Physics Past Paper - Runners' Positions (PDF)

1. Suppose the graphs presented is that of the position of two runners, Mars (red line) and Venus (in blue) recorded within 20 seconds - with different scenarios. started at the started...

1. Suppose the graphs presented is that of the position of two runners, Mars (red line) and Venus (in blue) recorded within 20 seconds - with different scenarios. started at the started the but Mars is 5 · same point · at same time , · blue is steeper ; VENUS is faster meters ahead of Venus · Venus stopped at a point 5m away from Mars · red is steeper ; MARS is faster at the end · they met the run of. GRAPH A GRAPH B started the · started at the same time , but Mars is 5 · at same time , I same point meters ahead of Venus · BLUE is steeper ; Venus is faster BOTH distance from O mark of position · some stopped are · parallel lines > -> same slope > - same speed · at the end , venus at pointa 5 m away from Mars GRAPH C GRAPH D Identify whether EACH of the following statements is TRUE or FALSE based on the graphs provided. Provide a justification to support EACH answer. a. There were no graphs (GRAPH A-D) shown that indicate that Mars and Venus traveled at the same speed. FALSE There is a graph that show they have the same speed -. > Graph C : parallel lines b. Both Mars and Venus ran for 20 seconds. TRUE All graphs are up to 20s-. same slopes -- c. Mars is relatively a slower runner than Venus in ALL scenarios. 4 d. GRAPH B shows that Mars was 5 meters away from Venus at the beginning of the run. e. GRAPH A & D show that Venus was farther away from Mars by 5 meters at the 20-second mark of the run. TRUE At the end Venus (blue line) , was at some point sm away , 2 from Mars TRUE Mars' at to - FALSE Mars is slower in Graph AYD. , is at 5m while Venus at to is at O m in at same pace for Graph Cy faster Graph B.  2. Suppose the graph below shows the velocity of a car in a 24-hour journey. 35km + 127 5 km Az by & total area = · & T 162 5 km. 2 = (e) (15) 2 A, = (bitbyn Az = 127 5. = (4 0((5) - b,+ A, = 35km 7 · speed h constanteasing V 9 9b Az risped I + As I -b - -62 Identify whether EACH of the following statements is TRUE or FALSE based on the graph provided. Provide a justification to support EACH answer. - FALSE.The untismoug 1 s T a. 8 The total distance and displacement traveled by the car is 162.5 meters. > TRUE. Magnitude is scalar b. The magnitude of the car’s acceleration at 0-3 h and 7-10h is equal. -. the In these time intervals c. The car was at rest for three (3) hours. TRUE. Seen at 10th-other , numerical values of the Od. The acceleration for the first 3 hours is the same as the acceleration on the 13th to 18th hour of slopes are -. the trip. purple - the same S e. The greatest velocity recorded is 15 km/h. - Segment ↳ point (13h) TRUE Highest 5 - 8 red : graph - - = a. - 3- our the a = 1 67. mis Yeoge -18th fi , : FALSE mthe While for S it's 3 m/s" S  EXERCISES Provide a complete solution. 1. A physics book slides off a horizontal tabletop that is 0.600 m tall and lands 0.385 meters from the foot of the table. Determine the initial horizontal velocity of the physics book. 5 points 2. A ball Is thrown horizontally from a height of 13 m with an initial speed of 15.0 m/s. How long will it take the ball to reach the ground? At what horizontal distance from the point of release will it strike the ground? 5 points 3. A football is kicked with an initial velocity of 12 m/s at an angle of 25-degrees. Determine the time of flight, the horizontal displacement, and the peak height of the football. 10 points constant of I unknown igt -unknown Ye ba - ① & dobrum V - - knowns - * legth This > can't be used -et Not = - Vox dx ? : + = ↓ Y unknown start on looking TIP : always unknowunknowns Eg + 2 - t dy = 9 8) +2 givene dx Vox t e - = +( - 60. = 0 10/5 -. 0 385m = Vox 2(40 60) z.. = t ? 8 355 see. #id e value issues our. 10m/s 1 = Vox · root with have two 355. The initial horizontal canbe usedtening Vox na : ↳ velocity (Vox) of the physic book is om/s HLP (wit) ! ② ball thrown hizontally > - Voxt 3) dx - = di = 13m = 15m/s/71 63)). 0 m/s may * dx+ Vox = 15. dahil 24 45m dx to =. At 3 t na. Time dx = ? take the ball to pres & will · The time it TIP : startop te Ifrom reach the ground is 1 635 >. by PRP for time , item to solve then, it will be 24 45 m away dy = t gt. one 8)t Leigh =( 9. - = (3m · t 1. 635  EXERCISES Provide a complete solution. 1. A physics book slides off a horizontal tabletop that is 0.600 m tall and lands 0.385 meters from the foot of the table. Determine the initial horizontal velocity of the physics book. 5 points 2. A ball Is thrown horizontally from a height of 13 m with an initial speed of 15.0 m/s. How long will it take the ball to reach the ground? At what horizontal distance from the point of release will it strike the ground? 5 points 3. A football is kicked with an initial velocity of 12 m/s at an angle of 25-degrees. Determine the time of flight, the horizontal displacement, and the peak height of the football. 10 points neither horizontal thusMUSTor them el FIRST z nor vertical ; 3 Vo = 12 m/ 12 cos25" 10 Stm/s Vox Vocos-. = = = - - = 250 > 12 sin 25 = 5 07 mis Voy Vosint. = = Height :? - to look for time aree dx =? Again, start 5 07. IgtV bakasame d Volino + = dymax -? L ↓ & C unknown ⑭known peak Vy 0 - hand notnationembi , = sa = or m/s 1 & 5.07-9 8 m/s. My = Vosino + gt Voy may t Vox , , , 0 5 07 + ( 9 8)t volsunknown -. =. dx = t = d eto 9 8 PWEDE ⑭known-. -. ↳ DI PA and + 8 525. na lang vertisin at -peak 0 525. + pero pohinahanap light it make 2 peak ! t flight = werk T halfway through 10 86m/s. Flight 210 52) Vocostt-1 04s Grajectory. dx. Flight = = 1 04S T E % m/y (1 04/). light 10 S.. = dx = 11 29m Reminder dx-treight. :. 07 m/s peak dy- 5 - dy Vosin E = DONE 5a 0 Your turn to. concludes [ na ! = 5 07m/s (0 523) + + (9 8) 10 523.... dy = 1 31. m 3-5 | VECTOR ADDITION: Two-dimensional LEARNING COMPETENCIES: The learners are expected to: 1. Associate corresponding directions to the axes and quadrants of the coordinate plane, 2. Determine ways that a vector can be presented (illustration, magnitude and direction, use of coordinates), 3. Transform vectors from magnitude and direction to coordinates, and vice-versa (𝜃 in standard and non-standard position), 4. Use graphical method in adding vectors (head-to-tail method), and 5. Use analytical method in adding vectors (component method). Introduction Vectors are physical quantities defined by a magnitude and direction. From the previous sessions, it has been known that it can also be represented in terms of coordinates. It is illustrated as a ray (in layman’s term, an arrow) with the length indicating the vector’s magnitude and the arrowhead indicating the vector’s direction. Similar to physical quantities, the fundamental operations can also be used in vectors. This time, we’re going to focus on adding vectors in one-dimension. Addition of vectors can be done either graphically or analytically. Objectives: 1. Draw vectors following a scale, 2. Rewrite an angle with bearing to an angle in standard position and vice-versa, 3. Add vectors using graphical and analytical method; and 4. Compare the results of analytical and graphical method of vector addition. Materials: Ruler Protractor Graphing paper (or Bond paper) for graphing Color pens or color pencil (fine point will be preferable for graphing) Procedure: 1. Complete the table using your knowledge in transforming vectors as well as locating them. 2. Add the vectors graphically by plotting them using the scaling of 5 m: 1 inch and connecting them head-to-tail (See the figure below). For the rest of the remaining vectors, continue doing steps 2-3 until the last vector. 3. When all the vectors have been drawn in head-to-tail arrangement to each other, you’re now ready to draw the resultant vector. This is a vector we can obtain by connecting the head of the first vector drawn to the tail of the last vector drawn (See vector R below). Resultant vector is just the sum of all the vectors you have drawn. 4. To wrap up the graphical method in adding vectors, we’ll measure the length and the angle (direction) of the resultant with the use of a ruler and protractor, respectively. Report values as measured. Do not forget that after measuring with the use of the ruler, you need to use the scale to retun it to the actual value. In the example on the left, it reads about 3.5 in. 3.5 1 Using the scale in ratio and proportion: =5 we 𝑋 get the R’s magnitude to be 17.5 meters. When measuring using a protractor, the 0° mark should be placed on an axis as shown. This somewhat reads 19°. Since the rotation started from the East, going to the North, the bearing of the resultant vector is North of East. In other words, R = 17.5 meters, 19° North of East 5. On this part, we are going to add the same set of vectors using analytical method. All you need to do is sort the coordinates according to dimensions: x & y. 6. After this, indicate the sum of these x- and y- components in the TOTAL column. 7. Use the Pythagorean theorem to solve for R’s magnitude and the tangent function to determine the angle (direction) of the vector. 8. Ultimately, report the computed values for magnitude and direction of the resultant vector.  use since use cosine 9 ↑ NAME: ____________________________________________ SECTION: __________ SCORE: __________ Vector Angle in Standard Coordinates Quadrant/ Axis Position A 10 m, North 90° (0, 10) + y – axis B 15 m, East ou (15 0) , + X-axis C 25 m, 30° East of South - 600 or 3000 (12 5 21 65). -. Q , 1650 & D 20 m, 15° North of West - 1950 or (19 32, 5 18).. Q & E 10 m, Southwest 2250 1350 (-7 07 7 Q 077 - or.. , zu For GRAPHICAL METHOD of vector addition, use the space WITHIN THIS BOX ONLY. Use the color coding you’ve set for the vectors from the table earlier. same anglesuse You will X-Y a in Soering / ordinates components RESULTANT VECTOR ____________________________________ ____________________________________ magnitude direction with bearing For ANALYTICAL METHOD of vector addition Vector x-component y-component 𝑅 = √(∑ 𝑥)2 + (∑ 𝑦)2 R(1 - 7/m magnitude A 0 10 ~ B 15 ↑ R = 3. C 25103300 = 12 5 25 sin 300 = 21 65 - ∑𝑦.. 𝜃 = tan−1 (∑ 𝑥 ) -0 3 (a) in D 20 c ins, =. E 31 TOTAL ∑ 𝑥= 1 1 ∑ 𝑦= - 13 54 ↓. addall. add all 3 71m RESULTANT VECTOR ____________________________________. 85 , ____________________________________. (1. 11 , 13 54) >. - Q based magnitude Drama direction with bearing 85 3) on the Signs!. NogWg 6 85 318 -. SofE &