Skema Fizik Kertas 2 peperiksaan akhir tahun sbp 2011 ting 4

SKEMA PEPERIKSAAN AKHIR TAHUN TING 4 (SBP) 2011 No 1

Answer

Mark

(a) (b)

Tail To measure the depth

1 1

(c) (d)

0.01 cm To measure internal diameter// to measure external diameter.

1 1

(a) (b) (c)

Impulsive force High force //does not fasten seatbelt

4 2

1 1

F 1 1 = - 3000 N // 3000 N Wear a seat belt // use seatbelt // use airbag

(d)

1 Total

3

(a) (b)

Reflection of light

5 1

(i)

1. The ray is reflected twice 2.The angle of incidence equals to the angle of reflection. (ii)

2

(c)

4

(i) (ii)

(a) (b)

(c)

A straight line is extrapolated from the reflected ray to the image below the object O. 15 cm 20 cm Total Volume increase // length increase To ensure net rate flow of heat is zero/ amount of heat transfer from the patient’s body to the thermometer = amount of heat transfer from thermometer to the patient’s body // To reach thermal equilibrium Principle of thermal equilibrium state that two bodies in thermal equilibrium when: -

It has reach the same temperature

-

Net rate of heat transfer between two bodies is zero

i)

(d)

1 1 1 6 1 1

1 1

x 100 ⁰C =

x 100 ⁰C

1 1

= 53.3 ⁰C ii) use thinner-walled glass bulb/reducing the diameter of

1

the capillary tube Total 5

(a) (b)

(c) (d)

6

(a) (b)

(i) (i) (ii) (iii) (i) (ii) (i)

Pressure Size of tube at P is bigger Velocity of air at P < velocity at Q // vice versa Pressure of air at P > Pressure of air at Q The higher the velocity of air the lower the pressure Bernoulli’s Principle.

7 1 1 1 1 1 1 1

(ii)

P P> PR> PQ Total Elastic potential energy Diagram 6.2(a) > Diagram 6.1(a) // vice-versa Diagram 6.2(a) > Diagram 6.1(a) // vice-versa

1 8 1 1 1

(i) (ii)

(c)

(i) (ii)

(d) (e) (f)

Diagram 6.2(b) > Diagram 6.1(b) // vice-versa Diagram 6.2(b) > Diagram 6.1(b) // vice-versa When the energy stored is big, the velocity is big Principle of conservation of energy No change Total

7

(a)

(i) (ii)

Pascal’s Principle Same // equal

(b)

(i)

FR = FP AR AP AR = FR x AP FP = (5000)(4) 50 = 400 cm2 Component: valves Reason : Liquid can flow in one direction and does not back flow Piston B is bigger. Reason: To produce large output force. Use released valves Reason: Liquid flows back to the storage reservoir Total Quantity of heat needed to change 1kg substance from liquid to gas at boiling point without any change in temperature.

(i) (c) (ii) (iii)

8

(a)

(i) (b)

(ii)

(c)

E = Pt = 400 x 200 = 8.0 x 104 J Q = mL L = Pt m = 8.0 x 104 J 0.1 kg = 8.0 x 105 J kg-1 Liquid A :

1 1 1 1 1 8 1 1

1 1 1 1 1 1 1 1 10 1

1 1

1 1

Pt = mL t = (3.0)390 600 = 1.950 s

1 1

Liquid B : Pt = mL 1

t = (2.5)900 1200

1

= 1.875 s Liquid C : Pt = mL t = (4.0)400

1

900 = 1.777 s (d)

9

(a) (b)

(i) (ii)

(i)

(ii)

(c)

Liquid C The time taken is the shortest/The things inside the refrigerator can be chilled faster Total Apparent weight is actual weight minus the buoyant force. 1. Apparent weight in 9.1(b) is less than 9.1(c) 2. The density of water is greater than density of oil 3. The buoyant force in 9.1(b) is greater than 9.1(c) 1. The higher the density the greater the buoyant force. 2. The greater the buoyant force the smaller the apparent weight. 1. The copper block sinks because its density is greater than the density of water. 2. The total density of air + bowl copper sheet is less than the density of water. 3. Weight equals to the buoyant force. 4. The surface area is also bigger that causes it to float easily

1 1 12 1 1 1 1 1 1 1 1 1 1

(d) Suggestion The submarine must be streamlined shape The material used must be strong Equip with ballast tanks

Reason To reduce water resistance

To withstand increasing pressure underwater To pump in water to submerge and pump out water to float Divide the submarine into To protect the crews from smaller compartments with drowning if leakage happens strong doors Equip the submarine with To provide air to the crew. oxygen tanks

10

Total

20

10

(a) (b)

(i)

(ii)

(c)

Extension is the length of the extended spring with load minus the length of spring without load. 1. The elasticity of the spring arrangement in Diagram 10.2 is higher 2. The pulling force in Diagram 10.2 is higher. 3. The period of oscillation in Diagram 10.2 is shorter. 3. The greater the elasticity the greater the force. 4. The greater the elasticity the shorter the period of oscillation. 5. Elastic potential energy converts to kinetic energy. 6. Make the spring shorter. 7. Increase the diameter of the spring wire. 8. Make the diameter of spring smaller.

1

3 2

4

(d) modification

Fit and light attire Jump with maximum strength/force Elastic spring board

Position of spring board is high // the height is more than 4 m Strong spring board

explaination

Less friction Reach maximum height to make 4 summersaults. to increase the elastic potential energy to jump higher.

10

Can make 4 summersaults during jumping. Not easily broken

Total

20

11

(a) (b)

(i)

(ii) (iii)

(c)

Force that acts against sliding motion of two surfaces sin 450 = W2 W W2 = W sin 450 W2 = 40 sin 450 W2 = 28.28 N Fn = 28.8 -20 = 20 N F = ma a=F m = 8.28 2 = 4.14 ms-2

1

1 1 1

1

1

Wooden block slides downwards because: - W2 > frictional force - Net force is not zero // forces are not in equilibrium

1 1

Wooden block is stationary - W2 = frictional force - Net force is zero // forces are in equilibrium

1 1

(d) Aspects Type of frame: - Aluminium Type of string: - nylon Characteristics of string: - inelastic Angle of inclination: - big Chosen frame: - Q

Total

explanation Light and easy to support. Strong and does not break easily// Long lasting

2

Does not extend easily due to weight of frame. Smaller tension of the string Because Q has aluminium frame, inelastic nylon string and bigger angle of inclination from horizontal line.

2

2

2 2

20

12

(a) (b)

(i)

(ii) (c)

(d)

(i)

(ii)

Quantity of heat required to raise the temperature of 1 kg object by 1 °C.  Heat flows from heater to the mercury in the thermometer 

The mercury expands / rises up



When thermal equilibrium state is achieved the temperature of thermometer is equal to the temperature of Aluminium block.

Oil has a better thermal conductivity than air // heat can flow through oil faster than air Characteristics Explanation Low specific heat capacity Easy to get hot With lid Reduced heat loss Low mass Easy to carry /manage Plastics High specific heat capacity // not easy to be hot S is chosen Low specific heat capacity, With lid, Low mass and the handle is made of plastic Mass,m =ρV = 800 x 0.0001 = 8 x 10-1 kg

1 1 1 1

1

10

1 1

pt= mcθ 1000(t) = (8 x10-1)x 2000 x 90 t = 144 s

1 1 1

Total

20