# GCE Edexcel GCE Mechanics M2 (6678)

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```					                                GCE
Edexcel GCE
Mechanics M2 (6678)

Summer 2005

Mark Scheme (Results)
Mechanics M2 (6678)
Edexcel GCE
June 2005
6678 Mechanics M2
Mark Scheme

Question                                                  Scheme                                         Marks
Number

P
1 (a)         Driving force =                                                 B1
v
21000
= 600 ⇒ v = 35 m s–1                                 M1 A1
v
(3)
P                  1
(b)         = 600 + 1200.g.                                           M1 A1
v                 14
( = 1440 N)
21000              21000
= 1440 ⇒ v =       ≈ 14.6 or 15 m s–1         M1 A1
v                1440
(4)

2 (a)   B     4               x       C

5             (x = 3)

A          4          D

M(AB): 7 × 3.5 + 5 × 5.5 + 4 × 2 = 20 × x                     M1 A2,1,0

⇒ 20 x = 24.5 + 27.5 + 8 = 60 ⇒ x = 3 cm                dep      M1 A1
(5)

(b)               X     3.5           kM   M(XY):

M × (3.5 − 3) = kM × 3.5           M1 A1 √
M                            ⇒ k = 1.
7
A1
(3)
Y

6678 Mechanics
3 (a)              v = (18 – 12t2)i + 2ctj                                                       M1 A1 A1

t=   3
2
: v = –9i + 3cj                                                         M1

⏐v⏐ = 15 ⇒ 92 + (3c)2 = 152                                            M1

⇒ (3c)2 = 144 ⇒ c = 4                                       A1
(6)

(b)                                a = – 24ti + 8j                                  M1

t=3:
2
a = –36i + 8j                                    M1
A1 √
(3)

4 (a)
12.6                            → 12.6t = x                            B1

0.1                                                         0.1 = 4.9 t2                         B1

x2
x              ⇒ 0 .1 = 4 .9 ×                                  M1
12.6 2

⇒ x = 1.8 m                                      A1
(4)

(b)                 u                                        → u cos α .t = 2.5                     M1 A1
)α                                              ↑ u sin α .t =   1
2
gt   2
M1 A1
2.5
24
u.   t = 2.5
25
7         2.5.25
u. = 4.9.
25           24u
4.9 × 2.5 × 25 2
u =
2
7 × 24
⇒ u ≈ 6.75 or 6.8 m s–1                          M1 A1
(6)

6678 Mechanics
5 (a)                  2u                     2u

3m                    2m

v             2u

CLM:               6mu − 4mu = 3mv + 4mu                                                         M1 A1
⇒ v = −2u3
A1
NLI:               2u − v = e.4u                                                                 M1 A1
⇒ 4eu = 8 u ⇒ e = 2 .
3       3
M1 A1
(7)
(b)                                     2u               0

2m               5m

x               y
5my + 2mx = 4mu                                                               M1 A1
y − x = 5 .2u = 6 u
3
5
A1
Solve:             x = −7u
2
M1 A1
2
7
u      〈 2 u so B does not overtake A
3
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
30g
P = 90 g.   2
3
≈ 1020 N (1000N)          A1
(4)
(b)                              →            X = P cos 60 = P  1
2
M1 A1
( ≈ 509 N (510N) )

↑            Y + P cos 30 = 30 g                       M1 A1
(⇒ Y = –588 N)
resultant =       ( X + Y ) = (509 2 + 588 2 ) ≈ 778 N
2        2
M1 A1
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                                             M1

Hence reaction is horizontal                                               A1

6678 Mechanics
7 (a) PE lost = 3 × g × 8 sin 30 = 3 × g × 8 × 0.5 = 117.6 J ≈ 118J                 M1 A1
or 120J          (2)

(b)      KE gained =   1
2
× 3 × 5 2 = 37.5 J                                     M1 A1
Work-energy:     F × 8 = 117.6 − 37.5 = 80.1                               M1 A1√
⇒ F = 10.0125 ≈ 10 N                 `                     A1
(5)
(c)            R = 3g cos 30 (= 25.46 N)                                          B1
10
F = µR ⇒ µ =           ≈ 0.393 or 0.39                             M1 A1
25.46
(3)
(d)          Work done by friction = 80.1 as before                        M1
Work-energy: 1 × 3 × v 2 = 1 × 3 × 2 2 + 117.6 − 80.1
2              2
M1 A2,1,0√
⇒ v ≈ 5.39 or 5.4 m s–1                  A1
(5)

6678 Mechanics