t. (a) Find the speed of the car, in km h−1 in terms of t.
(2) The speed limit is 40 km h−1. (b) Find the range of times during which the car is breaking the speed limit. Give your answer in
minutes. (4)
(c) Find the average speed of the car over the whole journey.
(5)
(Total 11 marks) _____________________________________________________________________________
TOTAL FOR PAPER IS 100
UVI Further Maths Vacation Work
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UVI Further Maths Vacation Work
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1 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
States that:
M1 2.2a 5th
Decompose algebraic
fractions into partial fractions − two linear factors.
Equates the various terms.
Equating the coefficients of x:
Equating constant terms:
M1* 2.2a
Multiplies both of the equations in an effort to equate one of the two variables.
M1* 1.1b
Finds A = 8 A1 1.1b
Find B = −2 A1 1.1b
(5 marks)
Notes
Alternative method
Uses the substitution method, having first obtained this equation:
Substitutes to obtain 272
− B = 27 (M1)
Substitutes to obtain 275
A = 43.2 (M1)
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2 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
Begins the proof by assuming the opposite is true.
‘Assumption: there do exist integers a and b such that ’
B1 3.1 7th
Complete proofs using proof by contradiction.
Understands that
‘As both 25 and 15 are multiples of 5, divide both sides by 5 to
leave ’
M1 2.2a
Understands that if a and b are integers, then 5a is an integer, 3b is an integer and 5a + 3b is also an integer.
M1 1.1b
Recognises that this contradicts the statement that ,
as 15
is not an integer. Therefore there do not exist integers a and
b such that ’
B1 2.4
(4 marks)
Notes
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3 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
(a) Finds d 2sin 2
dx tt= − and d cos
dy tt=
M1 1.1b 6th
Differentiate simple functions
defined parametrically
including application to tangents and
normals.
Writes −2sin 2t = − 4sin t cos t M1 2.2a
Calculates d cos 1 cosecd 4sin cos 4y t tx t t= = −−
A1 1.1b
(3)
(b) Evaluates d
dyx
at 56
t π= −
A1 ft 1.1b 6th
Differentiate simple functions
defined parametrically
including application to tangents and
normals. Understands that the gradient of the tangent is 12
, and then the
gradient of the normal is −2.
M1 ft 1.1b
Finds the values of x and y at 56
t π= −
5 1cos 26 2
x π = ×− =
and 5 1sin6 2
y π = − = −
M1 ft 1.1b
Attempts to substitute values into 1 1( )y y m x x− = −
For example, 1 122 2
y x + = − −
is seen.
M1 ft 2.2a
Shows logical progression to simplify algebra, arriving at: 122
y x= − + or 4 2 1 0x y+ − =
A1 2.4
(5)
(8 marks)
Notes
(b) Award ft marks for a correct answer using an incorrect answer from part a.
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4 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
States that
cos3cot 3sin3
xxx
= M1 2.2a 6th
Integrate using trigonometric
identities. Makes an attempt to find
cos3 dsin 3
x xx
∫
Writing ( )( ) ( )
f 'd ln f
fx
x xx
= ∫ or writing ln (sin x) constitutes an
attempt.
M1 2.2a
States a fully correct answer1 ln sin33
x C+ A1 1.1b
(3 marks)
Notes
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5 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
Demonstrates an attempt to find the vectors AB
, AC
and BC
M1 2.2a 5th
Find the magnitude of a
vector in 3 dimensions.
Finds ( )0,4, 2AB = −
, ( )5,4,8AC =
and ( )5,0,10BC =
A1 1.1b
Demonstrates an attempt to find | |AB
, | |AC
and| |BC
M1 2.2a
Finds ( ) ( ) ( )2 2 2| | 0 4 2 20AB = + + − =
Finds ( ) ( ) ( )2 2 2| | 5 4 8 105AC = + + =
Finds ( ) ( ) ( )2 2 2| | 5 0 10 125BC = + + =
A1 1.1b
States or implies in a right-angled triangle 2 2 2c a b= + M1 2.2a
States that 2 2 2| | | | | |AB AC BC+ =
B1 2.1
(6 marks)
Notes
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6 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
(a) States or implies that ( )2pq( ) 5 2x x= − M1 2.2a 5th
Find composite functions. States or implies that 2qp( ) 5 2x x= − M1 2.2a
Makes an attempt to solve ( )2 25 2 5 2x x− = − . For example, 2 225 20 4 5 2x x x− + = − or 26 20 20 0x x− + = is seen.
M1 1.1b
States that 23 10 10 0x x− + = . Must show all steps and a logical progression.
A1 1.1b
(4)
(b) ( )( )2 4 100 4 3 10 20 0b ac− = − = − < M1* 2.2a 5th
Find the domain and range of composite functions.
States that as 2 4 0b ac− < there are no real solutions to the equation.
B1* 3.2b
(2)
(6 marks)
Notes
(b) Alternative Method
M1: Uses the method of completing the square to show that 25 653 0
3 9x − + =
or
25 6533 9
x − = −
B1: Concludes that this equation will have no real solutions.
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7 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
Begins the proof by assuming the opposite is true.
‘Assumption: there is a finite amount of prime numbers.’
B1 3.1 7th
Complete proofs using proof by contradiction. Considers what having a finite amount of prime numbers means
by making an attempt to list them:
Let all the prime numbers exist be
M1 2.2a
Consider a new number that is one greater than the product of all the existing prime numbers:
Let
M1 1.1b
Understands the implication of this new number is that division by any of the existing prime numbers will leave a remainder of 1. So none of the existing prime numbers is a factor of N.
M1 1.1b
Concludes that either N is prime or N has a prime factor that is not currently listed.
B1 2.4
Recognises that either way this leads to a contradiction, and therefore there is an infinite number of prime numbers.
B1 2.4
(6 marks)
Notes
If N is prime, it is a new prime number separate to the finite list of prime numbers, .
If N is divisible by a previously unknown prime number, that prime number is also separate to the finite list of prime numbers.
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8 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
Attempts to write a differential equation.
For example, ddF Ft∝ or is seen.
M1 3.1a 7th
Construct simple differential equations.
States A1 3.1a
(2 marks)
Notes
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9 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
(a) Recognises that it is a geometric series with a first term 100=a and common ratio 1.05=r
M1 3.1a 6th
Use geometric sequences and
series in context. Attempts to use the sum of a geometric series. For example,
( )9
9
100 1 1.05
1 1.05S
−=
− or
( )9
9
100 1.05 1
1.05 1S
−=
− is seen.
M1* 2.2a
Finds 9 £1102.66S = A1 1.1b
(3)
(b) States
( )100 1.05 16000
1.05 1
n −>
− or
( )100 1 1.056000
1 1.05
n−>
−
M1 3.1a 5th
Use arithmetic sequences and
series in context. Begins to simplify. 1.05 4n > or 1.05 4n− < − M1 1.1b
Applies law of logarithms correctly log1.05 log 4n > or log1.05 log 4n− < −
M1 2.2a
States log 4log1.05
n > A1 1.1b
(4)
(c) Uses the sum of an arithmetic series to
state ( )29 100 28 60002
d + = M1 3.1a 5th
Use arithmetic sequences and
series in context. Solves for d. d = £11.21 A1 1.1b
(2)
(9 marks)
Notes M1 Award mark if attempt to calculate the amount of money after 1, 2, 3,….,8 and 9 months is seen.
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10 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
Selects 2cos 2 2cos 1x x≡ − as the appropriate trigonometric identity.
M1 2.2a 6th
Integrate using trigonometric
identities. Manipulates the identity to the question: 2cos12 2cos 6 1x x≡ − M1 1.1b
States that ( ) ( ) 2 1cos 6 d 1 cos12 d
2x x x x= +∫ ∫
M1 1.1b
Makes an attempt to integrate the expression, x and sin x are seen.
M1 1.1b
Correctly states 1 1 sin122 12
x x C + +
A1 1.1b
(5 marks)
Notes
Student does not need to state ‘+C’ to be awarded the third method mark. Must be stated in the final answer.
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11 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
(a) Writes tanx and secx in terms of sinx and cosx. For example,
sin 1tan sec cos cos
1 sin1 sin1
xx x x x
xx
− − =−−
M1 2.1 5th
Understand the functions sec, cosec and cot.
Manipulates the expression to find sin 1 1cos 1 sin
xx x− × −
M1 1.1b
Simplifies to find A1 1.1b
(3)
(b) States that or B1 2.2a 6th
Use the functions sec, cosec and cot
to solve simple trigonometric
problems.
Writes that or M1 1.1b
Finds A1 1.1b
(3)
(6 marks)
Notes
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12 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
(a) Rearranges ( )8 10x t= + to obtain 80
8xt −
= M1 1.1b 8th
Use parametric equations in
modelling in a variety of contexts.
Substitutes 808
xt −= into 2100y t= −
For example,280100
8xy − = −
is seen.
M1 1.1b
Finds 21 564 2
y x x= − + A1 1.1b
(3)
(b) Deduces that the width of the arch can be found by substituting 10t = ± into ( )8 10x t= +
M1 3.4 8th
Use parametric equations in
modelling in a variety of contexts.
Finds x = 0 and x = 160 and deduces the width of the arch is 160 m.
A1 3.2a
(2)
(c) Deduces that the greatest height occurs when d 0 2 0 0dy t tt= ⇒ − = ⇒ =
M1 3.4 8th
Use parametric equations in
modelling in a variety of contexts.
Deduces that the height is 100 m. A1 3.2a
(2)
(7 marks)
Notes
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13 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
Makes an attempt to set up a long division.
For example: is seen.
M1 2.2a 5th
Divide polynomials by
linear expressions with a remainder.
Award 1 accuracy mark for each of the following: 2x seen, 2x seen, −21 seen.
For the final accuracy mark either D = 138 or 138
6x + or the remainder
is 138 must be seen. 2
3 2
3 2
2
2
2 216 8 9 12
6
2 9
2 12
21 12
21 126
138
x xx x x x
x x
x x
x x
x
x
+ −
+ + − +
+
−
+
− +
− −
A4 1.1b
(5 marks)
Notes
This question can be solved by first writing and then solving for A, B, C and D. Award 1 mark for the setting up the problem as described. Then award 1 mark for each correct coefficient found. For example:
Equating the coefficients of x3: A = 1
Equating the coefficients of x2: 6 + B = 8, so B = 2
Equating the coefficients of x: 12 + C = −9, so C = −21
Equating the constant terms: −126 + D = 12, so D = 138.
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14 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
Recognises the need to use the chain rule to find d
dVt
For example d d d dd d d dV V r St r S t= × × is seen.
M1 3.1a 8th
Construct differential
equations in a range of contexts.
Finds 2d 4dV rr= π and d 8
dS rr= π
M1 2.2a
Makes an attempt to substitute known values. For
example,2d 4 1 12
d 1 8 1V rt r
π −= × ×
π
M1 1.1b
Simplifies and states d 6dV rt= −
A1 1.1b
(4 marks)
Notes
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Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
15 Recognises the need to write ( )3 2sin sin sinx x x≡ M1 2.2a 6th
Integrate using trigonometric
identities. Selects the correct trigonometric identity to write
( ) ( )2 2sin sin sin 1 cosx x x x≡ − . Could also write 2sin sin cosx x x−
M1 2.2a
Makes an attempt to find ( ) 2sin sin cos dx x x x−∫ M1 1.1b
Correctly states answer 31cos cos3
x x C− + + A1 1.1b
(4 marks)
Notes
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16 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
(a) Finds and M1 3.1a 7th
Use numerical methods to solve
problems in context.
Change of sign and continuous function in the interval ⇒ root
A1 2.4
(2)
(b) Makes an attempt to differentiate h(t) M1 2.2a 7th
Use numerical methods to solve
problems in context.
Correctly finds A1 1.1b
Finds and M1 1.1b
Attempts to find 1x
01 0 1
0
h( ) 0.2903...19.35h ( ) 13.6792...
xx x xx
= − ⇒ = −′ −
M1 1.1b
Finds A1 1.1b
(5)
(c) Demonstrates an understanding that x = 19.3705 and x = 19.3715 are the two values to be calculated.
M1 2.2a 7th
Use numerical methods to solve
problems in context.
Finds and M1 1.1b
Change of sign and continuous function in the interval ⇒ root
A1 2.4
(3)
(10 marks)
Notes
(a) Minimum required is that answer states there is a sign change in the interval and that this implies a root in the given interval.
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17 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
(a) Demonstrates an attempt to find the vectors KL
, LM
and KM
M1 2.2a 6th
Solve geometric problems using
vectors in 3 dimensions.
Finds ( )3,0, 6KL = −
, ( )2,5,4LM =
and ( )5,5, 2KM = −
A1 1.1b
Demonstrates an attempt to find | |KL
, | |LM
and| |KM
M1 2.2a
Finds ( ) ( ) ( )2 2 2| | 3 0 6 45KL = + + − =
Finds ( ) ( ) ( )2 2 2| | 2 5 4 45LM = + + =
Finds ( ) ( ) ( )2 2 2| | 5 5 2 54KM = + + − =
A1 1.1b
Demonstrates an understanding of the need to use the Law of Cosines. Either 2 2 2 2 cosc a b ab C= + − × (or variation) is seen, or attempt to substitute into formula is made
( ) ( ) ( ) ( )( )2 2 254 45 45 2 45 45 cosθ= + −
M1 ft 2.2a
Makes an attempt to simplify the above equation. For example, 36 90cosθ− = − is seen.
M1 ft 1.1b
Shows a logical progression to state 66.4θ = ° B1 2.4
(7)
(b) States or implies that KLM∆ is isosceles. M1 2.2a 6th
Solve geometric problems using
vectors in 3 dimensions.
Makes an attempt to find the missing
angles 180 66.421...2
LKM LMK −∠ = ∠ =
M1 1.1b
States 56.789...LKM LMK∠ =∠ = ° . Accept awrt 56.8° A1 1.1b
(3)
(10 marks)
Notes
(b) Award ft marks for a correct answer to part a using their incorrect answer from earlier in part a.
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H9 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
a Use of Newton’s second law. M1 3.1b 8th
Understand general
kinematics problems with
vectors.
2=
Fa M1 1.1b
24 32 6
t t = + −
(m s−2) A1 1.1b
(3)
b Integrate a M1 1.1a 8th
Solve general kinematics
problems using calculus of
vectors.
2 32 11 2
t t = + + −
v c (m s−1) A1 1.1b
0=c because initially at rest. A1 2.4
Integrate v M1 1.1a
3 4
2 13 41 13 2
t t
= + + −
r c (m)
A1 1.1b
c = 0 because initially at origin. A1 2.4
(6)
c Subsititute t = 1 M1 1.1a 6th
Understand general
kinematics problems with
vectors.
2 11 2
= + − v
M1 1.1b
31
= −
(m s−1) A1 1.1b
(3)
(12 marks)
Notes
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a Moment from bus = 5000 × 2 × g M1 3.1a 5th
Find resultant moments by considering direction.
= 10 000g (N m) A1 1.1b
Moment from gold = 1000 × 12 × g M1 3.1b
= 12 000g (N m) A1 1.1b
Moment from people = 70 × 8 × n × g M1 3.1a
= 560ng (N m) A1 1.1b
Total moment = (22 000 − 560n)g (N m) A1 1.1b
(7)
b Forming an equation or inequality for n and solving to find (n = 39.28…)
M1 1.1b 5th
Solve equilibrium problems involving
horizontal bars.
Need 40 people. A1 3.2a
(2)
c New moment from gold and extra person is 1070 × 12 × g (N) M1 3.1a 5th
Solve equilibrium problems involving
horizontal bars.
New total moment = (22840 − 560n)g (N m) M1 1.1b
n = 40.78… A1 3.2a
42 people (including the extra) A1 2.4
(4)
(13 marks)
H11 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
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H10 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
a Use of suvat equations M1 1.1a 8th
Derive formulae for projectile
motion.
10 cosx t θ= A1 1.1b
2110 sin2
y t gtθ= − M1 1.1b
210 sin 5t tθ= − A1 1.1b
Substitute x = 10 and y = −5 M1 1.1a
Solve x equation for t M1 1.1b
1cos
tθ
= A1 1.1b
Substitute into y equation M1 1.1a
25 10 tan 5secθ θ− = − A1 2.1
Use of 2 2sec 1 tanθ θ= + M1 2.1
( )2tan 1 1θ − = legitimately obtained A1 2.1
(11)
b Solve for tan θ M1 1.1a 8th
Solve problems in unfamiliar
contexts using the concepts of friction and
motion.
tan θ = 0 or tan θ = 2 A1 1.1b
θ = 0 or 63.43…(°) (accept awrt 63) A1 1.1b
(3)
(14 marks)
Notes
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H12 Scheme Marks AOs
Pearson Progression Step
and Progress descriptor
a Integrate a w.r.t. t M1 1.1a 5th
Use integration to determine
functions for velocity and/or displacement.
2180 180a t t= − A1 1.1b
(2)
b 2180 180 40t t− > M1 3.1a 7th
Solve general kinematics
problems in less familiar contexts.
( )( )20 3 2 3 1 0t t− − < A1 1.1b
1 23 3
t< < A1 2.4
Breaking the speed limit between 20 and 40 minutes. A1 3.2a
(4)
c Integrate v w.r.t. t M1 1.1a 5th
Use integration to determine
functions for velocity and/or displacement.
( )2 390 60x t t C= − + A1 1.1b
When 1, 30t x= = A1 3.1b
Average speed =distance
time M1 1.1b
30 km h−1 A1 1.1b
(5)
(11 marks)
Notes
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