Mark Scheme (Results) January 2010

Mark Scheme (Results) January 2010

GCE

Mechanics M1 (6677)

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January 2010 Publications Code UA023034 All the material in this publication is copyright ? Edexcel Ltd 2010

Question Number

Q1. (a) (b) LM

Alternative to (b)

January 2010 6677 Mechanics M1

Mark Scheme

Scheme

I = 2?12 - 2? 3 = 18 ( N s)

2?12 - 8m = 2? 3 + 4m Solving to m = 1.5

I = m(4 - (-8)) = 18

Solving to m = 1.5

Marks

M1 A1 (2)

M1 A1 DM1 A1 (4)

[6]

M1 A1 DM1 A1 (4)

Q2. (a) s

8

First two line segments B1

Third line segment B1

8, 75 B1

(3)

O (b)

1 ?8? (T + 75) = 500

2 Solving to T = 50

75 t

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

[8]

GCE Mechanics M1 (6677) January 2010

Question Number

Q3.

(a)

A R( )

Scheme 30?

20 N

60 ?

B

T N

C mg

20 cos 30? = T cos 60? T = 20 3 , 34.6, 34.64,...

Marks

M1 A2 (1,0)

A1

(4)

(b)

R( ) mg = 20sin 30? + T sin 60?

m = 40 ( 4.1) ,4.08 g

Q4. (a)

X

Y

1.8 m

A

1.5 m W

1.5 m

20

M ( A)

W ?1.5 + 20? 3 = Y ?1.8 Y = 5 W + 100 ? 63

(b)

X + Y = W + 20 X = 1 W - 40

63

(c)

5W 6

+

100 3

=

8

1W 6

-

40 3

W = 280

Alternative to (b) M(C) X ?1.8 + 20 ?1.2 = W ? 0.3

X = 1 W - 40 63

M1 A2 (1,0)

A1

(4)

[8]

M1 A2 (1, 0)

cso

A1

(4)

or equivalent M1 A1

A1

(3)

M1 A1 ft

A1

(3)

[10]

M1 A1 A1

GCE Mechanics M1 (6677) January 2010

Question Number

Scheme

Marks

Q5. (a) (b)

s = ut + 1 at2 2.7 = 1 a ? 9

2

2

( ) a = 0.6 m s-2

M1 A1

A1

(3)

R

? R

R = 0.8g cos 30? ( 6.79)

B1

Use of F = ?R

B1

0.8g

0.8g sin 30? - ?R = 0.8? a

M1 A1

30?

(0.8g sin 30? - ?0.8g cos 30? = 0.8? 0.6)

? 0.51

accept 0.507 A1

(5)

(c)

R

X

? R

0.8g

30?

R cos 30? = ?R cos 60? + 0.8g

( R 12.8)

X = R sin 30? + ?R sin 60?

Solving for X, X 12

accept 12.0

M1 A2 (1,0)

M1 A1 DM1 A1

(7) [15]

Alternative to (c)

R = X sin 30? + 0.8 ? 9.8sin 60? ?R + 0.8g cos 60? = X cos 30?

Solving for X,

X

=

?0.8g sin 60? + 0.8g cos 60? cos30? - ? sin 30?

X 12

accept 12.0

M1 A2 (1,0) M1 A1

DM1 A1 (7)

GCE Mechanics M1 (6677) January 2010

Question Number

Q6. (a) N2L A:

(b) N2L B:

Scheme

5mg - T = 5m ? 1 g 4

T = 15 mg ? 4

T - kmg = km ? 1 g 4

k =3

Marks

M1 A1

cso

A1

(3)

M1 A1

A1

(3)

(c) The tensions in the two parts of the string are the same

(d) Distance of A above ground Speed on reaching ground

s1

=

1 2

?

1 4

g

?1.22

=

0.18g

( 1.764)

v = 1 g ?1.2 = 0.3g ( 2.94)

4

For B under gravity

(0.3g )2 = 2gs2

s2

=

(0.3)2

2

g

(

0.441)

S = 2s1 + s2 = 3.969 4.0 (m)

B1

(1)

M1 A1 M1 A1

M1 A1

A1

(7)

[14]

GCE Mechanics M1 (6677) January 2010

Question Number

Scheme

Marks

Q7. (a)

v = 21i +10j - (9i - 6j) = 3i + 4j

4

( ) ( ) speed is 32 + 42 = 5 km h-1

M1 A1 M1 A1 (4)

(b)

tan = 3 ( 36.9?)

4

bearing is 37, 36.9, 36.87, ...

M1

A1

(2)

(c)

s = 9i - 6j + t (3i + 4j)

M1

= (3t + 9)i + (4t - 6)j ?

cso A1

(2)

(d) Position vector of S relative to L is

(3T + 9) i + (4T - 6) j - (18i + 6j) = (3T - 9) i + (4T -12) j

M1 A1

(3T - 9)2 + (4T -12)2 = 100

M1

25T 2 -150T +125 = 0

(T 2 - 6T + 5 = 0)

or equivalent DM1 A1

T = 1, 5

A1

(6)

[14]

GCE Mechanics M1 (6677) January 2010

GCE Mechanics M1 (6677) January 2010

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