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Mark Scheme (Results) January 2008

GCE

GCE Mathematics (6678/01)

Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH


January 2008 6678 Mechanics M2 Mark Scheme Question Number 1.

Scheme (a)

KE lost is

(b)

Work energy

Marks

1 × 2.5 × 82 = 80 ( J ) 2 80 = R × 20 R=4

M1 A1

(2)

ft their (a)

M1 A1 ft A1 (3) [5]

ft their a

M1 A1ft (3) A1

Alternative to (b) 02 = 82 − 2 × a × 20 ⇒ a = ( − )1.6 R = 2.5 × 1.6 =4

N2L

2.

(a)

p& = ( 6t − 6 ) i + ( 9t 2 − 4 ) j

(ms ) −1

M1 A1

9t 2 − 4 = 0 t = 23

(b)

(c) (+/-)

M1 DM1 A1

t = 1 ⇒ p& = 5 j 2i − 6 j = 0.5 ( v − 5 j) v = 4i − 7 j

(2)

(m s ) −1

ft their p&

(3)

B1ft M1 M1 A1

(4) [9]


Question Number

3.

Scheme 20 000 = 16 F

(a)

( F = 1250 )

F = 550 + 1000 × 9.8sin θ 1 ¿ Leading to sin θ = 14

Ê

(b)

Marks

N2L Ê

ft their F cso

550 + 1000 × 9.8 × sin θ = 1000a ( 550 + 1000 × 9.8 × 141 = 1000a ) or 1250 = 1000a ( a = ( − )1.25)

M1 A1 M1 A1ft A1 (5)

M1 A1

v 2 = u 2 + 2as ⇒ 162 = 2 × 1.25 × y M1 y ≈ 102 accept 102.4, 100 A1 Alternative to (b) Work-Energy

4.

(a) Mass ratio

x y

1 2

(4) [9]

×1000 ×162 − 1000 × 9.8 × 141 y = 550 y M1 M1 A1 y ≈ 102 accept 102.4, 100 A1 (4)

Triangle Circle S 9π 126 − 9π 126 (28.3) (97.7) 7 5 x 4 5 y

B1 B1ft 4, 7 seen B1

126 × 7 = 9π × 5 + (126 − 9π ) × x ft their table values x ≈ 7.58 (

882 − 45π ) 126 − 9π

awrt 7.6

M1 A1ft A1

126 × 4 = 9π × 5 + (126 − 9π ) × y ft their table values M1 A1ft 504 − 45π y ≈ 3.71 ( ) awrt 3.7 A1 (9) 126 − 9π

(b)

y 21 − x θ ≈ 15°

tan θ =

ft their x , y

M1 A1ft A1

(3) [12]


Question Number

5.

Scheme

(a)

N

Marks B

2a 30°

a

mg

R 3mg

a A

Μ(A)

Fr

N × 4a cos 30° = 3mg × a sin 30° + mg × 2a sin 30° 5 5 mg = 7.07…m) N = mg tan 30° ( = 4 4 3 → Fr = N , ↑ R = 4mg Using Fr = µ R 5 mg = µ R for their R 4√3 5 µ= awrt 0.18 16 √ 3

M1 A2(1,0) DM1 A1 B1, B1 B1 M1 A1 (10)

[10] Alternative method: M(B): mg × 2a sin 30 + 3mg × 3a sin 30 + F × 4a cos 30 = R × 4a sin 30 11mga sin 30 + F × 4a cos 30 = R × 4a sin 30 11mg 4 3 +F = 2R 2 2 ↑ R = 4mg , Using Fr = µ R 8µ 3 =

5 , 2

M1A3(2,1,0) DM1A1 B1 B1

µ=

5 16 √ 3

M1 A1


6.

(a) ↑

→ 30 = 2ut −47.5 = 5ut − 4.9t 2 −47.5 = 75 − 4.9t 2 75 + 47.5 t2 = ( = 25) 4.9

eliminating u or t

DM1

t =5 ¿

cso

(b)

30 = 2ut ⇒ 30 = 10u ⇒ u = 3

(c)

↑ →

y& = 5u − 9.8t = −34 x& = 2u = 6 v 2 = 62 + ( −34 ) v ≈ 34.5

2 −1

A1 (6) M1 A1 (2)

M1 requires both x& and y&

(ms )

B1 M1 A1 DM1

accept 35

M1 A1 A1

DM1 A1 (5)

[13] Alternative to (c) 1 2

mvB2 − 12 mv A2 = m × g × 47.5 with

v A2 = 62 + 152 = 261

vB2 = 261 + 2 × 9.8 × 47.5 ( = 1192 )

vB ≈ 34.5

(ms ) −1

accept 35

BEWARE : Watch out for incorrect use of v 2 = u 2 + 2as

M1 A(2,1,0) DM1 A1 (5)


Question Number

7.

Scheme

(a)

Marks

2u

u

2m

3m

x

y

4mu + 3mu = 2mx + 3my y − x = 12 u

LM NEL

M1 A1 B1

Solving to y = 8 u ¿ 5

cso

x = 11 u

(b) Energy loss

1 × 2m 2

(( 2u ) − ( 2

10

11 10

= (c)

8 5

or equivalent

)

(

u ) + 12 × 3m u 2 − ( 85 u ) 2

2

)

9 mu 2 20

M1 A1

(5)

B1 M1 A(2,1,0) A1

(5)

u

3m

m

s 24 mu 5

LM

t = 3ms + mt

M1 A1

t − s = 85 eu

NEL

Solving to s = For a further collision

11 10 u

2 u 5

B1

(3 − e)

M1 A1

> 52 u ( 3 − e )

e>

1 4

M1 ignore e ≤ 1

A1

(7) [17]


M2 MS Jan 2008