M3 MS Jan 2006

January 2006

6679 Mechanics M3 Mark Scheme

Question Number 1.

Scheme (a)



Marks

 T cos 60  0.8g F  T sin 60 [or  F cos 60  0.8g cos30 ] F  0.8g tan 60  14 (N)

both M1 (M2) accept 13.6 M1 A1 (3)

(b)

0.8 g allow in (a) M1   15.68 sin 30 24  x HL 15.68   x  0.78 (cm) accept 0.784 M1 A1 1.2 T

(3) (c)

E

24  x  6.1  J  2 1.2 2

accept 6.15 M1 A1ft (2) Total 8 marks

2.

(a)

dv 1 1  2sin t  v  A  4cos t 2 2 dt v  4, t  0  4  A  4  A  8

M1 A1 M1

1 2

v  8  4cos t

A1 (4)

(b)



  8  4cos ...

...

1  t dt 2 

1 2

 8t  8sin t

...0 2  4    2  

ft constants awrt 6.9

M1 A1ft M1 A1 (4) Total 8 marks

1

January 2006

6679 Mechanics M3 Mark Scheme

Question Number 3.

Scheme

(a)

N2L

d 1 2 c  v  2 dx  2  x

ma  

Marks

cm x2

B1

1 2 c v  A 2 m 2c v2  B  m 

ignore A

2c R 1 1  Leading to v 2  U 2  2c      x R x  R, v  U

 B U2 

M1 A1

M1 cso

A1 (5)

1 1  1   1 1  mU 2   m U 2  2c     2 2  2   2R R  1 Leading to c  RU 2 2

(b)

M1 A1 A1 (3) Total 8 marks

4.

h 3   5M x  3M   2M  h  r  2 8   3h 3 7h 3 5x   2h  r   r 2 4 2 4 14h  3r  x 20

(a)

M1 A2(1,0)

cso

M1 A1 (5)

(b) 

x 20r 4  14h  3r 3 6 Leading to h  r 7

tan  

r



M1 A1 M1 A1 (4) Total 9 marks

2

January 2006

6679 Mechanics M3 Mark Scheme

Question Number

Scheme

5.

Marks

A l B 1 4

l

O x P (a) (b) N2L

T  mg 

HL

  14 l

   4mg

l

M1 A1 (2)

mg  T  mx 1 4mg  4 l  x  mg   mx l d2 x 4g  x  2 dt l

M1 M1 A1 cso

M1 A1 (5)

(c)

v2   2  a2  x2  

4g  l l     l  4 16  2

2

Leading to v    gl  2 1

M1 A1 M1 A1 (4)

2

gl 1 4mg. 16 1 2 3l or energy,  mv  mg. for the first M1 A1 in (c) 2 l 2 4 (d) P first moves freely under gravity, then (part) SHM.

B1 B1 (2) Total 13 marks

3

January 2006

6679 Mechanics M3 Mark Scheme

Question Number 6.

Scheme (a)

A



l

v C u    3gl 

B Energy



Marks



1 m u 2  v 2  mgl 1  cos   2

M1 A1

v 2  gl  2 gl cos   mv 2 T  mg cos   l mgl 1  2cos    l T  mg 1  3cos   

N2L

M1 A1 M1 cso A1 (6)

1 T  0  cos    3

(b)

2  gl  v  gl  gl  v    3  3 2

B1 1

2

M1 A1 (3)

  gl  2 2       .      3   gl 8 4l v 2  u 2  2 gh  2 gh  .  h  3 9 27 4l 40l H  l 1  cos     27 27 1

 gl  2  v y    sin   3

(c) vy

v

1

M1 M1 A1 M1 A1 (5) Total 14 marks

4

January 2006

6679 Mechanics M3 Mark Scheme

Question Number 7.

Scheme

(a)

N2L

Marks

 kg   T cos 30  m  2a cos 30     3a  2kmg  T 3

M1 A1 cso A1 (3)



(b)

R  mg  T sin 30  k  mg 1    3

M1 A1 A1 (3)

(R 0)

(c)



k 3

ignore k > 0, accept k < 3

M1 A1 (2)

(d)

A T 2a X N2L

mg

 2g   T cos   m  2a cos      a  T  4mg  T sin   mg Eliminating T 1 AX  2a sin   a 2



AO  2a sin 30  a 

AX 

M1 A1

M1 M1 A1

1 AO , as required  2

cso B1, A1 (7) Total 15 marks

5