Sheme 4531 SKEMA PEMARKAHAN PHYSICS PAPER 2. Notes. lea) 1 P. To avoid parallax error 2.6cm Vernier Callipers. 2( a )(i) 1 To more sensitive of heat

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1 453 SKEMA PEMARKAHAN marks PHYSCS PAPER 2 NO Sheme - Notes lea) P To avoid parallax error 2.6cm Vernier Callipers 2( a )(i) To more sensitive of heat - t can detect any small change oftemperature 5mm st 75 mm - 00 C (50-5) - 00 X =77.4 0C 3( a) The Rate of change ofdisplacement Total distance = 72 m Averge speed = 72 4 = 5.]4 ms" st 4(a) 5 (a) (bxi) ] ] Principle of conservation ofmomentum Momentum of the bottle is equal to momentum of water jet but in opposite directon Momentum of the water, 0.2 x 2 = 0.4 kgms" is equal to momentum ofthe rocket, 0.2 X V The speed ofthe rocket = 0.2 Starting block for sprinter' Rate of change of distance Air flows with high velocity Water level in the tube rises and the roof also rises Air pressure above is higher 453 HAK CPTA PKPSM TERENGGANU 9

2 453 SKEMA PEMARKAHAN (iii) (i) 6(a) As higher speed,the pressure ofair is lower Benoulli's priciple Air speed at R is higher Air pressure bat R is lower Volime of gas in Diagram 6.2 is smaller The rate of molecules collision is hgher Gas pressure in Diagram 6.2 is higher. f the volume ofthe gas decrease,then the pressure increase Boyle's Law (e) 00 X 6 = V X 40 7 (a) Pressure by molecules of gas To measure between two levels of water (e) 8 (axi) (bxi) & Measured the height oftwo level, x em Pressure of gas = Pat,+ x m water Gas Pressure = 0 m + 0. m = 0. m water Gas Pressuere = 0. X 0 X X 0 3 = 0. X 04 Pa Use mercury as liquid in the manometer Use longer tube Single force oftwo or more forrces acted to the point or object The resultant force is zero/the object at rest/the object move at constant velocity T = 0 -Sl'--' n---6=0 =.55 N (i) Diagram 8.3 Angle of string, smallest Diagram OAK CPTA PKPSM TERENGGANU 92

3 453 SKEMA PEMARKAHAN 2 00 The angle of string is largest (iii) (2) Diagram 8.2] Because the tension ofstring is smallest 9 (a) Elasticity of a material is its property that enables it to return to its original size and shape when the force that was acting on it is removed. 5 Force that acted on the springs are the same Both springs are extended Spring N is extended more than spring M 5 th Spring M has a bigger spring constant than spring N The shorter the spring the bigger the spring constant The longer the spring the smaller the spring constant k= F =.2 x 4 =3 Ncm' st - two kind offorces; attractive forces and repulsiveforce - stretching causes molecules to be displaced away from each other -Attractive intermolecular forces are acting to oppose the stretching -when the force is removed attractive intermolecular forces greater than repulsive force so that it can returns to its original form (exi) st - more comfortable - can carries more load - absorb bumper lsi 2 fdl 5 th 6 th Modifications ofthe spring. A medium spring constant Pemalar springyang sederhana 2. Small density Ketumpatan rendah 3. High strength kekuatan yang tinggi Reason Not too stiff and can oscillate with low frequen cy Tidak terlalu keras dan boleh bergetar denganfr kuensi rendah small mass/flight jisim kecil/mngan spring not easily broken tidok mudah rosok TOTAL HAK CPTA PKPSM TERENGGANU 93

4 453 SKEMA PEMARKAHAN 0 (a) (i) ls ls (Max 0 maries Measure ofthe degree of hotness of a body The temperature increase for first 7 minitues The temperature not change after 7 minutes For last 4 minute, no net heat flow between the block and the water/the net rate of heat flow between two bodies is zero The temperature ofthe block and the water are same Thermal equilibrium. (A thermometer placed below the thounge) - to contact between thermometer and the mouth of he patient Heat flow from the patient to bulb ofthermometer (The thermometer left under the tounge for few minutes) to thermal equilibrium reached The tennometer reading shows the body temperature of the patient. Used mercury as a liquid because it is sensitive to change in temperature Used mercury because it is easy to read Used small capillary tube to expand at high rate for the small change in temperature Constriction part ofcapillary tube To prevent the mercurry return to original positon whle reading the temperature Place the temperatuere in pure melting ice Then placed the temperature in steam from boyling water Marks two points i.e at the melting ice (O C) and boiling water/steam (l00 C). Divide the temperature range between two points into equal parts. TOTAL HAK CPTA PKPSM TERENGGA\'lJ 94

5 453 SKEMA PEMARKAHAN (a)(i) Buoyant Force Force experience when an object totally or partly immersed into the liquid st Density of the gas inside the balloon less dense then air Air displaced of balloon and produced buoyant force The buoyant force is equal to weight ofballoon and load st Used helium gas ts light/less dense then air Mass of load is 20 kg Total weight of balloon and the load equal to buoyant force 5 th Tension allowed ofthe rope is 300 N 6 th To ensure the rope not break 7 th SetC 8 th Because its used helium gas, mass of load is 20 kg and tension allowed is greater than 250 N 9 th A is not suitable because mass of the load causes weight of the load and the balloon less then buoyant force loth The balloon will rise up ( Accept any other set and the reason) st Resultant force = =245N (Reject without unit.) Use F=ma st 245 = 5 a a = 49 ms" (Reject without unit.) air resistance is zero 2 (a) Pressure = Force Surface area overall (at least show mano meter) gas tank _- --L _ 453 HAK Cl'TA PKPSM TERENGGANli 95

6 453 SKEMA PEMARKAHAN i ", t Pressure of gas P atm + h cmhg The liquid used is mercury High density Diameter small To sensitive to change of gas pressure Glass bore stem is thick Not easy to brake The length is approximately m The atmospheric pressure and gas pressure less than 00 cm Hg//the height of mercury column less than m Choose manometer Q Used mercury, has small diameter, thick bore stem and length 0.8 m P 2 = cmhg X 3.6 X X 0 4 Pa _JL ~_

Lecture Outline Chapter 15. Physics, 4 th Edition James S. Walker. Copyright 2010 Pearson Education, Inc.

Lecture Outline Chapter 15. Physics, 4 th Edition James S. Walker. Copyright 2010 Pearson Education, Inc. Lecture Outline Chapter 15 Physics, 4 th Edition James S. Walker Chapter 15 Fluids Density Units of Chapter 15 Pressure Static Equilibrium in Fluids: Pressure and Depth Archimedes Principle and Buoyancy