January 2006

6678 Mechanics M2 Mark Scheme

Question Number 1.

Scheme

Marks

Kinetic Energy = 12  3  82  96, J

(a)

B1 B1 (2)

F   3g

(b)

B1

3gx12 = 96   0.27 or 0.272

Work-Energy

M1 A1ft A1 (4)

Alternative for (b)

a N2L

8 0 8  2 12 3 3g 3g  3 x 2

2

B1 8 3

M1 A1 A1

  0.27 or 0.272

2.

r  (2t  4)i  (3  3t 2 ) j r3  10i  24 j

(a)

(4) 6 M1 A1 substituting t =3 M1

r3   102  242   26  ms1 

(b)

(5)

0.4  v  10i  24 j   8i  12 j v  30i  54 j

3.

M1 A1

 ms 

12000   800 15 800  R  1000  0.2 R  600 

1

ft their r3 M1 A1ft A1 (3) 8

Tr 

(a) N2L

M1 ft their 800 M1 A1ft cso A1 (4)

(b)

1000 g 

1  Tr  R 40 7000 Tr  U U  20

1

M1 A1 M1 accept 19.7

M1 A1 (5) 9

January 2006

6678 Mechanics M2 Mark Scheme

Question Number 4.

Scheme (a)

LM

NEL

Marks

3u

2u

2m

m

x

8u/3

8 6mu  2mu  2mx  mu 3 2   x  u 3   8 u  x  5ue 3 2 Solving to e  5

M1 A1

M1 A1 M1 A1 (6)

1 1 2 2 (b) Initial K.E.=  2m  3u    m  2u   11mu 2 2 2 2

2

1 2  1 8  Final K.E. =  2m  u    m  u   4mu 2 both M1 2 3  2 3  Change in K.E. = 7mu 2  M1 Subtracting and simplifying M1 A1 to kmu 2 A1cso (c)

8  14 m  u  v   mu 3  3  v  2u 

e

2 3  8 4 3

(3)

M1 A1

M1 A1 (4) 13

2

January 2006

6678 Mechanics M2 Mark Scheme

Question Number 5.

Scheme (a)

Marks

12mx  6m  9 x  4 12

M1 A1 M1 A1

12my  16m  8m

y = (b)

12  k 

2 3

m  4  12m  4 12

 km  3

(4) ft their x

k 6

(c)

(d)

(3)

18m    12m  23 ,  tan  

4 4

M1 A1ft A1

,    83.7

  94

M1 A1 (2) ft their  , cao

M1 A1ft A1

9 (3) 12

6.

(a)

N 4W

a W 2a R

R

R

 R  5W 4Wacos   W.2a cos  R4asin   R.4acos (B): Having enough equations & solving them for    0.35  S  (5  k)W Use of F  0.35S or F  0.35S (B ): kW4a cos  W.2acos   F4asin  S.4acos Having enough equations & solving them for k (b)

k  10 7

awrt 1.42

k 10 7

ft their k, accept > and decimals

3

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

(6)

(7) 13

January 2006

6678 Mechanics M2 Mark Scheme

Question Number 7.

Scheme

ux  11cos30

(a)  (b)

11cos30 t  10  t  1.05

(s)

cao

s  11sin 30 t  4.9t 2  0.37

 2  1  0.37  0.63 (c)

Marks

(3) B1 M1 A1 A1

(m)

(4)

10   t    V cos 30  10 4.9 100 s  V sin 30  2 1 V cos30 V cos 2  V 2  136.86 V cos30 t  10

V  12

B1 M1 A1

accept 11.7

M1 A1 M1 A1 M1 A1 (6)

(d) B and/or T are not particles ( They have extension giving a range of answers)

4

B1 (1) 14

M2 MS Jan 2006

    12000 800 15     10 24   r i j substituting t =3 M1 (2 4) (3 3 ) t t    r i j M1 A1 3. (a)   2. (a) 0.4 10 24 8 12   ...