GCE Edexcel GCE Mechanics M2 (6678)
Summer 2005
Mechanics M2 (6678)
Edexcel GCE
Mark Scheme (Results)
June 2005 6678 Mechanics M2 Mark Scheme Question Number
Scheme
1 (a)
Driving force =
Marks
P v
B1
21000 = 600 ⇒ v = 35 m s–1 v
M1 A1 (3)
P 1 = 600 + 1200.g. v 14
(b)
M1 A1
( = 1440 N)
21000 21000 = 1440 ⇒ v = ≈ 14.6 or 15 m s–1 v 1440
M1 A1 (4)
2 (a)
B
4
x
C 5
A
4
(x = 3)
D
M(AB): 7 × 3.5 + 5 × 5.5 + 4 × 2 = 20 × x ⇒ 20 x = 24.5 + 27.5 + 8 = 60 ⇒ x = 3 cm
M1 A2,1,0 dep
M1 A1 (5)
(b)
X
3.5
kM
M(XY):
M × (3.5 − 3) = kM × 3.5 ⇒ k = 17 .
M
M1 A1 √ A1 (3)
Y
6678 Mechanics June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
v = (18 – 12t2)i + 2ctj
3 (a)
t=
3 2
M1 A1 A1
: v = –9i + 3cj
M1
⏐v⏐ = 15 ⇒ 92 + (3c)2 = 152
M1
⇒ (3c)2 = 144 ⇒ c = 4
A1 (6)
(b)
t = 32 :
a = – 24ti + 8j
M1
a = –36i + 8j
M1 A1 √
(3)
4 (a) → 12.6t = x
12.6 0.1
B1
0.1 = 4.9 t2 x
B1
x2 ⇒ 0 .1 = 4 .9 × 12.6 2
M1
⇒ x = 1.8 m
A1 (4)
(b)
→ u cos α .t = 2.5
u
↑ u sin α .t =
)α
1 2
gt
2
M1 A1 M1 A1
2.5
24 t = 2.5 25 7 2.5.25 u. = 4.9. 25 24u 4.9 × 2.5 × 25 2 2 u = 7 × 24 ⇒ u ≈ 6.75 or 6.8 m s–1 u.
M1 A1 (6)
6678 Mechanics June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
5 (a)
2u
2u
3m
2m v
2u
6mu − 4mu = 3mv + 4mu ⇒ v = − 23 u 2u − v = e.4u ⇒ 4eu = 83 u ⇒ e = 23 .
CLM: NLI:
M1 A1 A1 M1 A1 M1 A1 (7)
(b)
2u
0
2m
5m
x
y
5my + 2mx = 4mu y − x = 53 .2u = 65 u
M1 A1 A1
x = − 72 u
Solve:
2 u 7
M1 A1
〈 23 u so B does not overtake A
M1
So no more collisions
A1 cso (7)
6 (a)
Y X
0.5 P
M(A):
P × 0.5 sin 60 = 30 g × 1.5
M1 A2
P = 90 g.
A1
30g
≈ 1020 N (1000N)
(4) M1 A1
X = P cos 60 = P
→
(b)
2 3
1 2
( ≈ 509 N (510N) )
Y + P cos 30 = 30 g
↑
M1 A1
(⇒ Y = –588 N) resultant =
( X + Y ) = (509 2 + 588 2 ) ≈ 778 N 2
M1 A1
2
or 780N
(6)
(c) In equilibrium all forces act through a point M1 P and weight meet at mid-point; hence reaction also acts through mid-point so reaction horizontal A1 cso (2) OR M(mid-point): Y x 1.5 = 0 ⇒ Y = 0 Hence reaction is horizontal 6678 Mechanics June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
M1 A1
7 (a) PE lost = 3 × g × 8 sin 30 = 3 × g × 8 × 0.5 = 117.6 J ≈ 118J or 120J (b)
KE gained = Work-energy:
(c)
1 2
M1 A1 (2)
× 3 × 5 2 = 37.5 J
F × 8 = 117.6 − 37.5 = 80.1 ⇒ F = 10.0125 ≈ 10 N
M1 A1 M1 A1√ A1
`
(5)
R = 3g cos 30 (= 25.46 N) 10 F = µR ⇒ µ = ≈ 0.393 or 0.39 25.46
B1 M1 A1 (3)
(d)
Work done by friction = 80.1 as before Work-energy: 12 × 3 × v 2 = 12 × 3 × 2 2 + 117.6 − 80.1 ⇒ v ≈ 5.39 or 5.4 m s–1
6678 Mechanics June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
M1 M1 A2,1,0√ A1 (5)
6678 Mechanics June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics