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PAPER 2(Section A) 8 Structure Question(Q1 – Q8) Total Marks for Paper 2 Section A (60 Marks) 1 hour and 30 minutes

Question 1 (Knowledge, understanding & application) a)Name the part label P b) What is the function of the part label P? P 1 Diagram 1 shows a micrometer screw gauge when the jaws are closed. Rachet 1 mark To ensure the pressure exerted does not exceed // undue pressure is not exerted 1 mark P

c)What is the value of zero error shown by the micrometer above? d)Give one reason why micrometer screw gauge is more accurate compare to a vernier calliper mm 1 mark The scale is smaller 1 mark Question 1 (Knowledge, understanding & application)

Diagram 2 shows a boy throwing a ball upwards at a velocity of 10 m s -1. The ball decelerates to a maximum height before accelerating downwards. Diagram 2 Question 2 (Knowledge, understanding & application)

c)Sketch a displacement, s against time,t graph to describe the motion of the ball. 2 marks Question 2 (Knowledge, understanding & application)

Diagram 3 shows an arrangement of apparatus used to determine the atmospheric pressure in a laboratory. The length of the glass tube is 100 cm and the atmospheric pressure in the lab is 75 cm Hg. Question 3 (Knowledge, understanding & application)

(a)Name the apparatus shown in Diagram 3. (b)What is in space X (c)(i)What is the value of H? (ii) What happens to height, H, when this apparatus is submerged in water. (d ) The number of the coil increases the strength of the magnetic field increases.. Mercury Barometer1 mark Vacuum 1 mark 75 cm/0.75 m/750 mm 1 mark Increase 1 mark Question 3 (Knowledge, understanding & application)

(iii) Give a reason for your answer in c(ii). (d)State one application of atmospheric pressure in everyday life. The pressure outside glass tube increases // pressure of water + atmospheric pressure 1 mark Vacuum cleaner// siphon // straw // syringe // water pump 1 mark Question 3 (Knowledge, understanding & application)

Diagram 4 Diagram 4 shows a truck pulling a car with a cable. The cable is at an angle of 60 0 to the horizontal. The force, F, of the cable is 1500 N. Question 4 (Knowledge, understanding & application)

(a ) What is meant by force?. Anything that can move a stationary object // stop a moving object // change direction / shape / speed of an object. 1 mark Question 4 (Knowledge, understanding & application)

(b)On Diagram 4, draw the horizontal component, F x and the vertical component F y for F. In your drawing show the direction of F x and F y. 2 mark Question 4 (Knowledge, understanding & application)

(c)Calculate the magnitude of the horizontal component, F x. (d)(i) What is the effect of the component of forces, F x and F y to the towed car? Fx = 1500 x cos 60 = 750 N Fx = 1500 x cos 60 ° = 750 N1 mark Fx : to make car move forward // overcome frictional force1 mark Fy: to lift the car off the ground // to move the car upwards// to overcome the weight of the car 1 mark Question 4 (Knowledge, understanding & application)

(a ) Diagram 5.1 and Diagram 5.2 show the hair shampoo is pressed out with the same force from a shampoo container.. Question 5 (KUA and Conceptualization Skill)

(a) What is meant by pressure ? (b) Based on Diagram 5.1 and Diagram 5.2, compare (i)the volume of shampoo in the container. Normal Force / surface area // Normal force per unit area1 mark Diagram 5.2 is bigger than Diagram 5.1/ greater // vice versa 1 mark Question 5 (KUA and Conceptualization Skill)

(b) Based on Diagram 5.1 and Diagram 5.2, compare (ii) the volume of shampoo that spurts out from the container (c ) Based on your answer in (b), (i) relate the volume of shampoo in the container with the volume of shampoo that spurts out from the container.. Diagram 5.2 is bigger than Diagram 5.1 / greater// vice versa 1 mark When the volume of shampoo in the container is less, the volume of shampoo spurted out from the container is less // vice versa 1 mark Question 5 (KUA and Conceptualization Skill)

(c ) Based on your answer in (b), (ii) relate the volume of shampoo that spurts out from the container with the pressure exerted by the shampoo. (1 mark) (iii) state the relationship between pressure and the volume of shampoo in the container. When the volume of shampoo spurted out from the container is big, the pressure exerted towards the shampoo is big // vice versa When the volume of the shampoo in the container increase the pressure also increases.1 mark Question 5 (KUA and Conceptualization Skill)

(d) State one characteristic of the liquid (e) Name the principle involved that enable the shampoo to spurt out from the container. Liquid used is non compressible / hard to compress1 mark Pascal’s principle1 mark Question 5 (KUA and Conceptualization Skill)

Diagram 6.1 and 6.2 show a virtual image produced by a plane mirror and a convex mirror respectively.. Diagram 6.1Diagram 6.2 Question 6 (KUA and Conceptualization Skill)

(a) What is meant by virtual image? (b) Based on Diagram 6.1 and Diagram 6.2, (i)compare the size of the image. Image that cannot be formed on screen1 mark Image in Diagram 6.1 is bigger than Diagram 6.2// vice versa 1 mark Question 6 (KUA and Conceptualization Skill)

(b)(ii)state one other similarity of the image formed besides virtual. Upright1 mark Question 6 (KUA and Conceptualization Skill)

(c ) In Diagram 6.3, draw a ray diagram to show how the image in Diagram 6.2 is formed. DIAGRAM 6.3 Question 6 (KUA and Conceptualization Skill)

(e) Name the light phenomenon that occurs. (f) State one advantage of using convex mirror as the side mirror of a car.. Reflection1 mark Wide angle of reflection // wider the vision1 mark Question 6 (KUA and Conceptualization Skill)

Diagram 7 Diagram 7 shows a hydrometer used to determine the density of a liquid. Question 7 (KUA and Problem Solving)

(a)(i) Name the physics principle involved. 1 mark (ii) Explain why the hydrometer floats on the surface of the liquid.. Archimedes’s principle // Forces in equilibrium // Principle of flotation The buoyant force = weight of the hydrometer // density of liquid > density of hydrometer // the volume of the liquid displaced by the hydrometer is large Question 7 (KUA and Problem Solving)

(a)(iii) The volume of the hydrometer under the surface of the liquid is 25 cm 3. The density of liquid measured is 0.8 g cm -3. Calculate the buoyant force exerted to the hydrometer. [ 2 marks ] (b)(i) he hydrometer in Diagram7 is unsuitable to measure the density of an acid solution which has smaller density. [ 2 marks ]. F =  V g = (8 x 10 2 ) x ( 25 x ) x 10 // = 0.2 N the plastic wall is replaced by a glass wall increase the volume of the air filled bulb // reduce the mass of lead shot Question 7 (KUA and Problem Solving)

(b)(ii) Give the reasons for your answer in b(i) (iii) Suggest one method to increase the stability of the hydrometer (c ) What happens to the hydrometer in Diagram 7 when it is placed in a higher density liquid?. to avoid corrosion 1 mark add more lead shot1 mark to increase the buoyant force // to float the hydrometer1 mark Less submerged //hydrometer floats higher 1 mark Question 7 (KUA and Problem Solving)

Diagram 8.1 and Diagram 8.2 show a block of iron and a block of aluminium, each of mass 250 g, are heated by an immersion heater. The power of the immersion heater is 50 W.. Diagram 8.1Diagram 8.2 Question 8 (KUA and Decision Making)

The temperature-time graph for the two experiments are shown in diagram 8.3. Diagram 8.3 Question 8 (KUA and Decision Making)

(a) What is meant by heat? (b)Based on Diagram 8.3, calculate the change in temperature per minute for; (i) Iron. Energy transfer from higher temperature body to lower temperature // type of energy that flow 1 mark 80 2 = 40 °C per minute 1 mark Question 8 (KUA and Decision Making)

(b)Based on Diagram 8.3, calculate the change in temperature per minute for; (ii) Aluminium = 80 7 = °C per minute 1 mark Question 8 (KUA and Decision Making)

(c) Based on your calculation, which metal gets hot faster? Explain your answer. (d) Determine the specific heat capacity for iron and aluminium.. Iron 1 mark the rate of change of temperature is higher1 mark C iron = 50 x 2 x x 80 = 300 J kg-1 °C -1 2 mark C Al = 50 x 8 x x 80 = 1200 J kg-1 °C -1 1 mark Question 8 (KUA and Decision Making)

(e) Table 8 shows the specific heat capacity of materials which could be used to make a frying pan. Material Based on your answer in (c) and (d), which material would be suitable to make a frying pan? Explain your answer.. Material Bahan Specific heat capacity/ J kg -1 o C -1 Muatan haba tentu A780 B1528 C1415 Material Bahan Specific heat capacity/ J kg -1 o C -1 Muatan haba tentu Question 8 (KUA and Decision Making)

Based on your answer in (c) and (d), which material would be suitable to make a frying pan? Explain your answer. A 1 mark it has lowest specific heat capacity // easily to get hot 1 mark Question 8 (KUA and Decision Making)