Mark Scheme (Results) Summer 2007
GCE
GCE Mathematics Mechanics M1 6677
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June 2007 6677 Mechanics M1 Mark Scheme Question Number
Scheme (a)
1.
T ≈ 35.1
Marks
→ T sin 20° = 12
M1 A1
(N ) awrt 35
A1
(3)
T 20° 12
↑ W = T cos 20° ≈ 33.0
(b) W
(N )
M1 A1 DM1 A1
(4)
awrt 33 [7]
2.
4 ms
−1
m
0.3
2 ms
(a)
A:
−1
2 ms
I = 0.3 ( 8 + 2 ) =3
(b)
LM
Alternative to (b)
−1
( Ns )
0.3 × 8 − 4m = 0.3 × ( −2 ) + 2m
B:
M1 A1 A1 M1 A1
m = 0.5
DM1 A1
m ( 4 + 2) = 3 m = 0.5
M1 A1 DM1 A1
The two parts of this question may be done in either order.
(3)
(4) [7]
(4)
Question Number 3.
Scheme (a)
(
)
(
Marks
M(C) 8g × 0.9 − 0.75 = mg 1.5 − 0.9
)
M1 A1
Solving to m = 2 ¿
cso
DM1 A1
(4)
(b) A
D
B
x
5g M(D)
8g
2g
5 g × x = 8 g × ( 0.75 − x ) + 2 g (1.5 − x ) Solving to x = 0.6
4.
M1 A2(1, 0) DM1 A1 (5) [9]
(AD = 0.6 m)
(a) v
2 horizontal
lines Joined by straight line sloping down 25, 10, 18, 30 oe
B1 B1 B1
(3)
25
O (b)
10
18
30
t
25 × 10 + 12 ( 25 + V ) × 8 + 12 × V = 526
M1 A1 A1
Solving to V = 11 (c)
" v = u + at " ⇒ 11 = 25 − 8a
a = 1.75
(m s ) −2
DM1 A1 ft their V
(5)
M1 A1ft A1
(3) [11]
Question Number 5.
Scheme (a)
Marks
R 1.2 40° F 0.25g ↑
± R + 1.2sin 40° = 0.25g Solving to R = 1.7
(b)
→
( N)
F = 1.2 cos 40°
Use of F = µ R 1.2 cos 40° = µ R
µ ≈ 0.55
M1 A1 accept 1.68
( ≈ 0.919 )
DM1 A1
(4)
M1 A1 ft their R accept 0.548
B1 DM1 A1ft A1 cao (6) [10]
Question Number 6.
Scheme (a) s = ut + 12 at 2
⇒ 3.15 = 12 a × a = 2.8
(b)
Marks
9 4
M1 A1
(m s ) ¿ −2
0.5 g − T = 0.5 × 2.8
N2L for P:
T = 3.5
(d) The acceleration of P is equal to the acceleration of Q. v = u + at ⇒ v = 2.8 × 1.5
( or
(3)
M1 A1
( N)
T − mg = 2.8m 3.5 5 ¿ = m= 12.6 18
(c) N2L for Q:
(e)
cso A1
A1
(3)
M1 A1 cso
DM1 A1
(4)
B1
(1)
M1 A1
v = u + 2as ⇒ v = 2 × 2.8 × 3.15 ) 2
2
(v
2
2
= 17.64, v = 4.2 )
v = u + at ⇒ 4.2 = −4.2 + 9.8t 6 t = , 0.86, 0.857 (s) 7
DM1 A1 DM1 A1
(6) [17]
Question Number 7.
Scheme
v=
(a)
Marks
(
)or any equivalent
8i + 11j − 3i − 4j
v = 2i + 6j
2.5
A1
b = 3i − 4j + vt ft their v
(b)
(
)
A1cao
i component: −9 + 6t = 3 + 2t t=3
(
)
(
M1 A1
( ))
or vC = √ 62 + −2
(3)
M1 M1 A1
j component: 20 + 3λ = −4 + 18 λ = −2 (d) v B = √ 22 + 62
(3)
M1 A1 ft
= 3i − 4j + 2i + 6j t
(c)
M1 A1
(5)
2
M1
Both correct The speeds of B and C are the same
A1 cso
A1
(3) [14]