Mark Scheme (Results) Summer 2010
GCE
GCE Mechanics M2 (6678/01)
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Summer 2010 Publications Code UA024472 All the material in this publication is copyright Š Edexcel Ltd 2010
Summer 2010 Mechanics M2 6678 Mark Scheme Question Number
Scheme
Q1
Marks
3t + 5
dv = 3t + 5 dt v = ∫ ( 3t + 5 ) dt
M1*
v = 32 t 2 + 5t ( + c) t =0 v=2 ⇒ c=2 v = 32 t 2 + 5t + 2
A1
t =T
DM1*
6 = 32 T 2 + 5T + 2
12 = 3T 2 + 10T + 4 3T 2 + 10T − 8 = 0 ( 3T − 2 )(T + 4 ) = 0
T=
2 3
(T = −4 )
∴T =
2 3
(or 0.67)
B1
M1
A1 [6]
GCE Mechanics M2 (6678) Summer 2010
Question Number
Scheme
Q2
Marks
0 m s-1
4 m s-1 R
F
12 m
30
0.6g (a)
K.E gained = 12 × 0.6 × 42 P.E. lost = 0.6 × g × (12 sin 30) Change in energy = P.E. lost - K.E. gained M1 A1 A1
= 30.48 Work done against friction = 30 or 30.5 J (b)
R (↑ )
A1
R = 0.6 g cos 30
B1
30.48 12 F = µR 30.48 µ= 12 × 0.6 g cos 30 µ = 0.4987 µ = 0.499 or 0.50
B1ft
F=
GCE Mechanics M2 (6678) Summer 2010
(4)
M1 A1
(4) [8]
Question Number
Scheme
Q3
Marks
A 10 cm
10 cm
B (a)
mass ratio dist. from BC
C
12 cm AB
AC
BC
frame
10 4
10 4
12 0
32 x
Moments about BC:
10 × 4 + 10 × 4 + 0 = 32x 80 x= 32 x = 2 12 (2.5)
(b)
B1 B1
M1 A1
A1
(5)
B
θ D A
C Mg Moments about B:
Mg
Mg × 6sin θ = Mg × ( x cos θ − 6sin θ )
12sin θ = x cos θ x tan θ = 12 θ = 11.768..... = 11.8º Alternative method : C of M of loaded frame at distance
x from D along DA 1 x tan θ = 2 6 θ = 11.768..... = 11.8º 1 2
M1 A1 A1
A1
(4)
B1 M1 A1 A1 [9]
GCE Mechanics M2 (6678) Summer 2010
Question Number
Scheme
Q4
Marks
a m s-2 20 m s-1
T
θ
R
750g (a)
15000 = 750 20 R(parallel to road) T = R + 750 g sin θ R = 750 − 750 × 9.8 × 151 R = 260 * T=
(b)
M1 M1 A1 A1
(4)
a m s-2 20 m s-1
260N
T’
θ 750g
T′ =
18000 = 900 20
900 − 260 − 750 × 9.8 × 151 a= 750 a = 0.2
GCE Mechanics M2 (6678) Summer 2010
M1 M1 A1
A1
(4) [8]
Question Number
Scheme
Marks
Q5 (a)
I = mv − mu = 0.5 × 20i − 0.5 (10i + 24 j) = 5i − 12 j 5i − 12 j = 13 Ns
(b)
θ
M1 A1 M1 A1
(4)
5 12
12 5 θ = 67.38 θ = 67.4°
tan θ =
(c)
K.E.lost = 12 × 0.5 (102 + 242 ) − 12 × 0.5 × 202
= 69 J
GCE Mechanics M2 (6678) Summer 2010
M1 A1
(2)
M1 A1 A1
(3) [9]
Question Number
Scheme
Q6
D
Marks
θ
2a T
F A
R
a
2a mg
(a)
M(A)
a mg M1 A1 A1
3a × T cos θ = 2amg + 4amg
B1
A1
* (b)
3a × T × cos θ = 2amg + 4aMg
(5)
M1 A1
*
cso
A1
(3) [8]
GCE Mechanics M2 (6678) Summer 2010
Question Number
Scheme
Marks
Q7 (a)
(b)
Vertical motion:
v 2 = u 2 + 2as
M1 A1
θ = 22.54 = 22.5° (accept 23)
A1
(3)
Vert motion P → R : s = ut + 12 at 2 M1 A1 A1
t = 4.694... Horizontal P to R:
A1 M1
= 173 m
( or 170 m)
(c) Using Energy:
A1
(6)
M1 A1
v 2 = 2 ( 9.8 × 36 + 12 × 40 2 )
v = 48.0….. v = 48 m s -1 (accept 48.0)
A1
(3) [12]
GCE Mechanics M2 (6678) Summer 2010
Question Number
Scheme
Q8
u
u
A v
3m
Marks
m
B w
e=
1 2
(a) (i) Con. of Mom: 3mu − mu = 3mv + mw 1 2
N.L.R: (1) – (2)
2u = 3v + w (u + u ) = w − v u = w−v u = 4v v = 14 u
(ii) In (2)
(1) (2)
u = w − 14 u w = 54 u
(b) B to wall: N.L.R:
5 4
M1# A1 M1# A1 DM1# A1
A1
u × 52 = V V = 12 u
(7)
M1 A1ft
(2)
(c)
u A 1 4
1 2
u B
5 16a time = 4a ÷ u = 4 5u 1 16a 4 Dist. Travelled by A = u × = a 4 5u 5 1 1 In t secs, A travels ut , B travels ut 4 2
B to wall:
Collide when speed of approach = 4a − 54 a 16a 4 64a = × = 3 5 3u 15u 4u 16a 64a 112a Total time = + = 5u 15u 15u * ∴t=
GCE Mechanics M2 (6678) Summer 2010
B1ft B1ft
, distance to cover =
M1$
DM1$ A1 A1
(6) 15
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