M3 MS Jun 2009

June 2009 6679 Mechanics M3 Mark Scheme Question Number Q1 (a)

Scheme 6

Marks

Resolving vertically: 2T cos θ = W

6

4.5 7. 5

Hooke’s Law:

T

T

80 × 3.5 4 W = 84N T=

M1A2,1,0 M1A1 A1

W

(b)

EPE = 2 ×

80 × 3.52 , = 245 (or awrt 245) 2× 4

M1A1ft,A1

80 × 7 2 (alternative = 245 ) 16

Q2

(a)

Object Cone Base Marker

Mass m 3m 4m

[9] B1(ratio masses) B1(distances)

c of m above base 2h+3h h d

m x 5h +3m x h = 4m x d

M1A1ft

d = 2h

A1

(b)

r 1 = d 12

M1A1ft

6r = h

A1

2h

r

8371_9374 GCE Mathematics January 2009

[8]

103

Question Number Q3

Scheme

(a)

r

θ

R

d

(b)

Marks

↔ R sin θ = mxω 2 x 3g R × = mx × r 2r 3mg R= 2

M1 A1

R cosθ = mg 3mg d × = mg 2 r 2 d= r 3

M1 A1

↨

x

M1 A1

M1 A1

mg

Q4

(a)

Volume =

1

∫

1 4

1

πy 2 dx = ∫1 π 4

[8]

1 dx x4

M1A1

1

−1 ⎤ ⎡ = ⎢π × 3 ⎥ 3x ⎦ 1 ⎣

A1ft

4

(b)

− 1 64 = π ( + ) = 21π 3 3 1 21πρx = ρ ∫ πy 2 xdx = ρ ∫ π 4 xdx x

A1

*

M1A1

1

⎡ −1 ⎤ 21πx = π ⎢ 2 ⎥ ⎣ 2x ⎦ 1

A1ft

4

5 1 − 1 16 ( + )= x= 14 2 21 2

or awrt 0.36

A1

y = 0 by symmetry

B1 [9]

8371_9374 GCE Mathematics January 2009

104

Question Number Q5 (a)

Scheme

cos- 1

1

60 °

4

T

mg

Marks

Energy: 1 1 1 ( mu 2 +)mgl (cosθ − ) = mv 2 2 4 2 Resolving: mv 2 T − mg cosθ = l Eliminate v2: 1 1 T = mg cosθ + (2mgl (cosθ − )) l 4 mg T = 3mg cosθ − 2

*

(b)

1 2

M1

l

16

A1

vertical motion under gravity: ↕ 0 = (v cos 30°) 2 − 2 gs

⇒ v2 =

3l

M1

M1

d 2

M1A1

1 4

θ = 60° ⇒ mv 2 = 2mgl ( − )

A

M1A1

0=

gl 2

gl 3 3l × − 2 gs ⇒ s = 2 4 16

Distance below A =

l 3l 5l − = 2 16 16

A1

M1A1 [11]

Alternative for end of (b) using energy

d 60 ° v cos 60

l v 60 °

8371_9374 GCE Mathematics January 2009

1 2 1 mv − mgl cos 60 = m(v cos 60) 2 − mgd M1A1 2 2 gl gl gl 1 M1 − = × − gd 4 2 4 4 1− 4 + 8 5l A1 d= l= 16 16

105

Question Number Q6 (a)

Scheme

At max v, driving force = resistance 80 Driving force = v

⇒

B1

80 1 = k × 202 ⇒ k = 20 100

80 8000 − v 3 − kv 2 (= ) v 100v dv 8000 − v 3 ⇒v = dx 10000v 2 D 8 10000v ∫4 8000 − v 3 dv = ∫0 1dx 8 ⎡ 10000 3 ⎤ D = ⎢− ln 8000 − v ⎥ 3 ⎣ ⎦4

(b)

= (− (c)

M1A1

M1

F = ma ⇒ 100a =

*

Marks

*

A1 M1A1 A1

10000 7488 ln ) = 193.7..... ≈ 194 m (accept 190) 3 7936

T 8 10000v dv 8000 − v 3 = ⇒ ∫ 1dt = ∫ dv 0 4 8000 − v 3 dt 10000v 1 2×6 8 ⎫ ⎧ 4 ⇒ T ≈ × 2 × 10000 × ⎨ + + ⎬ 2 ⎩ 7936 7784 7488 ⎭ ⇒ T (= 31.1409....) ≈ 31

M1 A1

M1A1 M1 A1 [14]

8371_9374 GCE Mathematics January 2009

106

Question Number Q7 (a)

Scheme m od=16 a=2

Marks

mod=12 a=1

A

B

5m

d

5-d

16(d − 2) 12(4 − d ) = 2 1 ⇒ d = 3 .2 so extensions are 1.2m and 0.8m. (b) If the particle is displaced distance x towards B then 16(1.2 + x) 12(0.8 − x) (= 20 x) − mx&& = − 2 1 20 ⇒ && x = −40 x or && x = − (⇒ SHM) m (c) 2π T= 40 10 a= their ω x = a sin ωt their a, their ω 1 1 = sin 40t 4 2 Hooke’s law: Equilibrium ⇒

40t =

π

Proportion

6

(⇒ t =

π

6 40

)

A1 A1 M1A1ft A1ft A1 B1ft B1ft M1 A1 M1

4t 4π 40 1 = × = T 6 40 2π 3

8371_9374 GCE Mathematics January 2009

M1A1A1

M1A1 [16]

107

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