M2 MS Jan 2005

January 2005 6678 Mechanics M2 Mark Scheme Question Number

Scheme (a) M(A) W  4a  T  8a sin  Using a value of sin  and solving cso T  56 W 

1.

Y

T 

X W

2.

(a) Mass ratios Centres of mass

Marks



(b)

circle 9 6

X  T cos = 23 W

rectangle 200; 10

M1 A1 A1 3

plate 200  9 x

(b)

5 10.7   25

tan  

7

B1; B1ft B1

9  6   200  9  x  200 10 x  10.7 (cm)

M1 A1 M1 A1 4

cao ft their x cao

M1 A1

5

M1 A1ft A1

3

8

Question Number 3.

Scheme

Marks

KE lost is 12  0.6  102  92 

B1

Total loss in energy is 41.0 (J)

B1 M1 A1 4

  5.7 J  PE lost is 0.6  9.8 12sin 30   35.28 J 

(a)

accept 41

R  0.6  9.8  cos30   5.09

(b) WE

B1

40.98    0.6  9.8  cos3012   0.67 or 0.671

ft their (a)

M1 A1ft M1 A1 5

9

Alternative for (b)

92  102  19      2 12  24  mg sin 30   mg cos30  m   19 24  a

N2L

awrt 0.79

B1

ft their a

M1 A1ft

  0.67 or 0.671

4.

(a)

M1 A1 5

r  6i   2t  3 j

B1

F  0.4  6i  11j F =   2.42  4.42   5.0

 r   3t

0.4something obtained by differentiation, with t = 4 modulus of a vector

M1

accept more accurate answers

A1

  t  t j  C  4t  3 i   t  t  4  j

r  3t 2  4t i 

(b) Using boundary values,



3

3 2

1 3 3

2

3 2 2

r  61i  49 13 j

t = 4,

OS   612  49 13

2

1 3

2

  78 (m)

accept more accurate answers

M1 4

M1 A1 A1 M1 A1 5 9

Question Number 5.

Scheme

Marks

50000  F  25  F  2000 

(a) 

or equivalent M1 M1 cso A1

F  R  750 R  1250 

N2L 1500  2000  2500a a  1.4  ms2 

(b)

ignore sign of a cao

(c) Trailer: T  R  1500 1.4 or Car: T 1500  750  1000  1.4 T  850  N 

252  2 1.4  s  s  223.2...

(d)

W  1500  s = 335 (kJ)

ft their s accept 330

(e) Resistances vary with speeds

6.

2u

u

3m

2m

x

y

LM 6mu  2mu  3mx  2my NEL y  x  3eu Solving to y  15 u  9e  4   cso

(b) Solving to x  52 u  2  3e  x0  e 2 3

(c)

M1 A1 A1

3

M1 A1

2

M1 M1 A1ft A1 4 B1

(a)

 e 1

oe

2 3

ft their e for glb

2m  15 u  9e  4   u   325 mu

Solving to e 

7 9

3

1

13

M1 A1 B1 M1 A1 5 M1 A1 M1 A1 A1ft 5 M1 A1

awrt 0.78

M1 A1 4

14

Question Number 7.

Scheme  u y  32  53   19.2 

(a)

20  19.2t  4.9t t  4.8 or 4.77 (s) 2

(d)

1 each error

 ux  32  54   25.6 d  25.6  4.77...  120 or 122  m 

(b)

(c)

Marks

B1 M1 A1

 vy2  19.22  2  9.8  4 vy2  447.04, vy  21.14 V 2  447.04  25.62 V  33 or 33.2  ms 1 

tan  

21.14 25.6

25.6   , ...   or cos   33.2  

  40 or 39.6

B1 M1 A2(1, 0) A1 5

ft their components

3

M1 M1 A1 A1

4

M1 A1ft

or resultant

Alternative for (c) 2 2 1 2 m V  32   mg  4 V 2  1102.4 V  33 or 33.2  ms 1 

There is a maximum penalty of one mark per question for not rounding to appropriate accuracy.

A1

3

M1 A1 M1 A1

4

15