Stpm 960/2 Physics 2007

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PN Wan Marzuni Bt Wan Mohamad

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STPM

960/2 PHYSICS 2007

Q1 /2007A cylinder of mass 10 kg and radius 0.2 m rolls down an inclined plane of height 1.0 m without slipping from rest. Calculate the linear velocity of the cylinder when it reaches the ground. [4] [ Moment of Inertia of a cylinder of mass m and radius is r ]

U is transform to K, Ui = Kf-total

2

222

22

4

3

2

1

2

1

2

1

2

1

2

1

mv

r

vmrmv

IwmvK total

1

2

2

2

6.3

08.13

)0.1)(81.9(3

44

3

msv

v

v

mghmv…1

……………1

…..1

……1

Q2/2007 (a) On the same axes, sketch graphs of displacement against time to show underdamped, critically damped and overdamped oscillation. Label your graphs. [3] (b) What is meant by resonance in forced oscillation? [2]

(b) Resonant occurs when the forced frequency equals the natural frequency and at this instant the amplitude of the oscillation is maximum.

underdamped

overdampedCritically damped

Q3/2007(a) In each of the following cases, state any change in the interference pattern of a two-slit arrangement.

(i) One slit is covered by an opaque material. [1](ii) The distance between the slit is decrease. [1]

(b) Explain how interference can be used to determine the flatness of lens surface. [2]

(a) (i) The interference pattern changes to a diffraction pattern of a single slit. …………….1(ii) d decreases, therefore ∆y increases. ……1

(b) Some have the same optical path difference………………………………….1

During the interference, the maximum bright and the minimum dark intensity pattern is seen distorted/irregular ……..1

Q4/2007. A cylindrical brass rod with Young’s modulus 9.7 x 1010 Pa and original diameter 10.0 mm experiences only elastic deformation when a tensile load of 200 N is applied. Calculate

(a) the stress that produces the deformation. [3]

(b) If the original length of the rod is 0.25 m, calculate the change in the length of the rod. [2]

Pa

A

F

6

23

0

105.2

4

)1010(

200

m

E

ll

strain

stressE

6

9

6

0

104.6

1097

)25.0)(105.2(

…………1

...1

…..1

…….1

….1

Q5/2007A 1.5 F capacitor and a 2.0F capacitor are

connected in series across a 24 V source. (a)Calculate the equivalent capacitance

across the source. [2](b)(b) Calculate the charge stored on each

capacitor [2]

(c) If a dielectric is added to each capacitor, explain what happens to the charge stored on each capacitor [2]

.

FC

CCC

eq

eq

86.0

0.2

1

5.1

1111

21

(a)

(b)Q = CeqV …………………………...1 = 0.86 x 24 = 20.6 μC …………………………1

(c) The charge will increase because dielectric increases the capacitance of each capacitor …………………………1,1

.........................1

......................................1

Q6/2007 An ideal solenoid consists of 1000 turns of wire per cm wound around an air filled cylindrical structure. The solenoid is 2.0 cm long and cross-sectional area 1.8 cm2. A current of 2.0 A passes through the wire.

(a) Calculate the magnetic flux in the solenoid. [3] (b)Calculate the self-inductance of the solenoid. [2]

l

INB

ABB

0

.

Wb

Al

IN

5

47

0

1052.4

108.102.0

20000.2104

…………………………..1

…………………………..1

……..…………….1

HL

L

I

NL B

2

5

1052.4

0.2

)1052.4()1000(

(b) Self-inductance

………………………….1

…………… …1

Q7/2007A light beam of wavelength 0.110 nm

collides with an atom. After the collision, an electron is emitted with kinetic energy 180 eV.

(a)Calculate the energy absorbed by the atom. [3]

(b)Calculate the velocity of the electron emitted. [2]

00

00

hc

E

hfE

eV

eV4

2

9

819

34

1013.1

1001.113

10111.0

)1000.3(1060.11063.6

(a) …………………………..….1

…………………………..….1

………………………1

16

31

19

2

1095.7

1011.9

1060.11802

2

1

msv

v

vmKE e(b) ………………..1

……………1

Q8/2007 In a nuclear reactor, a very slow moving neutron is absorbed by a stationary boron atom. The equation for the nuclear reaction is

HeLiBn 42

73

105

10

After the reaction, the speed of the helium atom is 9.10 x 106 ms-1 and the kinetic energy of the neutron is approximately zero. [ me = 1.0087 u, mB = 10.0130 u, mLi = 7.0160 u, mHe = 4.0026 u]

KE of Li = Δ mc2 – KE of He ………………………1

Δmc2 = (1.0087u – 7.0160u – 4.0026u + 10.0130u) c2

= 3.1 x 10-3 x 1.66 x 10-27 x (3.0 x 108)2

= 4.631 x 10-13 J ………………………………1

KE of Li = 4.631x10-13 –

½(4.0026 x1.66 x 10-27)(9.10x106)2 ………….1

= 1.88 x 10-13 eV

or = 1.88 MeV ………………………………..1

(a) Calculate the kinetic energy og Lithium atom after the reaction.

(b) Calculate the reaction energy. [2]

Reaction energy = (mn + mB – mLi - mHe ) c2

19

2827

106.1

)1000.3)(1066.1)(0026.40160.70130.100087.1(

Mev89.2

Q9 (a) An object is projected with initial speed v at an angle θ with the horizontal. Derive an equation for the path of the projectile. [3]

Horizontal displacement, x = (v cos θ)t …….1

Vertical displacement, y = (v sin θ)t – ½ gt2...1

Substitute for t,

……...1

222

2

cos2tan

cos2

1

cos)sin(

xv

gxy

v

xg

v

xvy

9(b) The diagram below shows a jeep travelling at constant speed 10 ms-1 towards a security post.

Security

post

bulletjeep

0.7 km

15 m

A security guard holding his rifle 15 m above the ground shoots horizontally when the jeep is 0.7 km from the security post. The bullet strikes the jeep. Calculate

(i) The time taken by the bullet to strike the jeep. [2]

Using s = ut + ½ gt2 ……………………………1

15 = 0 + ½ (9.81) t2

t = 1.75 s ……………………………….1

(ii) The distance of the jeep from the security post when it is struck, [2]

Distance of moving jeep is = 10 (1.75)

= 17.5 ……………1

Distance from security post = 700 -17.5

= 682.5 m ……...1

(iii) The initial speed of the bullet, [2]

Initial speed of the bullet,

u = distance from security post /time ..1

= ………………….1

139075.1

5.682 ms

2222 2.17390 yx vvv

(iv) The speed and direction of the bullet when it strike the jeep. [6]

The speed of the bullet when it strike the jeep’

vx = u = 390 ms-1

vy = v0 sin θ + gt ……………………………1

= 0 + 9.81 ( 1.75) = 17.2 ms-1

………….………1 .

V= 390.4 ms-1 …………………………..1

22

22

2.17390

v

vvv yx

x

y

v

vtan

390

2.17tantan 11

x

y

v

v

390

2.17tantan

tan

11

x

y

x

y

v

v

v

v

θ = 2.50 .…………………………1

The angle strike is 2.50 below the horizontal …………………1

………….1

10 (a) What is meant by Doppler effect? [2]

Doppler effect is the change in frequency …….1due to relative motion between the source and an observer. ……...………………………………1

(b) A toy car with the siren on is approaching a wall at a speed of 2.0 ms-1. The siren produces a sound wave in the form

y = 2.5 x 10-5 sin 2π ( 500 t – 1.4 x )

Where x is in metres and t in seconds. Calculate the speed and frequency of the sound wave. [4]

y = 2.5 x 10-5 sin 2π ( 500 t – 1.4 x )

1357

)4.1(2

)500(2

msv

v

k

wv

• ………..1

• …...1

W = 2 π f …...1

= 500 Hz …..1

(c)The sound wave in (b) is then reflected from the wall. A detector is fixed on the ground behind the source of the sound wave to observe any frequency changes. Determine (i) The frequencies detected by the detector. [4]

(i) The detector received the note from the reflected sound. Then

………………….1

………………….1Hzf

f

fvv

vf

s

8.502'

5002357

357'

'

The detector received the note from the reflected sound. Then

…………………1

…………………..1Hzf

f

fvv

vf

s

2.497"

5002357

357"

"

(ii) The beat frequency [2]

Beats, Δf =| f’ – f” | ……………………1

= 5.6 Hz ……………………1

(iii) The equation of the reflected wave [1]

y = 2.5 x 10-5 sin 2π ( 500 t + 1.4 x ) ……..1 (iv) The intensity ratio of the approaching wave to the reflected wave. [2]

Intensity (amplitude)2 .........................1

..…………………………….11

2

1 I

I

Q11 The following is a pV diagram for 0.2 mol ideal monoatomic gas.

p/Pa

3 x 105 PaA

C B

2.0 12.0V/(x10-3)m3

The gas expands isothermally from A to B and is compressed from B to C at constant pressure. It finally undergoes a constant volume process from C to A. Calculate,

(a)The temperature at A, B and C [5]

At point A,

…………………………..1

KT

nR

VpT

nRTVp

A

AAA

AAA

361

)31.8(2.0

)100.2)(100.3( 35

………………………………………1

At point B (Isothermal), therefore TA = TB

TB = 361 K ………………………….1

At point C constant pressure

KT

nR

VpT

Papp

PaV

nRTp

C

CCC

BC

BB

2.60

)31.8(2.0

)100.2)(100.5(

100.5

100.5

34

4

4

….……………………1

………………………………..1

(b) The nett work done during the cycle [5]

Work done

Along AB

JW

W

v

vnRTW

A

BA

3

3

3

1007.1

100.2

100.12)361)(31.8)(2.0(

ln

…………………..1

Along BC

JW

W

TTnRW

TnRW

BC

BC

CBBC

BC

500

)2.60361)(31.8(2.0

)(

.………………….1

………..1

ΔW = WAB – WBC ……………………1

= 1.07 X 103 – 5.00 X 102

= 5.7 X 102 J ………………….1

(c) The nett heat absorbed by the gas [5]

dQ = dU + dW ………………………….1

Since it is a complete cycle:

T is the same .......................................….1

dU = 0 …………………………1

dQ = dW ………………………..1

= 5.7 x 102 J ………………..1

Q 12 (a) A 0.3 μF capacitor is connected to a source of alternating current with output voltage V = 240 sin 120 πt.

(i) Calculate the reactance of the capacitor. [3]For a capacitor the reactance decreases as the frequency increases ………….1

wcX c

1

The frequency of the voltage source.

W = 120 π rad s-1 …...............1

8842

)103.0)(120(

16

c

c

X

X

……………….1

(ii) Determine the r.m.s. current flowing through the capacitor. [3]

AI

I

X

VI

VV

srm

rms

c

rmsrms

rms

2

0

1092.1

88422

240

2

……………….……1

……………….……1

………………1

(b)A rectifier is a device which conducts electric current in one direction only such as diode.

(i) Describe briefly a full wave rectifier.

A full-wave rectifier can be constructed by using 4 diodes which allow current flowing through a load (resistance R) in one direction only. ……………………………….…...............1

A full-wave rectifier can be constructed by using 2 diode with centre-tapped transformer which allow current flowing through a load (resistance R) in one direction only. ………………………………….1

When the voltage cycle is negative, diodes D2 and D4 conducts to allow current through the load from A to B. (Region 2) …………………………1

When the voltage cycle is positive, diode D1 and D3 conduct to complete the circuit and allow current to pass through the load from A to B. (region 1)………………1

(b) (ii) With the aid of diagrams describe briefly the process of smoothing rectified alternating current voltages. [5]

Smoothing is required to make the pulsating current constant. This is achieved by connecting a capacitor in parallel to the load resistance. ……………………………1

Smoothen voltage in bold reduces pulsating. ….1

The capacitor in parallel charges up when the current flows through AB. …………………….1

When the voltage across AB starts to drop ( source changes polarity), the capacitor discharges through the resistor, across AB. ……………1

The discharges keeps the voltage across R (or AB) up. By making the discharge long enough, i.e. used larger capacitance (τ = RC) the discharge through the resistor can continue until the source voltage become positively high again. ………………………..1

Q 13 (a) State Bohr’s postulate for an atom. [2]

Bohr’s postulate:The electron in an atom moves in a circular orbit about the nucleus ………1Only certain permissible/stable/quantised / stationary states circular orbit are allowed.or angular momentum = …………….1

Energy emitted only when electron transits from higher to lower orbit. …….1 any 2 marks

2nh

(b) The diagram below shows an electron of mass m and charge – e moving at speed v in a circular orbit of radius r around a nucleus.

v

m

+e

-e

r

If the force of attraction between the electron and the nucleus provides the centripetal acceleration of the electron, derive an expression for the radius of the nth orbit of the electron. [5]

(b) Centripetal force = Electrostatic force

r

kemv

r

ke

r

mv

22

2

22

04

1

k, …….( 1, 1)

……………(1)

Angular momentum for nth allowed orbit

nn

nn

mr

nhv

nhL

rmvL

2

2

……………………1

……………………1

………(2) n =1, 2, 3,…..

Substitute equation (2) in equation (1)

222

2

22

4

2

nmke

hr

r

ek

mr

nhm

n

nn

n = 1, 2, 3,……

………..1

(c) An electron in a Bohr orbit has kinetic energy 8.64 x 10-20 J. (i) Calculate the speed of the electron. [3]

Kn = ½ mv2 .............................................1

8.64 x 10-20 = ½ [9.11 x 10-31] v2 ……….1

15

31

20

1036.4

1011.9

1064.82

msv

v

………….1

(ii) Determine the allowed orbit. [3]

51063.61036.41085.82

)1060.1(

21036.4

2

34512

219

0

25

n

n

nh

ev

nhrmv

n

n

…………………………1

………………1

……………………………………1

(c) Calculate the radius of the orbit. [2]

mr

r

mv

er

r

e

r

mv

n

n

9

253112

219

20

2

20

22

1033.1

)1036.4)(1011.9)(1085.8(4

)1060.1(

4

4

……………………………..1

….1

Q 14 (a) The bombardment of a beryllium nucleus by an -particle produces a fundamental particle X, as follows:

XCBe 126

94

(i) Complete the equation above by giving the proton and nucleon numbers to the -particle and X. [2]

XCBe 10

126

94

42 …….2

(ii) What are -particle and X? [2]

= helium nucleus, X = neutron ……(1,1)

(iii) State two important properties which cause X difficult to be detected. [2]

-No charge / neutral / does not deflect by electric field or magnetic field ………………1

-Does not produce ionization effect / no interaction with matter ……………..1

(b) Determine the equivalent energy in MeV of a mass of 1 u. [5]

1 u = 1.66 x 10-27 kg ……………………….........1

From E = mc2 ………………………………………1

= (1.66 x 10-27) (3.00 x 108)2 (1.60 x 10-19)-1 ….1

= 9.34 x 108 eV …………………………………1

= 934 MeV …………………………………1

(c)An element of unknown atomic mass is mixed with a atom in a mass spectrometer. The radii of curvature of the element and are 26.2 cm and 22.4 cm respectively.

What is the possible element?

State any assumption you make. [4]

C126

C126

144.22

2.26

12

m

m

rm ……………………..1

……………………1

Nitrogen / carbon 14 ………………………….1

Assumption: Both are equally charged …….1

THANK YOU