Mark Scheme (Results) January 2010
GCE
Mechanics M2 (6678)
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January 2010 Publications Code UA022965 All the material in this publication is copyright Š Edexcel Ltd 2010
January 2010 6678 Mechanics M2 Mark Scheme Question Number Q1.
Scheme
dv = 6t − 4 dt 6t − 4 = 0 ⇒ t =
Marks
M1 A1 M1 A1
2 3
s = ∫ 3t 2 − 4t + 3 dt = t 3 − 2t 2 + 3t ( + c)
t = 23 ⇒ s = − 16 27 +2 so distance is
M1 A1 38 27
m
M1 A1 [8]
2u
u
Q2. 2m
m
v1
v2
CLM: 4mu − mu = 2mv1 + mv2 M1 A1
i.e. 3u = 2v1 + v2
3eu = −v1 + v2
NIL:
M1 A1
v1 = u(1 − e)
DM1 A1
v2 = u(1 + 2e)
A1 [7]
Q3.
1 2
x 0.5 x 20 2 ;
10R =
1 2
0.5g x 10
x 0.5 x 20 2 − 0.5g x 10
⇒ R = 5.1
B1 B1 M1 A1 DM1 A1
[6]
GCE Mechanics M2 (6678) January 2010
Question Number Q4.
Scheme
(i)
I ↑ = 0.25 × 40 sin 60 = 5 3 I ← = 0.25(−20 + 30) = 2.5 |I|=
Marks
(8.66)
one component both
75 + 6.25 = 9.01 (Ns)
M1 A1
M1 A1
(ii)
sinθ sin60 o = 40 1300 o θ = 106 (3 s.f.)
or tan θ = ±
(4)
M1 A1
5 3 2.5
oee θ = 106°
M1 A1
(4) [8]
Alternative to 4(i) Use of I = m(v − u)
30 2 + 40 2 − 2 x 30 x 40cos60 o
M1
( = 1300)
I = 0.25 1300 = 9.01 N s (3 s.f.)
M1 A1 A1
2nd Alternative to 4(i) u = 30i , v = 40 cos 60i + 40 sin 60 j = 20i + 20 3 j 1 I = (−10i + 20 3 j) = −2.5i + 5 3 j 4
GCE Mechanics M2 (6678) January 2010
M1 A1 etc
Question Number Q5.
Scheme
Marks
(a) 490 −R=0 3.5
B1 M1 A1
R = 140 N
A1
24 1 + 70 g. − 40u = 0 14 u
B1
(4)
(b)
40u 2 − 49u − 24 = 0
M1 A2,1,0
(5u − 8)(8u + 3) = 0
DM1
u = 1 .6
DM1 A1
(7) [11]
Q6.
m( B) : R × 4 cos α = F × 4 sin α + 20 g × 2 cos α Use of F =
1 R 2
M1 A2 M1
Use of correct trig ratios
B1
R = 160N or 157N
DM1 A1 [7]
GCE Mechanics M2 (6678) January 2010
Question Number Q7.
Scheme
(a)
Rectangle
Marks
Semicircles
Template, T
24x
4.5π
4.5π
24x + 9π
B2
x
4 x3 3π
4 x3 3π
x
B2
⎛ 4 x 3⎞ ⎛ 4 x 3⎞ − 4.5π x ⎜ = (24 x + 9π ) x 24 x 2 − 4.5π x ⎜ ⎟ ⎝ 3π ⎠ ⎝ 3π ⎟⎠
distance = x =
4 2x 2 − 3 (8x + 3π )
**
M1 A1
A1
(7)
(b) When x = 2, tan θ =
6 4− x
x=
20 16 + 3π 6
=
4−
=
B1
20 16 + 3π
48 + 9π . 22 + 6π
M1 A1
A1
(4) [11]
GCE Mechanics M2 (6678) January 2010
Question Number Q8.
Scheme
Marks
x = ut
(a)
B1
y = cut − 4.9t 2 eliminating t and simplifying to give
y = cx −
M1 A1
4.9 x 2 ** u2
4.9 x 2 0 = cx − 2 u
(b)(i)
0 = x (c − (ii)
4.9 x u 2c ) ⇒ R = = 10c 4.9 u2
(5c ) 2 = 2.5c 2 10
dy 9.8 x x =c− 2 =c− 5 dx u
(c)
When x = 0, So, c −
(5)
M1
M1 A1 M1
When x = 5c, y = H = 5c 2 −
DM1 A1
M1 A1
(6)
M1 A1
dy =c dx
B1
x −1 = 5 c
DM1 A1
1 x = 5(c + ) c
A1
(6) [17]
Alternative to 8(c) u
θ uc = 7c
θ
GCE Mechanics M2 (6678) January 2010
u 1 v = = cu c u u 7 ⇒v= = c c 7 v = u + at ; − = 7c − 9.8t c 7 1 (c + ) t= c 9.8 1 x = ut = 7t ; x = 5(c + ) c tan θ =
u v
B1 M1 A1 M1 A1 A1
GCE Mechanics M2 (6678) January 2010
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