1.	(a)		s = ut + 	M1 A1
		a = 0.6 (m s–2)	A1	3
 
(b)

		 R = 0.8g cos30°(≈ 6.79)	B1
		Use of F = μR	B1
	 0.8g sin 30° – μR = 0.8 × a	M1 A1
	(0.8g sin 30° – μ 0.8g cos30° = 0.8× 0.6)
		μ ≈ 0.51	accept 0.507	A1	5
 
(c)

	↑	R cos30° = μR cos60° + 0.8g	M1 A2 (1,0)
		(R ≈ 12.8)
		→	X = R sin 30°+μR sin 60°	M1 A1
	Solving for X,	X ≈ 12	accept 12.0	DM1 A1	7
Alternative
	 R = X sin 30° + 0.8× 9.8sin 60°	M1 A2 (1,0)
	 μR + 0.8g cos60° = X cos30°	M1 A1
		
	Solving for X,	X ≈12	accept 12.0	DM1 A1
[15]

2.		0.5g sinθ – F = 0.5a	M1 A1 A1
		 seen	B1
		R = 0.5g cosθ	M1 A1
	Use of sinθ= or cosθ=  or decimal equiv or
decimal angle e.g 53.1° or 53°	B1
		 5.88
	m s–2 or 5.9 m s–2	DM1 A1
[9]

3.	(a)

	M1 A1
angle is 153.4°	A1	3
 
(b)		(4 + p)i + (q – 5)j	B1
	(q – 5) = –2(4 + p)	M1 A1
	2p + q + 3 = 0 * 	A1	4
 
(c)			B1
		 R = 2i – 4j	M1
		 	M1 A1 f.t.
			M1 A1 f.t.
		 	A1 cao	7
[14]

4.	(a)	

	R = 0.5g cos α = 0.4g	M1 A1
	4 = F + 0.5g sin α	M1 A1
	F = μR used	M1
	4 = 0.4g.μ + 0.3g
	Þ μ » 0.27(0)	M1 A1	7
1st two M1’s require correct number of the
correct terms,with valid attempt to resolve
the correct relevant term (valid ‘resolve’ = x sin / cos).
4th M1 (dept) for forming equn in μ, + numbers only
 
(b)	

	0.5a = 0.3g - 0.27 × 0.4g	M1 A2,1,0ft
	Þ a » (+) 3.76 m s-2 (or 3.8)	A1	4
In first equn, allow their R or F in the equation for full marks.
A marks: f.t. on their R, F etc. Deduct one A mark (up to 2) for each wrong term.
[11]

5.	

(a)	R( perp to plane):
P sin 30 + 10 cos 30 = 18	M1 A1
	Solve:     P ≈ 18.7 N	M1 A1	4

(b)	R( // plane):
P cos 30 = 10 sin 30 + F	M1 A1
F = 18m used	M1
	Sub and solve:    m = 0.621 or 0.62	M1 A1	5
 
(c)	Normal reaction now = 10 cos 30	M1 A1
	Component of weight down plane = 10 sin 30   (= 5 N)   (seen)	B1
	Fmax = mRnew ≈ 5.37 N    (AWRT 5.4)	M1
	     5.37 > 5 Þ does not slide	A1 cso	5
[14]

6.	(a)	

R (perp to plane): R = 2g cos 20	M1 A1
» 18.4 or 18 N	A1	3
 
(b)	R (// to plane): 18 – 2g sin 20 – F = 2a	M1	A1
F = 0.6 R used		B1
	¯
Sub and solve: a = 0.123 or 0.12 m s–2	M1	A1	5
[8]

7.	(a)	
R = 400g cos 15° (» 3786 N)	B1
F = 0.2R used	B1
T + 0.2R = 400g sin 15°	M1 A1
	¯
T » 257 or 260 N	M1 A1	6

(b)	400g sin 15° – 0.2 × 400g cos 15° = 400a	M1 A1
a = 0.643(…)	A1
50 =  × 0.643 × t2	M1 A1f.t.
t = 12.5 or 12 s	A1	6
[12]

8.	(a)	
R R = mg cos 30	B1
R ma = mg sin 30 – F	M1 A1
F = 0.4 R used	B1
	¯
Eliminate R    ma = mg sin 30 – 0.4 mg cos 30	M1
	¯
Solve: a = 4.9 – 0.4 × 9.8 × /2	M1
» 1.5 or 1.51 m s–2	A1	7
 
(b)	v2 = 2 × 1.51 × 3 Þ v = 3 or 3.01 m s–1	M1 A1	2
 
(c)	1.5/1.51 m s–2 (same as (a))	B1 ft	1
[10]

9.	(a)	
R(↖): R = 3g cos 30° (= 25.46 N)	M1 A1
F = 0.4R » 10.2 N (accept 10 N)	M1 A1	4
 
(b)	R(↗): -F + 3g sin 30° = 3a	M1 A2
(-1 eeoo)
	Þ a » 8.3 m s-2	M1 A1
“v2 = u2 + 2as” : 62 = 2 × a × s	M1
Þ s » 2.17 m (accept 2.2 m)	A1	7
[11]

10.	(a)	

	R() N  + 24 cos 60°  =  2g cos 30°	M1 A1 A1
	Þ N = 16.97 – 12 = 4.97 N	M1 A1
	Þ F = 0.4 . 4.97 = 1.99 N	M1 A1
	R() 2a  =  24 cos 30°  –  2g cos 60°  –  1.99	A1	8
 
(b)	Þ  a  »  4.5 m s–2

	R()	N¢ = 2g cos 30° = 16.97	M1 A1
	Þ F¢max  =  0.4 . 16.97 = 6.79 N
	Component of weight down plane = 2g sin 30° = 9.8 N	M1
(c)	9.8 > F¢max Þ  net force down plane Þ  parcel moves	A1	4
	2f  =  9.8 – 6.79,  Þ    f  »  1.5 m s–2	M1 A1, A1
[12]