Ans Phy Paper 2 Sgor 2008

1 PROGRAM PENINGKATAN PRESTASI MATA PELAJARAN SAINS DAN MATEMATIK TAHUN 2008 JABATAN PELAJARAN SELANGOR MARKING SCHEME SECTION A

MARKS

MARKING CRITERIA No. 1.(a) Displacement // distance // speed // velocity // acceleration // deceleration // time

SUB 1

(b) Tick the correct answer refer to 1(a) in the correct box

1

(c) Average velocity = 80 30 = 2.67 ms -1

1

2. (a) Pascal’s principle (b)

P =

(c) 200

F = A =

20 N 0.1 m 2 200 Nm -2 (answer with correct units)

=

F2

F2 0.5 = 100 N (answer with correct units)

1

1 1 1 1

Constant velocity

1

(ii)

deceleration

1

(b)

a =

10 - 5 2 2.5 ms -2 (answer with correct units)

1

5 + 10 X 2 2 15 m (answer with correct units)

1

3.(a)(i)

= (c)

s = =

5

1

1

4(a). Unstable nuclei that disintegrate to become more stable by emitting radiation

1

(b)(i)

25

1

beta particles

1

(ii)

4

1

Output force increases

(d)

TOTAL

6

2 (c)

Mark at ( 0, 4000 )

1

Mark ( 2 000, 6000 ) ( 12 000, 1000 ) ( 18 000, 500 ) ( 24 000, 250 )

2

Plot graph with smooth curve 1 5(a)(i) (ii)

Chemical energy to kinetic energy // Mechanical energy to kinetic energy Electrical energy to kinetic energy

1 1

(b) Energy can be changed from one form to another

1

(c) When work is done

1

(d) Conservation of energy

1

(e)(i)

sin 30

=

5 AB 10 m

1 1

= 800 X 10 = 8000 J

1

AB = (ii) 6(a)(i) (ii) (iii) (b)(i) (ii) (ii) (c) 7(a)(i) (ii) (b)

W

Depth at P is less than depth at Q Pressure of water at P is less than pressure of water at Q The volume of the bubble at P is bigger than at Q

1 1 1

Depth increases, pressure increases Pressure increases, the volume of bubble decreases Boyle’s Law

1 1 1

From Q to P, pressure decreases Distance between molecules increase / expand

1 1

Interference Constructive interference

1 1

λ

1

= =

340 2000 0.17 m

1

c (i)

Increase x α 1 a

1 1

(ii)

Increase

1

f

α

1 f

and

x α

λ //

7

x α

1 f

1

8

8

3

(d)(i) (ii)

1

destructive interference

1

+

10 8.(a) Input

Output

A

B

C

0

0

1

0

1

1

1

0

1

1

1

0

2

(b)(i) Input

Output

P

Q

R

S

T

0

0

1

1

0

0

1

1

1

0

1

0

1

1

0

1

1

0

0

1

AND GATE

(ii)

3 1

(iii) 1

(c)(i)

X = Y =

1 1

P 0 0 1 1

(ii)

(iii)

NOT AND

R =

Q 0 1 0 1 P

•

Q

R 0 0 1 0

2 1

12

4 FIZIK KERTAS 2 ( Section B ) No. 9(a)

b(i) (ii) (iii) (iv)

(c)

Marking Criteria/Answers

Marks

The distance between the focal point and the optical centre. 1. C is centre of a circle or CM is radius of circle 2. The light strike perpendicular to the mirror. 3. Incident angle is 00. 4. reflection angle is 00 or reflection law is obeyed

1 1 1 1 1

same diagram 10.2(b) is further than 10.2(a) diagram 10.2(b) is smaller than 10.2(a) the further the object from the mirror, the smaller the image formed the further the object, the smaller the linear magnification

1 1 1 1 1

Convex mirror

The image formed is virtual, upright an diminished

Large diameter

More object can be seen

More curve the mirror

Wider field view

Less thickness

Avoid multiple image formed

Top corner

Can look easily by observer

10

20 No. 10.(a)

Marking Criteria / Answers The current produced when the magnetic flux is cut across by a conductor// Changing of flux at conductor

(b)(i) 10.2 - no relative motion between the magnet and the coil // 10.3 - there is relative motion (ii) Number of turns in 10.3 is less than number of turns in 10.4 (iii) Number of turns increases, The change in magnetic flux increases Induced current increases (iv) Faraday’s Law

Marks 1

1 1 1 1

5

(c) Magnetic flux is cut , induced current is produced When Θ = 90 o maximum current produced // Θ = 0o minimum current produced Commutator is used to ensure the direction of the current that flows through the external circuit is in one direction

1 1 1

1

(d) 1.

Suggestion Step down transformer

Rationale To reduce voltage

2.

Ns : Np = 240 : 6 = 40 : 1

To reduce 240V to 6V

3.

Use diode

To change AC to DC

4.

Use capacitor

To smooth the output current

5.

Use laminated soft iron core

To reduce heat loss due to eddy current

10 20

No. 11.(a) (i)

Marking Criteria/Answers

An upwards force on an object placed in a liquid which comes from the liquid itself and makes the object appear to lose weight.

h2

(ii)

Marks

1

h1

P = hρg The pressure difference in the fluid = P2 - P1 = h2ρg - h1ρg P = F/A, A = area of the object (bottom and top) The difference between the two forces (at bottom and top) = P2A - P1A = A(P2 - P1) = A ρ g(h2 - h1) Buoyant force = Vρ g = mg = weight of the fluid displaced (m = ρV; volume of the object = volume of the liquid displaced)

1 1 1 1

6 (b) W air (i) Total weight

= =

0.8 X 1200 X 10 9600 + 4000

= 9600 N = 13 600 N

(ii) Buoyant force =

1.3 X 1200 X 10

= 15 600 N

1 1 1

(iii) Net force

= 15 600 N - 13 600 N = 2000 N 1

(c) ENVELOPES – CONSTRUCT FROM NYLON (i) • Lightweight material // • Reduce the total weight of the balloon. Resultant force is bigger. Lift up the balloon higher // • Strong, not easily torn, can withstand the strong winds which could easily wreck the balloon. BALLOON SHOULD BE LARGE SIZE • Displace more volume of air // • Weight of air displaced is greater // • To create sufficient buoyant force. Lift up the balloon higher.

1

1 1

USE TWO BURNERS • To warm up the air in the balloon quickly // • To keep the balloon rising

1

THE BASKET MUST BE MADE OF RATTAN • Light and flexible/safe material // • To stop the balloon gradually AND minimizing the impact // • Prolong the collision time between basket and ground// reduce impulsive force when basket hits the ground

1

1

1

1 (ii) S Material of envelope is nylon, large size of the envelope, use two burners and the basket is made of rattan

1 1 20

No.

Marking Criteria/Answers

12 (a)(i) Low melting point // low resistance // thin wire. (ii) High voltage surge or short circuit produces high current. High current produces high heat (I2Rt). Heat raise temperature to melting point. Fuse melt and break the circuit. (iii) If fuse is connected after the load, the live wire and the load still has high potential difference after the fuse blow unless the switch is off (iv) Earth the chassis by connecting it to the earth wire.

Marks 1

3 1 1

7 (b) Thin fuse has less mass hence lower heat capacity.

1 1

shorter time to heat up to melting point and blow.

1 1 ic catridge thstand higher temperature because sparks created by high voltage, can be huge.

l current of device is 10 A.

1 1 1 1 1 1

mum rating must be higher than normal current 13 A

c)(i)

g point must be low

2

(ii) blow. 2 osen because ower heat capacity, ceramic catridge, normal current 10 A and melting point

Power of A = 2 W Power of B = 4 W Note: A has lower resistance than B produce lower power because of same current. Power of A = 18 W Power of B = 9 W Note: A has lower resistance than B produce higher power due to higher current.

20