January 2006 6677 Mechanics M1 Mark Scheme Question Number




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January 2006 6677 Mechanics M1 Mark Scheme




Question Number


Scheme

Marks

1.

(a) Distance after 4 s = 16 x 4 – ½ x 9.8 x 42
= – 14.4  h = (+) 14.4 m
(b) v = 16 – 9.8 x 4
= –23.2  speed = (+) 23.2 m s–1

M1 A1
A1

(3)

M1 A1
A1



(3)

6

2.

(a) CLM: 3 x 4 + 2 x 1.5 = 5 x v
v = 3 m s–1
(b) (i) CLM: 3 x 4 – m x 4 = –3 x 2 + m (x 1)
m = 3.6
(ii) I = 3.6(4 + 1) [or 3(4 + 2)]
= 18 Ns

M1 A1
A1

(3)

M1 A1
A1



(3)

M1
A1

(2)

8

Question Number


Scheme

Marks

3.

(a) M(C): 25g x 2 = 40g x x
x = 1.25 m
(b) Weight/mass acts at mid-point; or weight/mass evenly distributed (o.e.)
(c) y 1.4

M(C):

25g 15g 40g 40g x 1.4 = 15g x y + 25g x 2
Solve: y = 0.4 m

M1 A1
A1

(3)

B1

(1)



M1 A1

M1 A1



(4)
8

4.



R = 103/2 i – 5j
Using P = 7j and Q = RP to obtain Q = 53i – 12j
Magnitude = [(53)2 + 122]  14.8 N (AWRT)
angle with i = arctan (12/53)  64.2
bearing  144 (AWRT)

Alternative method


θ .Vector triangle correct

P Q

Q2 = 102 + 72 + 2 x 10 x 7 cos 60

120

Q  14.8 N (AWRT)



R

14.8 = 10

sin 120 sin θ


 θ = 35.8,  bearing 144 (AWRT)

M1 A1


M1 A1


 M1 A1
M1 A1


A1

(9)

B1
M1 A1
A1
M1 A1 


M1 A1, A1



9





Question Number


Scheme

Marks

5.



18 (a) R( perp to plane):

P P sin 30 + 10 cos 30 = 18

18μ Solve: P18.7 N

10 (b) R( // plane):

P cos 30 = 10 sin 30 + F
F = 18μ used
Sub and solve: μ = 0.621 or 0.62
(c) Normal reaction now = 10 cos 30
Component of weight down plane = 10 sin 30 (= 5 N) (seen)
Fmax = μRnew  5.37 N (AWRT 5.4)
5.37 > 5  does not slide


M1 A1


M1 A1


(4)
M1 A1

M1

 



M1 A1

(5)


M1 A1
B1

M1


A1 cso

(5)
14





Question Number


Scheme

Marks

6.

(a) Speed of A = (12 + 62)  6.08 m s–1
(b) tan θ = 1/6  θ  9.46

θ

6 Bearing  351



1

(c) P.v. of A at time t = (2 – t)i + (–10 + 6t)j


p.v. of B at time t = (–26 + 3t)i + (4 + 4t)j
(E.g.) i components equal  2 – t = –26 + 3tt = 7
j components at t = 7: A: –10 + 6t = 32
B: 4 + 4t = 32
Same, so collide at t = 7 s at point with p.v. (–5i + 32j) m
(d) New velocity of B = (3i + 4j) m s–1

P.v. of B at 7 s = –26i + 4j + 1.6(3i + 4j) x 7 = 7.6i + 48.8j


PB = bp = 12.6i + 16.8j (in numbers)
Distance = (12.62 + 16.82) = 21 m

M1 A1


(2)

M1 A1
A1

(3)

B1 (either)


M1 A1

M1


A1 cso

(5)


B1
M1 A1

M1



M1 A1


(6)
16




Question Number


Scheme

Marks

7.

(a) T

A: 3mg sin 30 – T = 3m.g

3mg

T = mg


(b) T R



F: R(perp): R = mg cos 30

mg

R(//): Tmg sin 30 – F = m.g


Using F = μR

 μ = 0.693 or 0.69 or
(c)

T T Magn of force on pulley = 2T cos 60 = mg
Direction is vertically downwards


M1 A1
A1

(3)
M1 A1
M1 A2, 1, 0
M1


M1

A1



(8)
M1 A1 
B1 (cso)

(3)


14









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