MATHEMATICS MM03 Unit Mechanics 3 - Douis.net

Jun 11, 2007 - eed of (2.38) Speed of (0.6) collides again with. B. A. B. A. ∴. X. M1A1. M1. A1 m1. A1F. M1. A1F. E1. 6. 3. For both (1) and (2). Dependent on ...
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General Certificate of Education June 2007 Advanced Level Examination

MATHEMATICS Unit Mechanics 3 Monday 11 June 2007

MM03

1.30 pm to 3.00 pm

For this paper you must have: * an 8-page answer book * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.

Time allowed: 1 hour 30 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MM03. Answer all questions. Show all necessary working; otherwise marks for method may be lost. 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. * *

* * *

*

Information The maximum mark for this paper is 75. The marks for questions are shown in brackets.

* *

Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. *

P94382/Jun07/MM03 6/6/6/

MM03

2

Answer all questions.

1 The magnitude of the gravitational force, F, between two planets of masses m1 and m2 with centres at a distance x apart is given by F¼

Gm1 m2 x2

where G is a constant. (a) By using dimensional analysis, find the dimensions of G.

(3 marks)

(b) The lifetime, t, of a planet is thought to depend on its mass, m, its initial radius, R, the constant G and a dimensionless constant, k, so that t ¼ kma Rb Gg where a , b and g are constants. Find the values of a , b and g .

(5 marks)

2 The unit vectors i, j and k are directed due east, due north and vertically upwards respectively. Two helicopters, A and B, are flying with constant velocities of (20i  10j þ 20k) m s1 and (30i þ 10j þ 10k) m s1 respectively. At noon, the position vectors of A and B relative to a fixed origin, O, are (8000i þ 1500j þ 3000k) m and (2000i þ 500j þ 1000k) m respectively. (a) Write down the velocity of A relative to B.

(2 marks)

(b) Find the position vector of A relative to B at time t seconds after noon.

(3 marks)

(c) Find the value of t when A and B are closest together.

(5 marks)

3 A particle P, of mass 2 kg, is initially at rest at a point O on a smooth horizontal surface. The particle moves along a straight line, OA , under the action of a horizontal force. When the force has been acting for t seconds, it has magnitude ð4t þ 5Þ N . (a) Find the magnitude of the impulse exerted by the force on P between the times t ¼ 0 and t ¼ 3 . (3 marks) (b) Find the speed of P when t ¼ 3 .

(2 marks)

(c) The speed of P at A is 37.5 m s1 . Find the time taken for the particle to reach A. (4 marks)

P94382/Jun07/MM03

3

4 Two small smooth spheres, A and B, of equal radii have masses 0.3 kg and 0.2 kg respectively. They are moving on a smooth horizontal surface directly towards each other with speeds 3 m s1 and 2 m s1 respectively when they collide. The coefficient of restitution between A and B is 0.8 . (a) Find the speeds of A and B immediately after the collision.

(6 marks)

(b) Subsequently, B collides with a fixed smooth vertical wall which is at right angles to the path of the sphere. The coefficient of restitution between B and the wall is 0.7 . Show that B will collide again with A.

(3 marks)

5 A ball is projected with speed u m s1 at an angle of elevation a above the horizontal so as to hit a point P on a wall. The ball travels in a vertical plane through the point of projection. During the motion, the horizontal and upward vertical displacements of the ball from the point of projection are x metres and y metres respectively. (a) Show that, during the flight, the equation of the trajectory of the ball is given by y ¼ x tan a 

gx 2 ð1 þ tan2 aÞ 2 2u

(6 marks)

(b) The ball is projected from a point 1 metre vertically below and R metres horizontally from the point P. (i) By taking g ¼ 10 m s2 , show that R satisfies the equation 5R2 tan2 a  u2 R tan a þ 5R2 þ u2 ¼ 0

(2 marks)

(ii) Hence, given that u and R are constants, show that, for tan a to have real values, R must satisfy the inequality R2 4

u2 ðu2  20Þ 100

(2 marks)

(iii) Given that R ¼ 5 , determine the minimum possible speed of projection. (3 marks)

P94382/Jun07/MM03

s

Turn over

4

6 A smooth spherical ball, A, is moving with speed u in a straight line on a smooth horizontal table when it hits an identical ball, B, which is at rest on the table. Just before the collision, the direction of motion of A makes an angle of 30° with the line of the centres of the two balls, as shown in the diagram.

Line of centres 30° A

B

u

The coefficient of restitution between A and B is e. pffiffiffi 3 , show that the speed of B immediately after the collision is (a) Given that cos 30° ¼ 2 pffiffiffi 3 uð1 þ eÞ (5 marks) 4 (b) Find, in terms of u and e, the components of the velocity of A, parallel and perpendicular to the line of centres, immediately after the collision.

(3 marks)

2

(c) Given that e ¼ 3 , find the angle that the velocity of A makes with the line of centres immediately after the collision. Give your answer to the nearest degree.

P94382/Jun07/MM03

(3 marks)

5

7 A particle is projected from a point on a plane which is inclined at an angle a to the horizontal. The particle is projected up the plane with velocity u at an angle y above the plane. The motion of the particle is in a vertical plane containing a line of greatest slope of the inclined plane.

u

y

a

(a) Using the identity cosðA þ BÞ ¼ cos A cos B  sin A sin B , show that the range up the plane is 2u2 sin y cosðy þ aÞ g cos2 a

(8 marks)

(b) Hence, using the identity 2 sin A cos B ¼ sinðA þ BÞ þ sinðA  BÞ , show that, p a as y varies, the range up the plane is a maximum when y ¼  : (3 marks) 4 2 (c) Given that the particle strikes the plane at right angles, show that 2 tan y ¼ cot a

END OF QUESTIONS

P94382/Jun07/MM03

(4 marks)

MM03 - AQA GCE Mark Scheme 2007 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.

June 07

3

MM03 - AQA GCE Mark Scheme 2007 June series

MM03 Q

Solution 1(a) [G ]MM MLT −2 = L2 [G ] = L3 M −1T −2 (b)

Marks M1 A1

Total

A1F

3

Comments

t = kmα R β G γ T = Mα Lβ M −γ L3γ T −2γ −2γ = 1



α −γ = 0 ⇒ β + 3γ = 0

γ =−

L, M, T for G are needed to gain M1

m1 m1 A1F

Getting 3 equations Solution Finding α , β , γ

1 2

α = −

⇒ β=

M1 A1F

1 2

3 2 Total

5 8

4

MM03 - AQA GCE Mark Scheme 2007 June series

MM03 (cont) Q Solution 2 (a) B v A = v A − v B

= (20i − 10 j + 20k ) − (30i +10 j +10k ) = –10i − 20 j +10k (b)

r = (8000i +1500 j + 3000k )

B 0A

−(2000i + 500 j +1000k ) = 6000i +1000 j + 2000k B rA = (6000i +1000 j + 2000k )

Marks

Total

M1A1

2

Simplification not necessary

3

Simplification not necessary

M1

M1 A1F

+(−10i − 20 j +10k )t B rA = (6000 – 10t )i + (1000 − 20t ) j +(2000 +10t )k (c)

r

2

B A

= (6000 − 10t ) 2 + (1000 − 20t ) 2 +(2000 +10t ) 2

The helicopters are closest when is minimum. y = (6000 – 10t ) 2 + (1000 – 20t ) 2

r

M1 A1F

2

B A

+(2000 + 10t ) 2 dy = 2(–10)(6000 – 10t ) dt +2(–20)(1000 – 20t ) + 2(10)(2000 +10t ) = 0 t = 100 Alternative: ⎛ 6000 – 10t ⎞ ⎛ –10 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1000 – 20t ⎟ • ⎜ –20 ⎟ = 0 ⎜ 2000 + 10t ⎟ ⎜ 10 ⎟ ⎝ ⎠ ⎝ ⎠ −60000 + 100t − 20000 + 400t

m1 A1F A1F

5

(M1) (A1F) (m1) (A1F)

+20000 + 100t = 0 600t = 60000 t = 100

Comments

(A1F)

Total

(5)

10

5

MM03 - AQA GCE Mark Scheme 2007 June series

MM03 (cont) Q 3(a)

Solution

Marks

Total

Comments

3

I = ∫ (4t + 5)dt

M1

0

3

= ⎡⎣ 2t 2 + 5t ⎤⎦ 0

m1

= 33 Ns

A1

Or evaluation of constant 3

Alternative : I = Area under the Force–Time graph 17 + 5 ×3 2 = 33 Ns =

(b)

(M1) (m1) (A1)

(3)

M1 A1F

2

I = mv − mu 33 = 2v − 2(0) v

= 16.5 ms

−1

t

(c) I = (4t + 5)dt = 2(37.5) − 2(0) ∫

M1

0

2t 2 + 5t − 75 = 0

A1

−5 ± 25 + 8 × 75 4 t =5 t=

m1 A1F Total

4 9

For one value of t identified only

4(a) Conservation of momentum : 0.3(3) − 0.2(2) = 0.3v A + 0.2vB

M1A1

3v A + 2vB = 5 ----------------(1) Newton's experimental law : vB − v A 5 vB − v A = 4 -----------------(2)

0.8 =

M1 A1 m1

Solving (1) and (2) vB = 3.4

A1F

v A = −0.6

(b)

v 3.4 v = 2.38

0.7 =

For both (1) and (2) Dependent on both M1s 6

For both solutions

3

Cannot be gained without A1F

M1 A1F

Speed of B (2.38) ! Speed of A (0.6) ∴ B collides again with A

E1

Total

9

6

MM03 - AQA GCE Mark Scheme 2007 June series

MM03 (cont) Q 5(a)

Solution

y = ut sin α −

Marks M1 A1

1 2 gt 2

x = ut cos α t=

(ii)

x u cos α

A1 2

M1

A1

10 R 2 (1 + tan 2 α ) 2 2u 2 2 5R tan α − u 2 R tan α + 5R 2 + u 2 = 0

1 = R tan α −

6

Answer given

2

Answer given

2

Answer given

M1 A1

For real solutions of the quadratic :

u 4 R 2 − 20 R 2 (5R 2 + u 2 ) ≥ 0 4

M1

2

u − 20u 100 2 ( u u 2 − 20) R2 ≤ 100 R2 ≤

(iii)

Comments

M1

1 ⎛ x ⎞ ⎛ x ⎞ y = u⎜ ⎟ sin α − g ⎜ ⎟ 2 ⎝ u cos α ⎠ ⎝ u cos α ⎠ gx 2 y = x tan α − 2 u cos 2 α gx 2 y = x tan α − 2 (1 + tan 2 α ) 2u

(b)(i)

Total

A1

u 2 (u 2 − 20) 100 4 u − 20u 2 − 2500 ≥ 0

M1

umin 2 = 61.0

A1

52 ≤

(or 10 + 2600)

umin = 7.81

A1F

Total

Condone equation 3

13

7

3 sf required

MM03 - AQA GCE Mark Scheme 2007 June series

MM03 (cont) Q 6(a)

Solution

usin30°

Marks

Total

Comments

0

Before:

A

ucos30° usin30°

B

0

0

After:

A

vA

B

vB

Con. of Mom. along the line of centres: mu cos30° = mv A + mvB

3 u ----------(1) 2 Newton's experimental law : vB − v A e= u cos30° − 0 3 vB − v A = ue ----------(2) 2 Solving (1) and (2) : v A + vB =

vB =

(b)

⊥ ||

(c)

A1

M1 A1

3 u (1 + e) 4

A1

1 u sin 30° = u 2 3 3 vA = u− u (1 + e) 2 4 3 vA = u (1 − e) 4

α = tan

−1

α = tan −1

M1

5

usin30 accepted

B1 M1 A1F

1 u 2 3 ⎛ 2⎞ u ⎜1 − ⎟ 4 ⎝ 3⎠ 6 3

Answer given

3

Simplification not needed

3

To the nearest degree required

M1 A1F

α = 74°

A1F Total

11

8

MM03 - AQA GCE Mark Scheme 2007 June series

MM03 (cont) Q 7(a)

Marks

Solution j

Total

Comments

u

i

θ

α

α g 1 2 gt cos θ 2 2u sin θ y=0 ⇒ t = g cos α 1 x = ut cos θ − gt 2 sin α 2 y = ut sin θ −

M1A1 A1F M1A1 2

2u sin θ 1 ⎛ 2u sin θ ⎞ R=u cos θ − g ⎜ sin α g cos α 2 ⎝ g cos α ⎟⎠ 2u 2 sin θ cos(θ + α ) R= g cos 2 α (b)

1 2u 2 × [sin(2θ + α ) + sin(−α )] 2 R= g cos 2 α R is maximum when sin(2θ + α ) = 1 or 2θ + α = ∴

(c)

θ=

π 4



m1 A1

8

Dependent on M1s Answer given

3

Answer given

B1 M1

π 2

α

A1

2

2u sin θ g cos α u cos θ x" = 0 ⇒ t = g sin α 2u sin θ u cos θ = g cos α g sin α 2 tan θ = cot α y=0

M1

⇒ t=

For using y=0 and x" = 0 A2 for both correct

M1 A2,1

A1

4

Answer given N.B. A problem arose which ultimately affected the marking of part 7(c). Please see the Report on the Examination for details.

Total TOTAL

15 75

9

klm

Scaled mark component grade boundaries - June 2007 exams GCE Component Code Component Title ICT3 ICT4 ICT5 ICT6 LAW1 LAW2 LAW3 LAW4 LAW5 LAW6 MD01 MD02 MFP1 MFP2 MFP3 MFP4 MM03 MM04 MM05 MM1A/C MM1A/W MM1B MM2A/C MM2A/W MM2B MPC1

Maximum Scaled Mark

Scaled Mark Grade Boundaries B C D

A

GCE INFO AND COMM TECH UNIT 3 GCE INFO AND COMM TECH UNIT 4 GCE INFO AND COMM TECH UNIT 5 GCE INFO AND COMM TECH UNIT 6

60 90 90 90

43 69 62 59

37 63 55 51

31 57 49 43

26 51 43 36

21 46 37 29

GCE LAW UNIT 1 GCE LAW UNIT 2 GCE LAW UNIT 3 GCE LAW UNIT 4 GCE LAW UNIT 5 GCE LAW UNIT 6

65 65 65 85 85 70

47 47 43 56 56 47

43 42 38 51 52 43

39 37 33 47 48 39

35 32 28 43 45 35

31 27 23 39 42 32

GCE MATHEMATICS UNIT D01 GCE MATHEMATICS UNIT D02 GCE MATHEMATICS UNIT FP1 GCE MATHEMATICS UNIT FP2 GCE MATHEMATICS UNIT FP3 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT M03 GCE MATHEMATICS UNIT M04 GCE MATHEMATICS UNIT M05 GCE MATHEMATICS UNIT M1A - COURSEWORK GCE MATHEMATICS UNIT M1A - WRITTEN GCE MATHEMATICS UNIT M1B GCE MATHEMATICS UNIT M2A - COURSEWORK GCE MATHEMATICS UNIT M2A - WRITTEN GCE MATHEMATICS UNIT M2B GCE MATHEMATICS UNIT PC1

75 75 75 75 75 75 75 75 75 25 75 75 25 75 75 75

63 59 64 56 60 62 64 62 58 20 60 58 20 63 65 60

56 51 56 49 53 54 56 54 50 18 51 49 18 54 57 52

49 44 48 42 46 46 48 46 42 15 43 40 15 46 49 44

42 37 41 35 39 38 40 39 34 13 35 31 13 39 41 37

35 30 34 29 32 31 33 32 27 10 28 23 10 31 33 30

E