* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download smjk yu hua kajang marking scheme peperiksaan
Survey
Document related concepts
Equivalence principle wikipedia , lookup
Negative mass wikipedia , lookup
Pioneer anomaly wikipedia , lookup
Artificial gravity wikipedia , lookup
Electromagnetism wikipedia , lookup
Coriolis force wikipedia , lookup
Centrifugal force wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Lorentz force wikipedia , lookup
Matter wave wikipedia , lookup
Fictitious force wikipedia , lookup
Woodward effect wikipedia , lookup
Weightlessness wikipedia , lookup
Transcript
SMJK YU HUA KAJANG MARKING SCHEME PEPERIKSAAN PERTENGAHAN TAHUN 2014 PHYSICS PAPER 2 FORM 4 NO MARKING CRITERIA No 1 1 (a) (b) (i) (b) (ii) (b) (ii) Systematic + 0.03 cm To measure the internal diameter Actual Reading: 4.38 cm – 0.03cm = 4.35cm MARKS SUB TOTAL 1 1 1 1 TOTAL 4 No 2 (a) (i) Momentum is product of mass and velocity or Momentum = mass x velocity (b) (i) Momentum before collision: 2.5 + 0 = 2.5 kgms-1 (ii) Momentum after collision: 0.4 + 2.1 = 2.5 kgm-1 (c) Compare: Total momentum before collision equals to total momentum after collision. SUB 1 (d) (i) Principle of conservation of momentum 1 1 1 1 TOTAL No 3 (a)(i) (ii) (iii) (b) (c) (d) No 4 (a) (b)(i) (ii) (iii) -1 Initial Velocity u = s/t = 0.5/0.02 = 25 cms Final velocity: v = s/t = 0.8/0.02 = 40 cms-1 a = (v-u)/t = (40-25)/0.02x3 = 250 cms-2 = 2.50 ms-2 Uniform acceleration or constant acceleration Net Force F=ma 1.5 x 2.5 = 3.75 N New distance of ST = 1.6 cm Force is defined as the rate of change of momentum Horizontal Component: F cosθ = 20 x cos 40 = 15.32 N Verticle component : F sinθ = 20 x sin 40 = 12.86 N Reaction force acting on stroller R = mg + Fy = 8x10 + 12.86 = 92.86 N (No need to draw Free Body Diagram) (c) Acceleration : F = ma a = Fx/m = 15.32 /8 = 1.92 ms-2 (d) Decreases MARKS 5 SUB TOTAL 1 1 1 1 1 1 TOTAL 6 SUB 1 1 1 1 TOTAL 2 1 TOTAL 7 No 5 (a)(i) A to B : Constant velocity //uniform velocity // acceleration = 0 (ii) B to C: Uniform acceleration //constant acceleration (iii) C to D : Uniform deceleration // constant deceleration (b) Acceleration in 1st 3 seconds a = 5/5 = 1 ms-2 (c) Total distance travelled s = 12.5m + 35m + 22.5m + 15m = 85m (d) Average velocity = total displacement / total time v = 85/18 = 4.72 ms-1 SUB 1 1 1 2 1 1 1 TOTAL No 6 (a)(i) Fw = FR If in words also accept answer (ii) Fw > FR If in words also accept answer (iii) Fw < FR (b) F = ma FW – FR = ma 20000 – FR = 1500 x 5 FR = 20000 – 7500 = 12,500N (c) Impulsive Force = (mv – mu)/t Note: 110 kmh-1 is converted to 30.556 ms-1 Impulsive Force = (0 – 1500 x 30.556) / 0.5 = 91668 N or 91667 N TOTAL SUB 1 8 TOTAL 1 1 2 1 1 (d) 2 ways to reduce the impulsive force Way 1: Soft front bumpers Way 2: Crumple zone at the front bonnet 1 1 TOTAL 9 No 7 (a) Forces in equilibrium means Net force or resultant force acting is zero // Force vectors when added can form a closed polygon. (b) T1 String T2 o SUB 2 TOTAL 1 o 45 45 W T1 and T2 can be interchanged. If direction of T1 and T2 is not correct, less 1 mark No labels : no marks (c) 1 2 T2 W T1 Triangle is correctly drawn. Angle not important Labels are all correctly placed (d) Since forces are in equilibrium Resultant Force = 0 1 (e) Tension of the string T sin 45 + T sin 45 = 2 x 10 2T sin 45 = 20 T = 14.14 N (f) String will snap/break 1 1 1 TOTAL 10 No 8 (a) X: Meter ruler Y: Vernier Caliper Z: Micrometer Screw gauge (b) Average X : 1.3 cm ( If 1.26 or 1.27 is wrong) Y: 1.23 cm Z: 1.204 cm (c) Accurate instrument is : Z (d)(i) Passengers are thrown forward (ii) Inertia (iii) 1. Anti inertia Seatbelts 2. Air bags 3. ABS Brakes 4. Crumple zone 5. Head rest 6. Bumpers 7. Shatter proof windscreen (Select any 3) (iv) Higher speed, higher frictional force Value: > 30,000 N (Any value greater than 30000N accept) Both answers right, then award 1 mark SUB 1 1 1 TOTAL 3 2 1 1 1 3 1 TOTAL 12 No 9 (a) (i) Momentum is the product of mass and velocity or p=mv where m: mass (kg) and v is the velocity (ms-1) (ii) o Total momentum in diagram 9.1 is zero o Total in Diagram 9.2 m1v1 + m2v2 = 60x5 + 50(-6) = 0 o Total momentum before and after boy jumped to jetty are equal o Principle of conservation of momentum (b) (i) SUB 1 TOTAL 1 1 1 4 1 1 3 36 48 t/s Straight line through the (0,0) Values are marked correctly on axis labels on axis and units of physical quantity 1 1 1 No units in axis label – less 1 mark (ii) Acceleration of train: Gradient of graph a = 36/48 = 0.75 ms-2 (c)(i) 2 1 1 2 V / ms-1 36 48 120 t/s (ii) Total distance travelled is the area under the v-t graph Area = ½ x (120 + 72) x 36 = 3456 m (d) (i) (i) Area of shaded part is 432m V / ms-1 432 = ½ (t-120) x 36 (t-120) = 24 sec 36 (ii) 120 t t/s 6 2 2 Area= 432m 48 2 (ii) Deceleration = gradient a = 36/24 = - 1.5 ms-2 (iii) Acceleration is the rate of change of velocity 2 TOTAL 20 No 10 (a) (i) Impulsive Force (ii) 1. The surface of metal block is hard 2. The time of impact is shorter 3. The impulsive force is bigger (iii) Drop on a soft surface // drop on a pile of sand/mud SUB 1 1 1 1 TOTAL 1 3 1 (b) (i) 10 Specification Mass of pile driver Height of pile driver Pile Material Characteristic Bigger mass of load Higher position of pile drive Steel Pile is used Shape of the Shape must base pile pointed Choose Q is chosen Explanation Produce a bigger force during impact Produce a higher velocity when striking the pile Steel is stronger and stiffer. Is not brittle and does not crack on impact be Produce a higher pressure // easy to push the pile into the ground Because piling system Q has big mass of pile driver, position of pile driver is high and steel piles are use with sharp pointed base 2 2 2 2 2 (c)(i) W = mg = (50) (10) = 500 N (ii) Velocity of metal pile just before it hits the pile 1 2 v2 = u2 + 2gh = 0 + 2 (10) (20) = 400 v = 20 ms-1 (iii) Impulsive Force F = mv – mu t = 50 (0-20) 0.5 1 = 2000 N 1 1 2 1 TOTAL Prepared by Pradeep Kumar Chakrabarty SMJK Yu Hua Kajang 2014 20