M2 CENTRE OF MASS PAST QUESTIONS
1.
A triangular frame is formed by cutting a uniform rod into 3 pieces which are then joined to
form a triangle ABC, where AB = AC = 10 cm and BC = 12 cm, as shown in the diagram above.
(a)
Find the distance of the centre of mass of the frame from BC.
(5)
The frame has total mass M. A particle of mass M is attached to the frame at the mid-point of
BC. The frame is then freely suspended from B and hangs in equilibrium.
(b)
Find the size of the angle between BC and the vertical.
(4)
(Total 9 marks)
City of London Academy
1
2.
[The centre of mass of a semi-circular lamina of radius r is
4r
from the centre]
3
A template T consists of a uniform plane lamina PQROS, as shown in the diagram above. The
lamina is bounded by two semicircles, with diameters SO and OR, and by the sides SP, PQ and
QR of the rectangle PQRS. The point O is the mid-point of SR, PQ = 12 cm and QR = 2x cm.
4 2x 2  3
(a)
Show that the centre of mass of T is a distance
8 x  3
cm from SR.
(7)
The template T is freely suspended from the point P and hangs in equilibrium.
Given that x = 2 and that θ is the angle that PQ makes with the horizontal,
(b)
show that tan θ =
48  9
.
22  6
(4)
(Total 11 marks)
City of London Academy
2
3.
A shop sign ABCDEFG is modelled as a uniform lamina, as illustrated in the diagram above.
ABCD is a rectangle with BC = 120 cm and DC = 90 cm. The shape EFG is an isosceles triangle
with EG = 60 cm and height 60 cm. The mid-point of AD and the mid-point of EG coincide.
(a)
Find the distance of the centre of mass of the sign from the side AD.
(5)
The sign is freely suspended from A and hangs at rest.
(b)
Find the size of the angle between AB and the vertical.
(4)
(Total 9 marks)
City of London Academy
3
4.
A uniform lamina ABCD is made by joining a uniform triangular lamina ABD to a uniform
semi-circular lamina DBC, of the same material, along the edge BD, as shown in the diagram
above. Triangle ABD is right-angled at D and AD = 18 cm. The semi-circle has diameter BD and
BD = 12 cm.
(a)
Show that, to 3 significant figures, the distance of the centre of mass of the lamina ABCD
from AD is 4.69 cm.
(4)
Given that the centre of mass of a uniform semicircular lamina, radius r, is at a distance
4r
3
from the centre of the bounding diameter,
(b)
find, in cm to 3 significant figures, the distance of the centre of mass of the lamina ABCD
from BD.
(4)
The lamina is freely suspended from B and hangs in equilibrium.
(c)
Find, to the nearest degree, the angle which BD makes with the vertical.
(4)
(Total 12 marks)
City of London Academy
4
5.
y
C (3m)
B (5 m)
5
O
8
A (km )
x
The diagram above shows a rectangular lamina OABC. The coordinates of O, A, B and C are (0,
0), (8, 0), (8, 5) and (0, 5) respectively. Particles of mass km, 5m and 3m are attached to the
lamina at A, B and C respectively.
The x-coordinate of the centre of mass of the three particles without the lamina is 6.4.
(a)
Show that k = 7.
(4)
The lamina OABC is uniform and has mass 12m.
(b)
Find the coordinates of the centre of mass of the combined system consisting of the three
particles and the lamina.
(6)
The combined system is freely suspended from O and hangs at rest.
(c)
Find the angle between OC and the horizontal.
(3)
(Total 13 marks)
City of London Academy
5
6.
A
5cm
3cm
12 cm
O
5cm
B
C
21cm
A set square S is made by removing a circle of centre O and radius 3 cm from a triangular piece
of wood. The piece of wood is modelled as a uniform triangular lamina ABC, with ABC = 90°,
AB = 12 cm and BC = 21 cm. The point O is 5 cm from AB and 5 cm from BC, as shown in the
diagram above.
(a)
Find the distance of the centre of mass of S from
(i)
AB,
(ii)
BC.
(9)
The set square is freely suspended from C and hangs in equilibrium.
(b)
Find, to the nearest degree, the angle between CB and the vertical.
(3)
(Total 12 marks)
City of London Academy
6
7.
A
B
X
a
2a
C
D
a
F
2a
E
A uniform lamina ABCDEF is formed by taking a uniform sheet of card in the form of a square
AXEF, of side 2a, and removing the square BXDC of side a, where B and D are the mid-points
of AX and XE respectively, as shown in the diagram above.
(a)
Find the distance of the centre of mass of the lamina from AF.
(4)
The lamina is freely suspended from A and hangs in equilibrium.
(b)
Find, in degrees to one decimal place, the angle which AF makes with the vertical.
(4)
(Total 8 marks)
City of London Academy
7
8.
A
24 cm
8 cm
X
O
B
The diagram above shows a template T made by removing a circular disc, of centre X and radius
8 cm, from a uniform circular lamina, of centre O and radius 24 cm. The point X lies on the
diameter AOB of the lamina and AX = 16 cm. The centre of mass of T is at the point G.
(a)
Find AG.
(6)
The template T is free to rotate about a smooth fixed horizontal axis, perpendicular to the plane
1
of T, which passes through the mid-point of OB. A small stud of mass m is fixed at B, and T
4
and the stud are in equilibrium with AB horizontal. Modelling the stud as a particle,
(b)
find the mass of T in terms of m.
(4)
(Total 10 marks)
City of London Academy
8
9.
B
C (6m)
A
D (2m)
2a
3a
The figure above shows four uniform rods joined to form a rigid rectangular framework ABCD,
where AB = CD = 2a, and BC = AD = 3a. Each rod has mass m. Particles, of mass 6m and 2m,
are attached to the framework at points C and D respectively.
(a)
Find the distance of the centre of mass of the loaded framework from
(i)
AB,
(ii)
AD.
(7)
The loaded framework is freely suspended from B and hangs in equilibrium.
(b)
Find the angle which BC makes with the vertical.
(3)
(Total 10 marks)
City of London Academy
9
10.
y
A
4m
4
B
O
9
6m
x
4
C
2m
The figure above shows a triangular lamina ABC. The coordinates of A, B and C are (0, 4) (9, 0)
and (0, –4) respectively. Particles of mass 4m, 6m and 2m are attached at A, B and C
respectively.
(a)
Calculate the coordinates of the centre of mass of the three particles, without the lamina.
(4)
The lamina ABC is uniform and of mass km. The centre of mass of the combined system
consisting of the three particles and the lamina has coordinates (4, λ).
(b)
Show that k = 6.
(3)
(c)
Calculate the value of λ.
(2)
The combined system is freely suspended from O and hangs at rest.
(d)
Calculate, in degrees to one decimal place, the angle between AC and the vertical.
(3)
(Total 12 marks)
City of London Academy
10
11.
B
C
4 cm
5 cm
A
4 cm
D
A thin uniform wire, of total length 20 cm, is bent to form a frame. The frame is in the shape of
a trapezium ABCD, where AB = AD = 4 cm, CD = 5 cm and AB is perpendicular to BC and AD,
as shown in the diagram.
(a)
Find the distance of the centre of mass of the frame from AB.
(5)
The frame has mass M. A particle of mass kM is attached to the frame at C. When the frame is
freely suspended from the mid-point of BC, the frame hangs in equilibrium with BC horizontal.
(b)
Find the value of k.
(3)
(Total 8 marks)
12.
20 cm
A
B
5 cm
3 cm
O
10 cm
6 cm
D
City of London Academy
C
11
This diagram shows a metal plate that is made by removing a circle of centre O and radius 3 cm
from a uniform rectangular lamina ABCD, where AB = 20 cm and BC = 10 cm. The point O is 5
cm from both AB and CD and is 6 cm from AD.
(a)
Calculate, to 3 significant figures, the distance of the centre of mass of the plate from AD.
(5)
The plate is freely suspended from A and hangs in equilibrium.
(b)
Calculate, to the nearest degree, the angle between AB and the vertical.
(3)
(Total 8 marks)
13.
A
6 cm
P
M
S
B
C
3 cm
Q
2 cm
R
4 cm
The diagram above shows a decoration which is made by cutting the shape of a simple tree from
a sheet of uniform card. The decoration consists of a triangle ABC and a rectangle PQRS. The
points
P and S lie on BC and M is the mid-point of both BC and PS. The triangle ABC is isosceles with
AB = AC, BC = 4 cm, AM = 6 cm, PS = 2 cm and PQ = 3 cm.
(a)
Find the distance of the centre of mass of the decoration from BC.
(5)
City of London Academy
12
The decoration is suspended from Q and hangs freely.
(b)
Find, in degrees to one decimal place, the angle between PQ and the vertical.
(4)
(Total 9 marks)
14.
A uniform ladder AB, of mass m and length 2a, has one end A on rough horizontal ground. The
coefficient of friction between the ladder and the ground is 0.6. The other end B of the ladder
rests against a smooth vertical wall.
A builder of mass 10m stands at the top of the ladder. To prevent the ladder from slipping, the
builder’s friend pushes the bottom of the ladder horizontally towards the wall with a force of
magnitude P. This force acts in a direction perpendicular to the wall. The ladder rests in
equilibrium in a vertical plane perpendicular to the wall and makes an angle  with the
horizontal, where tan  = 32 .
(a)
Show that the reaction of the wall on the ladder has magnitude 7mg.
(5)
(b)
Find, in terms of m and g, the range of values of P for which the ladder remains in
equilibrium.
(7)
(Total 12 marks)
15.
4m
A
O
B
D
5a
C
m
2a
2m
A loaded plate L is modelled as a uniform rectangular lamina ABCD and three particles. The
sides CD and AD of the lamina have lengths 5a and 2a respectively and the mass of the lamina
is 3m. The three particles have mass 4m, m and 2m and are attached at the points A, B and C
respectively, as shown in the diagram above.
City of London Academy
13
(a)
Show that the distance of the centre of mass of L from AD is 2.25a.
(3)
(b)
Find the distance of the centre of mass of L from AB.
(2)
The point O is the mid-point of AB. The loaded plate L is freely suspended from O and hangs at
rest under gravity.
(c)
Find, to the nearest degree, the size of the angle that AB makes with the horizontal.
(3)
A horizontal force of magnitude P is applied at C in the direction CD. The loaded plate L
remains suspended from O and rests in equilibrium with AB horizontal and C vertically below
B.
(d)
Show that P =
5
4
mg.
(4)
(e)
Find the magnitude of the force on L at O.
(4)
(Total 16 marks)
16.
A
3a
B
2a
D
City of London Academy
C
a
E
14
A uniform lamina ABCD is made by taking a uniform sheet of metal in the form of a rectangle
ABED, with AB  3a and AD  2a , and removing the triangle BCE, where C lies on DE and
CE = a, as shown in the diagram above.
(a)
Find the distance of the centre of mass of the lamina from AD.
(5)
The lamina has mass M. A particle of mass m is attached to the lamina at B. When the loaded
lamina is freely suspended from the mid-point of AB, it hangs in equilibrium with AB
horizontal.
(b)
Find m in terms of M.
(4)
(Total 9 marks)
17.
Three particles of mass 3m, 5m and m are placed at points with coordinates (4, 0), (0, 3) and
(4, 2) respectively. The centre of mass of the system of three particles is at (2, k).
(a)
Show that  = 2.
(4)
(b)
Calculate the value of k.
(3)
(Total 7 marks)
City of London Academy
15
18.
A
8a
B
6a
G
X
E
4a
C
D
The diagram above shows a uniform lamina ABCDE such that ABDE is a rectangle, BC = CD,
AB = 8a and AE = 6a. The point X is the mid-point of BD and XC = 4a. The centre of mass of
the lamina is at G.
(a)
Show that GX =
44
15
a.
(6)
The mass of the lamina is M. A particle of mass M is attached to the lamina at C. The
lamina is suspended from B and hangs freely under gravity with AB horizontal.
(b)
Find the value of .
(3)
(Total 9 marks)
19.
A
E
B
D
M
30 cm
C
40 cm
A uniform plane lamina is in the shape of an isosceles triangle ABC, where AB = AC.
The mid-point of BC is M, AM = 30 cm and BM = 40 cm. The mid-points of AC and AB
are D and E respectively. The triangular portion ADE is removed leaving a uniform plane
lamina BCDE as shown in the diagram above.
(a)
Show that the centre of mass of the lamina BCDE is 6 23 cm from BC.
City of London Academy
16
(6)
The lamina BCDE is freely suspended from D and hangs in equilibrium.
(b)
Find, in degrees to one decimal place, the angle which DE makes with the vertical.
(3)
(Total 9 marks)
-----------------------------------------------------------------------------------------------------------------------1.
(a)
AB
AC
BC
frame
mass ratio
10
10
12
32
B1
dist. from BC
4
4
0
x
B1
Moments about BC:
10  4  10  4  0  32x
80
x
32
x  2 12 (2.5)
M1 A1
A15
(b)
Moments about B:
Mg  6sin   Mg   x cos   6sin  
M1 A1 A1
12sin   x cos
x
tan  
12
City of London Academy
17
θ = 11.768..... = 11.8°
A14
Alternative method :
C of M of loaded frame at distance
1
2
x from D along DA
tan θ =
1
2
x
B1
M1 A1
6
θ = 11.768..... = 11.8°
A1
[9]
2.
(a)
Rectangle
Semicircles
24x
4.5π
4.5π
x
43
3
43
3
Template, T
24x + 9π
B2
x
B2
 43
 4 3 
24x2 – 4.5π × 
 – 4.5π × 
 = (24x + 9π) x
 3 
 3 
M1 A1
distance = x 
(b)
8 x  3 
x
When x = 2,
tanθ =
4 2x 2 – 3
6
4– x


**
20
16  3
6
20
4–
16  3
48  9
.
22  6
A17
B1
M1 A1
A14
[11]
3.
(a)
Ratio of areas triangle:sign:rectangle = 1 : 5 : 6
(1800:9000:10800)
B1
Centre of mass of the triangle is 20cm down from
AD (seen or implied)
B1
 6 × 45 –1× 20 = 5× y
City of London Academy
M1A1
18
(b)
y  50cm
A15
Distance of centre of mass from AB is 60cm
B1
Required angle is tan
60
50
–1
(their values)
M1A1ft
= 50.2° (0.876 rads)
A14
[9]
4.
(a)
B1
MR
108
18π
108 + 18π
x1 (→)
4
6
x
B1
from AD
yi (↓)
6

from BD
8

y
AD(→): 108(4) + 18π (6) = (108 + 18π) x
x
M1
432108
 4.68731... 4.69(cm) (3sf ) AG
10818
A14
(b)
yi (↓)
from BD
6

8


BD  : 108(6) 18 (– 8 )  (10818 ) y
y
504
 3.06292... 3.06(cm)(3sf)
108 18
y
B1 oe
M1
A1ft
A14
(c)
City of London Academy
19
θ = required angle
y
tan 
12 – 4.68731..
3.06392..

12 – 4.68731..
θ = 22.72641... = 23 (nearest degree)
dM1
A1
A14
[12]
5.
(a)
(b)
M(Oy)
(8 + k)m × 6.4 = 5m × 8 + km × 8
1.6k = 11.2  k = 7 *
M(Oy)
27m x = 12m × 4 + 5m × 8 + 7m × 8
16
x
3
M(Ox)
(c)
tan  =
  26°
27m y = 12m × 2.5 + 8m × 5
70
y
27
y
x

cso
M1A1
DM1A1
M1A1
5.3 or better
A1
M1A1
2.6 or better
35
72
A1
M1A1ft
awrt 25.9 °
A1
[13]
6.
(a)
Mass ratio
Triangle
126
x
y
7
4
City of London Academy
6
Circle
9
(28.3)
5
5
S
126 – 9
(97.7)
x
y
B1 B1ft
4, 7 seen
B1
20
3
(b)
126 × 7 = 9 × 5 + (126 – 9) × x ft their table values
882  45
)
awrt 7.6
x  7.58 (
126  9
M1 A1ft
126 × 4 = 9 × 5 + (126 – 9) × y
504  45
)
y  3.71 (
126  9
M1 A1ft
tan 
  15°
A1
ft their table values
awrt 3.7
y
A1
ft their x, y
21  x
M1 A1ft
A1
[12]
7.
–
=
(a)
M(AF) 4a2.a – a2.3a/2 = 3a 2 .x
x  5a / 6
(b)
Symmetry  y = 5a/6, or work from the top to get 7a/6
tan q =
5a / 6
2a  5a / 6
(
M1A2,1,0
A14
x
2a  y
)
q  35.5°
M1
Taking moments about AF or a parallel axis, with mass proportional
to area. Could be using a difference of two square pieces, as above,
but will often use the sum of a rectangle and a square to make the L shape.
Need correct number of terms but condone sign errors for M1.
A1
A1 all correct
A1
A0 At most one error
A1
5a/6, (accept 0.83a or better)
B1ft
M1A1ft
A14
Condone consistent lack of a’s for the first three marks.
NB: Treating is as rods rather than as a lamina is M0
City of London Academy
9
21
3
B1ft
x  y = their 5a/6, or y = distance from AB = 2a – their 5a/6.
Could be implied by the working. Can be awarded for a clear
statement of value in (a).
M1
Correct triangle identified and use of tan
2a  5a / 6
is OK for M1.
5a / 6
Several candidates appear to be getting 45° without identifying a correct
angle. This is M0 unless it clearly follows correctly from a previous error.
A1ft Tan expression correct for their 5a/6 and their y
A1
35.5 (Q asks for 1 d.p.)
NB: Must suspend from point A. Any other point is not a misread.
[8]
8.
(a)
Mass ratios
Large
242
(c.1810
Small
82,
c.200
Template
512
anything in ratio 9: 1 : 8
c.1610)
M(A) 9 × 24 = 16 × 1 + 8 x
x = 25 (cm) exact
(b)
1
m
4
1 

 (36  x) M  12  m 
4 

3
m (o.e.e.)
M=
11
M(axis) 1LM = 12 ×
B1,B1ft
M1*A1
DM1*A16
ft their x
M1†A1ft
DM1†A14
[10]
9.
(a)
Total mass = 12m (used)
M1
(i)
M(AB): m.3a/2 + m.3a/2 + m.3a + 6m.3a + 2m.3a = 12m.x
indep M1 A1
5
x a
2
A1
M(AD): m.a + m.a + m.2a + 6m.2a = 12m.y
indep M1 A1
4
y a
3
A1
(ii)
City of London Academy
22
7
(b)
tan 
2a  4a / 3
5a / 2
M1 A1 f.t
   14.9°
A1cao 3
[10]
10.
(a)
(b)
12mx = 6m  9
M1
x = 4 12
A1
12my = 16m  8m
M1
y
A1
2
3
(12 + k) m × 4 = 12m × 4 12 + km × 3
ft their x
4
M1 A1ft
k=6
(c)
18m   = 12m ×
(d)
tan θ = 4 , 
4
9
A1
2
3
,
 =
4
9
M1A1 2
ft their  , cao
θ ≈ 83.7
3
M1 A1ft A1
3
[12]
11.
(a)
B
4
x
C
5 (x = 3)
A
4
D
M(AB): 7 × 3.5 + 5 × 5.5 + 4 × 2 = 20 × x
 20 x = 24.5 + 27.5 + 8 = 60  x = 3 cm
City of London Academy
M1 A2,1,0
dep M1A1
23
X
3.5
kM
(b)
M
Y
M(XY):
M × (3.5 – 3) = kM × 3.5
1
k= .
7
M1 A1ft
A1
3
[8]
12.
(a)
Mass ratios
Centres of mass
circle
9
6
rectangle
plate
200;
200 – 9
10
x
9 × 6 + (200 × 9) x = 200 × 10
x  10.7 (cm)
B1; B1ft
B1
M1
A1
5
cao
(b)
tan =
5
10.7
M1 A1ft
ft their x
  25°
A1
3
[8]
cao
13.
(a)
Rectangle Triangle Decoration
Mass Ratio 6 12 18 Ratio 1:2:3
1
CM from BG (–)1
2 x
2
18 × x = –6 × 1
x
1
+ 12 × 2
2
5
accept exact equivalents
6
City of London Academy
B1
B1
M1 A1
A1
24
5
G
–x
(b)
0
3
Q
Identification and use of correct triangle
tan  
1
ft their x
3 x
θ 14.6° cao
M1
M1 A1ft
A1
4
[9]
B
N
a
10mg
mg
R
a
14.
A
(a)
(b)
P
M(a) N × 2asin = mg × a cos + 10mg × 2a cos
2N tan = 21mg
N = 7mg (*)
M1 A2(1, 0)
 R = 11mg
Fr = 0.6 × 11mg = 6.6mg
For min P Fr  Pmin = 7mg – 6.6mg = 0.4mg
For max P Fr  Pmax = 7mg + 6.6mg = 13.6mg
0.4mg  P  13.6mg
B1
B1
M1 A1
M1 A1
cso A1 7
cso M1 A1
[12]
Note: In (a), if moments are taken about a point other than A, a complete set of equations for
finding N is needed for the first M1. If this M1 is gained, the A2(1, 0) is awarded for the
moments equation as it first appears.
City of London Academy
25
15.
(a)
(b)
(c)
AD: 10m x = 3m
5a
+ 3m  5a
2
M1 A1
x = 2.25a *
A1
AB: 10m y = 2m  2a + 3m  a
M1
y = 0.7a
A1
tan  =
2.5a  x
2
M1 A1 f.t.
y
 = 20 (nearest degree)
A1
R
A
3
3
B
O S
G
(d)
2a
10mg
P
5a
D
C
a
= P  2a
4
5a
5a
(OR: 4mg 
 3 mg 
= P  2a)
2
2
5mg
P=
* (exact)
4
M(0), 10mg 
(e)
M1 A1 A1
A1
5mg
; R = 10mg
4
5mg 65
F = S 2  R2 =
(10.1 mg)
4
S=
B1; B1
M1 A1 4
[16]
16.
(a)
B1
Area:
CM from AD:
6a 2
3a
2
6×
City of London Academy
a2
3a
2
2a 

 2a 

3 

8a
 1×
=5
3
4
5a 2
(Ratio)
x
B1 B1
x
M1
26
x
19a
15
A1
5
AWRT 1.27a
x
(b)
mg
Mg
M(X),
3a
 3a 19a 
Mg  
 = mg ×
2
15 
 2
7M
m=
, 0.156M, 0.16M
45
M1 A1 ft A1
A1
4
[9]
17.
(a)
Use of (8 + )m
i: 3m × 4 + m × 4 = (8 + )m × 2
Solving to  = 2 (*)
B1
M1
M1 A1 4
(b)
j: 5m × (3) + 2m × 2 = 10m × k
k = 1.1
M1 A1
A1
3
[7]
18.
(a)
MR
48a2
CM
4a
48a2 × 4a – 12a2 ×
(b)
4
a = 60 x
3
12a2
()
1
× 4a
3
60a2
B1, B1 ft
x
B1
M1 A1
Solving to x =
44
a (*)
15
A1
M × 4a = M ×
44
a
15
M1 A1
City of London Academy
27
6
=
11
15
A1
[9]
19.
(a)
ABC
Relative mass
4
Distance of centre of mass
10
from BC
(1 each error or omission)
ADE
1
20
BCDE
3
y
B3
(4 × 10) – (1 × 20) = 3 y
20
3
6 23 =
(b)
tan  =
= y
M1 A1
(T)
A16
15  y
20
E
M1
D
G
=
15  203
20
=
5
12
 = 22.6 (1 d.p.)
A1
A13
[9]
1.
Some candidates struggled with this question. Despite the question being explained clearly with reference to rods it was not
uncommon to see the triangle treated as a lamina. Another common error was to treat the rods as being of equal mass.
The geometry of the symmetrical triangular figure was appreciated by nearly all candidates with the height of the triangle correctly
calculated as 8 cm, although it was disappointing to find several candidates not recognising the 3-4-5 triangle and engaging in more
work than expected to find the height of the triangle.
For part (a) those candidates who answered the question as set and worked with three rods had little difficulty in producing a
relevant moments equation and arriving at the correct result. However it was disappointing to find a significant number of
candidates treating the triangle as a lamina, and they were happy simply to write down the answer as 83 cm.
For part (b) most candidates could either write down or calculate the distance of the new centre of mass from BC and proceed to find
the required angle. 3 out of 4 marks were available for those who had treated the shape as a lamina. A number of candidates ignored
the extra particle added to the framework and answered their own question. Very few students used the method of taking moments
about B to find the angle.
2.
(a)
The method was understood by most candidates and there was no problem in forming a moments equation for the centre of
mass. Common errors included simplifying
43
3
to 4π rather than
4

, using the area of a circle rather than a
semicircle, and the use of 6 for the radius of the semicircle. From a correct table, accuracy marks were often lost in the
moments equation because of a sign error. In general, those candidates who set out the masses and distances in a table
tended to make fewer errors.
Many candidates made it more difficult to obtain the given answer by taking their measurements from PQ and attempting
City of London Academy
28
3
to subtract their result from 2x, although this was often successfully completed. One advantage of this approach was that
they were less likely to make a sign error in their moments equation.
Candidates very rarely justified the modulus sign at all, with most candidates simply writing the final answer after their last
2
line of working. Students who had a negative coefficient for the x term in the numerator were more likely to deal with this.
(b)
3.
The fact that the answer was given did guide some candidates to the correct result, indicating that they clearly appreciated
the ‘show that’ nature of the question. Most candidates substituted x = 2 correctly into the given result and went on to find
the tangent of an angle. Many candidates did identify the correct triangle although some went to great lengths to find the
distance of the centre of mass from SP as an expression in x, not realising that it could be found by symmetry. Furthermore,
they often did not then realise that their expression cancelled to 6.
Many of those who made progress with this part found the angle to the vertical, with quite a few unconvincingly converting
to the given result or simply leaving it as the reciprocal.
This question was generally well answered by most candidates. It was pleasing to see many correct evaluations of the centre of mass
of the triangle and many completely correct solutions to part (a). A few candidates attempted breaking up the lamina into rectangles
and triangles rather than subtracting the moment of the triangle from the moment of the rectangle and so made the question much
more difficult. Of those who calculated the areas of the rectangle and triangle correctly some failed to subtract these and added them
instead. Most seemed to be happy with the use of large numbers for areas and only a few reduced these to a ratio. A very small
number of candidates tried to replace the lamina by a framework of rods.
In part (b) a significant number of candidates failed to recognise that the lamina was symmetrical and wasted time in finding the
distance to the centre of mass from AB using the same method, rather than using symmetry to write it down.
Many candidates lost the last two marks by finding the angle between AD and the vertical instead of the angle required.
4.
The majority of candidates applied the correct mechanical principles to solve this problem. Most were able to find the relative
masses and the centres of mass of the semi-circle and the triangle and obtain a correct moments equation. Many candidates did not
show sufficient working to demonstrate that their equation led to the given result in part (a).
In part (b) the most common error was to fail to realise that the two centres of mass were on opposite sides of the line BD and they
hence had a sign error in their expression. Those who decided to take moments about a line through A, perpendicular to AD avoided
this problem.
Candidates were generally able to use the given result to find the centre of mass of the semicircle, although it was quite common to
see it written incorrectly as 8π.
A clear diagram tended to lead candidates to identify the correct angle in part (c) and the correct ethod for finding it.
5.
This question was often answered very well. Most candidates took moments about Oy or Ox, although alternatives were seen.
Weaker candidates did not seem to be very confident in dealing with a lamina plus a set of particles.
(a)
Full marks were usually seen. The alternative method of taking moments about axes through the centre of mass was seen
and was usually implemented successfully.
A small number were too casual in claiming the printed result from a correct moment’s equation – candidates need to
remember that with a given answer a little more detail is required.
(b)
This was usually correct. Many candidates calculated the y coordinate of the centre of mass of the three particles as
8
3
and
then used that to calculate the centre of mass of the system.
Any errors were usually in the total mass (e.g. taken to be 40m or equal to the total mass of only those which contributed to
the moment). A few responses did not include the lamina and scored nothing. Some candidates treated OABC as a set of
connected rods, rather than as a lamina.
City of London Academy
29
(c)
6.
7.
Candidates who answered part (b) incorrectly often picked up two marks here from the follow through. The majority of
candidates recognised the correct triangle, although many of them then calculated the incorrect angle. A number of
responses used 5 and / or 8, which scored nothing. A few candidates lost the final mark by using rounded values from part
(b) and arrived at an incorrect value for the angle.
Many candidates achieved full marks on this question, whilst others just missed out on the final mark because they did not notice the
instruction to give their final answer correct to the nearest degree.
(a)
Where there were difficulties these usually arose when a candidate tried to work with the geometry of the triangle, finding
lengths of medians, etc in order to find the location of the centre of mass of the triangle – many seemed to be completely
unaware of the simple result they could apply and invariably made algebraic errors in their work. Most candidates
understood that the circle had been cut out of the triangle, but quite a few added the circle to the triangle in their working. A
few candidates treated the triangle as if it were just three rods, and others confused the centre of mass of the triangle with
the centre of mass of the set square in the course of their working.
(b)
Most candidates correctly identified the required angle. A few did not use their answers from part (a) at all, they simply
found the angle BCA.
In part (a) the first stage of this question requires the candidate to divide the given lamina into appropriate sections. There are
several options, large square minus small square, three small squares, a rectangle plus a square, or even dividing the shape into three
right angled triangles.
Most errors in finding the distance of the centre of mass from AF were due to sign errors in setting up the moments equation, errors
in finding the distance from AF of centres of mass of the constituent parts, or equations which were dimensionally incorrect (usually
missing a’s).
Some candidates saw the object as a collection of connected rods despite its being described as a lamina formed from a sheet of
card.
Part (b) Although some candidates used the symmetry of the figure to determine the distance of the centre of mass from FE (or AX),
many repeated the working from part (a). Some candidates did not appear to realise that they needed to find the distance of the
centre of mass from one of the horizontal lines, or they simply assumed that it was a. Many candidates did not draw diagrams,
making it difficult to determine exactly what they were trying to do. A few diagrams suggested that candidates did not understand
what would happen if the lamina were suspended from A. It was expected that candidates would use the right angled triangle with
5
7
a and a , but it is possible to find the distances of the centre of mass from A and F and use the cosine rule, as some
6
6
candidates attempted to demonstrate.
8.
Many candidates reached for calculators in this question – it was unusual to see the simplified form (9 : 1 : 8) for the ratio of areas.
Despite this, many solved the problem successfully. The most common error was in failing to subtract the area of the disc removed
in the moments equation or in failing to subtract the moment of the disc removed. Candidates did not always choose to take
moments about A, but many correct solutions were seen. A minority of candidates were either searching the formula booklet for
inspiration or confused by more advanced work that they have studied, and attempted to use the formula for centre of mass of a
sector of a circle. Candidates generally scored either full marks or no marks for (b). Some tried to bring in the areas from (a) and
ended up with dimensionally incorrect equations that earned no marks. Several chose to take moments about a different axis and
then usually neglected the reaction at the pivot.
9.
This question was well answered by many candidates and the method in part (a) was well known. Some, however, used “m” as mass
per unit length for the framework, or counted the masses of the particles more than once in an attempt to consider each rod
separately. Common sense often failed to prevail, with the mass of the whole system sometimes appearing as different values in the
two equations. Most were able to attempt part (b), but many failed to use (2a – y) in their ratio. The very few who decided to use
sine instead of tangent were usually successful.
City of London Academy
30
10.
In part (a) many saw intuitively that
x  4.5
because of the distribution of the masses. Some made life more difficult by taking
moments about different axes – especially to find the
y-coordinate. Many completed the second part easily, but weaker candidates tried to find the area of the triangle and somehow got
the required answer by devious means thereafter. A worrying number thought that 6 × 9 was 36. A small minority failed to complete
this part through not knowing the position of the centre of mass of a triangular lamina . Part (c) was usually completely correct or
else they had absolutely no idea. In the final part there were the usual problems of identifying the angle correctly with some
attempting to find sine or cosine, using Pythagoras or the cosine rule.
11.
Many treated the framework as a lamina and lost several marks in the first part, although most knew the correct method. Of those
who correctly treated the problem as a framework, a few tried to use three sides of a square and two sides of a triangle and made
errors locating the appropriate centres of mass. Many candidates wasted time finding the distance of the centre of mass from BC. In
part (b) there were some good solutions but many were unable to deal correctly with the introduction of M into the question.
12.
This question was also well done. A few added the masses of the circle and rectangle, rather than subtract them, and a few errors of
sign in the moments equation were seen. Some candidates ignored the obvious symmetry of the diagram and took moments to find
the distance of the centre of mass from AB and this, besides being unnecessary work, caused further difficulties if the distance found
was incorrect. When an accuracy is specified in a question, the candidate is expected to give their answer to that accuracy to gain
full marks.
13.
This was another question in which full marks were common. In part (a), some candidates, when taking moments, ignored the fact
that the centre of mass of the rectangle was below BC and the centre of mass of the triangle was above BC. Many candidates
produce solutions in which it is not at all clear what coordinate system they are using and what axis they are taking moments about.
In part (b) most could identify a suitable triangle and trigonometric manipulation was usually completed correctly and to the degree
of accuracy requested. However, the length of one side to the triangle was often given as 2 cm when it was 1 cm.
14.
Part (a) was generally well done. The majority of candidates found the most efficient method, which is to take moments about A.
The use of a calculator is not appropriate in questions like this and, for full marks, the candidate is expected to obtain the exact
answer, as printed, using a simple trigonometric identity or equivalent argument. Part (b) was rarely answered fully. Most could find
one limiting value of P, usually 0.4mg, the value associated with friction acting towards the wall, but very few realised that, if the
builder’s friend pushed hard enough, the direction of friction reversed. The commonest way of obtaining a range was to say that the
maximum value of P was associated with a frictional force of zero. However some candidates, having obtained their first value
efficiently by resolving, thought they might get a second value by taking moments about one or more points and it was possible to
waste much time this way.
15.
Many candidates found the paper too long and relatively few attempts at this question went beyond part (c) and the majority of
candidates scored less than half marks. The method for (a) and (b) was well known and there were many correct solutions. The
principal error was failure to include the mass of the lamina, producing equations which were missing a term, and divisions by 7m
instead of 10m. Some candidates tried to include the lamina by using its area 10a2 instead of its given mass. Marks were often lost
unnecessarily in part (c) by candidates, who knew exactly what to do, doing it carelessly.
The required angle was found probably less frequently than its complement and those who found the correct answer often failed to
round it to the nearest degree. Huge numbers of candidates lost easy marks by not making it clear that they knew the basic method.
Re-drawing the diagram to show the centre of mass below the point of suspension is a far less popular option than it used to be. It is
therefore essential that candidates indicate that the line they draw through G represents the vertical.
It is difficult to know the extent to which (d) and (e) were omitted due to difficulties rather than lack of time but good attempts were
rare. Of those who tried (d) a significant proportion were attempting to answer a different question about attaching an extra particle
at C. Some of those who attempted moments took them about points other than O but failed to realise that the unknown force at O
would contribute.
City of London Academy
31
Far too many attempted to arrive at the printed answer of 1.25mg by number juggling, concocting any old equation which looked
half-way convincing. The same was true to a lesser extent in part (a). It is sad to see candidates losing marks they possibly deserve
by inventing extra terms to produce the right numerical answer and so turning what may have been a minor numerical slip into a
wrong method.
Part (e) was rarely attempted and even more rarely correctly. A common misconception was that the force at C was vertical and this
led to some worthless moments attempts about C. Some candidates correctly combined the known 10mg and 1.25mg forces without
any reference to the point of suspension, and others simply added up everything which had been mentioned so far.
16.
(a)
The method here was generally well known but there were often mistakes in the mass ratios and the distance for the
triangle.
(b)
This proved to be a lot more demanding with some candidates not knowing where to start. Most who took moments about
AD missed out the third term.
17.
Questions of this type have not often been set in recent years but it was found straightforward, although a few did leave the question
till last. Full marks were scored by many candidates. Those candidates who chose to measure their distances other than from the
axes would have been well advised to state their method more clearly than many did, as it was not always easy to distinguish a nonstandard method from an erroneous one. It was not uncommon in (b) for distances to be measured from the line y = -3 with the
candidates then either forgetting to adjust their answer of 1.9 or failing to do so correctly.
18.
Part (a) was done very well although some lost unnecessary marks by the partial omission of a from their solution or from the use of
approximate decimals on the way to the required exact answer. One error commonly seen was to find the distance of the centre of
2
mass of the triangle from AE as 10 3 a. Part (b) proved more difficult. Many were unclear how they should be using the given
masses, attempting, often unsuccessfully, to write an equation involving both masses and areas. The simplest correct solution is to
take moments about the X, obtaining the equation
 M  4a  M 
44
a , which leads quickly to the answer. This part
15
proved effective in distinguishing candidates who gained high grades.
19. No Report available for this question.
City of London Academy
32