M2 MS Jun 2009

Page 1

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

97

M1A1A1 A1

(4) [6]


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M2 MS Jun 2009 by The de Ferrers Academy - Issuu