M1 MS Jun 2008

Mark Scheme (Results) Summer 2008

GCE

GCE Mathematics (6677/01)

June 2008 6677 Mechanics M1 Final Mark Scheme Question Number

1.

Scheme

Marks

I = mv ⇒ 3 = 0.4 × v

(a)

M1 A1

( )

v = 7.5 ms −1

(b)

0.4

0.6

v

5

0.4 × 7.5 = 0.4v + 0.6 × 5 0 = 0.4v ⇒ v = 0 ¿

cso

(a) v 2 = u 2 + 2as ⇒ 17.52 = u 2 + 2 × 9.8 × 10 Leading to u = 10.5

(b)

(3)

7.5

LM

2.

A1

v = u + at ⇒

17.5 = −10.5 + 9.8T 6 T =2 (s) 7

M1 A1 A1 (3) [6]

M1 A1 A1

M1 A1 f.t.

DM1 A1 (4)

Alternatives for (b)

[7]

s=(

OR

(3)

17.5 + −10.5 u+v )T ⇒ 10 = ( )T 2 2 20 =T 7

s = ut + 12 at 2

⇒ − 10 = 10.5t − 4.9t 2

M1A1 f.t.

DM1A1 (4)

M1 A1 f.t.

6 ⎛ 5⎞ Leading to T = 2 , ⎜ − ⎟ Rejecting negative DM1 A1 (4) 7 ⎝ 7⎠ (b) can be done independently of (a) s = vt − 12 at 2 ⇒ − 10 = −17.5t + 4.9t 2 M1 A1

6 5 Leading to T = 2 , 7 7 5 For final A1, second solution has to be rejected. leads to a negative u.

7

DM1 A1

(4)

Question Number

3.

Scheme

Marks

8 6 θ ≈ 53°

tan θ =

(a)

(

M1 A1

) (= 2.4i + 3.2j)

F = 0.4 6i + 8j

(b)

(

(2)

M1

)

F = √ 2.4 2 + 3.2 2 = 4

M1 A1

(3)

The method marks can be gained in either order.

(

)

v = 9i − 10j + 5 6i + 8j

(c)

( )

M1 A1

= 39i + 30 j ms −1

A1

(3) [8]

4.

(a)

v 25 shape 25, 10, 30, 90

B1 B1

(2)

10

O (b) 30 × 25 +

90 t

30

1 25 + 10 t + 10 60 − t = 1410 2 7.5t = 60 t=8 s

(

)

(

)

M1 A1 A1

()

a=

DM1 A1

25 − 10 = 1.875 ms −2 8

( )

1 78

M1 A1

(7) [9]

Question Number

5.

Scheme

Marks

(a) 15

R 30°

(↑)

50°

X

15sin 30° = R sin50°

M1 A1

( )

R ≈ 9.79 N

DM1 A1 (4)

(→) X − 15cos30° = R cos50° X ≈ 19.3 (N )

(b)

ft their R

M1 A2 ft DM1 A1 (5) [9]

Alternatives using sine rule in (a) or (b); cosine rule in (b) (a) R

15

50°

30°

R 15 = sin30° sin50°

M1 A1

( )

R ≈ 9.79 N

(b)

X 15 R = = sin100° sin50° sin 30°

DM1 A1 (4)

M1 A2 ft on R

X

( )

X ≈ 19.3 N

DM1 A1 (5)

X 2 = R 2 + 152 − 2 x 15 x Rcos100o OR: cosine rule; any of R 2 = X 2 + 152 − 2 x 15 x X cos30o

M1 A2 ft on R

152 = R 2 + X 2 − 2 x X x Rcos50o

( )

X ≈ 19.3 N

DM1 A1 (5)

Question Number

6.

Scheme

Marks

(a) X 2.4 0.8 A

B 8g

()

12g

8g × 0.8 + 12g × 1.2 = X × 2.4

M A

( )

X ≈ 85 N

M1 A1 accept 84.9,

26g 3

DM1 A1 (4)

(b) X + 10

X 2.4 0.8 x

A

B 8g

12g

( ) (X + 10)+ X = 8g + 12g (X = 93)

R ↑

()

M A

M1 B1 A1

8g × 0.8 + 12g × x = X × 2.4

( )

x = 1.4 m

M1 A1 accept 1.36

A1

(6) [10]

Question Number

7.

Scheme

(a)

Marks

R 45 N

µR

50°

4g

30° R = 45cos 40° + 4g cos30° R ≈ 68 (b)

accept 68.4

Use of F = µ R

M1 M1 A2 (1, 0)

F + 4g sin 30 = 45 cos 50° Leading to µ ≈ 0.14

M1 A2 (1, 0) DM1 A1 (5)

accept 0.136

DM1 A1 (6) [11]

Question Number

8.

Scheme

Marks

(a) T

T

µ 2g

30

µ 3g

s = ut + 12 at 2

⇒ 6 = 12 a × 9

M1

( )

a = 1 13 ms −2

30 − µ5g = 5a

(b) N2L for system symbol

µ= (c)

T−

ft their a, accept

14 10 = 3g 21

awrt 0.48 DM1 A1 (4) M1 A1 ft

14 4 × 2g = 2 × 3g 3

( )

Leading to T = 12 N Alternatively

or

(2)

M1 A1ft

ft their µ , their a, accept symbols

T − µ 2 g = 2a

N2L for P

A1

N2L for Q 30 − T − µ 3g = 3a

( )

Leading to T = 12 N

awrt 12 DM1 A1 (4)

M1 A1 awrt 12 DM1 A1

(d) The acceleration of P and Q (or the whole of the system) is the same.

4 ×3= 4 3 N2L (for system or either particle) −5µ g = 5a a = −µg v = u + at ⇒ 0 = 4 − µ gt (e) v = u + at ⇒ v =

Leading to t =

6 s 7

()

B1

(1)

B1 ft on a or equivalent

M1 DM1

accept 0.86, 0.857

A1

(4) [15]