93/01(a) Semester 1, 2009 Page 1 of 1

THE UNIVERSITY OF SYDNEY

PHYS 1001

PHYSICS 1 (REGULAR)

Solutions

JUNE 2010

Time allowed: THREE Hours

MARKS FOR QUESTIONS ARE AS INDICATED TOTAL: 90 MARKS

INSTRUCTIONS All questions are to be answered. Use a separate answer book for section A and section B. All answers should include explanations in terms of physical principles. DATA Density of fresh water = 3 31.000 10 kg.m

Free fall acceleration at earth's surface g = 29.80m.s

Gravitational constant G = 11 2 26.67 10 N.m .kg

Speed of light in a vacuum c = 8 13.00 10 m.s

Speed of sound in air v = 1344 m.s

Avogadro constant NA = 1236.023 10 mol

Universal gas constant R = 1 18.314 J.mol .K

Boltzmann constant k = 1231.380 10 J.K

Stefan-Boltzmann constant = 8 25.67 10 W.m .K 4

ADV_Q01=REG_Q01 Question 1 A mass of 2m is connected to two masses of mass m and 3m as shown below. The strings connecting the masses are light. The pulleys and the slopes are frictionless.

(a) Draw separate free body diagrams for each of the masses, taking care to identify all forces

acting.

(b) If the angles of the slopes are 30 , = and the system is released from rest, does the hanging mass 2m remain where it is, accelerate upwards or accelerate downwards? Justify your answer.

(5 marks)

Solution (a)

(1 mark for left diagram; mark each for similar right diagrams) (b) Take vertically upwards as the positive direction for the 2m mass and downwards along the slope as the positive direction for the m and 3m masses Apply Newtons Second Law to the 2m mass as follows:

1 2

1 2

2 22 2

T T m g m aT T m a m g+ =

+ = + (1)

Apply Newtons Second Law to the m mass as follows: 1

1

sinsin

T m g m aT m g m a

= =

(2)

Apply Newtons Second Law to the 3m mass as follows: 2

2

3 sin 33 sin 3

T m g m aT m g m a

= =

(3)

Equation (1) Equation (2) Equation (3)

1 2 1 2( )( 2 2 ) ( sin ) (3 3 sin )6 2 4 sin0

6 2 4 sin( 2 sin ) / 3

T T T Tm a m g ma mg m a m g

m a m g m g

m a m g m ga g g

+ = + + = = =

For 30 = we have sin 0.5 =

( 2 (0.5)) / 3 0a g g= = . So the system is in equilibrium. Alternate Derivation With insight, assume system is equilibrium and calculate tensions 1T and 2T finding that 4 sin 2m g m g =

(1 mark for answer; 2 marks for valid justification)

REG_Q02 Question 2

A projectile of mass 2 m is fired with a speed u at an angle above the horizontal. At the highest point of its trajectory the projectile explodes into two fragments of equal mass, one of which falls vertically with zero initial speed. The range of the projectile if it had not exploded is R . Ignore air resistance.

(a) Is momentum conserved during the explosion? Briefly explain your answer.

(b) How far from the point of firing does the other fragment land, assuming that the terrain is level? Express your answer in terms of R .

(b) By what factor has the kinetic energy of the projectile changed just after the explosion when compared with its value just before the explosion?

(5 marks)

Solution (a) Momentum is conserved during the explosion because the forces acting on the two fragments as a result of the explosion are internal to the projectile system. (1 mark) (b) The backward fragment falls at distance 1 / 2x R= from point of firing since at the highest point the projectile has travelled half the range. The centre of mass falls at distance R since the centre of mass is unaffected by the explosion, which is internal to the projectile system. Taking the origin at the launch point and from the centre of mass formula we derive the value for

2x , the point at which the forward fragment lands.

1 1 2 2

1 21

22

122

32 2

22

cmm x m xx

m mm R mxR

mR R x

x R

+=

++

=

= +

=

The second fragment lands at 3 / 2R from the launch point. Alternate method using conservation of momentum Take xu as the horizontal velocity of the projectile just before the explosion and xv as the velocity of the forward fragment just after the explosion. Using conservation of momentum we have 2 (0)

2x x

x x

mu m mvv u

= + =

We can treat the horizontal and vertical motions separately. The project would have travelled a further distance / 2R from the peak of the path until it hit the ground. The forward fragment travels at twice the horizontal speed and hence goes a distance 2 / 2R R= until it hits the ground. The total distance that it has travelled from the origin is therefore / 2 3 / 2R R R+ = (2 marks) (c) Kinetic energy before explosion: 2 212 2 x xmu mu= Kinetic energy after explosion: 212 0xm v + From conservation of momentum at explosion:

2 02x x

x x

mu mvv u

= +=

Therefore kinetic energy after explosion: ( )22 21 12 2 2 2x x xm v m u mu= =

Therefore kinetic energy has increased by a factor 2. (2 marks)

REG_Q03 Question 3

Albert stands on a rotating platform with his arms outstretched holding lead weights in each hand. When Alberts arms are pulled radially inwards his angular velocity increases.

(a) Justify the application of conservation of angular momentum to this situation.

(b) Apply the conservation of angular momentum to this situation to explain the increase in Alberts angular velocity.

(c) If instead of pulling his arms inwards, Albert dropped the weights, describe what happens to his motion. Justify your answer.

(5 marks)

Solution (a) There is no external angular torque acting on the system of Albert, the rotating platform, and the weights. Therefore the angular momentum of the system will be conserved. (1 mark) (b)

1 1 2 2I I = ; the act of pulling arms inwards reduces the moment of inertia of the system. Hence

2 1I I< , therefore from the conservation law 2 1 > and the angular velocity increases. (1 mark) (c) Nothing happens. Alberts angular velocity remains unchanged. (1 mark) The system includes the weights. The angular momentum that they had before being dropped is the same as after. Also, Alberts angular momentum before the weights were dropped is the same as after. The fact that the weights horizontal component of motion is in a straight line after dropping does not change the angular momentum. Hence nothing changes. (2 marks)

ADV_Q04=REG_Q04 Question 4 (a) Before giving you an injection, a physician swabs your arm with isopropyl alcohol, a

volatile liquid at room temperature.

(i) Why does this make your arm feel cold?

Because of a fear of needles, you sit on a chair gripping the metal armrests with your hands.

(ii) Why do the metal armrests feel cool to the touch?

(b) Use the concepts of the kinetic-molecular model to explain why the pressure of an ideal gas in a container increases as heat is added to it.

(c) Explain in terms of entropy and the Second Law of Thermodynamics why air molecules in a sealed room do not spontaneously collect into one half of the room, occupying half the volume and leaving a vacuum in the other half. Your answer should include consideration of the number of microscopic states of the molecules in the room.

(5 marks)

Solution (a) (i) Evaporative cooling To evaporate the alcohol requires heat (its latent heat of

vaporization). This comes from the persons skin leaving it cooler than before. (1 mark) (ii) Conduction metal is a good thermal conductor, so it effectively transfers away the heat

from your hands. (1 mark). (b)

pV N k T= and 23 12 2p avg

k T K m v= = which is the average translational kinetic energy of the

gas molecules. When heat is added to the gas, the average molecular speed increases and so does T and .p (1 mark) (c) Answer 1 The entropy statement of the Second Law says that the entropy of a closed system can never decrease, so a closed system can never spontaneously undergo a process that decreases the number of possible microstates.

lnS k w= where, S is entropy and, w is the number of possible microstates. The Second Law states that

0S and so the number of microstates cannot decrease (which would be the case if all molecules were in one half of the volume). Answer 2 The Second Law states that the entropy or degree of disorder of a closed system increases or remains constant. Putting all molecules into a half of the volume of the room corresponds to greater order (i.e. decreased disorder). It is therefore inconsistent with The Second Law. Answer 3 If N air molecules spontaneously collected in half the volume of the room, then the number of possible microstates would decrease by a factor 2N . The probability of finding a given molecule in one half of the room is so the probability of finding all N molecules in half the room is

12

N

which for a large number of molecules is very small.

(2 marks for reasonable explanation)

REG_Q05 Question 5 The vertical displacement of a wave as a function of horizontal displacement x and time t is given by

( , ) 0.5cos(6.3 3.1 ),y x t x t= + where ,x y are in metres, t is in seconds.

(a) Which of the following types of mechanical waves describes the wave: transverse, longitudinal, neither, or both?

(b) Calculate the wavelength of the wave.

(c) Calculate the speed of the wave.

(d) Another wave is generated in the same medium with a vertical displacement given by

( , ) 0.5cos( 6.3 3.1 ).y x t x t= +

Give a brief, qualitative description of this second wave compared to the one above.

(e) Consider both waves travelling in the medium simultaneously. Give a brief, qualitative description of the resultant wave behaviour.

(5 marks)

Solution (a) The wave is a transverse wave (1 mark) (b) The wavelength is given by

2 1.0 m.6.3 = =

(1 mark) (c) The speed is given by

( )10.5 m.s .

3.1/ 2v

T

= = =

(1 mark) (d) The second wave is identical to the first, but travelling in the opposite direction. (1 mark) (e) Using the principle of superposition, the resultant wave is a standing wave. (1 mark)

REG_Q06 Question 6 (a) Two equal masses are attached to separate identical vertical springs next to one another

(see diagram below). One mass is pulled so its spring stretches 20 cm and the other pulled so its spring stretches only 10 cm. The masses are released simultaneously. Which mass passes the equilibrium position first? Justify your answer in physical terms.

(b) The string of a guitar with fundamental frequency, 256 Hz, is plucked while two tuning forks (two-pronged forks which can vibrate with a pure musical tone, see picture below) are on a table nearby. The natural frequencies of the two tuning forks are 512 Hz and 384 Hz, respectively, and they are silent before the string is plucked. Discuss whether you think either of the tuning forks will start to vibrate and why.

(5 marks)

Solution (a) The spring-mass systems are identical, and the period of oscillation depends only on the force

constant and the mass kTm

= which is therefore also the same for both systems. As they are

released at the same time from maximum displacement they reach equilibrium at the same time (they are always in phase). Some students might mention that the extra displacement by one of the masses is exactly matched/compensated by the extra force (and hence acceleration) because F k x= (2 marks) (b) The harmonics of the guitar string are 256, 512, 768, 1024 Hzetc. When plucked the harmonic content of the sound produced by the guitar will include components of all harmonics. The tuning fork with natural frequency 512 Hz will therefore be able to resonate with the guitar string while the other fork will remain silent. (1 marks for correct answer; 2 marks for suitable justification)

FND_Q09=REG_Q07 Question 9 or 7

A ballistic pendulum can be used to determine the speed of a bullet fired into it. The above diagram shows an example of such a pendulum.

A bullet of mass 0.10 kg is fired horizontally at a speed v into the block of wood of mass 1.0 kg, which is suspended motionless from the ceiling by a string. The distance from the ceiling to the point of impact at the centre of the block is 0.50 m . The bullet stops in the block.

(a) Write an expression for the following quantities in terms of the initial speed of the bullet v which is not yet known:

(i) the kinetic energy of the bullet just before the impact;

(ii) the momentum of the bullet just before the impact;

(iii) the speed of the block (with the bullet embedded in it) just after the collision;

(iv) the kinetic energy of the block (with the bullet embedded in it) just after the collision.

(b) Derive an expression for the maximum height h above the test position that the block (with bullet embedded in it) reaches after the collision.

Suppose that a bullet is fired into the block with a speed such that the block rises until the string is horizontal.

(c) Calculate the value of the initial speed of the bullet.

(10 marks)

Solution Deduct mark one time for missing units. No units are required when no specific values have been substituted. (a)

(i) 2 21 0.050 J2

K m v v= =

(1 mark) (ii) 10.10 kg.m.sp mv v = =

(1 mark) (iii) All forces are internal to the system, momentum is therefore conserved, and so after beforep p= Therefore

( )

10.10 0.0909 m.s1.10

after after

after

p m M v mv

mv vm M

v

= + =

=+

= =

(1 mark for method, 1 mark for correct result) (iv) The kinetic energy after the collision is

( ) 2

2

2

12(0.5)(1.10)(0.0909 )0.004545 J

afterK m M v

vv

= +

=

=

(1 mark for method, 1 mark for correct result) (b) The block rises to a height h until all the kinetic energy is transformed into potential energy:

( )( )1.10 (9.80) 10.78 J

K m M g h

h h

= +

= =

(2 marks for energy transformation: numerical value not required) (c) If the string is horizontal, then 50cm 0.50 mh = = so

2

2

1

0.004545 (10.78)(0.50) 5.395.39 / 0.004545 1185.8

34.4 m.s

vvv

= =

= =

=

(2 marks)

REG_Q08 Question 8

A 70 kg skier starts from rest at the top of a ski slope that is inclined at an angle of 10 degrees to the horizontal. The slope is 200 m long and the coefficient of kinetic friction between the slope and the skier is 0.075k = .

(a) Draw a diagram showing the forces acting on the skier as she slides down the hill (Assume that she does not use her ski poles).

(b) Calculate her acceleration down the slope.

(c) Calculate her speed at the bottom of the slope.

(d) Calculate the work done by each of the forces shown in your answer to part (a), during the time that the skier comes down the slope.

(e) Work is a measure of energy transfer. What are the energy transformations associated with the work done by each of the forces in your answer to part (d)?

(10 marks)

Solution 70 kgm = 10 = 200 mx = 0.075k =

(a)

W m g= sinXW m g = cosYW m g = kf N= (3 marks) (b) Apply Newtons Second law to the X and Y directions.

+Y direction: 0 cosY YF N W N m g = = = +X direction: sin cosX X X X kF W f ma ma mg m g = = =

So therefore

( )2

sin cos

(9.80)(sin10 .(0.075)cos10 ) 0.98m.sX ka g

=

= =

(2 marks) (c) Assume constant acceleration, speed at the bottom is therefore

2 20 2v v a x= +

where 0 0.0v = ; 20.98m.sa = 200mx =

and therefore 1(2)(0.98)(200) (19.8) 20m.s .v = = =

(2 marks)

(d) Work done by the normal force 0= since the angle between the force and the displacement is 90 . Work done by weight 4sin (70)(9.80)(200)(sin10 J.) 2.4 10m g x = = = Work done by friction 4cos (0.075)(70)(200)(9.80)(cos10 ) 1.0 10 J.kf x mgx = = = (2 marks) (e) Work done by weight causes an increase in the kinetic energy of the skier. Work done by friction increases the internal energy of (heats) the snow and skier. (1 mark)

REG_Q09 Question 9

A 1.00 mole sample of monatomic ideal gas contained in a cylinder with a movable piston goes through a thermodynamic cycle composed of the following four thermodynamic processes:

(1) The system starts from state A and after an adiabatic expansion process it reaches state B;

(2) The system starts from state B and after an isobaric compression process it reaches state C;

(3) The system starts from state C and after an adiabatic compression it reaches state D;

(4) The system starts from state D and returns to its initial state A after an isochoric process.

The table below gives data for pressure, volume and temperature at various (but not all) points in these processes.

Pressure (Pa) Volume

(m3) Temperature

(K)

state A 48.00 10 460

state B 44.24 10 27.00 10 357

state C 25.80 10 296

state D 24.78 10

(a) What is the volume of the ideal gas at the state A?

(b) What is the pressure of the ideal gas at state D?

(c) What is the temperature of the ideal gas at state D?

(d) From state B to state C, what is the work done by the gas?

(e) From state B to state C, how much heat leaves the gas?

(f) Sketch a pV diagram to show the whole cyclic process A B C D .

(10 marks)

Solution

Slightly different results are obtained for 1.67 = or 5 1.66.....3

= = Mark either as correct.

(a) Ideal gas equation gives:

1 12 3

4

(1.00mol)(8.314J.mol .K )(460K) 4.78 10 m(8.00 10 Pa)

AA

A

n RTVp

= = =

.

(1 mark) (b) State B C is isobaric so

44.24 10 PaC Bp p= = (1 mark) State C D is adiabatic so

CD C

D

Vp pV

=

53

= since there are 3 degrees of freedom for an ideal monatomic gas.

(1 mark) 5/32 3

42 3

4 5/3 4

(5.80 10 m )(4.24 10 Pa)(4.78 10 m )

(4.24 10 Pa)(1.213) 5.85 10 Pa

= = =

(1 mark) (c) Ideal gas equation gives:

4 2 3

1 1

(5.85 10 Pa)(4.78 10 m ) 336K(1.00mol)(8.314 J.mol .K )

D DD

p VTn R

= = = .

(1 mark) (d)

2

1

V

V

W p dV=

State B C is isobaric so pressure is constant and equal to 44.24 10 Pa.Bp = So

( ) ( )4 2 3 2 34 2 3

(4.24 10 Pa) 5.80 10 m 7.00 10 m

(4.24 10 Pa)(1.20 10 m ) 509 JB C BW p V V

= =

= =

(1 mark) (e) Method 1 State B to State C is an isobaric (constant pressure) change for a monatomic gas. So

( ) ( )1 15(1.00 mol) (8.314 J.mol .K ) 296K 357 K2

1270 J

BC p C BQ nC T T = =

=

Method 2 Change in internal energy for a monatomic gas is given by:

( ) ( )1 1

32

(1.5)(1.00mol)(8.314J.mol .K )(296K 357 K)761J

bc V c b c bU nC T T n R T T

= =

= =

Heat added to gas is given by

761 5091270 J

Q U W= += =

In either case, the amount of heat that leaves the gas is 1270 J. (2 marks) (f)

(only sketch required 2 marks)

ADV_Q10=REG_Q10 Question 10 (a) An engineer has devised two heat engines, A and B, the detailed energy-flow diagrams of

which are shown below.

In the diagrams, HT is the temperature of a hot reservoir and CT is that of a cold

reservoir. HQ and CQ represent the quantities of heat absorbed and rejected by the engine during one cycle. Check and explain whether both engines were designed by the engineer using correct thermodynamic principles.

(b) A man enters a sauna where the air temperature is o47 CairT = . His skin temperature is

o36 CskinT = . Assuming he is completely naked and his skin surface area is 21.5m ,A =

calculate the net rate (in watts) at which the mans skin is heated by:

(i) Conduction. Assume that the heat is conducted to the skin through a 10 mm thick layer of air (thermal conductivity 1 10.024 W.m .Kk = );

(ii) Radiation. Assume that the emissivity of the skin is given by 1.0e = .

(iii) In addition to the energy transferred to his skin, the man is also generating heat by internal metabolic processes. If the total energy reaching his skin is at a rate of 300 W, at what rate (in litres per hour) must perspiration evaporate from the mans skin in order to maintain a constant skin temperature? The heat of vaporisation of water at o36 C is 6 12.4 10 J.kg .

(10 marks)

Solution (a)

The First Law of Thermodynamics Work Done First Law

H CW Q Q=

Consistent with First Law

Heat Engine A

600 J

400 J

no

Heat Engine B

500 J

500 J

yes

Engine B is consistent with the First Law but Engine A is not consistent with the First Law. The Second Law of Thermodynamics (engine efficiency) Efficiency

H

WeQ

=

Maximum Efficiency

max 1 CH

TeT

=

Efficiency maxe e

Heat Engine A

0.60

0.50

no

Heat Engine B

0.50

0.40

no

Engine B violates the Second Law according to the maximum efficiency formula (which incorporates the First Law in its derivation). Engine B is therefore not a viable engine.

The Second Law of Thermodynamics (entropy change) Entropy Change

0

total hot engine cold

CH

H C

S S S S

QQT T

= + +

= + +

Consistent with Second Law

0S

Heat Engine A

11000 6000 0.33 J.K

600 300 + + = +

yes

Heat Engine B

11000 5000 0.33 J.K

500 300 + + =

no

According to this analysis of the Second Law, Engine B is not viable but Engine A is. However, Engine A is not compatible with the First Law and so is not a viable engine. (Engine A: 1 mark for analysis and 1 mark for conclusion. No need to use Second Law if stated as not viable because of violation of First Law.) Engine B (1 mark for analysis, 1 mark for conclusion) (b)(i) Rate of heating by conduction:

1 1 2

( )

47 K 36 K(0.024 W.m .K )(1.5m )0.01 m

39.6 40 W

air skinT TH k AL

=

=

= =

(2 marks) (ii) Rate of radiative heating:

( ) ( )

4 4

4 42 8 1 4

2 8 1 4 9 4

( )

(1.5m )(1.0)(5.67 10 W.m .K )( 320K 309K )

(1.5m )(1.0)(5.67 10 W.m .K )(1.37 10 K )116 120 W

air skinH Ae T T

=

=

= = =

(2 marks)

(iii)

/ totvv v

Hdm dQ dtQ m Ldt L L

= = =

(1 mark) Rate of perspiration (mass) loss is given by:

1

4 16 1

300J.s 1.25 10 kg.s2.4 10 J.kg

tot

v

HL

= =

( mark) Rate of perspiration (volume) loss is given by:

4 3

7 3 11.25 10 m 1.25 10 m .s1000s

dVdt

= =

(this assumes fresh water with density 31000kg.m ) This is converted to 1L.hr as ( )( )7 11.25 10 1000 (60)(60) 0.45L.h (litres per hour) = ( mark) Some students will add answers to part (i) and (ii) to the 300 W . Give full marks if otherwise correct.

REG_Q11=ADV_Q11 Question 11

Figure A shows a 1.0 m long pipe which is closed at one end and open at the other. The air in the pipe is vibrating as a standing wave. Figure B shows a representation of the displacement of an air molecule relative to its equilibrium position as a function of position x along the pipe at one instant of time.

(a) Copy Figure B and add to it a dashed line showing the displacement at a time one half period later.

(b) Make a new figure in your answer book, similar to Figure B but showing the pressure variation as a function of position x along the pipe at the same instant of time as in Figure B.

(c) The speed of sound in air is 1344 m.s . Calculate the wavelength and frequency of the standing wave mode shown in Figure B.

(d) What is the fundamental frequency of the pipe?

The pipe is now filled with helium (speed of sound in helium is 1965m.s ).

(e) What are the wavelength and frequency for the mode shown in Figure B in this case?

(10 marks)

Solution (a) The curve / 2T later is the inverted curve as shown below (dashed line)

(2 marks) (b) The correct pressure curve should show pressure nodes where there are displacement anti-nodes and vice versa. The pressure varies around atmospheric pressure 0p as in the diagram below.

(1 mark for this or same but upside down) The correct phase would have the curve being less than atmospheric pressure at 0x = as above. The correct phase can be seen from the positive to negative transitions on the displacement curve. While positive the displacement of molecules is in the x+ direction and when then negative the displacement is towards the x direction creating a higher than atmospheric pressure around the transition point. (1 mark for correct phase) (c)

4 m 0.8m5

= =

(1 mark) 1344m.s 430Hz

0.8mvf

= = =

(1 mark)

(d) For fundamental

4.0 m = (1 mark)

1344 m.s 86Hz4 m

vf

= = =

(1 mark) (e)

0.8m = (1 mark)

1965m.s (1206)1210Hz0.8m

vf

= = =

(1 mark)

REG_Q12 Question 12

A block is attached to a horizontal spring with a spring constant of 25.0kg.s . The block is displaced 0.5 m from equilibrium and released. The block exhibits simple harmonic motion with a period of 4.0s . The block is moving on a frictionless surface, and the spring is of negligible mass (see Figure A).

(a) What is the mass of the block?

(b) What is the velocity of the block 1.0 seconds after it is released? (Hint: Consider the energy of the oscillating system).

The block is again displaced 0.5 m from its equilibrium position. At the instant it is released, it is given an initial velocity of 11.0m.s in the direction toward its equilibrium position.

(c) What is the amplitude of the blocks motion? (Hint: Consider the energy of the oscillating system).

The block is now vertically suspended by a 1.0 m long unstretchable string of negligible mass (see Figure B). The block is displaced 5 degrees from its equilibrium position and oscillates as a simple pendulum under the influence of Earths gravity.

(d) What is the period of oscillation of the block?

The block is now displaced 70 degrees from equilibrium and is released.

(e) Can the motion of the block be described as simple harmonic motion? Why or why not?

(10 marks)

Solution (a)

Using 2 2 kTm

= =

solve for m

22 4.05.0 2.03= 2.0 kg

2 (2)(3.14)Tm k

= = =

(2 marks) (b) The block is at the equilibrium position ( )0x = after 1.0s which is a quarter of the period of the oscillation. (1 mark)

Energy at initial point 0x (entirely elastic potential energy) is: 2

012

k x .

Energy at equilibrium point (entirely kinetic energy) is: 212

m v .

Energy is conserved and therefore

105.00.5 0.78 0.8m.s .2.03

kv xm

= = = =

(1 mark, give mark for using the mass from part (a) even if it is wrong) (c) Use energy equation and 11.0m.sv = , 0 0.5mx = . Initial energy is:

2 20

1 12 2

k x m v+ .

Energy at the amplitude of the oscillation is: 21

2k A .

These are equal so

2 2 20

22

0

22

1 1 12 2 2

(2.03)(1.0)(0.5)(5.0)

0.81 0.8m

k A k x mv

mvA xk

= +

= +

= +

= =

(1 mark for setting up problem with energy equation; 1 mark for answer)

(d)

1.02 (2)(3.14) 2.0s.9.80

LTg

= = =

(2 marks) (e) The motion is not described as simple harmonic motion. (1 mark) A pendulum exhibits simple harmonic motion only when the displacement from equilibrium is small (e.g. sin ). (1 mark)

REG_2010_headerREG_Q01=ADV_Q01REG_Q02REG_Q03ADV_Q04=REG_Q04REG_Q05REG_Q06FND_Q09=REG_Q07REG_Q08REG_Q09ADV_Q10=REG_Q10REG_Q11=ADV_Q11REG_Q12