GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME PHYSICAL SCIENCES GRADE 12 SESSION 1 (LEARNER NOTES)

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1 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) TOPIC 1: MECHANICS PROJECTILE MOTION Learner Note: Always draw a dagram of the stuaton and enter all the numercal alues onto your dagram. Remember to SELECT A DIRECTION AS POSITIVE OR NEGATIVE. SECTION A: TYPICAL EXAM QUESTIONS QUESTION 1: 6 mnutes (Taken from the WC Prelm. paper 008) A cross-bow (bow and arrow) s used to shoot an arrow ertcally upwards nto the ar from the top of an 80 m hgh platform. The arrow reaches a heght of 15 m aboe the platform and then falls to the ground below. Ignore the effects of ar frcton. B 15 m A C 80 m D 1.1 Calculate the magntude of the elocty of the arrow at the nstant t s shot up nto the ar from the top of the platform. (4) 1. Calculate the tme t takes for the arrow to reach the ground from the moment t s shot upwards (4) [8] Page 1 of 10

2 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) QUESTION : 4 mnutes (Taken from Gauteng Prelm paper 009) Sandle throws a small metal ball of mass 10 g ertcally up nto the ar. The ball accdentally lands n the gutter of a buldng. It remans n the gutter for 0.5 s durng whch tme t rolls a few centmetres n the gutter, and then falls through a hole n the gutter back to the orgnal poston n Sandle s hand. The upward elocty wth whch the ball left Sandle s hand was 8 m s -1. When the ball fnally falls back nto hs hand, the elocty s 7 m s -1 downward. Ignore frcton as well as all horzontal moement and answer the followng questons:. Maxmum heght aboe hand = 3.7 m Heght of gutter aboe the hand =.5 m Upward elocty at the start 8 m s -1 Downward elocty at the end 7 m s -1.1 At what speed would the ball hae fallen nto Sandle s hand f the ball had not fallen nto the gutter? (1). The maxmum heght that the ball reaches aboe Sandle s hand s 3.7 m. Proe that ths s correct by usng an equaton of moton and not energy prncples. (4) [5] Page of 10

3 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) QUESTION 3: 0 mnutes A helcopter s rsng ertcally at constant elocty. When the helcopter s at a heght of 100 m aboe the ground, a grl accdentally drops her camera out of the wndow of the helcopter. The elocty-tme graph below represents the moton of the camera from the moment t s released from the helcopter untl t strkes the ground. Ignore arresstance. (m s -1 ) 6 0 a 4 t (s) 3.1 What s the alue of the slope (gradent) of the graph? () 3. Use the gradent to calculate the tme a on the tme axs. (5) 3.3 Whch pont on the path of the camera corresponds to tme a? (1) 3.4 Use an equaton of moton to calculate the magntude of the elocty of the camera as t reaches the ground at 4 s. (4) 3.5 Use the graph to calculate the maxmum heght reached by the camera. (5) 3.6 Draw a rough dsplacement-tme graph and an acceleraton-tme graph to represent the moton of the camera from the moment t was released untl t ht the ground. Tme alues must be shown but y-axs alues need not be shown. (8) [5] Page 3 of 10

4 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) SECTION B: ADDITIONAL CONTENT NOTES Equatons of moton are equatons that are used to descrbe the moton of a body whle experencng a force as a functon of tme. These equatons apply only to bodes mong lnearly (unlaterally).e. n one drecton wth a constant acceleraton. The body s moton s consdered between two tme ponts: that s, from one ntal tme pont and ts fnal pont n tme. Moton can be descrbed n dfferent ways. Words: Dagrams: Graphs: When your frend explans hs frst experence n drng a car and tells you n detal how he struggles to pull off and stop. When you draw a sketch to explan a specfc moement. We use three dfferent graphs. 1. elocty tme graph. acceleraton tme graph 3. poston tme graph. METHOD FOR ANSWERING THESE QUESTIONS: STEP 1: Wrte down all the nformaton gen from the queston STEP : Identfy whch formula to use,.e. dentfy the known and unknown quanttes STEP 3: Substtute nto the equaton STEP 4: Interpret the answer EXAMPLE 1: A car s traellng at 5 m s -1 and t starts to accelerate at m s - for 3 s. What dstance wll the car coer n the 3 s? ANSWER: STEP 1: = 5 m s -1 a = m s - t = 3 s x =? STEP : Δx = Δt + ½ aδt STEP 3: Δx = (5)(3) + ½ ()(3) STEP 4: = 135 m The car wll trael 135 m n the drecton of the moton. Page 4 of 10

5 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) Introducton to the use of equatons of moton n the ertcal drecton A projectle s an object that s gen an ntal elocty by shootng or throwng etc, and once launched, the only force actng on t s the force due to graty. In the absence of ar resstance, the object s free fallng. Termnal elocty s reached when the downward force of graty and the upward force of ar resstance are equal, and now the object falls at a constant elocty as a result of there beng no resultant force actng n on the object. The Equatons of Moton for Vertcal Projectle Moton LINEAR MOTION VERTICAL PROJECTILE MOTION f = + a t f = + g t x = Δt + ½ a t y = Δt + ½ g t f = + a x f = + g y substtute g for a and Δy for Δx n ertcal projectle moton problems All objects are attracted to the earth wth a gratatonal force called WEIGHT. All objects wll accelerate towards the earth wth a constant acceleraton called the gratatonal acceleraton (g) whch has a alue of 9,8 m s -. g = 9,8 m s - whch s found on the nformaton sheet. Gratatonal acceleraton (g) s ALWAYS downwards no matter whether the object s beng thrown up or fallng down. Page 5 of 10

6 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) Important facts concernng Vertcal Projectle Moton: At the greatest heght of the upward moton, f = 0 m s -1 The object wll take the same tme to reach ts greatest heght from pont of upwards launch as the tme taken to fall back to pont of launch ( t up = t down ) f = 0 m s -1 = 0 m s -1 up down t up t down = max f = max Can hae moton descrbed by a sngle set of equatons of moton for the upward and downward moton. If the object s beng released from rest or beng dropped, ts ntal elocty s 0 m s -1. If the object s beng thrown upwards, t must start wth a maxmum elocty and as t moes up, the elocty decreases untl t stops. These are ectors thus drecton s mportant. Page 6 of 10

7 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) SECTION C: HOMEWORK QUESTION 1 A hot-ar balloon s rsng upwards at a constant elocty of 5 m s -1. When the balloon s 100 m aboe the ground, a sandbag s dropped from t (see FIGURE 1). FIGURE shows the path of the sandbag as t falls to the ground. Ignore ar resstance. 1.1 What s the acceleraton of: The hot-ar balloon whle the sandbag s n t? (1) 1.1. The sandbag the moment t s dropped from the hot-ar balloon? () 1. Determne the maxmum heght P, aboe the ground, reached by the sandbag after t s released from the hot-ar balloon. (3) 1.3 Calculate the tme taken for the sandbag to reach ths maxmum heght after t has been released. (3) 1.4 Calculate the total tme taken for the sandbag to reach the ground after t has been released. (4) 1.5 Wll the elocty of the hot-ar balloon INCREASE, DECREASE or REMAIN THE SAME mmedately after the sandbag has been released? Explan fully. (4) [17] Page 7 of 10

8 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) SECTION D: SOLUTIONS AND HINTS TO SECTION A QUESTION Take downwards as the poste drecton / upward s negate (4) f = 0 g = 9,8 m s - y = -15 m 1. x = 95 m g = 9,8 m s - = - 17,15 m s -1 t =? 0 f ay (9,8)( 15) 17,15 m s 1 17,15 m s 1 y = t + g t 80 = (-17,15) t + (9,8) t 4,9 t 17,15 t 80 = 0 t = = = 6,15 s OR (4) AB f 17,15s 0 a 9,8m s y?? t?? f BCD 0 a 9,8 m s y 95m t?? 1 f at y t 1 at 0 ( 17,15) (9,8) t t 1,75 s 95 0 t 4,40s 1 (9,8) t t total 1,75 4,40 6,15s [8] Page 8 of 10

9 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) QUESTION.1 8 m s -1 (1). aδy f 0 8 Δy m (4) [5] QUESTION m s - down () g = gradent = change n elocty /change n tme = 9, 8 a 0 a Therefore a = 0,61 s (5) 3.3 At the pont of maxmum heght reached where = 0.e. pont at the top of the moton. (1) 3.4 f = + gδt = 0 + (-9,8)(3,39) = 33, m s -1 down (4) 3.5 x = area under graph = ½ b h = ½ (0,61)(6) = 1,8 m Maxmum heght reached = ,8 = 101,8 m (5) Page 9 of 10

10 PHYSICAL SCIENCES GRADE 1 SESSION 1 (LEARNER NOTES) 3.6 x(m) t(s) shape a (m s - ) 0-9,8 shape t (s) (8) [5] Learner Note: Copy the equaton off the nformaton sheet as gen. Only then substtute nto the equaton, and then manpulate the equaton to make the unknown the subject of the formula. The SSIP s supported by Page 10 of 10