2015.1 L.41-42 ms -11 ha - M. Selkirk Confey College · 2015.1 L.41/42_MS 8/52 Page 8 of 48 exams DEB 3. A particle is projected from a point O on horizontal ground with an initial

Page 1 of 44

L.41/42

Pre-Leaving Certificate Examination, 2015

Applied Mathematics

Marking Scheme

Ordinary Pg. 4 Higher Pg. 21

2015.1 L.41/42_MS 2/52 Page 2 of 48 examsDEB


Applied Mathematics

Ordinary & Higher Level

Table of Contents

Ordinary Level Higher Level Q.1 ..................................... 04 Q.1 ..................................... 21

Q.2 ..................................... 06 Q.2 ..................................... 24

Q.3 ..................................... 08 Q.3 ..................................... 27

Q.4 ..................................... 10 Q.4 ..................................... 30

Q.5 ..................................... 12 Q.5 ..................................... 33

Q.6 ..................................... 14 Q.6 ..................................... 36

Q.7 ..................................... 16 Q.7 ..................................... 38

Q.8 ..................................... 18 Q.8 ..................................... 41

Q.9 ..................................... 20 Q.9 ..................................... 44

Q.10 ..................................... 46

examsDEB



Applied Mathematics

Ordinary & Higher Level

Explanation

1. Penalties of three types are applied to students’ work as follows:

Slips - numerical slips S(–1) Blunders - mathematical errors B(–3) Misreading - if not serious M(–1)

Serious blunder or omission or misreading which oversimplifies: - award the attempt mark only.

Attempt marks are awarded as follows: 5 (Att. 2), 10 (Att. 3)

2. Mark all answers, including excess answers and repeated answers whether cancelled or not, and award the marks for the best answers.

3. Mark scripts in red unless a student uses red. If a student uses red, mark the script in blue or black.

4. Number the grid on each script 1 to 9 in numerical order, not the order of answering.

5. Scrutinise all pages of the answer book.

6. The marking scheme shows one correct solution to each question. In many cases, there are other equally valid methods.

Current Marking Scheme

Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students’ work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice.

examsDEB



Applied Mathematics

Ordinary Level Marking Scheme (300 marks)

Six questions to be answered. All questions carry equal marks. (6 × 50m)

1. Four points P, Q, R, S lie on a straight level road. A train, travelling with uniform acceleration, passes point P with a constant speed of

10 m s–1 and 5 seconds later passes Q with a speed of 20 m s–1. The time taken by the train to travel from Q to R is twice the time it takes to travel from

R to S. Its speed at S is 26 m s–1. Find (i) the uniform acceleration of the train (10)

For 5,20,10, tvuPQ atuv )5)((1020 a ... (5m) a510 2sm2 a ... (5m) (ii) PQ , the distance from P to Q (10)

For PQ , 2,20,10 avu

asuv 222 ))(2(2)10()20( 22 s ... (5m) s4100400 3004 s m75s ... (5m)

examsDEB


(iii) QS , the distance from Q to S (10)

For QS , 2,26,20 avu

asuv 222 ))(2(2)20()26( 22 s ... (5m) s4276 m69s ... (5m) (iv) RS , the distance from R to S. (20)

For QS , atuv ))(2(2026 t t26 s3t ... (5m)

Time from Q to R is s.2)3(3

2 ... (5m)

For QR , 2,2,20 tau atuv )2)(2()20( v

1sm24 v ... (5m)

For RS , 2,26,24 avu

asuv 222 ))(2(2)24()26( 22 s s4576676 1004 s m25s ... (5m)


2. At a certain instant, a boat A is 96 m due north of another boat B.

A is travelling at a constant speed of 6 m s–1

in the direction 30° west of south. B is travelling due west at a constant speed

of 8 m s–1.

Find (i) the velocity of A in terms of i

and j

(10)

jivA

30cos630sin6 ... (5m)

jivA

2053 ... (5m)

(ii) the velocity of B in terms of i

and j

(5)

ivB

8 ... (5m)

(iii) the velocity of A relative to B in terms of i

and j

(10)

BAAB vvv

)8()2053( ijivAB

... (5m)

jivAB

2055 ... (5m)

(iv) the magnitude and direction of the velocity of A relative to B (10)

2 25 (5 20)ABv

17 21 msABv ... (5m)

and

5 20

tan 1 045

1tan 1 04 46 12 ... (5m)

A

B

96 m

6 m s−1

8 m s−1

30°

5

5 20⋅ABvr

θ


(v) the shortest distance between them in their subsequent motion. (15) Let M be the point of closest

approach on the path of A relative to B. Shortest distance: p

BM ... (5m)

96cos 96cos 46 12 ... (5m) 66 54 m ... (5m)

96

ABvr

A

B

M

p

θ

θ


3. A particle is projected from a point O on horizontal ground with an initial speed of u m s–1 at an angle of 60° to the horizontal.

One second after being projected, the

particle passes a point A, which is 20 m horizontally from the point of projection.

Find (i) the value of u (10)

When, 1, 20xt s

cos 60 (1) 20u ... (5m)

1

202

u

40u ... (5m) (ii) the height of A above the horizontal ground (10) The height of A above the horizontal ys at 1t ... (5m)

2140sin 60 (1) (10)(1)

2

29 64 m ... (5m) (iii) the maximum height of the particle above the horizontal ground (15) The maximum height occurs when 0yv

40sin 60 10 0t 34 64 10t 3 464t ... (5m) The maximum height ys at 3 464t ... (5m)

2140sin 60 (3 464) (10)(3 464)

2

60 m ... (5m)

O60°

20 m

uA


(iv) the other time that the particle has the same vertical height above the plane as at A. (15)

Let t be the times when the particle is at the height of m.6429

29 64ys

6429)10(2

160sin40 2 tt ... (5m)

234 64 5 29 64t t

20 5 34 64 29 64t t

25 34 64 29 64 0t t

234 64 (34 64) 4(5)(29 64)

2(5)t

... (5m)

34 64 24 64

10t

(We want the greater time.)

34 64 24 64

10t

5 928 st ... (5m)


4. A particle of mass 4 kg is connected to another particle of mass 3 kg by a taut, light, inextensible string which passes over a smooth light pulley at the edge of a rough horizontal table.

The coefficient of friction between the 4 kg

particle and the table is . The system is released from rest. (i) Show on separate diagrams the forces acting on each particle. (10)

T

a

4 kg a3 kg

T

3g

N

μN

4g

... (5m) ... (5m)

(ii) If = 3

2, find the common acceleration of the particles. (20)

For the 4 kg mass: 4N g

2

43

T N a

For the 3 kg mass: 3 3g T a … 1 ... (5m) Then:

2

(4 ) 43

T g a

8

43

gT a … 2 ... (5m)

Adding 1 and 2,

8

3 73

gg a ... (5m)

73

ga

2 210ms 0 476 ms

21a ... (5m)

4 kg

3 kg


(iii) Find the tension in the string. (10) From 1,

10

30 321

T

... (5m)

7

1030T

N57.28N7

200 T ... (5m)

(iv) Calculate the time it takes the 3 kg particle to fall 2 metres. (10) 0, 0 476, 2u a s

21

2s ut at

212 (0)( ) (0 476)

2t t ... (5m)

22 0 238t

2 8 403t 2 90 st ... (5m)


5. A smooth sphere A, of mass 5 kg, collides directly with another smooth sphere B, of mass 2 kg, on a smooth horizontal surface.

A and B are moving in opposite

directions with speeds of 3 m s–1 and 4 m s–1, respectively.

The coefficient of restitution for the collision is 3

2.

Find (i) the speed of A and the speed of B after the collision (30)

PCM: 1 2(5)(3) (2)( 4) (5) (2)v v ... (10m)

1 27 5 2v v

1 25 2 7v v … 1

NEL: )43(3

221 vv ... (10m)

1 2

14

3v v

1 2

282 2

3v v … 2

Adding 1 and 2,

1

77

3v

11

1ms

3v

and 2

52 7

3v

2

262

3v

12

13ms

3v ... (10m)

(ii) the percentage loss in kinetic energy due to the collision (15)

KE before collision 2 21 1(5)(3) (2)( 4) 38 5

2 2 ... (5m)

KE after collision 2 2

1 1 1 13(5) (2) 19 06

2 3 2 3

... (5m)

Loss in KE 38 5 19 06 19 44

4 m s−13 m s−1

A B5 kg 2 kg


Percentage loss in KE 19 44

100%38 5

50 50% ... (5m) (iii) the magnitude of the impulse imparted to B due to the collision. (5)

Impulse 13 50

(2) (2)( 4) Ns 16 67 Ns3 3

... (5m)


6. (a) Particles of weight 3 N, 1 N, 5 N and 2 N are placed at the points (4, –5), (p, –1), (p, q) and (3, 2q), respectively.

The co-ordinates of the centre of gravity of the system are (6, 1). Find (i) the value of p (10)

3(4) 1( ) 5( ) 2(3)

63 1 5 2

p p

... (5m)

66 12 5 6p p 66 6 18p 6 48p 8p ... (5m) (ii) the value of q. (10)

3( 5) 1( 1) 5( ) 2(2 )

13 1 5 2

q q

... (5m)

11 15 1 5 4q q q91611 27 9q 3q ... (5m)


(b) A quadrilateral lamina has vertices A, B, C and D. The co-ordinates of the vertices are

A(–1, 3), B(2, 9), C(8, 3) and D(5, –3). Find the co-ordinates of the centre of gravity

of the lamina. (30) Area Centre of gravity

ABC 1

(9)(6) 272

1 2 8 3 9 3

, (3,5)3 3

... (10m)

ACD 1

(9)(6) 272

1 5 8 3 3 3

, (4,1)3 3

... (10m)

Let (x,y) be the centre of gravity of the lamina. Then

(27)(3) (27)(4)

27 27x

7

3 52

x ... (5m)

and

(27)(5) (27)(1)

27 27y

3y ... (5m) Thus the co-ordinates of the centre of gravity are ).3,53(

CA

B

D

( 1,3)− (8,3)

(2,9)

(5, 3)−

6

6

9

B

A C

D


7. A uniform rod, [AB], of weight 30 N is smoothly hinged at end A to a rough vertical wall.

The rod is held in a horizontal position by a

light, inextensible string attached to point B on the rod and to the point C on the wall vertically above A.

The string [ BC ] makes an angle with the

rod [ AB ], as shown in the diagram. (i) Show on a diagram all the forces acting on the rod [ AB ]. (10)

C

AF

N 30l

lT

Bθ

... (10m) (ii) Write down the two equations that arise from resolving the forces horizontally

and vertically. (10) cosTN ... (5m) 30sin TF ... (5m) (iii) Write down the equation that arises from taking moments about the point B. (10)

Let 2AB l . Then

(30)( ) ( )(2 )l F l ...(5m, 5m)

B

C

θA


(iv) If the rod is in equilibrium and is on the point of slipping when 4

3tan , find the

coefficient of friction between the rod and the wall. (20) From the moments equation: 30 2F 15F ... (5m)

If 4

3tan , then

5

3sin and

5

4cos .

Then

3

15 305

T

3

155

T

25T ... (5m) and

4

255

N

20N ... (5m) If the rod is on the point of slipping, then

F

N

15

20

3

4 ... (5m)


8. (a) A particle describes a horizontal circle of radius r metres with uniform angular velocity radians per second.

Its speed is 8 m s–1 and its acceleration is 32 m s–2.

Find (i) the value of (15)

Speed 8 8r ...(5m)

and Acceleration 32

2 32r ...(5m)

then ( ) 32r 8 32

14 rads ...(5m)

(ii) the value of r. (5)

Then (4) 8r 2 mr ...(5m)

(b) A right circular cone is fixed to a horizontal surface. Its semi vertical angle is , where

15

8tan and its axis is vertical.

A smooth particle of mass 4 kg describes a horizontal circle of radius r cm on the smooth inside surface of the cone.

The plane of the circular motion is 30 cm above the horizontal surface.

(i) Find the value of r. (5)

15

8αtan

30r

m160cm16 r ... (5m)

30 cm

r

α


(ii) Show on a diagram all the forces acting on the particle (5)

… (5m) (iii) Find the reaction between the particle and the surface of the cone. (10) gR 4αsin ... (5m)

4017

8

R

N85R ... (5m) (iv) Calculate the angular velocity of the particle. (10)

2cosR mr

2ω)160)(4(17

15 R ... (5m)

2ω6408517

15

2ω64075

16

1875ω2

1srad4

325ω ... (5m)

R

4g

α


9. (a) State the Principle of Archimedes. (5)

Statement ... (5m)

A solid piece of metal has a weight of 50 N. When it is completely immersed in water, the metal weighs 35 N.

Find (i) the volume of the metal (10)

B = weight of water displaced 15=1000(V)(10) ... (5m) 33 m105.1 V ... (5m)

(ii) the relative density of the metal, correct to two decimal places. (5)

Weight of metal Vgρ

)10)(1051(ρ50 3 3333ρ

3331000

ρ s ... (5m)

(b) An object consists of a hemisphere of diameter 12 cm surmounted by a cone of diameter 12 cm and height 8 cm.

The relative density of the object is 60 and it is completely immersed in a tank or liquid of relative density 11 .

The object is held at rest with its axis vertical by a light, inextensible vertical string which is attached to the base of the tank.

Find the tension in the string.

[Density of water = 1000 kg m–3] (30)

3 22 11100 (0 06) (0 06) (0 08) (10)

3 3B

... (5m)

8 29B ... (5m)

3 22 1600 (0 06) (0 06) (0 08) (10)

3 3W

... (5m)

4 52W ... (5m)

and T W B ... (5m) 8 29 4 52T 3 77T N ... (5m)



Applied Mathematics

Higher Level Marking Scheme (300 marks)

Six questions to be answered. All questions carry equal marks. (6 × 50m)

1. (a) A train travels a distance d from rest at one station to rest at another station. The train travels for the first part of its journey with a constant acceleration 1f . It then immediately decelerates to rest at the second station with a constant deceleration 2f .

Show that the total time taken is

21

112

ffd .

(25) Let v be the maximum speed attained. Then

1: 11

vf

t

11

vt

f ... (5m)

2: 22

vf

t

22

vt

f ... (5m)

3: 1 2T t t

1 2

v vT

f f

1 2

1 1T v

f f

... (5m)

1 2

1 1

Tv

f f

Further answers overleaf

examsDEB

v

2t1t

d

T


4: 1

2d Tv ... (5m)

1 2

2 .1 1

Td T

f f

2

1 2

1 12d T

f f

1 2

1 12T d

f f

... (5m)

(b) A particle, P, starts from rest at a point A and moves with constant acceleration

f in a straight line. A time T, after P starts from A, a second particle, Q, starts from A and moves in the same direction along the same straight line as P.

Q moves with a constant speed of u. (i) Prove that Q will overtake P if fTu 2 . (20)

Let t T . Let 1s and 2s be the distances travelled by P and Q in time t. Then

21

1(0)( ) ( )( )

2s t f t

21

1

2s ft

and 2 ( )( )s u t T ... (5m) The particles will be level when 1 2s s ... (5m)

21

2ft ut uT

2 2 2 0ft ut uT

22 ( 2 ) 4( )(2 )

2

u u f uTt

f

22 2 2

2

u u fuTt

f

2 2u u fuT

tf

... (5m)


For Q to overtake P, this equation must have real solutions, i.e. 2 4 0b ac

2 2 0u fuT

2 2u fuT 2u fT ... (5m) (ii) Assuming Q does overtake P, i.e. that fTu 2 , express in terms of

u, f and T the length of time for which Q is ahead of P. (5)

Let 2

1

2u u fuTt

f

and

f

fuTuut

22

2

.

The length of time Q is ahead of P is

2 2

2 1

2 2u u fuT u u fuTt t

f f

22 2u fuT

f

... (5m)


2. (a) A man can swim at 3 m s –1 in still water. He swims across a river of width 60 metres. The river flows with a constant speed of 5 m s –1 parallel to the straight banks.

He swims at an angle to the upstream

direction but ends up going at an angle to the downstream direction.

(i) Show that tan θ =

cos35

sin3

. (10)

Let i

and j

be unit vectors in the direction of the river flow and perpendicular to the banks respectively.

Then 5Rv i

3cos 3sinMRv i j

and M MR Rv v v

3cos 3sin 5i j i

5 3cos 3sini j

... (5m) If this makes an angle with the

downstream direction, then

3sin

tan5 3cos

... (5m)

θα

3 m s–1

60 m5 m s–1

3 sin α

5−3 cos α θ

5

3

α

α

θ

θβ


(ii) Find the time taken for the man to cross by the shortest path. (15)

Let be the angle between MRv

and Mv

. Then by the Sine rule,

sin sin

3 5

3sin

sin5

For the shortest path, we require the largest possible value of . This is when 90 and

3

sin5

... (5m)

then

2 25 3 4Mv

and

4

sin5

... (5m)

Then the speed directly across the river is

4

3sin 3 2 45

Thus the time to cross the river by the shortest path is

60

25 s2 4

... (5m)


(b) Ship A is travelling with a constant speed of 10 m s –1 in the direction 30° north of east. At midday, ship B is 10 km due east of ship A, and is travelling in a straight line with a constant speed of v m s –1.

(i) Calculate the minimum possible value of v if B is to intercept A. (20)

(10cos30 ) (10sin 30 )Av i j

5 3 5Av i j

... (5m) Let B travel in the direction north of west. Then ( cos ) ( sin )Bv v i v j

Thus BA B Av v v

( cos sin )v i v j

5 3 5i j

( cos 5 3)v i

( sin 5)v j

... (5m)

For interception, BAv

must point due west, i.e.

sin 5 0v

5

sinv

... (5m)

For this to be possible, sin 1

5

1v

5v Thus the minimum value of v for interception is 5. ... (5m) (ii) If v = 6, show that B can travel in either of two directions to intercept A,

and find these directions, correct to the nearest degree. (5) If 6v , then interception occurs when 6sin 5 0

5

sin6

1 5sin 56

6 or 180 56 124 ... (5m)

10

300

v

10 kmA BBAv

r

Bvr

Avr

α


3. (a) A particle is projected from a point on a horizontal plane with speed 21 m s –1 at an angle to the horizontal. The particle then strikes a small target whose horizontal and vertical distances from the point of projection are 30 m and 10 m respectively.

Find (i) the two possible values of tan (15)

Given: for the same value of t, 30xs and 10ys

30xs : 21cos . 30t

10

7cost

... (5m)

10ys : 2

10 10021sin 10

7cos 2 49cos

g ... (5m)

230 tan 10sec 10

23 tan (1 tan ) 1

2tan 3tan 2 0 (tan 1)(tan 2) 0 tan 1 or tan 2 ... (5m) (ii) the two possible times taken to strike the target. (10)

If tan 1 , then 45 and 1

cos2

. The time taken is.

10 10 2

2 02 s1 7

72

... (5m)

If tan 2 , then 1

cos5

. The time taken is

10 10 5

3 19 s1 7

75

. ... (5m)


(b) A particle is projected up an inclined plane from a point O, with initial speed of 35 m s –1. The line of projection makes an angle with the inclined plane and the plane is inclined at an angle of 45 to the horizontal.

The plane of projection is vertical and contains the line of greatest slope. The particle is moving horizontally when it strikes the inclined plane at Q.

(i) Show that 3

1tan . (15)

xv

O

45°Q

35

45°

yv−vr

θ

35cos sin 45 .xv g t and 35sin cos 45 .yv g t

35cos2

x

gtv and 35sin

2y

gtv

Also, time of flight:

0ys

2

35sin . 02 2

gtt

70 2 sin

tg

... (5m)

At the time of flight:

tan 45 y

x

v

v

x yv v ... (5m)

70 2 sin 70 2 sin

35cos 35sin2 2

g g

g g

35cos 70sin 35sin 70sin 35cos 105sin

35 sin

105 cos

1

tan3

... (5m)


(ii) Find OQ . (10)

1

tan3

Then

1

sin10

and 3

cos10

Then time of flight

70 2 1 10 5

710g

... (5m)

and xOQ s at time of flight

23 10 5 10 5

357 710 2 2

gOQ

2

50275

50 2OQ

70 71 mOQ ... (5m)


4. (a) Two particles of masses 2 kg and 6 kg are connected by a light inextensible string passing over a fixed smooth pulley.

Initially the two particles are at rest at the

same horizontal level. The system is released from rest. The 6 kg particle takes 2 seconds to

strike horizontal ground. Find (i) the initial height of the particles

above the ground (15) By Newton’s 2nd Law: 2 2a T g and

4 4a g T ... (5m) adding 6 2a g

3

ga ... (5m)

Let h be the height of the particles above the ground. Then

21

2s ut at

21(0)(2) (2)

2 3

gh

2

3

gh ... (5m)

2 kg 6 kg


(ii) the greatest height above the ground to which the 2 kg mass rises. (10)

After 2 seconds, v u at

2

(0) (2)3 3

g gv

When the 4 kg mass strikes the ground, the 2 kg mass is at a height of

4

23

gh above the ground. ... (5m)

Let 1h be the further height the 2 kg mass rises to before coming to rest.

Then

with a g and 2

3

gu

2 2 2v u as

2

21

2(0) 2( )

3

gg h

2

1

42

9

ggh

1

2

9

gh

The maximum height of the 2 kg mass is

4 2 14

3 9 9

g g g

15 24 m ... (5m)

(b) A smooth wedge, of mass 2m and slope 45°, rests on a smooth horizontal plane. A particle of mass m is placed on the inclined face of the wedge.

The system is released from rest.

Find the speed of the mass m relative to the wedge, when the speed of the wedge is 50 m s–1. (25)

Wedge (2 ) sin 45m a N

22

Nma

2 2

Nma

2 2N ma ... (5m)


m

2m

45°

N

a

N1 2mg45°


Particle

bN

2mg

2a

2a2

mg

mg

45º 45º a

Forces Accelerations

2 2

a mgm N

... (5m)

2 22 2

a mgm ma

4ma mg ma 5a g

5

ga ... (5m)

2 2

a mgm b

5 2 2

g gb

3 2

5

gb ... (5m)

Wedge

0, 0 5,5

gu v a .

Then v u at

1

02 5

gt

5

2t

g

Particle: v u at

3 2 5

(0)5 2

gv

g

13 22 12 ms

2v ... (5m)


5. (a) A smooth sphere A, of mass 1m , collides directly with a smooth sphere B,

of mass 2m which is at rest on a smooth horizontal table. The coefficient

of restitution between the spheres is 1e .

The line of centres of the spheres is at right angles to a smooth vertical cushion at the edge of the table. Sphere B then strikes the cushion and rebounds.

The coefficient of restitution between sphere B and the cushion is 2e .

Show that there will be no further impact between the spheres if 212121 1 meeeem . (25)

Collision 1: Two spheres

PCM: 1 2 1 1 2 2(0)m u m m v m v ... (5m) 1 1

2 20

u m v

u m v

m v

1 1 2 2 1m v m v m u

NEL: 1 2 1v v e u ... (5m)

Then 1 1 2 2 1m v m v m u

2 1 2 2 1 2m v m v e m u

1 2 1( )m m v 1 1 2( )m e m u

1 1 21

1 2

m e mv u

m m

also 1 1 2 2 1m v m v m u

1 1 1 2 1 1m v m v e m u

1 2 2 1 1 1( ) ( )m m v m e m u

1 12

1 2

(1 )m ev u

m m

... (5m)

Collision 2: Spheres and cushion 3 2 2v e v

2 1 13

1 2

(1 )e m ev u

m m

... (5m)

There will be no further collision if 1 3v v

1 1 2 2 1 1

1 2 1 2

(1 )m e m e m eu u

m m m m

1 1 2 2 1 1 1m e m e m e m

1 2 1 1 2 1 1 2m e m e e m e m

1 2 1 2 1 2(1 )m e e e e m ... (5m)


(b) Two smooth spheres, each of mass m and radius r, collide while travelling on a smooth horizontal plane. Before impact, the speeds of the spheres are u and 4u respectively, and the spheres are moving in the same direction along parallel lines, a distance r21 apart.

The coefficient of restitution between the spheres is 2

1.

Find the angle between their directions of motion after impact, correct to the

nearest degree. (25) From the diagram

1 2 3

sin2 5

r

r

... (5m)

Thus 4

cos5

and 3

tan4

.

Let and be the directions of motion of

the two spheres after impact.

4uu

x y

m m

12u

12u

3u

3u

4u16u5

5

5

5

5

5

α α

βθ

jr

irBefore

After

PCM ( )i

: 16 4

5 5

u umx my m m

... (5m)

4x y u ...1

u

1⋅2r4u

2r1⋅2r

α


NEL ( )i

: 1 16 4

2 5 5

u ux y

... (5m)

6

5

ux y ...2

Adding 1 and 2,

14

25

ux

7

5

ux

and

7

45

uy u

13

5

uy ... (5m)

then

12125tan

7 75

u

u

59 74 and

335tan

13 135

u

u

12 99 Then the angle between their directions of motion 59 74 12 99 47 ... (5m)


6. (a) A particle P is moving at a constant speed on the inner surface of a smooth sphere of radius r.

The particle is describing horizontal circles r2

1 below the centre of

the sphere.

Prove that the speed of the particle is gr62

1. (25)

From the diagram,

112sin2

r

r

30 ... (5m) Then 1 cos30r r

1

3

2

rr ... (5m)

Then

: 2

Rmg ... (5m)

2R mg Circular motion:

2

1

3

2

mv R

r ... (5m)

2 3 3(2 )

2 2

rmv mg

2 3

2

grv

2 6

4

grv

1

62

v gr ... (5m)

°30

2330cos RR =°

O

rR

1r C

r21

mg

230sin RR =° R

θ


(b) A particle moves with simple harmonic motion in a straight line. It has velocities of 4 m s–1 and 2 m s–1 when it is at distances of

1 m and 2 m respectively from the centre of the motion. (i) Find the amplitude and the periodic time of the motion. (15) 4v when 1x :

2 2v a x

24 1a ...1 2v when 2x :

22 4a ...2 ... (5m) Dividing 1 by 2:

2

2

4 1

2 4

a

a

2

2

12

4

a

a

2

2

14

4

a

a

2 24 16 1a a

23 15a

2 5a

5 ma ... (5m) From 2,

2 5 4 2 and

2 2

s2

T

... (5m)

(ii) Calculate the least time taken for the particle to travel from a position

of rest to a position where its velocity is 2 m s–1. (10)

0v when 5x a Let cosx a t

5 cos 2x t 2v when 2x

2 5 cos 2t ... (5m)

2

cos 2 0 89445

t

2 0 4636t 0 23 st ... (5m)


7. (a) A uniform ladder, of weight W and length 2l, rests with its lower end, P, on rough horizontal ground. Its upper end, Q, is in contact with a rough vertical wall.

At both P and Q, the coefficient of friction is 3

1.

The ladder makes an angle of 2tan 1 to the horizontal.

Express in terms of l, the distance that a person of weight W can safely climb

before the ladder begins to slip. (25)

1tan 2 tan 2 Then : 1 2N F W

: 1F N ... (5m)

Limiting friction: 1

3F N

1 1

1

3F N

Let x be the distance that the person can ascend. Taking moments about P: 1 1.2 sin .2 cosN l F l

. cos . cosW l W x ... (5m) 1 12 tan 2lN lF Wl Wx

1 14 2lN lF Wl Wx

then 1 1

1 1 1

3 3 9F N F N

thus 1

29

N N W

9 18N N W 10 18N W

9

5

WN ... (5m)

and 1 1

3,

5 5

W WF N ... (5m)

then 3

4 25 5

W Wl l Wl Wx

12 2 5 5l l l x 9 5l x

9

5

lx ... (5m)

Q

Ptan−1 2

x

P

N

1F

F

W

Wl

l

1N Q

θ


(b) Two uniform rods, AB and BC, each of length 2l are smoothly jointed at B. The weight of AB is 3W and the weight of BC is 5W. The rods stand in equilibrium with the ends A and C on rough horizontal ground, with each rod making an angle with the vertical.

The coefficient of friction between A and the ground is 3

1, while the

coefficient of friction between C and the ground is . The angle is increased until both rods are on the point of slipping.

Find (i) the value of . (15)

Structure ABC : 1 8N N W

: 1

1

3N N ... (5m)

Moments about A: (3 )( ) (5 )(3 )W k W k

1( )(4 )N k

118 4W N

1

9

2N W

Then

7

2N W ... (5m)

and

1 7 9

3 2 2W W

7

27 ... (5m)

B

A C

α α

N

A

k k k k

N31

3W

B

5W

1μNC

1N


(ii) the value of when the rods are about to slip. (10)

Rod AB Moments about B

1

(3 )( sin ) (2 cos )3

W l N l

( )(2 sin )N l ... (5m)

7 7

(3 )(tan ) (2) (2 tan )6 2

W WW

7

3 tan 7 tan3

7

4 tan3

7

tan12

1 7tan

12 ... (5m)

30 3 .

B

Nl 3W

A

l

N31

α


8. (a) Prove that the moment of inertia of a uniform rod of mass m and length 2l

about an axis through its centre perpendicular to the rod is 2

3

1ml . (20)

Standard Proof Moment of mass element ... (5m) Moment of body ... (5m) Integral ... (5m) Deduce ... (5m)

(b) A uniform rod AB of mass m and length 2l has a particle of mass m attached at a distance 0x from A. The system is free to rotate about a horizontal axis through A perpendicular to the rod.

When the system makes small oscillations about the horizontal axis through A, the length

of the equivalent simple pendulum is 3

4l.

(i) Express x in terms of l. (15)

Let h be the distance from A to the centre of mass of the system.

( )( ) ( )( )m x m l

hm m

2

x lh

Let I be the moment of inertia of the system about the

horizontal axis through A.

Then rod particleI I I

2 24

3I ml mx ... (5m)

Given

4

3

I l

Mh

2 24433(2 )

2

ml mx lx l

m

... (5m)


A

B

mm =1

mm =2

xx =1

lx =2

A

B

m

x


2 24 4( )

3 3

ll x x l

2 2 24 3 4 4l x lx l

23 4x lx 3 4x l

4

3

lx ... (5m)

(ii) If the system is released from rest with AB horizontal, find the speed of

B when it is vertically below A. (15) When AB is horizontal, 1 0KE

1

43(2 )

2

ll

PE m g

7

3

lmg

7

3

mgl

When B is vertically below A,

21

1

2KE I

2

2 21 4 4

2 3 3

lml m

2 214

9ml

2 0PE

PCE: 1 1 2 2KE PE KE PE

2 27 14

3 9

mglml ... (5m)

2 3

2

gl

2 3

2

g

l

3

2

g

l ... (5m)


Let v be the velocity of B at its lowest point. Then v r

3

(2 )2

gv l

l

2 34

2

gv l

l

6v gl ... (5m)


9. (a) 3cm275 of a liquid of relative density 12 is mixed with 3cmV of another liquid of relative density 53 .

If there is no contraction of volume, and the relative density of the mixture is

952 find the value of V. (15)

6 6

1 2 1 and 2

275 10 10

2 1 3 5 2 95

m

V V

s

− −× ×ρ

⋅ ⋅ ⋅ Then

1 1 2 2

1 2

s V s Vs

V V

6 6

6 6

(2 1) (275 10 ) (3 5) ( 10 )2 95

(275 10 ) ( 10 )

V

V

... (5m)

577 5 3 5

2 95275

V

V

... (5m)

811 25 2 95 577 5 3 5V V 233 75 0 55V V = 425 ... (5m) (b) A block of mass 12 kg, and relative density 24 , is

held suspended by a string attached to a scale A. The block is completely immersed in 3cm1500 of a liquid of relative density 21 contained in a cylindrical beaker of mass 70 kg. The beaker sits on another scale B.

(i) If scale A registers x kg, find the value of x. (15) Let W be the weight and B be the buoyancy. 2 1W g and

Ls WB

s

T

B

W


1 2(2 1 )

4 2

g

... (5m)

0 6g

In equilibrium: T B W 0 6 2 1T g g ... (5m) 1 5T g If scale A registers a mass of x kg, then 1 5.x ... (5m)

(ii) If scale B registers y kg, find the value of y. (10)

Weight of liquid:

61000(1 2) 1500 10 g

1 8g

Let W be the total weight of the beaker and the liquid. Then

0 7 1 8W g g ... (5m) 2 5g

In equilibrium: R W B 2 5 0 6g g 3 1g

If scale B registers a mass of y kg, then 3 1.y ... (5m)

(iii) The radius of the beaker is 10 cm. Find the height, in cm, of the liquid in the beaker, correct to two decimal places. (10)

Let V be the volume of the block. Then 1000m sV 2 1 1000(4 2)V

30 0005 mV

3500 cm ... (5m)

Total volume of liquid and block 1500 500

32000 cm

Let h cm be the height of the liquid. Then

2(10) 2000h 6 37 cm.h ... (5m)

R

WB


10. (a) A particle moving in a straight line experiences an acceleration of

t6

1cos4– cm s–2 at time t seconds. At time 0t the particle is at rest and has

a displacement of 144 cm relative to a fixed point O on the line. (i) Find the first positive time that the particle reaches the point O. (20)

2

2

d 14cos

6d

xt

t

d 1

4cosd 6

vt

t

1

d 4 cos d6

v t t

1

4 6sin6

v t c ... (5m)

1

24sin6

v t c

When 0, 0t v : 0 24sin 0 c 0c The unique solution is

tv6

1sin24 ... (5m)

Then

d 1

24sind 6

xt

t

1

d 24 sin d6

x t t

1

24 6cos6

x t d

1

144cos6

x t d

When 0, 144t x 144 144cos0 d 144 144 d 0d The unique solution is:

1

144cos6

x t ... (5m)

When 0x :

1

0 144cos6

t


1

cos 06

t

1

6 2t

3 st 425.9t ... (5m) (ii) Show that the particle is moving with simple harmonic motion. (5)

acc 1

4cos6

t

acc 4144

x

acc 1

36x

As this is in the form

acc 2x ... (5m) the particle is moving with simple harmonic motion. (b) A particle moving in a straight line of mass m is acted upon by a force of

magnitude 5

2

x

m directed away from a fixed point O on the line, where x is the

distance of the particle from O. The particle starts from rest at a distance d from O.

Show that the velocity of the particle tends to a limit of 2

1

d. (25)

2

2 5

d 2

d

x mm

t x

5

d 2

d

vv

x x ... (5m)

5d 2 dv v x x

2 41 12

2 4 2

cv x

24

1 1 1.

2 2 2

cv

x



24

1v c

x ... (5m)

0v when x d :

4

10 c

d

4

1c

d

Unique solution:

24 4

1 1v

x d

4 4

1 1v

d x ... (5m)

In the limit as x ,

4 4

1 1lim limx x

vd x

... (5m)

4 4

1 1limxd x

4

10

d

2

1

d . ... (5m)


Notes:


Notes:


Notes:


Documents

2015.1 L.41-42 ms -11 ha - M. Selkirk Confey College · 2015.1 L.41/42_MS 8/52 Page 8 of 48 exams DEB 3. A particle is projected from a point O on horizontal ground with an initial