M2 MS Jan 2006

Page 1

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


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