M2 CENTRE OF MASS PAST QUESTIONS - …...City of London Academy 1 M2 CENTRE OF MASS PAST QUESTIONS 1. A triangular frame is formed by cutting a uniform rod into 3 pieces which are

City of London Academy 1

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)


2. [The centre of mass of a semi-circular lamina of radius r is 3

4r from the centre]

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.

(a) Show that the centre of mass of T is a distance 38

324 2

x

x 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 = .622

948

(4) (Total 11 marks)


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)


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 3

4r

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)


5.

y

C (3m) B (5m)

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)


6.

A

3cm

21cm

O

5cm

CB

5cm

12 cm

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)


7.

A B X

C D

EF2a

2aa

a

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)


8.

A B8 cm 24 cm

X O

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

of T, which passes through the mid-point of OB. A small stud of mass 4

1m is fixed at B, and T

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)


9.

A

B C m(6 )

D m (2 )

3a

2a

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)


10.

A

B

C

O x

y

4

2m

9

4m

6m

4

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)


11.

A

B C

D4 cm

4 cm5 cm

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.

A B

CD

O

3 cm

5 cm

20 cm

10 cm

6 cm


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

CBP

MS

Q R

3 cm

6 cm

2 cm

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)


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 = 23 .

(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

2m

m

2a

5a

A B

CD

O

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.


(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 = 4

5 mg.

(4)

(e) Find the magnitude of the force on L at O. (4)

(Total 16 marks)

16.

2a

3a

a

A B

CD E


A uniform lamina ABCD is made by taking a uniform sheet of metal in the form of a rectangle

ABED, with 3AB a and 2AD a , 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)


18.

G

A

E D

C

B

4aX

8a

6a

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 = a1544 .

(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

B C

DE

M

40 cm

30 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 632 cm from BC.


(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 M1 A1

80

32x

1

22 (2.5)x

A15

(b)

Moments about B:

6sin cos 6sinMg Mg x M1 A1 A1

12sin cosx

tan12

x


θ = 11.768..... = 11.8° A14

Alternative method :

C of M of loaded frame at distance x21 from D along DA B1

tan θ = 6

21 x

M1 A1

θ = 11.768..... = 11.8° A1 [9]

2. (a) Rectangle Semicircles Template, T

24x 4.5π 4.5π 24x + 9π B2

x 3

34

3

34 x B2

24x2 – 4.5π ×

3

34 – 4.5π ×

3

34 = (24x + 9π) x M1 A1

distance =

**38

3–24 2

x

xx A17

(b) When x = 2, 316

20

x B1

tanθ =

316

20–4

6

–4

6

x

M1 A1

.622

948

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 M1A1


cmy 50 A15

(b) Distance of centre of mass from AB is 60cm B1

Required angle is tan–1 50

60

(their values) M1A1ft

= 50.2° (0.876 rads) A14 [9]

4. (a)

MR

108

18π

108 + 18π

B1

x1 (→)

from AD

yi (↓)

from BD

4

6

6

8

x

y

B1

AD(→): 108(4) + 18π (6) = (108 + 18π) x M1

AG)sf3()cm(4.694.68731...18108

108432x

A14

(b)

yi (↓)

from BD

6

8 y B1 oe

yBD )18108()(–18)6(108: 8

M1

A1ft

(3sf)3.06(cm) 3.06292...18108

504

y A14

(c)


θ = required angle

4.68731..–12tan

y dM1

4.68731..–12

3.06392.. A1

θ = 22.72641... = 23 (nearest degree) A14 [12]

5. (a) M(Oy) (8 + k)m × 6.4 = 5m × 8 + km × 8 M1A1

1.6k = 11.2 k = 7 * cso DM1A1 4

(b) M(Oy) xm27 = 12m × 4 + 5m × 8 + 7m × 8 M1A1

3

16x 5.3 or better A1

M(Ox) ym27 = 12m × 2.5 + 8m × 5 M1A1

27

70y 2.6 or better A1 6

(c) tan = 72

35

x

y M1A1ft

26° awrt 25.9 ° A1 3 [13]

6. (a) Triangle Circle S

Mass ratio 126 9 126 – 9 B1 B1ft

(28.3) (97.7)

x 7 5 x

y 4 5 y 4, 7 seen B1


126 × 7 = 9 × 5 + (126 – 9) × x ft their table values M1 A1ft

x 7.58 (

9126

45882

) awrt 7.6 A1

126 × 4 = 9 × 5 + (126 – 9) × y ft their table values M1 A1ft

y 3.71 (

9126

45504

) awrt 3.7 A1 9

(b) x

y

21tan ft their yx, M1 A1ft

15° A1 3 [12]

7.

– =

(a) M(AF) 4a2.a – a

2.3a/2 = xa .3 2 M1A2,1,0

6/5ax A14

(b) Symmetry y = 5a/6, or work from the top to get 7a/6 B1ft

tan q = )2

(6/52

6/5

ya

x

aa

a

M1A1ft

q 35.5° A14

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)

Condone consistent lack of a’s for the first three marks.

NB: Treating is as rods rather than as a lamina is M0


B1ft yx = 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 tan6/5

6/52

a

aa is OK for M1.

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) Large Small Template

Mass ratios 242 8

2, 512 anything in ratio 9: 1 : 8 B1,B1ft

(c.1810 c.200 c.1610)

M(A) 9 × 24 = 16 × 1 + x8 M1*A1

x = 25 (cm) exact DM1*A16

(b) M(axis) 1LM = 12 × m4

1 ft their x M1†A1ft

mMx

4

112)36(

M = m11

3 (o.e.e.) 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

ax2

5 A1

(ii) M(AD): m.a + m.a + m.2a + 6m.2a = 12m.y indep M1 A1

ay3

4 A1 7


(b) 2/5

3/42tan

a

aa M1 A1 f.t

14.9° A1cao 3 [10]

10. (a) xm12 = 6m 9 M1

x = 214 A1

ym12 = 16m 8m M1

32y A1 4

(b) (12 + k) m × 4 = 12m × 214 + km × 3 ft their x M1 A1ft

k = 6 A1 3

(c) 18m = 12m × 32 , =

94 M1A1 2

(d) tan θ =

944 , θ ≈ 83.7 ft their , cao M1 A1ft A1 3

[12]

11. (a)

A

B C

D

x

4

4

5 (x = 3)

M(AB): 7 × 3.5 + 5 × 5.5 + 4 × 2 = 20 × x M1 A2,1,0

20 x = 24.5 + 27.5 + 8 = 60 x = 3 cm dep M1A1 5


(b)

X 3.5 kM

M

Y

M(XY):

M × (3.5 – 3) = kM × 3.5 M1 A1ft

k = 7

1. A1 3

[8]

12. (a) circle rectangle plate

Mass ratios 9 200; 200 – 9 B1; B1ft

Centres of mass 6 10 x B1

9 × 6 + (200 × 9) x = 200 × 10 M1

x 10.7 (cm) A1 5

cao

(b) tan = 7.10

5 M1 A1ft

ft their x

25° A1 3 [8]

cao

13. (a) Rectangle Triangle Decoration

Mass Ratio 6 12 18 Ratio 1:2:3 B1

CM from BG (–)12

1 2 x B1

18 × x = –6 × 12

1 + 12 × 2 M1 A1

6

5x accept exact equivalents A1 5


(b)

Q

G

3

–x0

Identification and use of correct triangle M1

1tan

3 x

ft their x M1 A1ft

θ 14.6° cao A1 4 [9]

14.

R

A

NB

P

a

a

10mg

mg

(a) M(a) N × 2asin = mg × a cos + 10mg × 2a cos M1 A2(1, 0)

2N tan = 21mg

N = 7mg (*) cso M1 A1 5

(b) R = 11mg B1

Fr = 0.6 × 11mg = 6.6mg B1

For min P Fr Pmin = 7mg – 6.6mg = 0.4mg M1 A1

For max P Fr Pmax = 7mg + 6.6mg = 13.6mg M1 A1

0.4mg P 13.6mg cso A1 7 [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.


15. (a) AD: 10m x = 3m 2

5a + 3m 5a M1 A1

x = 2.25a * A1 3

(b) AB: 10m y = 2m 2a + 3m a M1

y = 0.7a A1 2

(c) tan = y

xa 5.2 M1 A1 f.t.

= 20 (nearest degree) A1 3

(d) G O S

RB

P

CD

A

5a

2a10mg

M(0), 10mg 4

a = P 2a M1 A1 A1

(OR: 4mg 2

5a 3 mg

2

5a = P 2a)

P = 4

5mg* (exact) A1 4

(e) S = 4

5mg; R = 10mg B1; B1

F = 22 RS = 4

655mg (10.1 mg) M1 A1 4

[16]

16. (a)

Area: 6 5 (Ratio)a a a 2 2 2

B1

CM from AD: 2

3a

3

22

aa x B1 B1

6 × 2

3a 1 ×

3

8a = 5 x M1


15

19ax A1 5

AWRT 1.27a

(b)

x

mg

Mg

M(X),

Mg

15

19

2

3 aa = mg ×

2

3a M1 A1 ft A1

m = 45

7M, 0.156M, 0.16M A1 4

[9]

17. (a) Use of (8 + )m B1

i: 3m × 4 + m × 4 = (8 + )m × 2 M1

Solving to = 2 (*) M1 A1 4

(b) j: 5m × (3) + 2m × 2 = 10m × k M1 A1

k = 1.1 A1 3 [7]

18. (a)

MR 48a2 12a

2 60a

2 B1, B1 ft

CM 4a ()

3

1 × 4a x B1

48a2 × 4a – 12a

2 ×

3

4a = 60 x M1 A1

Solving to x = 15

44a (*) A1 6

(b) M × 4a = M × 15

44a M1 A1


= 15

11 A1 3

[9]

19. (a)

ABC ADE BCDE

Relative mass 4 1 3

Distance of centre of mass

from BC

10 20 y

(1 each error or omission) B3

(4 × 10) – (1 × 20) = 3 y M1 A1

63

2 = 3

20 = y (T) A16

(b) tan = 20

15 y M1

E D

G = 20

153

20 =

12

5 A1

= 22.6 (1 d.p.) 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 38

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 3

34 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


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

line of working. Students who had a negative coefficient for the x2

term in the numerator were more likely to deal with this.

(b) 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.

3. 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 3

8 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.


(c) 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.

6. 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.

7. 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

aa6

7 and

6

5, but it is possible to find the distances of the centre of mass from A and F and use the cosine rule, as some

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.


10. In part (a) many saw intuitively that 5.4x 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.


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 non-standard 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

mass of the triangle from AE as 1032

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 44

415

M a M a , which leads quickly to the answer. This part

proved effective in distinguishing candidates who gained high grades.

19. No Report available for this question.

Documents

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