Version 1.0
General Certificate of Education (A-level)
June 2011
Mathematics
MM03
(Specification 6360)
Mechanics 3
Final
Mark Scheme
Mark schemes are prepared by the Principal Examiner and considered, together with the relevant
questions, by a panel of subject teachers. This mark scheme includes any amendments made at the
standardisation events which all examiners participate in and is the scheme which was used by them
in this examination. The standardisation process ensures that the mark scheme covers the
candidates’ responses to questions and that every examiner understands and applies it in the same
correct way. As preparation for standardisation each examiner analyses a number of candidates’
scripts: alternative answers not already covered by the mark scheme are discussed and legislated for.
If, after the standardisation process, examiners encounter unusual answers which have not been
raised they are required to refer these to the Principal Examiner.
It must be stressed that a mark scheme is a working document, in many cases further developed and
expanded on the basis of candidates’ reactions to a particular paper. Assumptions about future mark
schemes on the basis of one year’s document should be avoided; whilst the guiding principles of
assessment remain constant, details will change, depending on the content of a particular examination
paper.
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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