Name:. CG : Date: Feb 2011

Transcription

1 Teacher s Cop JC Phsics H & H Name:. CG : Date: Feb Topic Kinematics B Mr. Ng Soo Kok (Extension : ) ng_soo_kok@moe.edu.sg Main Reference: College Phsics Vuille & Serwa (8 th Ed) Content Rectilinear motion Non-linear motion Learning Outcomes Candidates should be able to: (a) define displacement, speed, velocit and acceleration. (b) use graphical methods to represent distance travelled, displacement, speed, velocit and acceleration. (c) find displacement from the area under a velocit-time graph. (d) use the slope of a displacement-time graph to find the velocit. (e) use the slope of a velocit-time graph to find the acceleration. (f) derive, from the definitions of velocit and acceleration, equations which represent uniforml accelerated motion in a straight line. (g) solve problems using equations which represent uniforml accelerated motion in a straight line, including the motion of bodies falling in a uniform gravitational field without air resistance. (h) describe qualitativel the motion of bodies falling in a uniform gravitational field with air resistance. (i) describe and explain motion due to a uniform velocit in one direction and a uniform acceleration in a perpendicular direction. Internet Resources: (good resource) (good resource) (graphic simulation) (graphic simulation) (video) (graphic simulation) (graphic simulation) (video resources) Kinematics / JC / NgSK

2 . Introduction Kinematics [derived from the Greek word kīnēma: motion] is that branch of mechanics that studies the motion of objects without due consideration given to the force/s that cause them to move. We attempt to describe the motion using words, diagrams, graphs, and equations. Here we shall assume that the object is a small (point) object capable of translation motion onl (ie no rotational motion). The mass of the object is immaterial as we shall soon learn. To describe the motion and location of an object, we need to have a frame of reference. A frame of reference is a choice of coordinate sstems that defines the starting point or a point of reference so that the location of an object can be precisel determined relative to it. Spatial Dimension* Coordinates Needed Describes location of an object on -Dimensional x A Line (eg straight highwa, ruler ) -Dimensional x, A Plane (eg chess board, map) 3-Dimensional x,, z Space (eg hologram, 3D model) Worked Example Giving instructions to our friend to our home. Post Office (pt of origin) Main Street x Fig. Examples of spatial dimension * We will focus on the motion of an object in D and D. Kinematics Definitions Motion Graphs Kinematics Equations Fig. Rectlinear Motion (-D) Projectile Motion (-D) Rectilinear motion is the motion of an object on a straight line eg a stone dropped verticall. Projectile motion is the motion of an object on a plane (usu. a vertical plane) eg a stone thrown at an angle from the horizontal line. The object must not be propelled b some inside power engines eg rocket or missile. The flight of these powered objects is NOT characteristic of a projectile motion. Kinematics / JC / NgSK

3 Using the Cartesian sign convention When describing the location or motion of an object, we need to be mindful of scalar and vector qualities. (-x,) + (x,) The Cartesian sign convention is used to describe the directions of vector quantities. The arrows indicate the direction of the location or the velocit of the object. - x (-x,-) - (x,-) + x Motion on a vertical plane Fig. 3 Cartesian sign convention Diagrams + v= m s - + Examples Saing it - x 3 o A stone is thrown up at an angle from the horizontal surface. The stone is projected with a speed m s - at an angle 3 o above the positive x -direction. + x 4 o - v=3 m s - + x A stone is thrown down at an angle from the edge of a cliff. The stone is projected with a speed 3 m s - at an angle 4 o to the right of the negative -direction. For motions on a vertical plane, one cannot use the terms - north, west, east and south. Wh? Motion on a horizontal plane N, E, W & S are the main directions. N and S take precedence as the meridian directions, thus the intermediate directions are NE, NW, SE & SW. The are at 45 o to the main directions. Describing directions of A B C Short Form N5 o E S o E N75 o W Saing it 5 o East of North o East of South 75 o West of North Kinematics / JC / NgSK 3

4 . Definitions (a) define displacement, speed, velocit and acceleration.. Distance and Displacement (smbol : s ; S.I. unit: m) Distance is the total length of the actual path traveled b an object irrespective of the direction of motion. It is a scalar quantit. Displacement is the distance between the final position of an object and its initial position in a specific direction. It is a vector quantit with its direction starting from the initial position to the final position. Self-Assessment Difference between distance and displacement (-Dimensional) The diagram shows the motions of four particles. A number line is used to determine the respective initial and final positions of the particles. Complete the table below. Particle Distance / cm Displacement / cm ( right) 4-4 ( left) ( right) (right) The signs (+ or -) for displacement indicates direction (taking rightward as positive) The magnitude for the displacement is relative to the respective initial positions. Worked Example Difference between distance and displacement (-Dimensional) In an Orientation programme, a treasure map has the following instructions. The student has to follow the instructions in turn to locate the treasure. Find. Stand at the centre of the school field facing northwards.. Walk 7. m to the North. 3. Walk. m to the East. 4. Walk 5. m in the southwest direction (or S45 o W). 5. Walk 5. m in the direction of N3 o W (or 3 o West of North). (a) the total distance travelled. Total distance = = 39. m (b) the net displacement Kinematics / JC / NgSK 4

5 Method (Using a scaled diagram and vector addition) (You will need a protractor and a ruler for this). Scale. cm :. m. cm 7. cm 5. cm Starting point 5. cm Final Displacement (b scale drawing) Measurement =.3 cm (using ruler) Displacement =.3 m Angle = 6 o W of N (using protractor) Method (Using Resolution Method) The displacement and direction is resolved ( broken down ) into the x- and - components. Steps Displacement in x- direction / m Displacement in - direction / m () Stand at the centre. () Walk 7. m to the North +7. (3) Walk. m to the East. +. (4) Walk 5. m in the southwest direction (5) Walk 5. m in the direction of N3 o W Net Displacement Magnitude Net displacement = =.3 m Direction. m. tan = 58 o θ. 7.7 m The net displacement is.3 m in the direction of N58 o W (58 o West of North). You can find the treasure sooner and with less effort if ou appl a bit of math. Kinematics / JC / NgSK 5

6 . Speed and Velocit (smbol v ; S.I. unit: m s - ) Speed is the rate of change of distance travelled b an object (or a change of distance per unit time). It is a scalar quantit. () Average Speed <v> is defined as the total distance travelled b the object divided b the total time taken. Total Distance Travelled s Average Speed = < v >= Total Time Taken t () Instantaneous Speed is the distance travelled b an object per unit time at a particular instant of time. The instantaneous speed of a car is indicated b its speedometer. Velocit is the rate of change of displacement b an object (or a change of displacement per unit time). It is a vector quantit. () Average velocit, <v> is defined as the net displacement ( s) of the object divided b the time interval ( t). s < v >= t () Instantaneous velocit, v is the rate of change of displacement at a particular instance of time. v = ds dt The value ds is given b the gradient of the tangent of the displacement-time graph at a dt particular instant of time. The average velocit ( s/ t) of an object during the time interval is equal to the slope of the straight line joining the initial (A) and final (B) points on a displacement-time graph s / m A t B s (ds/dt) The instantaneous velocit (ds/dt) of an object at point A is given b the gradient of the tangent at that point during that instant of time. t / s Kinematics / JC / NgSK 6

7 Self-Assessment Difference between average speed and average velocit (-Dimensional) An athlete runs 5 m along a straight track and then changes direction racing back along the same track to the starting point in. s. Determine for that time interval, (a) the total distance travelled. (b) the total displacement. (c) the average speed. (d) the average velocit. 5 m (a) Total distance traveled = m (b) Total displacement = m (c) Average speed = Total distance / Total Time = m/. s = 9. m s - (d) Average velocit = Total displacement / Total time = m s - Worked Example 3 Difference between average speed and average velocit (-Dimensional) A plane flies 45 km east from cit A to cit B in 45. min and then 9 km south from cit B to cit C in 9. min. A 45 km B (a) What is the total distance (in km) traveled? (b) What is the net displacement (in km) for the whole trip? (c) What is the average speed (in km h - )? (d) What is the average velocit* (in km h - )? 9 km (a) Total distance traveled = = 35 km C (b) Magnitude of net displacement = = 6 km = km (3 s.f.) Direction 9 tanθ = = 45 θ = 63.4 o The net displacement of the plane is km, 63.4 o South of East. (c) Average speed = total distance 35 = = 6 km h- total time ( ) 6 (d) Magnitude of average velocit = Net displacement / Total time = 6 km /.5 h = 447 km h - (3 s.f.) Direction 9 tanθ = = 45 θ = 63.4 o The average velocit* of the plane is 447 km h -, 63.4 o South of East. * Note: Average velocit is a concept seldom used in practice. Kinematics / JC / NgSK 7

8 .3 Acceleration (smbol a ; S.I. unit: m s - ) Acceleration is defined as the rate of change of velocit (or change of velocit per unit time). It is a vector quantit. () Average acceleration <a>, is defined as the net change in velocit ( v), divided b the total time taken t. v < a > = t () Instantaneous acceleration a, is the rate of change of velocit at a particular instant of time. dv a = dt The value dv is given b the gradient of the tangent of the velocit-time graph at a dt particular instant of time. The average acceleration ( v/ t) of an object during the time interval is equal to the slope of the straight line joining the initial (A) and final (B) points on a velocit-time graph v / m s - A t B v (dv/dt) The instantaneous acceleration (dv/dt) of an object at point A is given b the gradient of the tangent at that point during that instant of time. t / s Watch video How to survive at high velocit : Worked Example 4 Acceleration (-D & -D) A F racing car travels through points ABCD with the respective instantaneous velocities as indicated in the diagram. The time intervals between AB, BC and CD are t, t and t 3 respectivel. A km h - t =. s B 8 km h - 8 km h - t =.5 s C (a) Express the speeds in m s -. (b) What is the acceleration (in m s - ) between points A and B? (c) What is the acceleration (in m s - ) between points C and D? (d) What is the average acceleration (in m s - ) between points B and C? N t 3 =.6 s km h - D Kinematics / JC / NgSK 8

9 (a) km h - = 36 3 = 7.8 m s - 8 km h - = =. m s - (b) Magnitude v v u a = = = = 5. 6 m s t t. The car decelerates at a rate of 5.6 m s -. The negative sign indicates the acceleration is acting to the west opposite to the initial velocit. Since the direction of the acceleration is opposite to the initial velocit, car slows down. Direction v B final v A initial v change in velocit Change in velocit is in the -x direction (ie west). Thus, direction of acceleration is negative. (-x) The acceleration between A and B is -5.6 m s - westward. (c) Magnitude v v u a = = = = 9. 3m s t t. 6 The car accelerates at a rate of 9.3 m s -. Direction v C initial Alternativel v v u 7. 8 (. ) a = = = = 9. 3m s t t. 6 The car accelerates at a rate of 9.3 m s - in the negative -direction (ie south). The negative sign here does not mean the car is decelerating. It means the acceleration is acting in the negative direction, same as the velocit. Thus, the car speeds up. v change in velocit v D final Change in velocit is in the - direction (ie south). Direction of acceleration is negative (-). Since the direction of the acceleration is the same as the initial velocit, the car speeds up. The acceleration between C and D is 9.3 m s - southward. (d) Magnitude v =. +. = 3. 4m s v 3. 4 a = = =. 6m s t. 5 Direction v C final v B initial v change in velocit There is acceleration even though the speed is constant. This is because there is a change in direction. This tpe of acceleration is called centripetal acceleration. Change in velocit is in the southwest direction. Thus, the average acceleration is in the southwest direction. Note that the speed is unchanged, but the velocit has change from eastward at point B to southward at point C. Note:. The acceleration on an object can be due to a change in the magnitude of the velocit or a change in the direction or both.. Negative acceleration does not necessaril mean an object is slowing down. If the acceleration and the velocit are both negative, the object is speeding up in the negative direction. 3. Negative acceleration is not the same as deceleration which means a reduction in speed. The negative here refers to the direction of the acceleration which can be in the x or direction. (See Page3 of Vuille and Serwa) Kinematics / JC / NgSK 9

10 3. Graphical Representations (b) use graphical methods to represent distance travelled, displacement, speed, velocit and acceleration. The motion of an object along a straight line can be described b three tpes of graphs namel: Graph Slope Area under the graph () Displacement time (s-t) () Velocit time (v-t) (3) Acceleration time (a-t) Instantaneous velocit ds v = dt Instantaneous acceleration dv a = dt Change in Displacement ds v = s = v dt dt Change in velocit dv a = v = a dt dt The velocit-time graph is the most versatile because each main feature holds special significance: the area beneath the v-t graph represents the change in displacement while the slope at an point on the v-t graph gives the acceleration. To gain a better understanding of the motion through graphical representations, we will do the following examples. 3. Displacement-time graph (d) use the slope of a displacement-time graph to find the velocit. Self-Assessment 3 The following s-t graph shows a 4. km race among three runners: A : the kiasu runner B : the stead runner C : the delaed runner. s / km A. (a) At t =. min, what were the placing and displacements of the three runners?. B C st place : A (.7 km) nd place : B (.8 km) 3 rd place : C (.3 km) t / min (b) Who eventuall came in first, second and third in the race? Give their respective finishing times? st is C (4. min), nd is B (5. min), 3 rd is A (6. min) Kinematics / JC / NgSK

11 (c) Estimate the initial (ie at t = min) velocities of the three runners in m per min. Initial velocit of A v = = 5 m min. (assume const. velocit during first. s) Initial velocit of B 8 v = = 8m min. Initial velocit of C 3 v = = 3m min. (assume const. velocit during first. s) (d) When did the eventual winner overtake the other two runners? C overtook B at t = 34. min and then it overtook A at t = 38. min (e) How can we tell from the graph that a runner is slowing down or speeding up? The slope of the graphs reveals the change in velocities. A is progressivel slowing down as the slope decreases with time. C is progressivel speeding up as the slope increases with time. B is moving with constant velocit as the slope is constant. Points to note in a displacement-time (s-t) graph The sign for s indicates the location of the object with respect to the point of reference. The slope of the tangent at a point on the graph gives the instantaneous velocit. An increase in the slope means that the velocit is increasing (ie object is accelerating) A decrease in the slope means that the velocit is decreasing (ie object is decelerating) A constant slope means that the velocit is constant (ie object is not accelerating) Where an intersection between two graphs occurs, it means that the two moving objects are alongside each other (same location) at the same point in time. 3. Velocit-time graph (c) find displacement from the area under a velocit-time graph. (e) use the slope of a velocit-time graph to find the acceleration. Worked Example 5 N93/II/ (modified) The graph shows the speed of two cars A and B which are travelling in the same direction over a period of 4 s. Car A, travelling at a constant speed of 4 m s -, overtakes car B at time t = s. In order to catch up with the car A, car B immediatel accelerates uniforml for s to reach a constant speed of 5 m s -. (a) Calculate the acceleration of car B for the first s. (b) At what time did car B catch up with car A? v / m s - 5 Car B 4 Car A 5 t 4 t / s Kinematics / JC / NgSK

12 Note: The point of intersection of the two graphs does not represent the point of overtaking. At this point in time, both cars share the same speed (ie 4 m s - ) ; the are NOT side b side of each other. (a) Acceleration of B = slope of graph during first s = (5-5)/ =.5 m s - (b) For car B to catch up with car A, both cars would need to share the same displacement. (i.e. s A = s B ) at the same point in time. Thus, the areas under each graph must be made the same. In the first s, displacement of each car: s A = 4 s B = ( ) = 8 m = 75 m When car B catches up with car A, s A = s B t = t t = 5. s Time taken = + 5. = 5 s 3.3 Relations between s-t, v-t and a-t graphs A good understanding of the relationships between the 3 graphs describing the same object in motion is ver important. In the following example, we will see wh negative acceleration does not necessaril means deceleration. Likewise, neither does positive acceleration necessaril means acceleration. Worked Example 6 A driver drives his car from one point to another point (, m awa) and reverse back on its track with the applications of the two pedals: Gas pedal (a.k.a. accelerator) which controls the acceleration of the car. Brake pedal which controls the deceleration. t = s v = m s - t =. s v = m s - t = 53. s v = m s - t = 57. s v = m s - START v = + a = + v = + a = - m 86 m 4 m negligible time for the turn FINISH t = 4 s v = m s - 4 m 86 m m v = - v = - a = + a = - t = s v = - m s - t = 67. s v = - m s - t = 57. s v = m s - Kinematics / JC / NgSK

13 s / km. Slope = Inst. velocit time / s v / m s - Area = of displacement Slope = Inst. acceleration time / s a / m s - 5. Area = of velocit time / s (a) (b) (c) (d) (e) (f) (a) (b) (c) (d) (e) (f) s v + constant + - constant - a Pedal applied gas brake gas brake Motion speeds up constant slows down speeds up constant slows down Note: The signs (+ or -) for v and a indicate the directions. The do not represent location except for s. A negative acceleration means the direction of a is negative while a positive acceleration means the direction of a is positive. When v and a are of the same sign, the car speeds up. When v and a are of opposite signs, the car slows down. Kinematics / JC / NgSK 3

14 Worked Example 7 The graphs in Example 9 are oversimplification of what actuall happens to the car. We assume that the change in acceleration occurs immediatel. If this was the case, the ride would be a jerk one as the change of velocit is ver abrupt. The following graphs are better representations of what happen in practice when the gas or brake pedal is applied graduall. Here, we are considering the motion of the car in one direction onl. v The shape of this graph often leads to misconception. The car is not making a turn around. The car is initiall speeding up, maintain constant speed, then slows down in the same direction. As the velocit is alwas positive, the displacement of car is alwas increasing. a t / s t / s (i) (ii) (iii) (iv) (v) (vi) (vii) (i) (ii) (iii) (iv) (v) (vi) (vii) Velocit increasing Constant velocit Velocit decreasing v a Increasing positive acceleration Constant positive acceleration Decreasing positive acceleration Zero acceleration Increasing negative acceleration Constant negative acceleration Decreasing negative acceleration Gas pedal down Gas pedal maintains Gas pedal retracts - Brake pedal down Brake pedal maintains Brake pedal retracts Note: Decreasing acceleration does not mean deceleration. It just means that the car is accelerating at a decreasing rate; the car is still speeding up albeit at a lesser rate. For part (iv), we assume that there is no friction acting against the motion of the car. Thus, the velocit of the car is maintained. In practice however, friction is alwas present and to maintain its constant motion, the gas petal needs to be pressed b a certain fixed amount to overcome friction. The area under the a-t graph gives the change in velocit of the car. Kinematics / JC / NgSK 4

15 Points to note in a velocit-time (v-t) graph The slope of the tangent at a point on the graph gives the instantaneous acceleration. An increase in the slope means that the acceleration is increasing. A decrease in the slope means that the acceleration is decreasing. A constant slope means that the acceleration is constant. The area under the graph represents the displacement. Self-Assessment 4 (N4//3) Modified A train travelling at. m s - passes through a station. The variation with time t of the speed of the v of the train after leaving the station is shown below. v / m s t/s (a) What is the acceleration of the train? v u a = = =. m s - t 5 (b) What is the speed of the train when it is 5 m from the station? Method Area under graph = displacement 5 = ( + v)t B inspecting the graph, the area is 5 m, when v = 8. m s -, t = 3 s Method v u v B substituting t = = into the above equation, v = 8. m s - a. Alternativel (to appl after learning Section 4) v = u + as v = + (. )( 5) = 64 v = 8. m s - Kinematics / JC / NgSK 5

16 4. Equations of Motion 4. Derivation of equations for linear motion (f) derive, from the definitions of velocit and acceleration, equations which represent uniforml accelerated motion in a straight line. Consider the motion of an object that accelerates from an initial velocit u to a final velocit v with constant acceleration a over a time interval t. From the definition of acceleration, v v-u a = = t t v = u + at ---() Since displacement is the area under the velocit-time graph, s = ( ) u + v t Substitute v = u + at into equation (), s = s = ut + ( ) u + u + at t at Rewriting equation () for --- () --- (3) v / m s - v - u t = a and substituting into equation () v u ½ at ut t at t / s v - u s = ( u + v) a v = u + as ---(4) Note:. These derived equations of motion appl onl to uniforml accelerated motion in a straight line.. Even in one dimensional problems s, u, v and a are vector quantities! It is important to be consistent in the usage of the signs. If the positive direction is taken to be that to the right, then i. points on the left of the point of reference have negative values for s ii. velocities directed to the left have negative values iii. accelerations directed to the left have negative values Equations v = u + at ( ) Relations v and t s = u + v t s, v and t s = ut + at s and t v = u + as s and v Kinematics / JC / NgSK 6

17 Worked Example 8 [J3/I/8] A motorist travelling at m s - can bring his car to rest in a distance of m. If he had been travelling at 3 m s -, in what distance could he bring the car to rest using the same braking force? Same breaking force implies that the car is undergoing the same deceleration. Case Case s m * u m s - 3 m s - v a?? t Not given Not given Method v = u + as = + a = 5. m s - a ( ) ( ) = s s = 9 m Method v / m s - 3? m t Self-Assessment 5 [J85//3] The velocit of a car which is decelerating uniforml changes from 3 m s - to 5 m s - in 75 m. After what further distance will it come to rest? A 5 m B 5 m C 75 m D m Solution Step : Find the acceleration v = u + as 5 = 3 + a(75) a = -4.5 m s - Step : Find the extra distance to come to rest v = u + as = 5 s = 5 m + ( 4. 5)s Kinematics / JC / NgSK 7

18 5. -D free fall in a uniform gravitational field (without air resistance) (g) solve problems using equations which represent uniforml accelerated motion in a straight line, including the motion of bodies falling in a uniform gravitational field without air resistance. If there is no air resistance, all objects irrespective of mass, shape or size will fall towards the Earth with the same acceleration known as acceleration of free fall or the acceleration due to gravit, g. Such objects are said to be in free fall if the onl force acting on it is the gravitational pull of the Earth. Near the surface of the Earth, the value of g = 9.8 m s - and it is directed towards the centre of the Earth. A bod falling from rest under the action of gravit without air resistance will increase in its velocit constantl at a rate of 9.8 m s - per s. Consider an object dropped from rest. Given that the acceleration due to free fall g is 9.8 m s -, we use MS Excel to obtain the following table and graphs s-t, v-t and a-t. Here, we assume that there is no air resistance and we take downward direction as positive. s = ½ g t v = gt Displacement time graph t / s s / m v / m s Velocit time graph.. Note. The s-t graph is parabolic. s t. The v-t graph is linear. v t velocit is dir. proportional to time velocit / m s time / s 3. The a-t graph is a flat line. a = g = constant a is independent of t Acceleration time graph Kinematics / JC / NgSK 8

19 Worked Example 9 [motion of bodies falling in a uniform gravitational field without air resistance.] At the top of a cliff m high, a student throws a rock verticall upwards with an initial velocit m s -. How much time later should he drop a second rock from rest so that both rocks arrive simultaneousl at the bottom of the cliff? Ignore air resistance. Assume upwards as positive. st rock nd rock s - m - m u + m s - m s - v Not given Not given a -9.8 m s m s - t t t Use b ± b 4ac a st rock nd rock m For the st rock s = ut + ½ at - = t + ½ (-9.8)t 4.9 t t = ( ) ± ( ) 4( 4. 9)( ) t = = 6.7 s ( 4. 9) For the nd rock s = ut + ½ at - = ()t + ½ (-9.8)t 4.9 t = t = 4.5 s Time interval = t - t = =.9 s Self-Assessment 6 [J85//4] The acceleration of free fall is determined b timing the fall of a steel ball photo-electricall as shown below. The ball passes P and Q at times t and t after release. point of release What is the acceleration of free fall? light beam P A h /( t t) B h /( t t ) C h /( t t ) D h /( t t ) E h / ( t t ) light beam Q h Photo-cells Solution ground Let s = distance dropped when at P s = gt.() s + h = gt () ()-() h = g( t t ) = h / t t Ans : D g ( ) Kinematics / JC / NgSK 9

20 Worked Example (Serwa Pg 45 Example.) A rocket moves straight upwards, starting from rest with an acceleration of +9.4 m s -. It runs out of fuel at the end of 4. s and continues to coast upward, reaching a maximum height before falling back to Earth. (a) Find the rocket s velocit and position at the end of 4.s. (b) Find the maximum height the rocket reaches. (c) Find the velocit the instant before the rocket crashes on the ground. Solution The rocket goes through two phases: Phase when the rocket is powered b the fuel. Phase when the rocket is at free fall. (a) Rocket is in Phase Using v = u + at v = + (+9.4)(4.) = 7.6 m s - = 8 m s - Using s = ut + ½ at t is given, so use v = u + at s = ut + ½ at s = + ½ (+9.4) (4.) = 35. m = 35 m (b) Rocket is in Phase v = u + as t is not given, so use v = u + as () = v + (-g) s = (+7.6) + (-9.8) s s = 74.9 m Maximum height Y max = s + s = = 94. = 94 m (c) Rocket is descending free fall v = u + as v = + (-9.8)(-94.) v = ±35.8 m s - v = 36 m s - in downward direction t is not given, so use v = u + as velocit is a vector quantit, so must give its magnitude and direction. In general, express our answers in or 3 s.f. However, if the value is required for further calculation, use the value that is in 4 s.f. to avoid further deviation from the true value due to compound error. Kinematics / JC / NgSK

21 Worked Example [N5/P/3] A tennis ball is dropped from a distance above a rigid horizontal surface and is allowed to bounce. The s-t graph is as shown below (note it is a -D problem not -D). Draw the corresponding v-t and a-t graphs for the motion. Assume no air resistance. s Ball is undergoing free fall when in the air even when it is rising up. Path is parabolic. (s = ½ at ) Slope gives the instantaneous velocit. Rebound heights decrease progressivel. t v Slopes are linear & parallel. Negative slopes = -9.8 m s - Positive slopes are almost vertical due to rebound. Peaks decrease progressivel. t a a-t graphs are flat when ball is in the air ie free fall. Acceleration = -9.8 m s - Positive acceleration is large due to sharp rebound. Peaks decrease progressivel. a = -9.8 m s - even when v = at the max height. t Self-Assessment 7 A tennis plaer on serve tosses a ball straight up. While the ball is in free fall, its acceleration A increase B decrease C increase and then decrease D decrease to zero and then increase E remain constant Ans : E Kinematics / JC / NgSK

22 Self-Assessment 8 [N//4] A steel ball is released from rest a distance above a rigid horizontal surface and is allowed to bounce. Which graph best represents the variation with time t of acceleration a? Ans: D Self-Assessment 9 [J3//9] A ball is released from rest above a horizontal surface. The graph shows the variation with time of its velocit. Areas X and Y are equal. velocit st impact nd impact 3 rd impact release X Y time Ball falling Ball rising Ball falling Ball rising Ball falling This is because A the ball s acceleration is the same during its upward and downward motion B the speed at which the ball leaves the surface after an impact is equal to the speed at which it returns to the surface for the next impact. C for one impact, the speed at which the ball hits the surface equals the speed at which it leaves the surface. D the ball rises and falls through the same distance between impacts. Ans : D Kinematics / JC / NgSK

23 6. Effect of air resistance on motion of objects falling under the action of gravit (h) describe qualitativel the motion of bodies falling in a uniform gravitational field with air resistance. When an object is dropped from rest, the onl force acting is its weight W. (Point A). As it descends, it picks up speed leading to a corresponding increased in air resistance R acting on the object. Since air resistance opposes the downward motion, it causes the net force acting on the object to decrease and its net acceleration to reduce. (Point B). Note that the object is not slowing down just because its net acceleration is decreasing. The velocit still increases but at a decreasing rate. Eventuall, when the air resistance acting upwards equals to the weight of the object, the resultant force is zero and there is zero acceleration from this point onwards. (Point C). The object then falls with a constant maximum velocit, called terminal velocit. Onset of terminal speed R R s No air resistance W W W With air resistance Point A Point B Point C v / m s - A The gradient at each point on the curve represents the object's instantaneous acceleration. The greatest acceleration occurs when the object is just released and the air resistance is negligible. In general, a heavier object with a small surface area such as steel ball-bearing takes a longer time to reach its high terminal velocit, as compared to a raindrop. Self-Assessment [N4//4] B C No resistance Slope= 9.8 m s - v = + gt steel ballbearing raindrop t / s v a t No air resistance With air resistance t No air resistance With air resistance t A steel sphere is released from rest at the surface of a deep tank of viscous oil. A multiple exposure photograph is taken of the sphere as it falls. The time interval between exposures is alwas the same. Which of the following could represent this photograph? For a sphere moving in viscous oil, its velocit increases from zero to a constant value (terminal velocit). The spacing between the dots has to slowl increase till a constant value. Ans B Kinematics / JC / NgSK 3

24 7. Projectile Motion (-D) in a Uniform Gravitational Field (i) describe and explain motion due to a uniform velocit in one direction and a uniform acceleration in a perpendicular direction. A projectile is an object propelled through space b the exertion of a force which ceases after launch (eg a basketball, tennis ball, arrow). If air resistance is negligible, the onl force acting on the object is its weight, acting downwards. Projectile motion is a -D curved motion of a particle subject to a constant acceleration, e.g. a ball thrown obliquel into the air. The gravitational acceleration g affects the vertical velocit onl and not the horizontal velocit, i.e. a x = and a = g. The velocit of the projectile along ever point of the trajector therefore changes all the time. The horizontal and vertical motions are completel independent of each other. We can easil solve projectile motion problems if we treat the horizontal motion and vertical motion separatel. The two are independent of each other. In other words, we break (resolve) a two-dimensional problem into one-dimensional problems (x- and -directions) and treat each separatel. After an object has been projected into space, the onl force acting on it is its weight. Thus, there is acceleration in the -direction while there is none in the x-direction. As can be seen in the diagram, the horizontal separations (s x t) for regular time intervals are constant while the vertical separations ( s t ) are changing. The result is a parabolic path. A ball dropped verticall will fall at the same rate as another ball projected horizontall. Check out (graphic simulation) This excellent simulation gives the trajector of a fired cannon ball. Effect of air resistance is illustrated. Kinematics / JC / NgSK 4

25 7. Equations of motion for projectile motion The diagram on the right is a tpical parabolic trajector. In solving projectile problems, we must first resolve the initial velocit u. In the x-direction, we have In the -direction, we have u x = u cosθ u = u sinθ s Horizontal motion (a x = ) x = v = u t x x u x Vertical motion (a = - g ) *Taking upwards as positive s = u t + a t v = u a t + v = u + a s u sin θ Projectile Motion (-D) u θ u cos θ x s x = ( u cosθ )t v x = u cosθ s v = θ gt = u sin θ gt ( u sin ) t ( usin ) gs v = θ Problem-Solving technique for Projectile Motion. Select a coordinate sstem and sketch the path of the projectile, including the initial and final positions and velocities.. Resolve the initial velocit vector into the x- and -components. 3. Treat horizontal motion and the vertical motion independentl. 4. Use the equation of motions for solving problems with constant velocit to analze the horizontal motion of the particle. 5. Use the equation of motions for solving problems with constant acceleration to analze the vertical motion of the projectile. Self-Assessment [ J3//7] Ans : C Kinematics / JC / NgSK 5

26 Worked Example [N97/III/b(ii)] [Context : Full parabolic path] A ball is thrown from the horizontal ground with an initial velocit of 5 m s - at an angle of 6 to the horizontal. Assuming that air resistance can be neglected, determine. the maximum height to which the ball rises,. the time of flight and 3. the range. v = u + a s. 3. ( ) ( ) = 5 sin s s = 8.6 m s = u t + a t = ( 5 sin 6 ) t + ( 9. 8) t t = (reject) or t =.65 s s u t a t x = x + x s x = 5cos = 9.9 m * value to be determined ( ) ( )( ) s * u + 5 sin 6 v a t? s u + 5 sin 6 v? a t * x s * u + 5 cos 6 v + 5 cos 6 a t.65 Worked Example 3 [Modified J8/II/3] [Context : Half parabolic path] m s - 5 m An aeroplane, fling in a straight line at a constant height of 5 m with a speed of m s -, drops an object. The object takes a time t to reach the ground and travels a horizontal distance d in doing so. Ignoring air resistance, (a) determine the velocit of the object just before it hits the ground. (b) calculate the values of t and d. (a) In x-direction v x= u x = m s - In -direction v = u + a s v = + ( 9. 8)( 5) v = 99. m s - x s d 5 u v * * a t t t Kinematics / JC / NgSK 6

27 Resultant velocit v = v x + v v = v = 3 m s - v tanθ = = v x θ = 6.3 The velocit of the object just before it hits the ground is 3 m s - at 6.3 below the horizontal. (b) Taking upwards as positive, s = u t + a t 5 = t + ( 9. 8) t t =. s s = u d = ()(.) = m t s -5 u v -99. a t * x s * u v a t. Worked Example 4 [Context 3: Other parabolic path] A ball is thrown from the top of one building towards a tall building 5 m awa. The initial velocit of the ball is m s - at m above the horizontal. How far above or below its original level will the ball strike the opposite wall? m s - 4 m Solution: 5 m u x = cos 4 = 5.3 m s -, u = sin 4 =.9 m s - Step : Find time taken to reach the wall. s x = u x t 5 = ( cos 4 ) t t = 3.64 s Step : Find the drop from original height. Take upwards as positive. s = u t + a t s = (sin 4)(3.64) + ( 9.8)(3.64) s =.3 s = m (below original level) The ball will strike.3 m below the original level. Kinematics / JC / NgSK 7

28 8. The Monke Gun Challenge A banana is aimed at the monke and projected at the precise moment when the monke releases his grip from the branch. Will the banana hit the monke before the monke reaches the ground? This is provided the banana does not hit the ground first before the monke. H The answer is alwas YES. Show mathematicall, that this is the case. θ x h Hint: The time of drop for the monke must be the same as the time of flight for the banana. The heights for the banana and the monke must be the same (h), when the banana crosses the path of the monke. Use the equations of motions rectilinear for the monke and projectile for the banana. me with our solution or drop it in m pigeon hole with our name and CG. The one with the best solution will win a prize! In addition, his/her solution will be presented on the LMS. Deadline: Mar. Kinematics / JC / NgSK 8

29 Kinematics : Summar Distance is the total length of the actual path traveled irrespective of the direction of motion. Displacement is the distance between the final position of an object and its initial position in a specific direction. Speed is the rate of change of distance travelled (or a change of distance per unit time). () Average Speed is defined as the total distance travelled b the object divided b the total time taken. () Instantaneous Speed is the distance travelled b an object per unit time at a particular time instant. Velocit is the rate of change of displacement. () Average velocit, is defined as the net displacement of the object divided b the time interval. () Instantaneous velocit, v is the rate of change of displacement at a particular time instant. Acceleration is the rate of change of velocit (or change of velocit per unit time). () Average acceleration, is defined as the net change in velocit, divided b the total time taken. () Instantaneous acceleration a, is the rate of change of velocit at a particular time instant. In the s-t graph, the slope gives the velocit of the object. In the v-t graph, the slope gives the acceleration of the object while the area under the graph gives its displacement. Rectilinear Motion: One dimensional motion with constant acceleration s = v = u + at s = ut + ( ) u + v t at v = u + as Projectile motion:two dimensional motion with constant acceleration in one direction. Horizontal motion (a x = ) How displacement changes with time? s x ( u cosθ )t Vertical motion (a = - g) *Taking upwards as positive = u sin θ t gt = ( ) s How velocit changes with time? How velocit changes with displacement? v x = u cosθ= constant = u sin θ gt v x = u cosθ= constant v = ( usin θ ) gs v End Kinematics / JC / NgSK 9