GCE

Edexcel GCE Mathematics Mechanics 2 M2 (6678)

June 2008

Mathematics

Edexcel GCE

Final Mark Scheme

General Marking Guidance •

All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last.

•

Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions.

•

Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie.

•

There is no ceiling on achievement. All marks on the mark scheme should be used appropriately.

•

All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.

•

Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited.

•

When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, the team leader must be consulted.

•

Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

June 2008 6678 Mechanics M2 Mark Scheme Question Number

Scheme

1600

1.

Resolve Ê : Tr + 2000 g × sin α = 1600

Tr

α

2.

(a)

LM NEL

Marks

(Tr = 816 ) P = 816 × 14 ( W ) ft their Tr ≈ 11 ( kW ) accept 11.4

2000g

3u

2u

4m

3m

x y = 4eu 12mu + 6mu = 4mx + 12meu 4eu − x = eu Eliminating x to obtain equation in e 3 Leading to ¿ e= 4

x = 3eu or

M1 A1ft A1 cso (6) [6]

B1 M1 A1 DM1

cso

9 u or 4.5u – 3eu seen or implied in (b) 4 2 1 1 1 2 2 2 ⎛9 ⎞ 1 Loss in KE = 4m ( 3u ) + 3m ( 2u ) − 4m ⎜ u ⎟ − 3m ( 3u ) 2 2 2 ⎝4 ⎠ 2 ft their x 5 3 = 24mu 2 − 23 mu 2 = mu 2 = 0.375mu2 8 8

(b)

M1 A1 A1

A1

(5)

B1 M1 A1ft

A1

(4) [9]

Question Number

Scheme

Marks

3.

1 (a ∆ KE = × 3.5 (122 − 82 ) ( = 140 ) or KE at A, B correct separately 2 ∆ PE = 3.5 × 9.8 × 14sin 20° ( ≈ 164.238 ) or PE at A, B correct separately ∆ E = ∆ KE + ∆ PE ≈ 304 , 300

(b) Using Work-Energy Fr = µ × 3.5 g cos 20° 304.238 ... = Fr × 14 304.238 ... = µ 3.5 g cos 20°×14 µ ≈ 0.674 , 0.67

ft their (a), Fr

B1 M1 A1 DM1 A1

(5)

M1 A1 M1 A1 ft

A1

(5) [10]

Alternative using N2L

µR

Fr = µ × 3.5 g cos 20°

R

mg

4.

(a)

N2L

2

2

2

( 6t − 5) i + ( t 2 − 2t ) j = 0.5a a = (12t − 10 ) i + ( 2t 2 − 4t ) j ⎛2 ⎞ v = ( 6t 2 − 10t ) i + ⎜ t 3 − 2t 2 ⎟ j ( +C ) ⎝3 ⎠ ⎛2 ⎞ v = ( 6t 2 − 10t + 1) i + ⎜ t 3 − 2t 2 − 4 ⎟ j ⎝3 ⎠

(b) When t = 3 ,

M1 A1

v = u + 2as ⇒ 8 = 12 − 2a × 14 20 ⎞ ⎛ ⎜ a = ⎟ (2.857 ….) 7 ⎠ ⎝ N2L R É : {their Fr}- mg sin 20° = ma M1 A1ft ft their Fr. Leading to µ ≈ 0.674 or 0.67 (5) A1 2

M1 A1

ft their a

A1

v 3 = 25i − 4 j

−5i + 12 j = 0.5 ( v − ( 25i − 4 j) )

v = 15i + 20 j

v = √ (15 + 20 2

2

) = 25 ( ms ) −1

M1 A1ft+A1ft (6)

M1 ft their v 3

M1 A1ft A1

cso M1 A1

(6) [12]

Question Number

5.

Scheme

(a)

Marks

P α

0.5a R 1.5a

α µR

W

R ( ↑ ) R + P cos α = W M ( A)

M1 A1

P × 2a = W × 1.5a cos α

3 ⎛ ⎞ ⎜ P = W cos α ⎟ 4 ⎝ ⎠ 3 R = W − P cos α = W − W cos 2 α 4 1 = ( 4 − 3cos 2 α ) W ¿ 4 (b) Using cos α =

2 , 3

R ( →)

M1 A1

DM1 cso

2 R= W 3

A1

(6)

B1

µ R = P sin α

3 Leading to µ = sin α 4 sin α = √ (1 − 94 ) = 35

(

5 4

µ=√

M1 A1

) awrt 0.56

DM1 A1

(5)

[11]

Question Number

6.

Scheme

(a)

M ( Oy )

(8 + k ) m × 6.4 = 5m × 8 + km × 8 1.6k = 11.2 ⇒ k = 7 ¿

(b)

M ( Oy )

cso

16 3

70 27 y 35 = x 72 θ ≈ 26°

DM1 A1

(4)

M1 A1 5.3 or better

27 my = 12m × 2.5 + 8m × 5

y=

(c)

M1 A1

27 mx = 12m × 4 + 5m × 8 + 7 m × 8

x= M ( Ox )

Marks

A1 M1 A1

2.6 or better

A1

(6)

M1 A1ft

tan θ =

awrt 25.9 °

A1

(3) [13]

Question Number

7.

Scheme

(a)

(↓)

Marks

u y = 25sin 30° ( = 12.5 )

12 = 12.5t + 4.9t 2 Leading to t = 0.743 , 0.74 (b)

either

( →)

−1 each error

⎛ 25 3 ⎞ u x = 25cos 30° ⎜ = √ ≈ 21.65 ⎟ 2 ⎝ ⎠ OB = 25cos 30°× t ( ≈ 16.094 58 )

( →)

TB ≈ 1.1 (m) (c)

B1

15 = u x × t ⇒ t =

(↓)

B1 ft their (a) awrt 1.09

15 2 3 (= ≈ 0.693 or 0.69) 5 ux

v y = 12.5 + 9.8t ( ≈ 19.2896 )

M1 A2 (1, 0) (5) A1

M1 A1ft A1 (4)

M1 A1 M1

V 2 = u x 2 + v y 2 ( ≈ 840.840 )

V ≈ 29 ( ms −1 ) , 29.0

or

(↓)

M1 A1

(5) [14]

2

sy = 12.5t + 4.9t (≈ 11.0) 1 1 m × 25 2 + mg × s y = mv 2 2 2 −1 V ≈ 29 ( ms ) , 29.0

M1

M1A1