General Certificate of Education Advanced Level Examination June 2011
Mathematics
MM03
Unit Mechanics 3 Wednesday 22 June 2011
9.00 am to 10.30 am
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For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
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Time allowed * 1 hour 30 minutes
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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.
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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.
P38108/Jun11/MM03 6/6/6/
MM03
2
A ball of mass 0.2 kg is hit directly by a bat. Just before the impact, the ball is travelling horizontally with speed 18 m s1 . Just after the impact, the ball is travelling horizontally with speed 32 m s1 in the opposite direction.
1
(a)
Find the magnitude of the impulse exerted on the ball.
(b)
At time t seconds after the ball first comes into contact with the bat, the force exerted by the bat on the ball is kð0:9t 10t 2 Þ newtons, where k is a constant and 0 4 t 4 0:09 . The bat stays in contact with the ball for 0.09 seconds. Find the value of k.
(2 marks)
(4 marks)
The time, t, for a single vibration of a piece of taut string is believed to depend on
2
the length of the taut string, l, the tension in the string, F, the mass per unit length of the string, q, and a dimensionless constant, k, such that t ¼ kl a F b q g where a , b and g are constants. By using dimensional analysis, find the values of a , b and g .
(5 marks)
(In this question, use g = 10 m s2 .)
3
A golf ball is hit from a point O on a horizontal golf course with a velocity of 40 m s1 at an angle of elevation y . The golf ball travels in a vertical plane through O. During its flight, the horizontal and upward vertical distances of the golf ball from O are x and y metres respectively. Show that the equation of the trajectory of the golf ball during its flight is given by
(a)
x 2 tan2 y 320x tan y þ ðx 2 þ 320yÞ ¼ 0 (b) (i)
(6 marks)
The golf ball hits the top of a tree, which has a vertical height of 8 m and is at a horizontal distance of 150 m from O. Find the two possible values of y .
(5 marks)
(ii) Which value of y gives the shortest possible time for the golf ball to travel from O
to the top of the tree? Give a reason for your choice of y .
(02)
(2 marks)
P38108/Jun11/MM03
3
The unit vectors i, j and k are directed due east, due north and vertically upwards respectively.
4
A helicopter, A, is travelling in the direction of the vector 2i þ 3j þ 6k with constant speed 140 km h1 . Another helicopter, B, is travelling in the direction of the vector 2i j þ 2k with constant speed 60 km h1 . (a)
Find the velocity of A relative to B.
(b)
Initially, the position vectors of A and B are ð4i 2j þ 3kÞ km and ð3i þ 6j þ 3kÞ km respectively, relative to a fixed origin.
(5 marks)
Write down the position vector of A relative to B, t hours after they leave their initial positions. (2 marks) Find the distance between A and B when they are closest together.
(c)
(8 marks)
A ball is dropped from a height of 2.5 m above a horizontal floor. The ball bounces repeatedly on the floor.
5
(a)
Find the speed of the ball when it first hits the floor.
(b)
The coefficient of restitution between the ball and the floor is e. (i)
(2 marks)
Show that the time taken between the first contact of the ball with the floor and the 10e seconds. (3 marks) second contact of the ball with the floor is 7
(ii) Find, in terms of e, the time taken between the second contact and the third contact
of the ball with the floor.
(1 mark)
(c)
Find, in terms of e, the total vertical distance travelled by the ball from when it is dropped until its third contact with the floor. (5 marks)
(d)
State a modelling assumption for answering this question, other than the ball being a particle. (1 mark)
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(03)
Turn over
P38108/Jun11/MM03
4
A projectile is fired from a point O on a plane which is inclined at an angle of 20 to the horizontal. The projectile is fired up the plane with velocity u m s1 at an angle of 30 to the inclined plane. The projectile travels in a vertical plane containing a line of greatest slope of the inclined plane.
6
The projectile hits a target T on the inclined plane.
u T
30 20
O
(a)
Given that OT ¼ 200 m, determine the value of u.
(b)
Find the greatest perpendicular distance of the projectile from the inclined plane. (4 marks)
(04)
(7 marks)
P38108/Jun11/MM03
5
Two smooth spheres, A and B, have equal radii and masses 4m and 3m respectively. The sphere A is moving on a smooth horizontal surface and collides with the sphere B, which is stationary on the same surface.
7
Just before the collision, A is moving with speed u at an angle of 30 to the line of centres, as shown in the diagram below. Before collision B
A Line of centres 30
3m
4m u
Immediately after the collision, the direction of motion of A makes an angle a with the line of centres, as shown in the diagram below. After collision
B
A a
Line of centres
4m
3m
The coefficient of restitution between the spheres is
5 . 9
(a)
Find the value of a .
(b)
Find, in terms of m and u, the magnitude of the impulse exerted on B during the collision. (3 marks)
(10 marks)
END OF QUESTIONS
Copyright ª 2011 AQA and its licensors. All rights reserved.
(05)
P38108/Jun11/MM03
Key to mark scheme abbreviations 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 c sf dp
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 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. 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.
MM03 Q Solution 1 (a) I = 0.2(32) + 0.2(18) I = 10 Ns
(b)
0.09
0
Marks M1 A1
k (0.9t-10t 2 )dt = 10
Total 2
Comments Condone +10
M1
Condone limits
A1F
Condone limits
m1
For substituting 0.09
0.09
10 k 0.45t 2 − t 3 3 0
= 10
1.215 × 10-3 k = 10 k = 8230
A1F
4 6
2
1
α
-2
β
-1
γ
T = L (MLT ) (ML )
α + β −γ = 0 β +γ = 0 −2 β = 1 β =− 1 γ= 2 α =1
1 2
M1 A1 m1
m1 A1F
Getting three equations
5
5
Solution
Q Solution 3 (a) x = 40cos θ .t
Marks
Total
M1
1 M1 A1 y = − (10)t 2 + 40sin θ .t 2 1 x x ) 2 + 40sin θ .( ) y = − (10)( m1 2 40cos θ 40cos θ x2 y=− + x tan θ 320cos 2 θ
320 y = − x 2 (1 + tan 2 θ ) + 320 x tan θ x 2 tan 2 θ − 320 x tan θ + ( x 2 + 320 y ) = 0 (b)(i)
1502 tan2 θ − 320(150)tanθ + (1502 + 320 ×8) = 0 1125tan θ − 2400tanθ +1253 = 0 2
tan θ =
(b)(ii)
2400 ± 24002 − 4(1125)(1253) 2(1125)
m1 A1
6
Answer Given (Condone missing brackets)
Correct quadratic
m1 A1F
θ = 50.7 , 42.4
A1F
θ = 42.4
B1F
150 and cos 42.4 > cos50.7 40cosθ
Dependent on both M1s
M1 A1
tan θ = 1.22 , 0.912
t=
Comments
E1
PI 5 For the smaller angle 2 13
OE
Q 4 (a)
Solution Marks Total (−2i + 3j + 6k )140 = −40i + 60 j + 120k M1 A1 uA = (2)2 + (3)2 + (6)2
uB =
(2i − j + 2k )60
= 40i − 20 j + 40k
A1
u = (−40i + 60j + 120k) − (40i − 20j + 40k)
M1 A1F
5
M1 A1F
2
(2)2 + (1)2 + (2)2
A B
= −80i + 80j + 80k (b)
r = (4i − 2 j + 3k ) − (−3i + 6 j + 3k ) +
A B
t (−80i + 80 j + 80k ) or (7i − 8 j) + t (−80i + 80 j + 80k )
(c)
r = (7 − 80t )i + (−8 + 80t ) j + (80t )k
A B 2
2
2
s = (7 − 80t ) + (−8 + 80t ) + (80t ) 2s
2
ds = 2(7 − 80t )(−80) + 2(−8 + 80t )(80) + dt 2(80t )(80) = 0
240t = 15
1 16 s2 = (7 − 80 × 0.0625)2 + (−8 + 80 × 0.0625)2 +
t = 0.0625
or
(80 × 0.0625)2 s = 6.16 km
or
38 km
Simplification not needed
Simplification not needed Subtracting B from A A difference of initial p.v. + t× A u B
B1F B1F M1 A1F
Differentiation
m1
Solving
A1F M1 A1F
Alternative (Not in the specification) A and B are closest A rB . A v B = 0 [(7 − 80t )i + (−8 + 80t ) j + (80t )k] . [−80i + 80j + 80k] = 0 −80(7 − 80t ) + 80(−8 + 80t ) + 80(80t ) = 0 240t = 15 t = 0.0625
Comments
B1 M1 A1 A1 M1 A1
8 15
Q 5(a)
(b)(i)
Solution v 2 = u 2 + 2as
Marks
v 2 = 02 + 2(9.8)(2.5) v=7
M1 A1
w =e 7 w = 7e
M1
9.8 2 t or ( 0 = 7e − 9.8t ) 2 10e 7e t= ( t = 2× ) 7 9.8 0 = 7et −
(ii)
t′ =
2
M1 A1
3
Answer given
B1
1
OE
9.8 2 t′ 2
10e 2 7
02 = (7e) 2 + 2( −9.8)h2
h2 = 2.5e
Or for correct method to find h4
M1
2
A1
h3 = 2.5e 2
02 = (7e2 ) 2 + 2(−9.8)h4 h4 = 2.5e4 h5 = 2.5e
A1
4
Total distance = 2.5 + 2(2.5e 2 ) + 2(2.5e 4 )
= 2.5 + 5e 2 + 5e 4
m1 A1
5
Alternative (not in the specification) K.E. after each bounce = e2 × K.E. before the bounce
P.E. at max. height after each bounce = e2 × P.E. at max. height before the bounce Height after first bounce = 2.5e 2
Height after second bounce = 2.5e 2
4
Total = 2.5 + 2(2.5e + 2(2.5e ) 2
= 2.5 + 5e + 5e
(d)
Comments
w′ = 7e2 0 = 7e2 t '−
(c)
Total
4
Motion in vertical line, No air resistance, No energy loss, Instantaneous bounce
4
(M1) (A1) (A1) (m1) (A1)
B1
1
12
Q Solution 6 (a) Perpendicular to the plane: 1 y = − gt 2 cos 20 + ut sin 30 2 0 = −4.9t 2 cos 20 + ut sin 30 2u sin 30 t = 0.108589568u or g cos 20
Marks
Total
Comments
M1 M1 A1
Parallel to the plane: 1 x = − gt 2 sin 20 + ut cos30 2
M1
200 = −4.9(0.108589568u ) 2 sin 20 + u (0.108589568u ) cos30
m1
u 2 = 2693
A1F
u = 51.9
(b)
or 51.894
A1F
y = − gt cos 20 + u sin 30 = 0
M1
51.9sin30 t = 2.817899 or 2.817580214 or g cos20
A1F
7
Accept 3 significant fig.
The greatest ⊥ distance = 1 − 9.8(2.817899)2 cos20 + 51.9(2.817899)sin30 or 2
m1
1 51.894sin30 2 51.894sin30 − 9.8( ) cos20+51.9( )sin30 2 9.8cos20 9.8cos20
= 36.5622 m or 36.5538 = 36.6 3sf
A1F
4 11
6 (a)
(b)
Alternative: x = 200cos 20 y = 200sin 30 200cos 20 = u cos50t 292.4 t= u 1 292.4 2 292.4 200sin30 = (−9.8)( ) + u sin50( ) 2 u u u 2 = 2693 u = 51.9 Alternative: 0 = (u sin 30) 2 − 2 g cos 20.s
B1 B1 M1 A1 M1 A1 A1
M1
2
(51.9sin 30) 2(9.8)cos 20 s = 36.6
s=
m1A1 A1
Do not accept 2693
Q 7 (a)
Solution Momentum of A is unchanged ⊥ to the line of centres 4mu sin 30 = 4mvA sin α u vA = ...............................(1) 2sin α C.L.M.: 4mu cos30 = 4mv A cos α + 3mvB
2 3u = 4vA cosα + 3vB .................(2) Restitution along the line of centres: vB − v A cos α 5 = 9 u cos30 5 3u vB = vA cos α + ...................(3) 18 2 3u = 4
u u 15 3u cosα + 3 cosα + 2sinα 2sinα 18
Marks
Total
M1 A1 M1A1 OE
A1F M1A1
Or equivalent, could be in part (b)
B1
Solving (1), (2) and (3) Dependent on three M1s
m1
7 3 7 = 6 2 tan α tan α = 3 π 3
α = 60 or (b)
A1F
Impulse on B = Change in momentum of B along the line of centres u 5 3u vB = cos 60 + 2sin 60 18 u 5 3u 4 3 vB = + (= ) 18 9 2 3
I = 3m(
u
+
5 3u ) − 3m(0) 18
10
M1 M1
2 3 4mu I= or 2.31mu 3
A1F TOTAL
Comments
3 13 75
Scaled mark unit grade boundaries - June 2011 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 62 68 62 88 69 67 68 68 68 63 58 67 62
59 64 55 62 54 54 76 64 59 59 62 61 57 51 60 55
52 56 48 55 47 46 66 55 51 52 54 53 51 46 52 48
46 49 41 48 41 38 57 46 43 46 46 46 45 41 44 41
40 42 34 41 35 30 48 38 36 40 39 39 39 36 37 34
34 35 28 34 29 23 39 30 29 34 32 32 33 31 30 28
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
70 74
55 63 61 68
47 54 50 56
40 45 39 45
33 36 28 34
26 28 18 23
MHEB1 MHEB2
GCE MODERN HEBREW UNIT 1 GCE MODERN HEBREW UNIT 2
100 100
80
61 71
54 62
47 54
40 46
34 38
MUSC1 MUS2A MUS2B MUS2C
GCE MUSIC UNIT 1 GCE MUSIC UNIT 2A GCE MUSIC UNIT 2B GCE MUSIC UNIT 2C
80 60 60 60
-
57 45 49 49
51 40 43 44
45 35 37 39
39 30 32 34
34 26 27 29
