June 2009 6678 Mechanics M2 Mark Scheme Question Number Q1
Scheme
I =mv-mu 1 1 5i – 3j = v - (3i + 7j) 4 4 v = 23i - 5j |v| =
Marks
M1A1 A1 M1A1
232 + 5 2 = 23.5
[5]
Q2
(a)
M1
dv = 8 − 2t dt 8 − 2t = 0 Max v = 8 × 4 − 4 2 = 16 (ms-1)
M1 M1A1 (4)
(b)
1 dt = 4t 2 − t 3 (+ c) 3 (t=0, displacement = 0 ⇒ c=0)
∫ 8t − t
2
M1A1
1 4T 2 − T 3 = 0 3 T T 2 (4 − ) = 0 ⇒ T = 0, 12 3
DM1 DM1 A1
T = 12 (seconds)
Q3
(a)
(b)
(5) [9]
Constant v ⇒ driving force = resistance ⇒ F=120 (N) ⇒ P=120 x 10 = 1200W
M1 M1 (2)
Resolving parallel to the slope, zero acceleration: P = 120 + 300 g sin θ (= 330) v 1200 ⇒v= = 3.6 (ms-1) 330
8371_9374 GCE Mathematics January 2009
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M1A1A1 A1
(4) [6]
Question Number Q4
Scheme
(a)
Taking moments about A: T 3g × 0.75 = × 0.5 2 7 .5 9 2 g T = 3 2g × = (= 62.4 N ) 5 2
D B
A
Marks
M1A1A1 A1 (4)
C
(b)
T
← ±H =
↑ ±V +
2 T 2
(=
9g ≈ 44.1N ) 2
= 3g (⇒ V = 3 g −
B1
9 g − 3g = ≈ −14.7 Ν) 2 2
M1A1
g ≈ 46.5( N ) 2 1 at angle tan −1 = 18.4o (0.322 radians) below the line of BA 3 161.6o (2.82 radians) below the line of AB (108.4o or 1.89 radians to upward vertical)
⇒ R = 81 + 9 ×
Q5
(a)
M1A1 M1A1 (7) [11]
Ratio of areas triangle:sign:rectangle = 1 : 5 : 6 (1800:9000:10800) Centre of mass of the triangle is 20cm down from AD (seen or implied)
B1 B1
⇒ 6 × 45 − 1 × 20 = 5 × y y = 50cm
M1A1 A1
(b) Distance of centre of mass from AB is 60cm
(5)
B1
60 50 = 50.2o (0.876 rads)
Required angle is tan −1
(their values)
M1A1ft A1 (4) [9]
8371_9374 GCE Mathematics January 2009
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Question Number Q6
(a)
(b)
Scheme
→ x = u cosαt = 10 1 ↑ y = u sin αt − gt 2 = 2 2 10 ⇒t= u cos α 10 100 g 2 = u sin α × − × 2 u cos α 2 u cos 2 α 50 g (given answer) = 10 tan α − 2 u cos 2 α
Marks M1A1 M1A1
M1 A1 (6)
100 g × 2 2 = 10 × 1 − 2u 2 × 1 100 g 100 g = 11.1 (m s-1) ,u= u2 = 8 8 1 1 mu 2 = m × 9.8 × 2 + mv 2 2 2 −1 v = 9.1ms
M1A1 A1 M1A1 A1 (6) [12]
8371_9374 GCE Mathematics January 2009
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Question Number Q7
Scheme
(a)
Marks
KE at X = Y
-1
14 m s
1 2 1 mv = × 2 × 14 2 2 2
GPE at Y = 7 ⎞ ⎛ mgd sin α ⎜ = 2 × g × d × ⎟ 25 ⎠ ⎝
B1
Normal reaction R = mg cos α
M1
24 1 Friction = µ × R = × 2 g × 8 25
M1A1
X
α
Work Energy:
B1
B1
1 mv 2 - mgd sin α = µ × R × d or equivalent 2
196 =
14 gd 6 gd 20gd = + 25 25 25
A1 (7)
d = 25 m (b) Work Energy
First time at X:
1 1 mv 2 = m14 2 2 2
24 1 Work done = µ × R × 2d = × 2 g × × 2d 25 8
Return to X:
1 1 1 24 mv 2 = m14 2 - × 2 g × × 50 2 2 8 25
v = 8.9 ms-1
M1A1
(accept 8.85 ms-1)
DM1A1 (4)
OR: Resolve parallel to XY to find the acceleration and use of v 2 = u 2 + 2as 2a = 2 g sin α − Fmax = 2 g ×
7 6 g 8g − = 25 25 25
M1A1
(accept 8.85 ms-1)
v 2 = (0+ )2 × a × s = 8 g ; v = 8.9
DM1;A1 [11]
8371_9374 GCE Mathematics January 2009
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Question Number Q8
Scheme
Marks
(a)
A
B
C 3m
4m u
m
v
0
kv
Conservation of momentum: 4mu − 3mv = 3mkv 3 (u + v) 4
Impact law:
kv =
Eliminate k:
3 4mu − 3mv = 3m × (u + v) 4
u = 3v (Answer given)
M1A1 M1A1
DM1 A1 (6)
(b)
kv =
3 (3v + v) , k = 3 4
M1,A1 (2)
(c) Impact law: (kv + 2v)e = vC − v B
( 5ve = vC − v B ) Conservation of momentum : 3 × kv − 1 × 2v = 3v B + vc ( 7v = 3v B + vc ) Eliminate vC : v B = v (7 − 5e) > 0 hence no further collision with A. 4
B1 B1 M1 A1 (4) [12]
8371_9374 GCE Mathematics January 2009
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8371_9374 GCE Mathematics January 2009
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