General Certificate of Education Advanced Level Examination June 2010
Mathematics
MM03
Unit Mechanics 3 Tuesday 22 June 2010
1.30 pm to 3.00 pm
d
For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
e
s
Time allowed * 1 hour 30 minutes
n
Instructions * Use black ink or black ball-point pen. Pencil should only be used for drawing. * Fill in the boxes at the top of this page. * Answer all questions. * Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. * You must answer the questions in the spaces provided. Do not write outside the box around each page. * Show all necessary working; otherwise marks for method may be lost. * Do all rough work in this book. Cross through any work that you do not want to be marked. * The final answer to questions requiring the use of calculators should be given to three significant figures, unless stated otherwise. * Take g ¼ 9.8 m s2 , unless stated otherwise.
e
d
n
o
C
Information The marks for questions are shown in brackets. * The maximum mark for this paper is 75. *
Advice * Unless stated otherwise, you may quote formulae, without proof, from the booklet.
P27938/Jun10/MM03 6/6/
MM03
2
A tank containing a liquid has a small hole in the bottom through which the liquid escapes. The speed, u m s1 , at which the liquid escapes is given by
1
u ¼ CV rg where V m3 is the volume of the liquid in the tank, r kg m3 is the density of the liquid, g is the acceleration due to gravity and C is a constant. By using dimensional analysis, find the dimensions of C.
(5 marks)
A projectile is fired from a point O on top of a hill with initial velocity 80 m s1 at an angle y above the horizontal and moves in a vertical plane. The horizontal and upward vertical distances of the projectile from O are x metres and y metres respectively.
2
(a) (i)
Show that, during the flight, the equation of the trajectory of the projectile is given by y ¼ x tan y
gx 2 ð1 þ tan2 yÞ 12 800
(5 marks)
(ii) The projectile hits a target A, which is 20 m vertically below O and 400 m
horizontally from O. 80 m s1 O
y
20 m A 400 m Show that 49 tan2 y 160 tan y þ 41 ¼ 0 (b) (i)
(2 marks)
Find the two possible values of y . Give your answers to the nearest 0.1°. (3 marks)
(ii) Hence find the shortest possible time of the flight of the projectile from O to A.
(2 marks) (c)
State a necessary modelling assumption for answering part (a)(i).
(1 mark)
P27938/Jun10/MM03
3
Three smooth spheres, A, B and C, of equal radii have masses 1 kg, 3 kg and x kg respectively. The spheres lie at rest in a straight line on a smooth horizontal surface with B between A and C. The sphere A is projected with speed 3u directly towards B and collides with it.
3
3u 1 kg
3 kg
x kg
A
B
C 1
The coefficient of restitution between each pair of spheres is 3 . (a)
Show that A is brought to rest by the impact and find the speed of B immediately after the collision in terms of u. (6 marks)
(b)
Subsequently, B collides with C. Show that the speed of C immediately after the collision is
4u . 3þx
Find the speed of B immediately after the collision in terms of u and x.
(6 marks)
(c)
Show that B will collide with A again if x > 9 .
(2 marks)
(d)
Given that x ¼ 5 , find the magnitude of the impulse exerted on C by B in terms of u. (2 marks)
The unit vectors i, j and k are directed east, north and vertically upwards respectively.
4
At time t ¼ 0 , the position vectors of two small aeroplanes, A and B, relative to a fixed origin O are ð60i þ 30kÞ km and ð40i þ 10j 10kÞ km respectively. The aeroplane A is flying with constant velocity ð250i þ 50j 100kÞ km h1 and the aeroplane B is flying with constant velocity ð200i þ 25j þ 50kÞ km h1 . (a)
Write down the position vectors of A and B at time t hours.
(3 marks)
(b)
Show that the position vector of A relative to B at time t hours is ðð20 þ 50tÞi þ ð10 þ 25tÞj þ ð40 150tÞkÞ km.
(2 marks)
(c)
Show that A and B do not collide.
(4 marks)
(d)
Find the value of t when A and B are closest together.
(6 marks)
s
Turn over
P27938/Jun10/MM03
4
A smooth sphere is moving on a smooth horizontal surface when it strikes a smooth vertical wall and rebounds.
5
Immediately before the impact, the sphere is moving with speed 4 m s1 and the angle between the sphere’s direction of motion and the wall is a . Immediately after the impact, the sphere is moving with speed v m s1 and the angle between the sphere’s direction of motion and the wall is 40°. 2
The coefficient of restitution between the sphere and the wall is 3 .
a
40°
6
4 m s1
v m s1
(a)
Show that tan a ¼ 2 tan 40°.
3
(3 marks)
(b)
Find the value of v.
(3 marks)
Two smooth spheres, A and B, have equal radii and masses 1 kg and 2 kg respectively. The sphere A is moving with velocity ð2i þ 3jÞ m s1 and the sphere B is moving with velocity ði 2jÞ m s1 on the same smooth horizontal surface. The spheres collide when their line of centres is parallel to the unit vector i, as shown in the diagram. i 2j j A
B Line of centres
i 2i þ 3j
P27938/Jun10/MM03
5 (a)
Briefly state why the components of the velocities of A and B parallel to the unit vector j are not changed by the collision. (1 mark)
(b)
The coefficient of restitution between the spheres is 0.5 . Find the velocities of A and B immediately after the collision.
(6 marks)
A ball is projected from a point O on a smooth plane which is inclined at an angle of 35° above the horizontal. The ball is projected with velocity 20 m s1 at an angle of 30° above the plane, as shown in the diagram. The motion of the ball is in a vertical plane containing a line of greatest slope of the inclined plane. The ball strikes the inclined plane at the point A.
7
A 20 m s1 30° 35°
O (a)
Find the components of the velocity of the ball, parallel and perpendicular to the plane, as it strikes the inclined plane at A. (7 marks)
(b)
On striking the plane at A, the ball rebounds. The coefficient of restitution between 4
the plane and the ball is 5 . Show that the ball next strikes the plane at a point lower down than A.
(6 marks)
END OF QUESTIONS
Copyright Ó 2010 AQA and its licensors. All rights reserved.
P27938/Jun10/MM03
MM03 - AQA GCE Mark Scheme 2010 June series
Key to mark scheme and abbreviations used in marking M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA
mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach
MC MR RA FW ISW FIW BOD WR FB NOS G c sf dp
mis-copy mis-read required accuracy further work ignore subsequent work from incorrect work given benefit of doubt work replaced by candidate formulae book not on scheme graph candidate significant figure(s) decimal place(s)
No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.
3
MM03 - AQA GCE Mark Scheme 2010 June series
MM03 Q
Solution 1
Marks B1 M1 A1
−1
LT LT −1 = M α Lβ T γ × L3 × ML−3 × LT −2
Total
Comments For dimensions of u M1 for equation with five components
1 = β +1 −1 = γ − 2 0 = α +1
Forming and solving equations (PI)
m1
β = 0, α = -1, γ = 1
2(a)(i)
The dimensions of C are M −1T
A1F
Alternative : LT −1 LT −1 = C × L3 × ML−3 × LT −2 LT −1 = C × L MT −2 The dimensions of C are M −1T
(B1) (M1A1) (m1) (A1F) Total
x = 80cosθ . t x t= 80cosθ
(ii)
(b)(i)
5 5
B1
gx 2 ( 1 + tan 2θ ) 12800
B1 M1 A1
5
Answer given
2
−20 = 400 tanθ − 9.8 × 400 ( 1 + tan 2θ ) 12800 2 122.5 tan θ − 400 tanθ + 102.5 = 0
M1
49 tan 2θ − 160 tanθ + 41 = 0
A1
160 ± 25600 − 4( 49 )( 41 ) 2 × 49 = 2.9850, 0.2803 θ = 71.5D , 15.7D tan θ =
(ii) For the shortest time 400 = 80cos15.7 D .t t = 5.19 (c)
For dimensions of u M1 for equation with five components
B1
y = 80sinθ .t − 1 gt 2 2 x x y = 80sinθ )2 − 1 g( 80cosθ 2 80cosθ
y = x tanθ −
5
• •
The projectile is a particle The air resistance is negligible Total
Condone + 20 2
Answer given
M1 A1 A1F
PI 3
M1 AIF
2
E1
1 13
4
MM03 - AQA GCE Mark Scheme 2010 June series
MM03 (cont) Q 3(a) C.L.M.
Solution
Marks
(1)3u = (1)v A + (3)vB
Total
M1 A1
Comments
M1 for three non-zero terms
Restitution :
(b)
(c)
(d)
1 × 3u = v − v B A 3 vB = u
M1 A1 m1
vA = 0
A1
C.L.M. 3u = 3wB + xwc Restitution : 1u = w − w C B 3 wC = 4u 3+ x u( 9 − x ) wB = 3( 3 + x )
Accept v A − v B Solution 6
A1 for both answers
M1 A1 M1 A1 m1 OE
A1
u( 9 − x ) 9
For further collision
Solution attempt, dep. on both M1s AG 6
A1 for both
2
AG
M1 A1
I = 5( 4u ) 3+5 5 u I= 2 Alternative: u( 9 − 5 ) I = 3u − 3 × 3( 3 + 5 ) I = 5u 2
M1 A1
2
(M1) Accept −
(A1F)
5u 2
Follow through on their wB Total
16
5
MM03 - AQA GCE Mark Scheme 2010 June series
MM03 (cont) Q Solution 4(a) rA = ( − 60i + 30k ) + (250i + 50 j − 100k )t
(b)
Marks
Total
M1
rB = ( − 40i + 10 j − 10k ) + (200i + 25 j + 50k )t
A1,2
r = [ ( − 60i + 30k ) + ( 250i + 50 j − 100k )t ] –
M1
B A
3
Comments For correct form
A1 for each
Attempt at the difference using their answers
[( − 40i + 10 j − 10k ) + (200i + 25 j + 50k )t ]
A1
r = ( − 20 + 50t )i + ( − 10 + 25t )j + (40 − 150t )k
B A
2
AG
(c) For collision ( − 20 + 50t )i + ( − 10 + 25t )j + (40 − 150t )k = 0 −20 + 50t = 0 ⇒ t=2 5 −10 + 25t = 0 ⇒ t=2 5 40 − 150t = 0 ⇒ t= 4 15
(d)
M1
m1 A1F
The relative position vector cannot be zero. Therefore A and B do not collide
E1
S 2 = ( − 20 + 50t )2 + ( − 10 + 25t ) 2 + (40 − 150t ) 2
M1A1
4
For minimum S dS 2 = 100( −20 + 50t ) + 50( −10 + 25t ) − dt 300( 40 − 150t ) = 0
M1 A1F m1 A1F
51250t − 14500 = 0 t = 0.283
Total Alternative: ⎛ −20 + 50t ⎞ ⎛ 50 ⎜ −10 + 25t ⎟ .⎜ 25 ⎜ 40 − 150t ⎟ ⎜ −150 ⎝ ⎠⎝
⎞ ⎟=0 ⎟ ⎠
−1000 + 2500t − 250 + 625t − 6000 + 22500t = 0 25625t − 7250 = 0 t = 0.283
(M1) (A1) (m1) (A1F) (A1F) (A1F)
6
Solution 6 15
MM03 - AQA GCE Mark Scheme 2010 June series
MM03 (cont) Q Solution 5(a) Parallel to the wall 4cosα = v cos40D Perpendicular to the wall v sin40D = 2 × 4sinα 3 3 tan α = tan 40D 2 (b)
Marks
Comments
M1
Correct trigonometric ratios
M1
Correct trigonometric ratios
A1
α = 51.5D
Total
3
AG
3 6 1
OE
M1 D
4cos51.5 cos40D v = 3.25 ms-1
v=
M1 A1
Total 6(a) The spheres are smooth, no force acting in j direction (b)
E1
vA = ai + bj
vB = ci + dj C.L.M. along i: 1(2) + 2(−1) = 1(a) + 2(c) a+ 2c= 0 Restitution along i : c − a = 0.5( 2 − ( −1 )) c − a = 1.5 c = 0 .5 a = −1
M1A1 M1A1
v A = −i + 3 j
A1F
vB = 0.5i − 2 j
A1F Total
7
6 7
Any valid reason
MM03 - AQA GCE Mark Scheme 2010 June series
MM03 (cont) Q
Solution On striking A: 7(a) 20sin30D .t − 1 (9.8)cos35D .t 2 = 0 2 t = 2.49
Marks
Total
Comments
M1A1 A1
AWRT
OE
Components of Velocity : u x = 20cos30D − 9.8sin35D (2.49)
M1
ux = 3.32
A1F D
D
u y = 20sin30 − 9.8cos35 (2.49)
M1
u y = −10
A1F
(or -9.99)
AWRT 7
(b) On Rebounding vx = 3.32 v y = 4 × 10 5 v y = 8 (or 7.99) The rebound angle = tan −1 8 3.32 = 67.5D ( or 67.4D ) D
D
For 4 × their u y 5
B1F
D
35 + 67.5 = 102.5
102.5D > 90D , therefore the second strike will be at a point lower down than A. Alternative: 4 × 10 = 8 5 0 = 8t − 1 g cos35.t 2 2 t = 1.9931 x = 3.32t − 1 g sin35.t 2 2 x = −4.55 or -4.56
The second strike will be at a point lower down than A. Total TOTAL
M1 A1F M1 A1F E1
6
(B1)
Dependent on the two M1s
Condone negative sign
(M1) (A1)
OE
(M1) (A1) (E1) 13 75
8
Scaled mark unit grade boundaries - June 2010 exams A-level
Code
Title
Max. Scaled Mark
Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E
MS1B MD02 MFP2 MM2B MPC2 MS2B XMCA2 MFP3 MM03 MPC3 MS03 MFP4 MM04 MPC4 MS04 MM05
GCE MATHEMATICS UNIT S1B GCE MATHEMATICS UNIT D02 GCE MATHEMATICS UNIT FP2 GCE MATHEMATICS UNIT M2B GCE MATHEMATICS UNIT PC2 GCE MATHEMATICS UNIT S2B GCE MATHEMATICS UNIT XMCA2 GCE MATHEMATICS UNIT FP3 GCE MATHEMATICS UNIT M03 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT S03 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT M04 GCE MATHEMATICS UNIT PC4 GCE MATHEMATICS UNIT S04 GCE MATHEMATICS UNIT M05
75 75 75 75 75 75 125 75 75 75 75 75 75 75 75 75
69 70 63 65 102 69 71 67 60 67 68 67 68 63
59 64 65 55 61 58 92 64 68 62 54 60 61 62 61 56
52 56 57 47 54 51 80 56 61 54 48 52 53 56 53 49
45 48 49 39 47 44 68 49 54 46 42 44 45 50 45 42
39 40 41 31 41 37 57 42 48 39 36 36 37 44 38 35
33 32 33 23 35 31 46 35 42 32 31 29 29 39 31 29
MEST1 MEST2 MEST3 MEST4
GCE MEDIA STUDIES UNIT 1 GCE MEDIA STUDIES UNIT 2 GCE MEDIA STUDIES UNIT 3 GCE MEDIA STUDIES UNIT 4
80 80 80 80
69 72
52 63 59 65
45 54 48 53
38 45 38 42
31 36 28 31
25 28 18 20
MHEB1 MHEB2
GCE MODERN HEBREW UNIT 1 GCE MODERN HEBREW UNIT 2
100 100
80
66 71
59 62
52 53
45 45
38 37
MUSC1 MUS2A MUS2B
GCE MUSIC UNIT 1 GCE MUSIC UNIT 2A GCE MUSIC UNIT 2B
80 60 60
-
57 46 48
51 41 42
45 36 36
39 31 31
34 26 26
