Save My Exams! – The Home of Revision
For more awesome GCSE and A level resources, visit us at www.savemyexams.co.uk/
Resultants and
Components of Forces
Mark Scheme 3
Level
A Level
Subject
Maths
Exam Board
OCR
Module
Mechanics 1
Topic
Force as a Vector
Sub Topic
Resultants and Components of Forces
Booklet
Mark Scheme- 3
60 minutes
Time Allowed:
Score:
/51
Percentage:
/100
Grade Boundaries:
A*
>85%
A
77.5%
B
C
D
E
U
70%
62.5%
57.5%
45%
<45%
Save My Exams! – The Home of Revision
For more awesome GCSE and A level resources, visit us at www.savemyexams.co.uk/
1
E − 400 = 6000 × 2
B1
M1
A1t
A1
Hence E = 1600
2
EITHER: R cosθ = 8 + 5cos 30°
R sin θ = 5sin 30°
Hence R 2 = (12.33...)2 + 2.52
R = 12.6
2.5
tan θ =
12.33...
θ = 11.5
OR:
Triangle of forces has 5, 8, R and 150°
R 2 = 82 + 52 − 2 × 5 × 8 × cos150°
Hence R = 12.6
5sin150°
sin θ =
= 0.1987...
12.58...
Hence θ = 11.5
3
(i)(a)
V = Pcos20o – 0.04g
R = Psin20o
M1
Magnitude is 0.143 N
A1ft
(i)(c)
0.143 = 0.04a
Acceleration is 3.57 ms-2
M1
A1ft
(ii)
R2 = 0.082 + (0.04g)2
Magnitude is 0.400 N (or 0.40 or
0.4 )
tan θ = +/-0.04g/0.08 or
tan(90 o - θ ) = +/-0.08/0.04g
M1
A1
Angle made with horizontal is
78.5o or 1.37 radians, or
angle made with vertical is 11.5 o
or 0.201 radians
Downwards or below
horizontal
For attempt at resolving || or ⊥ to 8 N force
For one completely correct equation
For a second correct equation
M1
A1t
For correct method for either unknown
For correct value
M1
For correct method for second unknown
A1t
For correct value
M1
A1
M1
A1
A1t
For considering any triangle with 5, 8, R
For correct triangle drawn or used
For use of cosine formula attempted
For correct expression for R 2
For correct value
M1
For use of sine formula with numerical R
B1
M1
A1
(i)(b)
M1
A1
A1
A1t
P = 0.417
For resultant force E − 400 stated or implied
For use of Newton II for the truck
For the correct equation
4 For correct answer 1600
4
7 For correct value
7
For setting V = 0
3
2
2
For using R = horizontal component
of P
ft value of P
For using Newton’s second law
ft magnitude of the resultant
For using R2 = P2 + W2
For using tan θ = Y/X or
tan(90 o - θ ) =X/Y
M1
A1
B1
5
Direction may alternatively be
shown clearly on a diagram or
given as a bearing
Save My Exams! – The Home of Revision
For more awesome GCSE and A level resources, visit us at www.savemyexams.co.uk/
4
M1
(i)
2.5 = 6.5sin θ
θ = 22.6o
A1
A1
M1
R = 6.5cos22.6o
R=6
A1
A1
(ii)
5 (i)
(ii)
i
6 (i)
(ii)
(iii)
X=5
Y = 12
= 52 + 122
Magnitude is 13 N
tan θ = 12/5
Angle is 67.4o
R + Tsin72o = 50g
T = 50g/sin72o
T = 515
T = mg
m = 52.6
X = Tcos72o
X = 159
7
3
[2]
M1
A1
M1
A1
[4]
3
An equation with R, T and 50 in linear combination.
R + 0.951T = 50g
M1
A1
[2]
M1
A1
B1
B1
[4]
B1
Using R = 0 (may be implied) and Tsin72o = 50(g)
Or better
Accept 52.5
Implied by correct
answer
Or better
B1
[2]
M1
(i)
2.5 = 6.5sin θ
θ = 22.6o
A1
A1
M1
R = 6.5cos22.6o
R=6
A1
A1
(ii)
AG Accept verification
For resolving forces in the j
direction or for using
Pythagoras or relevant
trigonometry.
X=-5 B0. Both may be seen/implied in (ii)
No evidence for which value is X or Y available from (ii)
award B1 for the pair of values 5 and 12 irrespective of
order
For using R2 = X2 + Y2
Allow 13 from X=-5
For using correct angle in a trig expression
SR: p=14.9 and Q=11.4 giving R=13+/-0.1 B2,
Angle = 67.5+/-0.5 B2
B1
B1
2
For resolving forces in the i
direction or for relevant use of
trigonometry
For resolving forces in the i
direction or for relevant use of
trigonometry
3
3
AG Accept verification
For resolving forces in the j
direction or for using
Pythagoras or relevant
trigonometry.