Mark Scheme (Results) January 2008
GCE
GCE Mathematics (6678/01)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
January 2008 6678 Mechanics M2 Mark Scheme Question Number 1.
Scheme (a)
KE lost is
(b)
Work energy
Marks
1 × 2.5 × 82 = 80 ( J ) 2 80 = R × 20 R=4
M1 A1
(2)
ft their (a)
M1 A1 ft A1 (3) [5]
ft their a
M1 A1ft (3) A1
Alternative to (b) 02 = 82 − 2 × a × 20 ⇒ a = ( − )1.6 R = 2.5 × 1.6 =4
N2L
2.
(a)
p& = ( 6t − 6 ) i + ( 9t 2 − 4 ) j
(ms ) −1
M1 A1
9t 2 − 4 = 0 t = 23
(b)
(c) (+/-)
M1 DM1 A1
t = 1 ⇒ p& = 5 j 2i − 6 j = 0.5 ( v − 5 j) v = 4i − 7 j
(2)
(m s ) −1
ft their p&
(3)
B1ft M1 M1 A1
(4) [9]
Question Number
3.
Scheme 20 000 = 16 F
(a)
( F = 1250 )
F = 550 + 1000 × 9.8sin θ 1 ¿ Leading to sin θ = 14
Ê
(b)
Marks
N2L Ê
ft their F cso
550 + 1000 × 9.8 × sin θ = 1000a ( 550 + 1000 × 9.8 × 141 = 1000a ) or 1250 = 1000a ( a = ( − )1.25)
M1 A1 M1 A1ft A1 (5)
M1 A1
v 2 = u 2 + 2as ⇒ 162 = 2 × 1.25 × y M1 y ≈ 102 accept 102.4, 100 A1 Alternative to (b) Work-Energy
4.
(a) Mass ratio
x y
1 2
(4) [9]
×1000 ×162 − 1000 × 9.8 × 141 y = 550 y M1 M1 A1 y ≈ 102 accept 102.4, 100 A1 (4)
Triangle Circle S 9π 126 − 9π 126 (28.3) (97.7) 7 5 x 4 5 y
B1 B1ft 4, 7 seen B1
126 × 7 = 9π × 5 + (126 − 9π ) × x ft their table values x ≈ 7.58 (
882 − 45π ) 126 − 9π
awrt 7.6
M1 A1ft A1
126 × 4 = 9π × 5 + (126 − 9π ) × y ft their table values M1 A1ft 504 − 45π y ≈ 3.71 ( ) awrt 3.7 A1 (9) 126 − 9π
(b)
y 21 − x θ ≈ 15°
tan θ =
ft their x , y
M1 A1ft A1
(3) [12]
Question Number
5.
Scheme
(a)
N
Marks B
2a 30°
a
mg
R 3mg
a A
Μ(A)
Fr
N × 4a cos 30° = 3mg × a sin 30° + mg × 2a sin 30° 5 5 mg = 7.07…m) N = mg tan 30° ( = 4 4 3 → Fr = N , ↑ R = 4mg Using Fr = µ R 5 mg = µ R for their R 4√3 5 µ= awrt 0.18 16 √ 3
M1 A2(1,0) DM1 A1 B1, B1 B1 M1 A1 (10)
[10] Alternative method: M(B): mg × 2a sin 30 + 3mg × 3a sin 30 + F × 4a cos 30 = R × 4a sin 30 11mga sin 30 + F × 4a cos 30 = R × 4a sin 30 11mg 4 3 +F = 2R 2 2 ↑ R = 4mg , Using Fr = µ R 8µ 3 =
5 , 2
M1A3(2,1,0) DM1A1 B1 B1
µ=
5 16 √ 3
M1 A1
6.
(a) ↑
→ 30 = 2ut −47.5 = 5ut − 4.9t 2 −47.5 = 75 − 4.9t 2 75 + 47.5 t2 = ( = 25) 4.9
eliminating u or t
DM1
t =5 ¿
cso
(b)
30 = 2ut ⇒ 30 = 10u ⇒ u = 3
(c)
↑ →
y& = 5u − 9.8t = −34 x& = 2u = 6 v 2 = 62 + ( −34 ) v ≈ 34.5
2 −1
A1 (6) M1 A1 (2)
M1 requires both x& and y&
(ms )
B1 M1 A1 DM1
accept 35
M1 A1 A1
DM1 A1 (5)
[13] Alternative to (c) 1 2
mvB2 − 12 mv A2 = m × g × 47.5 with
v A2 = 62 + 152 = 261
vB2 = 261 + 2 × 9.8 × 47.5 ( = 1192 )
vB ≈ 34.5
(ms ) −1
accept 35
BEWARE : Watch out for incorrect use of v 2 = u 2 + 2as
M1 A(2,1,0) DM1 A1 (5)
Question Number
7.
Scheme
(a)
Marks
2u
u
2m
3m
x
y
4mu + 3mu = 2mx + 3my y − x = 12 u
LM NEL
M1 A1 B1
Solving to y = 8 u ¿ 5
cso
x = 11 u
(b) Energy loss
1 × 2m 2
(( 2u ) − ( 2
10
11 10
= (c)
8 5
or equivalent
)
(
u ) + 12 × 3m u 2 − ( 85 u ) 2
2
)
9 mu 2 20
M1 A1
(5)
B1 M1 A(2,1,0) A1
(5)
u
3m
m
s 24 mu 5
LM
t = 3ms + mt
M1 A1
t − s = 85 eu
NEL
Solving to s = For a further collision
11 10 u
2 u 5
B1
(3 − e)
M1 A1
> 52 u ( 3 − e )
e>
1 4
M1 ignore e ≤ 1
A1
(7) [17]