1 Gain in KE = Loss of PE

2 (a)

(b) Resonance occurs when a system undergoing forced oscillation oscillates with the maximum amplitude when the driving frequency is slightly less than the natural frequency of the system.

3 (a) (i) A single slit diffraction pattern is formed.(ii) The fringe separation of the interference pattern increases.

[Use the formula : when a is decreased, y increases]

(b) The lens is placed on a flat piece of glass and a thin film interference pattern is formed by the reflected light.A uniform fringe separation will confirm the flatness of the lens surface.

4 (a)

(b)

5 (a)

C = 0.86 μF(b) For capacitors in series, the charge stored in each capacitor is equal to the total

charge stored.

Charge stored in each capacitor = CV = 0.86 μF x 24 =20.64 μC

(c) When a dielectric is added to each capacitor, the capacitance increases. The potential difference across the capacitor is constant.Therefore, the charge stored in the capacitor increases.

6 (a) Number of turns, N = 1000 x 2 = 2000

Magnetic flux,

(b) Self-inductance,

L = 4.52 x 10–2 H

7 (a) Energy absorbed by the atom = energy of a photon of the light beam

=

(b) KE of the electron, E = 180 eV = 180 x 1.60 x 10–19 J

.

Therefore,

8 (a) Mass defect, m = (1.0087 + 10.0130) – (7.0160 + 4.0026) = 0.0031 uTotal kinetic energy released, E = mc2

E = 0.0031 x 1.66 x10–27 x (3.00 x 108)2

E = 4.63 x 10–13 JThe total kinetic energy released consists of the KE of the Lithium atom and the KE of the Helium atom.Kinetic energy of the helium atom

=

Kinetic energy of the lithium atom = 4.63 x 10–13 – 2.75 x 10–13

= 1.88 x 10–13 JALTERNATIVE SOLUTIONUsing E = mc2,

Displacement

Time0

UnderdampedOverdampedCritically damped

Thin air flim

24 V

1.5 μF 2.0 μF

24 V

C

the KE equivalent to 1 u is, = 1.0 x 1.66 x 10–27 x (3.00 x 108)2

= 1.494 x 10–10 JIn a nuclear reaction, mass-energy is conserved.The conservation equation in terms of mass is:

The conservation equation in terms of kinetic energy is:

Let the kinetic of the Lithium atom = EL

[(1.0087 x 1.494 x 10–10) + 0 + (10.0130 x 1.494 x 10–10) + 0]= [(7.0160 x 1.494 x 10–10) + EL

+ (4.0026 x 1.494 x 10–10) + 2.75 x 10–13]EL = 1.88 x 10–13 J(b) The reaction energy is the energy released in the form of the total kinetic

energy of the decay products = 2.75 x 10–13 + 1.88 x 10–13 = 4.63 x 10–13 J

9 (a) The force in the object = – mg. Therefore, the acceleration of the object = –g.Let the time taken for the object to move from P to Q = tHorizontal displacement, …………..……. (1)

Vertical displacement,

………..... (2)

From (1),

Substitute into (2)

(b) (i) The time taken for the bullet to strike the jeep is equal to the time taken for the bullet to drop through a vertical height of 15 m.The initial vertical component of the velocity of the bullet = 0

Using the equation, , with u = 0, s = 15 m and g = 9.81 m s–2

(ii) Distance travelled by the jeep = 10 m s–1 x 1.75 s = 17.5 mDistance from the security post = 700 m – 17.5 m = 682.5 m

(iii) The bullet travels a horizontal distance of 682.5 m in 1.75 s

Initial speed of the bullet =

(iv) Vertical component of the velocity of the bullet = 0 + gt= 9.81 x 1.75= 17.17 m s–1

Speed of the bullet =

The speed of the bullet is 390.4 m s–1 at an angel 2.520 below the horizontal.

10 (a) Doppler effect is the apparent change in the observed frequency of a wave due to the relative motion of the source of the wave and the observer.

(b) . Compare with

Speed,

. Therefore frequency,

(Rest mass of n)+

(Mass equivalent to KE of n)+

(Rest mass of B)+

(Mass equivalent to KE of B)

(Rest mass of Li)+

(Mass equivalent to KE of Li)+

(Rest mass of He)+

(Mass equivalent to KE of He)

=

=

(KE equivalent to rest mass of n)+

(KE of n)+

(KE equivalent to rest mass of B)+

(KE of B)

(KE equivalent to rest mass of Li)+

(KE of Li)+

(KE equivalent to rest mass of He)+

(KE of He)

v sin θ

v cos θ

y

x

v

P

Q

390 m s–1

17.17 m s–1

θ

Wall Detector

v = 2 m s–1

(c) (i) Speed of sound waves = vSpeed of source of sound = uS

Frequency of sound given out by the sourceThe detector receives sound reflected from the wall and sound directly from the car.

Consider the Car and the WallThe car is a source moving towards the wall.

Apparent frequency of sound received by the wall,

Frequency of sound reflected by the wall and received by the detector = 503 HzConsider the Car and the Detector

The car is a source moving away from the detector.

Apparent frequency of sound received by the detector,

Frequency of sound from the car received directly by the detector = 497 Hz

(ii) Beat frequency = f1 – f2 = 503 – 497 = 6 Hz

(iii) The reflected travels in the opposite direction. The waves undergo a phase change of π due to reflection at the wall.Equation of the reflected wave is

(iv) Intensity is directly proportional to (amplitude)2

Since the approaching wave and reflected wave have equal amplitudes,the ratio of the intensities of the approaching wave and reflected wave is 1 : 1.

11 (a) Using , we have

At A,

TB = TA = 361 K since A to B is an isothermal expansion.

From B to C, pressure is constant. Therefore,

(b) From A to B, work done by the gas,

WAB = 1075 JFrom A to B, temperature is constant. Therefore, pAVA = pBVB

From B to C, work done on the gas, WBC = p(VC – VB)WBC = 5 x 104 (2.0 x 10–3 – 12.0 x 10–3)WBC = –500 J

From C to A, no work is done because the volume is constant.Net work done by the gas = 1075 – 500 = 575 J

(c) Use the First law of thermodynamics, From A to B, . Therefore, Q = W = 1075 JFrom B to C, pressure is constant. Therefore,

From C to B, volume is constant. Therefore,

Net heat absorbed by the gas = 1075 – 1251 + 750 = 574 J

12 (a) (i) Given, . Compare with, Therefore, V0 = 240 V and f = 60 Hz

Reactance, = 8.84 x 103 Ω

(ii) rms voltage,

rms current,

(b) (i) A full wave rectifier consists of 4 diodes connected in the form of a bridge circuit. W and X are connected to the AC supply.Y and Z are the outputterminals.In the first half cycle,diodes 1 and 3 are forwardbias; diodes 2 and 4 arereversed bias.The current flows through the load, R from A to B.In the next half cycle,diodes 2 and 4 are forward bias; diodes 1 and 3 arereversed bias.The current flows through the load, R in the same direction A to B.Therefore, in both half cycles, the current flows in thesame direction to produce a potential difference acrossthe load.(ii) Smoothing is achieved by connecting a capacitor

across the load. When the output voltage increases, the capacitor is charged.When the output voltage decreases, the capacitor dischargesthrough the load to maintain a steady voltage across the load.The capacitance should be large enough so that the time constant of the discharge

13 (a) Bohr's postulates1. There are certain allowed or permissible orbits for which the energy of the

electron is constant, that is, the electron does not transmit energy in the form of electromagnetic radiation.Energy is only transmitted when there is a transition of the electron from an orbit with higher energy to another orbit with lower energy.

2. In an allowed orbit, the angular momentum of the electron are integral

multiples of , where h = Planck’s constant.

(b) The electrostatic attraction between the nucleus (+e) and an electron (–e) acts as the centripetal force for the electron to orbit around the nucleus.

, that is, …….....................[1]

From Bohr’s postulate, angular momentum of the electron,

..……………….... [2] , n = 1, 2, 3, …….

From [2], .

Substitute into [1],

. Therefore,

(c) (i) .

(ii) From [1] ,

r = 1.33 x 10–9 m

From the answer to (b), , we have

n = 5The allowed orbit is n = 5.

(iii) The radius of the orbit = 1.33 x 10–9 m as calculated in (c)(ii)

Input

AC

R

Output

Varying DC

Z

W

Y

X

1

23

4

AB

AB

R

CSmoothed output

0Time

Outputvoltage

14 (a) (i)

(ii) The α-particle is the helium nucleus.X is a neutron.

(iii) 1. Carries no charge.2. Moves at high speed

(b) Using E = mc2 and 1 u = 1.66 x 10–27 kgE = (1 x 1.66 x 10–27) x (3.00 x 108)2 J1 J is equivalent to 1.60 x 10–19 eV of energy.

eV

E = 9.34 x 108 eV = 934 x 106 eV = 934 MeVThe energy equivalent to a mass of 1 u is 934 MeV.

(c) Let the element be X.When the ions of X and carbon-12 enter the magnetic field in the deflection chamber of the mass spectrometer, the magnetic force acts as the centripetal force. The ions move in a semicircle of radius r.

The radius of the circular path,

Assuming v, q and B to be constant,

Therefore, . That is, and mX = 14

The element could possibly be nitrogen. Assumptions: The ions of the element and carbon-12 carry the same charge and move with the same speed in the mass spectrometer.