Mark Scheme (Results) Summer 2008
GCE
GCE Mathematics (6677/01)
June 2008 6677 Mechanics M1 Final Mark Scheme Question Number
1.
Scheme
Marks
I = mv ⇒ 3 = 0.4 × v
(a)
M1 A1
( )
v = 7.5 ms −1
(b)
0.4
0.6
v
5
0.4 × 7.5 = 0.4v + 0.6 × 5 0 = 0.4v ⇒ v = 0 ¿
cso
(a) v 2 = u 2 + 2as ⇒ 17.52 = u 2 + 2 × 9.8 × 10 Leading to u = 10.5
(b)
(3)
7.5
LM
2.
A1
v = u + at ⇒
17.5 = −10.5 + 9.8T 6 T =2 (s) 7
M1 A1 A1 (3) [6]
M1 A1 A1
M1 A1 f.t. DM1 A1 (4)
Alternatives for (b)
[7]
s=(
OR
(3)
17.5 + −10.5 u+v )T ⇒ 10 = ( )T 2 2 20 =T 7
s = ut + 12 at 2
⇒ − 10 = 10.5t − 4.9t 2
M1A1 f.t. DM1A1 (4)
M1 A1 f.t.
6 ⎛ 5⎞ Leading to T = 2 , ⎜ − ⎟ Rejecting negative DM1 A1 (4) 7 ⎝ 7⎠ (b) can be done independently of (a) s = vt − 12 at 2 ⇒ − 10 = −17.5t + 4.9t 2 M1 A1
6 5 Leading to T = 2 , 7 7 5 For final A1, second solution has to be rejected. leads to a negative u. 7
DM1 A1
(4)
Question Number
3.
Scheme
Marks
8 6 θ ≈ 53°
tan θ =
(a)
(
M1 A1
) (= 2.4i + 3.2j)
F = 0.4 6i + 8j
(b)
(
(2)
M1
)
F = √ 2.4 2 + 3.2 2 = 4
M1 A1
(3)
The method marks can be gained in either order.
(
)
v = 9i − 10j + 5 6i + 8j
(c)
( )
M1 A1
= 39i + 30 j ms −1
A1
(3) [8]
4.
(a)
v 25 shape 25, 10, 30, 90
B1 B1
(2)
10
O (b) 30 × 25 +
90 t
30
1 25 + 10 t + 10 60 − t = 1410 2 7.5t = 60 t=8 s
(
)
(
)
M1 A1 A1
()
a=
DM1 A1
25 − 10 = 1.875 ms −2 8
( )
1 78
M1 A1
(7) [9]
Question Number
5.
Scheme
Marks
(a) 15
R 30°
(↑)
50°
X
15sin 30° = R sin50°
M1 A1
( )
R ≈ 9.79 N
DM1 A1 (4)
(→) X − 15cos30° = R cos50° X ≈ 19.3 (N )
(b)
ft their R
M1 A2 ft DM1 A1 (5) [9]
Alternatives using sine rule in (a) or (b); cosine rule in (b) (a) R
15
50°
30°
R 15 = sin30° sin50°
M1 A1
( )
R ≈ 9.79 N
(b)
X 15 R = = sin100° sin50° sin 30°
DM1 A1 (4)
M1 A2 ft on R
X
( )
X ≈ 19.3 N
DM1 A1 (5)
X 2 = R 2 + 152 − 2 x 15 x Rcos100o OR: cosine rule; any of R 2 = X 2 + 152 − 2 x 15 x X cos30o
M1 A2 ft on R
152 = R 2 + X 2 − 2 x X x Rcos50o
( )
X ≈ 19.3 N
DM1 A1 (5)
Question Number
6.
Scheme
Marks
(a) X 2.4 0.8 A
B 8g
()
12g
8g × 0.8 + 12g × 1.2 = X × 2.4
M A
( )
X ≈ 85 N
M1 A1 accept 84.9,
26g 3
DM1 A1 (4)
(b) X + 10
X 2.4 0.8 x
A
B 8g
12g
( ) (X + 10)+ X = 8g + 12g (X = 93)
R ↑
()
M A
M1 B1 A1
8g × 0.8 + 12g × x = X × 2.4
( )
x = 1.4 m
M1 A1 accept 1.36
A1
(6) [10]
Question Number
7.
Scheme
(a)
Marks
R 45 N
µR
50°
4g
30° R = 45cos 40° + 4g cos30° R ≈ 68 (b)
accept 68.4
Use of F = µ R
M1 M1 A2 (1, 0)
F + 4g sin 30 = 45 cos 50° Leading to µ ≈ 0.14
M1 A2 (1, 0) DM1 A1 (5)
accept 0.136
DM1 A1 (6) [11]
Question Number
8.
Scheme
Marks
(a) T
T
µ 2g
30
µ 3g
s = ut + 12 at 2
⇒ 6 = 12 a × 9
M1
( )
a = 1 13 ms −2
30 − µ5g = 5a
(b) N2L for system symbol
µ= (c)
T−
ft their a, accept
14 10 = 3g 21
awrt 0.48 DM1 A1 (4) M1 A1 ft
14 4 × 2g = 2 × 3g 3
( )
Leading to T = 12 N Alternatively
or
(2)
M1 A1ft
ft their µ , their a, accept symbols
T − µ 2 g = 2a
N2L for P
A1
N2L for Q 30 − T − µ 3g = 3a
( )
Leading to T = 12 N
awrt 12 DM1 A1 (4)
M1 A1 awrt 12 DM1 A1
(d) The acceleration of P and Q (or the whole of the system) is the same. 4 ×3= 4 3 N2L (for system or either particle) −5µ g = 5a a = −µg v = u + at ⇒ 0 = 4 − µ gt (e) v = u + at ⇒ v =
Leading to t =
6 s 7
()
B1
(1)
B1 ft on a or equivalent
M1 DM1
accept 0.86, 0.857
A1
(4) [15]