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]