1.	(a)	For whole system: 1200 – 400 – 200 = 1000a	M1 A1
		a = 0.6 m s^–2	A1	3
 
(b)	For trailer: T – 200 = 200 × 0.6	M1 A1 ft
		T = 320 N	A1
		OR:
	OR:	For car: 1200 – 400 – T = 800 × 0.6	M1 A1 ft
		T = 320 N	A1	3
 
(c)	For trailer: 200 + 100 = 200f or –200f	M1 A1
		f = 1.5 m s^–2 (–1.5)	A1
	For car: 400 + F – 100 = 800f or –800f	M1 A2
		F = 900	A1	7
	(N.B. For both: 400 + 200 + F = 1000f)
[13]

2.	(a)

s = ut + at^2 ⇒ 6 = a × 9	M1
a = 1 (m s^–2)	A1	2
 
(b)	N2L for system 30 – μ5g = 5a	ft their a, accept symbol	M1A1ft
 or awrt 0.48		DM1A1	4
 
(c)	N2L for P T – μ2g = 2a	ft their μ, their a, accept symbols	M1A1ft

Leading to T = 12 (n)	awrt 12	DM1A1	4
 
	Alternatively
N2L for Q
30 – T – μ3g = 3a		M1A1
Leading to T = 12 (n)	awrt 12	DM1A1

(d)	The acceleration of P and Q (or the whole of the system)
is the same.	B1	1
 
(e)	v = u + at ⇒ v = 		B1ft on a
N2L (for system or either particle)
–5μg = 5a	or equivalent	M1
a = –μg
v = u + at ⇒ 0 = 4 – μgt		DM1
Leading to t =  (s)	accept 0.86, 0.857	A1	4
[15]

3.	(a)	a =  =3i – 15j	M1A1	2
 
(b)	N2L F = ma = 6i – 3j	ft their a	M1A1
⎪F⎪ =√(6^2 + 3^2) ≈ 6.71 (N)	accept √45, awrt 6.7	M1A1	4
 
(c)	v_6 = (3i + 2j) + (3i – 1.5j)6 	ft their a	M1A1ft
= 21i – 7j (m s^–1)		A1	1
[9]

4.	(a)

Use of F = µR		B1
 P cos 20° = µR		M1A1
¡ R + Psin20° = 30g		M1 A1
P cos 20° = µ(30g – P sin 20°)		M1
P = 		M1
≈ 110 (N)	accept 109	A1	8

(b)	i	R + 150 sin 20° = 30 g		M1A1
	(R ≈ 242.7)
N2L 	150 cos 20° – μR = 30a		M1A1
a ≈ 		M1
= 1.5 (m s^–2)	accept 1.46	A1	6
[14]

5.	(a)	16^2  =  20^2  – 2 × a × 24  ⇒  a  =  3 m s^–2	M1 A1	2
(b)	v^2  =  20^2  –  2 × 3 × 30	M1 A1ft
	v  =  √220 or  14.8 m s^–1	A1	3
(c)	0.3  =  m × 3 ⇒  m  =  0.1 kg (*)	M1 A1	2
 
(d)	0.1(w + √220)  =  2.4	M1 A1ft
	w  =  9.17	A1
		↓
	0  =  9,17 – 3 × t	M1 A1ft
	t  ≈ 3.06 s	A1	6
[13]

6.	(a)	
R(↑) R + 150 sin 20 = 30g	M1 A1
⇒ R ≈ 243 N	A1	3
 
(b)	R(→): 150 cos 20 – 0.2R = 30a	M1 A1
⇒ a ≈ 3.08 m s^–2	A1	3

(c)	
S = 30g ⇒ F = 0.2 × 30g	M1 A1
30a’ = (–) 0.2 × 30g ⇒ a’ = (–) 0.2g (= 1.96)	M1 A1
0 = 12^2 – 2 × 0.2g × s (using new a’)	M1
⇒ s ≈ 36.7 m	A1	6
[12]

7.	(a)	a = [–14i + 21j – (6i – 27j)] ÷ 4	M1 A1
= (–5i + 12j) m s^–2	A1	3
 
(b)	|a| = √(5^2 + 12^2) = 13	M1
|F| = m|a| = 0.4 × 13 = 5.2 N	M1 A1	3
[6]
	Alt (b)
	F = 0.4(5i + 12j) = 2i + 4.8j	M1
	|F| = √(2^2 + 4.8^2) = 5.2 N	M1 A1	3

El Alsson British International School