M2 MS Jun 2005

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