Section A

1 U is transform to K, Ui = Kf-total

Ktotal =

=

=

Therefore = mgh or Ktotal = mgh or loss in P.E = gain K.E

v2 = (9.81)(1.0)

= 13.08

v = 3.6 m s1 or 4 m s1

2 (a)

(b) Resonant occurs when the forced frequency equals the natural frequency

or forced period equals the natural period

and at this instant the amplitude ( of the oscillation ) is maximum.

( 1 )overdamped

( 0 )

3 (a) (i) The interference pattern changes to a diffraction pattern (of a single slit.)

(ii) y =

d decreases, therefore y increases

or wider / further apart / less fringes

(b) The (optical) path / phase difference between some points on the imperfect lens

surface is different.

or

Some have the same optical path difference

During the interference,( the maximum / bright and the minimum / dark intensity)

pattern is seen distorted/uneven / irregular / not regular / width steadily

become smaller

4 (a) =

=

=

= 2.5 106 Pa or 2.546 106 Pa

(b)

=

= E or or

=

l =

= or or

2

= m or m or m

5 (a) =

=

Ceq = 0.86 F

(b) Q = Ceq V or

= 0.86 24

= 20.6 C or 21 C

(c) The charge will increase

because dielectric increases the capacitance of each capacitor

or

V initially decreases and further charging of capacitor

6 (a) B = B . A

B = or

=

=

= 4.52 105 Wb or 4.5 105 Wb

or

B = N B A = (nL) A µ0 n I

= n2 L A µ0 I

= (1000 102)2 (2 10-2) (1.8 10-4) (4 10-7) (2)

= 9.05 10-2 Wb or 9.0 10-2 Wb

(b) Self-inductance

L =

=

3

2

2

= 4.52 102 H

or

=

= 4.52 102 H or 4.5 102 H

7 (a) E0 = hfo

= or f0 =

=

= 113.01 102 eV

= 1.13 104 eV or 1.81 10-15 J

(b) KE = mev2

1

v =

= 7.95 106 m s1

8 (a) Since boron is at rest and neutron move very slowly, we can assume

the total momentum before reaction and after reaction is zero

or

mLi vLi = mHe vHe

Hence

Kinetic energy of the lithium

KELi = mLi vLi 2

= mLi

=

4

=

= 1.57 1013 J or

= 0.981 MeV

or

K.E of Li = – K.E of He

= 1.0087 + 10.0130 – 7.0160 – 4.0026

= 3.1 103 u

E =

= 3.1 103 1.66 1027 (3.0 108)2

= 4.631 1013 J

K.E of Li = 4.631 1013 – 4.0026 1.66 1027 (9.10 106)2

= 1.88 1013 J or

= 1.18 MeV

or

Conservation of momentum

= 1.57 1013 J

(b) Kinetic energy of the helium

KEHe = mHe vHe2

= (4.0026) (1.66 1027) (9.10 106)2

= 2.75 1013 J

= 1.72 MeV

The reaction energy = KELi + KEHe

= 0.981 MeV + 1.72 MeV

= 2.70 MeV

5

or

(b) Reaction energy = (mn + mB mLi mHe)c2

=

= 2.89 MeV

Section B

9 (a) Horizontal displacement x = (v cos) t

Vertical displacement y = (v sin) t

Substitute for t

(b) (i) Using s =

15 =

t = 1.75 s or 1.7 s

(ii) Distance of moving jeep is = 10(1.75) or = 10(1.7)

= 17.5 m = 17 m

Distance from security post = 700 – 17.5 = 700 – 17

= 682.5 m = 683 m

(iii) Initial speed of bullet u = or

= or =

= 390 m s1 = 402 m s-1

(iv) The speed of bullet when it strikes the jeep

vx = u = 390 m/s or = 402 m/s

vy = vo sin + gt or vy = vo sin ± gt or

= 0 + 9.81(1.75) = 0 + 9.81(1.7)

= 17.2 m s1 = 16.7 m s1

6

v = or

= (3902 + 17.22)0.5 = (4022 + 16.72)0.5

= 390.4 m s1 = 402 ms-1

tan =

= or =

= 2.5 = 2.4

the angle of strike is 2.5 below the horizontal or diagram

or 87.6 to the downward vertical

10 (a) Doppler effect is the change / shift in frequency due to relative motion

between a source and an observer.

(b) y1 = 2.5 10–5 sin 2 (500t – 1.4x),

= 357 m s-1

= 2f

f = 500 Hz

or

= 2f f = 500 Hz

m v =f = 500 0.714

= 357 m s-1 or 360 m s-1

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

7

f' =

f' =

= 502.8 Hz or 503 HzThe detector hears received the note directly from a sound source. Then,

f" =

f" =

= 497.2 Hz or 497 Hz

(ii) Beats, f = f – f"

= 5.6 Hz or 6.0 Hz

(iii) y2 = 2.5 10-5 sin 2 (500t + 1.4x)

(change sign)

(iv) Intensity (amplitude)2

11 (a) At point A pA VA = nRTA

TA =

=

= 361 K or 400 K

At point B (Isothermal) Therefore TA = TB

TB = 361 K or 400 K

At point C constant pressure

pV = nRT

= constant

8

TC = TB

= or =

= 60 K = 67 K or 70 K

or

= 5.0 104 Pa

K

(b) Work done

Along AB W = nRTA n

= (0.2)(8.31)(361) n

= 1.07 103 J

Along BC W = pC (V) or W = nR T

pC = or = nR ( TB - TC ) ( T= TB - TC )

=

= 5.0 104 Pa

W = pC (V)

= (5.0 104) (10.0 103)

= 5.0 102 J

W = WAB WBC

= 1.07 103 5.0 102

= 5.7 102 J or 6.0 102 J

(c) Net heat

Along BC Q = nCp (TC TB) ,

Q = nR (TC TB)

= (0.2) (8.31) (60 361)

= 1.25 103 J

9

Along CA Q = nCv (TA TC) , or

= nR (361 60)

= (0.2) (8.31) (301)

= 7.5 102 J

Along AB, Isothermal, therefore dU = 0

dQ = dW

= 1.07 103 J

Net heat = QAB + QBC + QCA

=1.07 103 1.25 103 + 7.5 102

= 5.7 102 J or 6.0 102 J

or

Net heat absorbed by gas = net work done by gas

= 5.7 102 J or 6.0 102 J

or

dQ = dU + dW

since it is a complete cycle, T is the same

dU = 0

dQ = dW

= 5.7 102 J or 6.0 102 J

12 (a) (i) (For a capacitor the reactance decreases as the frequency increases)

Xc =

The frequency of the voltage source

= 120 rad s1

Xc =

= 8842 or 9000

(ii) From Ohm’s law Io = 1

10

Io = or =

= 0.027 A

= 27 m A

The r.m.s. value Ir.m.s. =

=

= 0.019 A

= 19 m A or 20 m A

or

or

= 1.92 10-2 A

(b) (i) A full-wave rectifier can be constructed by using 4 diodes

which allow current flowing through a load (resistance R)

in one direction only.

or

or

A full-wave rectifier can be constructed by using 2 diodes

with centre-tapped transformer which allow current flowing

through a load (resistance R)

in one direction only.

or

11

A

~+

+Vi

D4

Load

B

D1

D2

D3

Vo

AB

When the voltage cycle is positive, diodes D1 and D3

conduct to complete the circuit and allow current

to pass through the load from A to B (region 1 ).

or

When the voltage cycle is positive, diode D1

conducts to complete the circuit and allow current

to pass through the load from A to B (region 1 ).

When the voltage cycle is negative, diodes D2 and D4

conducts to allow current through the load

across AB from A to B (region 2 ).

or

When the voltage cycle is negative, diode D2

conducts to allow current through the load from A to B (region 2 ).

(ii) Smoothing is required to make the pulsating current constant.

This is achieved by connecting a capacitor in parallel to

12

Vi

21

V0 or I0

the load resistance

or

(b) Smoothen voltage (in bold) reduces pulsating

The capacitor in parallel charges up when the current

flows through AB.

When the voltage across AB starts to drop (source changes

polarity), the capacitor discharges through the resistor, across AB

The discharges keeps the voltage across R (or AB) up.

By making the discharge long enough, i.e. used larger

capacitance ( = RC) the discharge through the resistor can

continue until the source voltage become positively high again.

or

For time 0 to A, capacitor is charged to maximum

From time A to B, capacitor discharged

13

+ +Load R Vo

Dischargecurrent

C

A

B

A CB

0

VAB without capacitor

VAB with capacitor

0

DischargingChargingV0

For smoother curve use larger capacitance

From B to C, capacitor is charged again

After C, the process repeats

13 (a) Bohr’s postulate

The electron in an atom moves in a circular orbit about the nucleus

/ proton.

Only certain / permissible / stable / quantised circular orbits

/ stationary states are allowed

or angular momentum =

Energy emitted only when electron transits from higher to lower orbit

(b) Centripetal force = Electrostatic force

= k =

mv2 = .............................. (1)

Allowed orbit means

Angular momentum for nth allowed orbit

L = mvnrn

=

vn = ........................ (2) n = 1, 2, 3, ...

Substitute equation (2) in equation (1)

= k

Solving for rn rn = (n = 1, 2, 3, ...)

or

(c) (i) Kn = mv2

14

8.64 1020 = [9.11 1031] [v]2

[v]2 = J

v =

v =

= 4.36 105 m s1

(ii) vn= or

vn = or vn =

4.36 105 =

n =

= 5

(iii) rn = or

r5 =

= 1.329 109 m or 1.33 109 m

or

r = 1.33 10-9 m

14 (a) (i) + +

15

(ii) = helium nucleus, X = neutron

(iii) No charge / neutral / does not deflect by electric field

or magnetic field

Does not produce ionisation effect / no interaction with matter

(b) 1 u = mass of an atom

= Mass of 1 mol carbon = 12 g

= 1.66 × 1027 kg

or

m = 1.66 × 1027 kg

From E = mc2

= (1.66 × 1027)(3.00 × 108)2 ×

= 9.34 × 108 eV

= 934 MeV

(c) m r

=

m = 14

Nitrogen / carbon-14Assumption: Both are equally charged or

m = 14 ( )

16