YEAR 10 SCIENCE 2023 “We're not actually going that fast- its just that the ink got smudged.” IMGHIUMER com by 1 Mecrocken Poultry in Motion Name: WORKBOOK Fun Fact: The word "Physics" Is taken from Greek saying "knowledge of nature". It is one of the oldest fields of science which dates back to 650 BC which was discovered by Greeks.Year 10 Physics Booklet: Forces and Motion Lesson 1 - 2: Quick Review of Fundamental Quantities and Units Physical Quantity Symbol Unit Name Unit Symbol Myoy m Ke loprau ka leugth (isto d ae Bi Tiwe iG second = volo] ped | V wolves per cooaf| 6" accohorehion Ow autee er saoal? ms? fore F Meuston M (gas) E weve & oule on Vocabulary By the time you finish this unit, you should be familiar with many new words. The list below contains some of the more important ones. Write their meanings in your own words and/or withhelp from your teacher. St unit: Galea Taternetionals (Latimotially agreed wt) vectorquantiy: Aaguitude 4 clivection Scalar quantity: Maguitode oly Distance: Som 4 alt par 4 body woves A Displacement: shracglt Line dichaa Cur pid f ong © tad pork Speed: Velocity: wousasA whhive lo tine ox ony Hirwhion dipuousid velebiwe to Pines ehrosgl line!ea Physics Booklet: Forces and Motion 2023 REVIEW: FUNDAMENTAL QUANTITIES AND UNITS All forms of motion have certain basic or fundamental things in common. They involve an objectthat moves some distance in a certain time in a certain direction. The mass of an object controls how easily or quickly we can change its speed. It takes more effort to catch a massive medicine ball than to catch a tennis ball! So, there are three important variables to consider when you investigate motion. These variables are the fundamental quantities of: + mass + length, or distance time. Note that a physical quantity is something you can measure, directly or indirectly. We call the units that most countries have agreed to use to measure these fundamental quantities fundamental u The fundamental units are: + kilogram (kg) for measuring mass metre (m) for measuring length second (s) for measuring time. Kilogram: the standard for mass Originally, the gram was defined as the mass of exactly one millilitre of water at 4° C. Users found that this unit was too small, and too difficult to set up, so replaced the water standard with the standard kilogram; the mass of a cylindrical piece of platinum-iridium alloy kept at S€vres in France. Other countries have copies of this mass, in the form of their own metal object with a ‘massas similar as possible to the standard kilogram. Metre: the standard for length Originally, the metre was defined as one ten-millionth of the distance between the North Pole andthe equator. This was later replaced by the distance between lines engraved on a special metal bar. Until 1960 the standard metre was defined as the distance between two lines on a platinum- iridium bar kept at 0° C (freezing point of water) in Sevres, France. However, modern technology needs extremely accurate measurement, so physicists devised a new international standard, Since 1983 the metre is defined as the distance that light travels in a vacuum in a time of 1/299,792,458 second. This provides a way for anyone with the right ‘equipment to produce a length of exactly one metre. This also means that the speed of light in a vacuum is 299 792 458 metres per second! Note: for all practical purposes this is 300 000 000 metres per second, or 300 000 km per second (over seven times around the Equator in one sec!)Yeart0 Physics Booklet: Forces and Motion 2023 Second: the standard for time The second was originally defined as 1/ggqpoth of an average day. 1 day = 24h = 24 = 60 min = 24 x 60 x 60s = 86 400s However, this is not accurate enough for modern science and technology as the length of a day, which is the time the Earth takes to rotate from one noon to the next, varies significantly over time. Since 1967 the second is defined as the time taken for a caesium atom to vibrate 9 192 631 770 times. This is hard to measure, but provides a way for everybody with the right equipment to measure a time of exactly one second with very great precision. Authorities therefore measure precise time with caesium atomic clocks. System Internationale (SI) Units These fundamental units for mass, length and time are part of the System Internationale Units.Physics uses the System Internationale Units, which is a standardised set of units used by scientists all over the world. Conversions are required when dati used that is not quoted in System Internationale U ~~ Physical Quantity | System Internationale Ui distance (d) metre (m) mass (m) kilogram (kg) time (t) ‘second (s) temperature(T) | kelvin) SCS electric current (I) ‘ampere (A) Force (F) Newton (N) velocity (v) metres per second (ms) QUESTIONS 4 What fundamental quantity do we measure in: kilograms?__/A(-2) seconds?__J (ag metres? uote 2 Indicate which of the following quantities are expressed in fundamental SI units (Yes) and which are not expressed in fundamental SI units (No) 130mg Mo 16s YO 073k Y9 93m > 181h Yo 8days “0 45Mg x0 idkm MoYear 10 Physics Booklet: Forces and Motion Baal Some other SI units The kilogram, metre and second are three of the Base Units of the International System (or, SI system) used around the world. Some derived SI units are: + square metre (m?) for area + metres per second (m/s or ms~) for speed + metres per second per second (m/s/s or ms~) for acceleration + ampere (A) for electric current + newton (N) for force + joule (J) for energy or work + watt (W) for power Working with symbols + If the unit is named after a person then the symbol starts with a capital letter, but the full name starts with a lowercase letter e.g. force is measured in newtons (N) * These are symbols, not abbreviations, so you never add a full stop after the symbol. + Always use the symbol, never an abbreviation, because they can be very confusing. For example, gms, as a symbol, means a gram metre second, a very strange unit that has no meaning! + When we refer to seconds, metres, kilograms or other plural units, the symbol remains the same - we don’t add an ‘s’. The symbol ‘s’ stands for ‘seconds’, and so adding an ‘s’ toa symbol adds ‘seconds’ to its meaning. The extra ‘s’ makes the unit nonsense, or causes confusion with other units. For example: 7 ss would mean ‘7 seconds seconds’, which doesnot make sense; 5 ms could be 5 metres or 5 milliseconds; and 14 kgs would mean ‘14 kilogram seconds’, which does not make sense. Multiples We often use multiples of SI units when referring to small or large quantities. Here are someexamples, + milligrams (mg) or grams (g) for small masses + tonnes for large masses. + millimetres (mm) or centimetres (cm) for small lengths + kilometres (km) for large distances. + milliseconds (ms) for very short times + minutes (min) and hours (h) for longer times. Note: the symbol for hour is h, not hr.Year 10 Physics Booklet: Forces and Motion a Sometimes you need to change the units of measurement. As a simple example, suppose that you want to compare the two length measurements, 3.47 m and 350 cm. Which is longer? The answer may be obvious to you, but to compare the lengths you either changed 3.47 m to 347 cm, or changed 350 cm to 3.50 m, maybe without thinking about it. CONVERSION FACTORS Conversion factors are any numbers that you multiply one unit by to get another unit. For example, to convert metres to centimetres we multiply by the conversion factor of 100. Using your data sheet, fill in the table below showing common metric prefixes: Prefix ‘Name Value ] Power of 10 u micro | 1/1000 000 Sa One millionth | a willl Ngo © | awk 4éo or” : kilo (000 1o° M Maps 000000 7 10% For the example below, label the arrows with the conversion factors for each. (HINT: use the tableyou just filled in to check what the prefixes mean.) © Converting distance POOQFS FY FY ym mm cm m km Mm OS OOSYear 10 Physics Booklet: Forces and Motion ae Example 1: Convert 654 grams into kilograms That is, 654 g = Pkg kilo means 1.000, so 1000g = 1kg 654g=? kg = 0-654 kg or, doing it another way, 1g =0-001 kg 654g= 654x 0-001 kg = 0-654 kg. There are often several ways to work out a problem. Use the method with which you are mostcomfortable. Example 2: Convert 45-6 kilometres to metres That is, 45-6 km. = ?m ‘kilo! means 1000 ie.km = 1000m 45:6km= 45-68 1000m = 45600m Example 3: Convert 1 hour 40 minutes to seconds That is, 1h 40 min = 2s 1h = 60min, andimin = 60s 1h = 60x60s = 3600s and 40min = 40x60s = 2400s © 1h40min= 3600s+2400s = 6000sYear 10 Physics Booklet: Forces and Motion 2023 WS1: Units and Measurements QUESTIONS Here are some problems for you to practise using these ideas. 4. Write in words, the meaning of each of the following, for example 4 mm = four millimetres, 6km 2em 5Mh .. 7ms 2. Complete the following conversions. a. 36-2kmtom —-34 20a b. 150cmtom bso c. 966mstos 0-46 “Why science teachers should not be given playground duty. d. 1600 gto kg |-boky e2zhts 71652008 f. 1755mmtom [-75s M 8 14500mgtog 14.5 ¢Year 10 Physics Booklet: Forces and Motion Brel Topic: VECTORS AND SCALARS Learning Objectives © State the difference between a vector and a scalar. «State examples of vectors and scalars (eg — distance v displacement; speed v velocity; acceleration (vector); time (scalar); force (vector); enerey/work (scalar); etc.) «Perform vector addition (displacement and velocity); specifically, parallel vectors and perpendicular vectors}. © Describe the differences between ‘distance’ and ‘displacement’ (including mathematically using vectors). Describe the differences between ‘speed’ and ‘velocity’ (including mathematically using vectors). «Describe opposite directions for vectors using positive and negative values ~ eg, describe the meaning of s > 0 and s <0; v> and v <0). magnitude head Girecti tail irection Parts of a vectorYeor 40 Physics Booklet: Forces and Motion 2023 DESCRIBING MOTION: VECTORS V SCALARS; DISPLACEMENT V DISTANCE; VELOCITY V SPEED 1. BRAINSTORM: In the space below, list things that could be measured to describe the motion of a moving object. 2. Define the following terms: + SCALAR (and provide three examples) Avawtily with atiguitudle awty ! Murr, elirloutt, tysut + VECTOR (and reve three examples) oat ly with bedh septa aul hohe! Ve vee rtton * DISTANCE (include some units that are used to measure this physical quantity) ea all gltllerou posclious/ laws a lodély tuover fo. wt, Ku, ofc - * DISPLACEMENT (include some units that are used to measure this physical quantity) a7 Hing dite frou poet 4 angts fo Leuwl durtiaaAinn . + SPEED (include some units that are used to measure this physical quantity) Movomsu An tea dieHion marta aly chaps cy ude * VELOCITY (include some units that are used to measure this physical quantity) Monorart ot steetgiet (te ot!Year 10 Physics Booklet: Forces and Motion 2023 Using the compass bearings shown below, draw vectors to represent the following physical quantities (see the example shown). Make sure you label them fully. The scale: 1 cm on the grid = 10 units on the quantity (eg - 40 m: 4 cm). tail (a) 10 metres NORTH 1Year 10 Physics Booklet: Forces and Motion real (c) 35 ms‘ NORTH WEST Fun Fact: The modern-day applications like GPS which literally show us the right paths of life is based on a physics formula of E= me 12Year 10 Physics Booklet: Forces and Motion 2023 3. When representing a vector quantity graphically, the two important features that need to be drawn are’ locred dasuidede aud direction 4, Physical quantities (vectors) can be added together. When adding vectors, you must follow the “head to tail” tule. See this method below ('a’ + 'b’ = ‘a + b') Tt atb He Tr fi The diagram below shows the addition of vectors in one dimension. aN 3N - 5N 1. magnitude: add 2. direction: same Here’s an example of adding two vector quantities that are perpendicular to each other. jtkm.E timn| + —HeBE , = tnmn Using the compass bearings shown below, draw vector addition graphs to determine the resultant physical quantities, Make sure you label the diagram fully. Measure both the magnitude and direction of the resultant. The scale: 1 cm on the grid = 1 unit on the map (eg— 4 m:4 cm), 13Year 10 Physics Booklet: Forces and Motion ones {a) 8 meast +5 meast = Bw E 14Year 10 Physics Booklet: Forces and Motion oes (c) 3msouth +4 meast = Sw $e (d) 8m west +6 m norttYear 10 Physics Booklet: Forces and Motion 2003 5. The map below shows part of the Perth CBD. The scale for this map is 1 cm: 50m (eg - 2 cm on the map = 1100 m in real life). As shown direction North is at the top of the map. NORTH WEST SOUTH From http://www.lookatwa.com.au/Zones/images/perthmap For this activity, use the following information: Average speed = Matt; y= ir rocity ~ {#isPlacement s werage velocity = “sPuacement, y_ $ ge velocity = “SPREE : (a) A person starts from the Barrack Street Jetty (intersection of Barrack St. and Riverside Drive). They walk up Barrack St to its intersection with Wellington St. The journey takes 300 s. Calculate: Distance = 350m Displacement =_ 3 SOu« -1 Speed = (7 Velocity LlTus Fun Fact: The astronauts who are at International Space Station experience the time slowly since both velocity and gravity have an effect on the speed of time. This is why astronauts kind of look younger to us. 16Year10 Physics Booklet: Forces and Motion 2023 {b) A person starts from the intersection of Milligan St. and Wellington St. They walk along Wellington St to its intersection with William St. They turn right and walk to the end of William St. where it intersects with Riverside Drive. The journey takes 500 s. Calculate: 15 x 50 (30 +950 = Displacement = Z7S a Distance velocity=___— OTS ws! Spee {c) A person starts from the intersection of Riverside Drive and Victoria Ave. They walk along Victoria Ave to its intersection with Hay St. They turn left and walk to the end of Hay St. where it intersects with King St. They turn right and walk to the end of King St. where it intersects with Wellington St. The journey takes 750. culate: Calculate 24 338 + [US qs xso = Distance = 68s Displacement = {7s 4 speed = U3 mS Velocty= 0-633 wes" (d) A person starts from the intersection of Riverside Drive and Mount St. They walk along Mount St. to its intersection with St Georges Tce. They turn right and walk along St Georges Tce, continue along that road when it turns into Adelaide Tce, and continue until it intersects with Victoria Ave. They turn right and walk to the end of Victoria Ave. where it intersects with Riverside Drive. They turn right and walk along Riverside Drive until they return to their starting point at the intersection of Riverside Drive and Mount St. The journey takes 1200s. Calculate: >>, 4 y2s 4 ISO% 3004 (G+ GO Distance = Sa) Displacement = Om speed = 1.03 wi! Velocity = OQus That's how I roll 7Year 10 Physics Booklet: Forces and Motion oy Lesson 4 - 6: KINEMATICS INERTIA Education Perfect Module: Forces and Motion: 1. Motion: 1. Distance and Time; 2. Displacement and Compass Directions Use the space below to take notes.Year 10 Physics Booklet: Forces and Motion ACTIVITY 2.2 Distance / Displacement Exercise 2023 Walk with a constant length pace as you follow these directions Start at the middle of the Basketball Court. Walk 20 paces NORTH Walk 20 paces WESTWalk 5 paces NORTH Walk 15 paces EAST Walk paces SOUTH Walk 5 paces EAST. This is the finish point Questions Choose an appropriate scal&and draw a scale diagram of your journey. 2. What distance have you walked? 3. What is your displacement? (straight line distance and direction from the start point). 19Year 10 Physics Booklet: Forces and Motion aes ACCELERATED MOTION Education Perfect Module: Forces and Motion Speed A roadrunner’s top speed is about 45 km/h while Coyotes can run as fast as 85 km/h 20Year 10 Physics Booklet: Forces and Motion 038, Lesson 6-7: SPEED Speed is the rate of change of distance of a body. ‘The average speed is the distance moved per unit time, Speed(av) = _distance/jime s Vav = > t Speed is a scalar quantity as it does not require any direction to be named. ‘The unit of speed is determined by dividing the units of length (m) by units of time (s), Metre/socond = "Ys = ms? (orm/s) why ms? Consider this pattern that you may have come across in mathematics. axaxa=a? axaza? aza’ a/a=a°=1 darat taxa Usain Bolt remains the fastest human on the planet: Following the pattern you will see that s*=1/s and mx 1/s or m/s That is, ms means ‘metres divided by seconds’, which tells you how many metres the object travelled in one second. Another common, but non SI, unit for speed is kilometres per hour Kilometre sour = My = kmh” (or km/h) Both km/h and km h~ are correct. However, scientists and engineers prefer to use the notation: == mstorkmh. Other units Some other units include: + km/s, or kms* for things like spacecraft + mm/s, or mm s* for things like snails. 21Year10 Physics Booklet: Forces and Motion ae CALCULATING AVERAGE SPEED Rearranging Formulae Mathematical formulae or equations can be rearranged so that the quantity you want is in theeasiest place for you to make the calculation. For example, suppose you want to calculate the distance travelled, and you know the time for thejourney was 2 minutes and the average speed was 2 ms” then the equation linking all three quantities is vy = Of, Here are two ways to rearrange the equation to find ‘ METHOD 1. Move the quantity you don’t want to the other side of the equation. To do this you have to ‘undo’ its mathematical operation. In this case, ‘d’ is divided by ‘t’, so to get rid of ‘t’ you need to multiplyboth sides by ‘t’ to ‘undo’ it. So vay = U/y vv xt = Of xt After cancelling both the ‘t's on the right side of the equation, Or OC! or d= Vw xt By putting the values for ‘t’ and Spjsy into the new equation, you can calculate ‘d’ METHOD 2 For formulae that have only three quantities, you can use the triangle formula diagram below. To find the formula you need: (i) cover the quantity you want to find (ii) the other two quanti find will be in the correct pattern to calculate the quantityyou want to e.g. if you want to find time t, cover t andsee that it equals d/ vav oe” rearranging below: i Va t 22Year 10 Physics Booklet: Forces and Motion pas Example 1: A girl can run 600 metres in 2 minutes. What is her average speed in metres per second? Distance = 600 m, Time = 2 min = 120s Average speed vv = 4p = 600M/y29, = Sm/s Ws2: Calculating Speed QUESTIONS 3.1 1. Atoy car travels 5 metres in 4 seconds. Calculate its average speed. Vv i aoe h2s wet 2. What is the average speed of a car that travels 250 m in 15 s? (6.705 3. By how much isa car exceeding a 60 kilometres per hour speed limi 1.5 hours? of d= roka d - 120 = 30 huh t= ish VE €° Te Cecordeg leit by eo hl 4. Calculate the average speed in metres per second of a rifle bullet that travels one kilometre in 0.5 seconds. d= tha= (oom i= ee = at 5) ee DaeS eae os 23Year 10 Physics Booklet: Forces and Motion poe 5. What is the average speed of a sprinter who runs 100 m in 12.5 s? i. t= ee gour 6. A body travels 36 min 4s. Whatis its average speed? yo 3b > q0ui 7. Determine the average speed (in km/h) of a plane that flies a distance of 720 km in 1.5 hours. vg 722 = 430 kwh re Answers: 1) 1.25m/s 2)16.7m/s 3) 20km/h 4) 2000m/s_ 5) 8m/s_ 6) 9m/s_—_7) 480km/h Example 2: How long would it take a car travelling a speed of 80 m/s to travel a distance of 25 m? Speed = 80 m/s, Distance = 25m vw = Ge so t= ovw = 25m/g0m/s = 0.31 seconds 24Year 10 Physics Booklet: Forces and Motion eal 9 f 1. How long would it take a car travelling at 65 m/s to travel a distance of 450. m? ‘WS3: Calculating Speed - QUESTIONS 3.2 . - dod _ 40. ve 2 lts £2 Bs ows 2. Acar travels a distance of 250 kilometres at 45 kilometres per hour. How long does it take to. cover this distance? d 230 = we = $s6h t vo as > TS 3. An aircraft travels at 500 m s“! between two towns. How long does it take to cover this distance if the towns are 30 km apart? 4 > BP ws 4. How far will a bird fiy in 25 s travelling with an average speed of 6 ms“? dz Vt = 6% % = /s0um a = 5. A person swims from one end of a pool to the other at an average speed of 1.5m What is the length of the pool? dz Vt= Isx w= Wm: 25Year 10 Physics Booklet: Forces and Motion aoe 6. Determine the distance travelled by a plane in 1 hour if it flies due south with a uniform speed of 90 m/s. teth _ 3600s ve ous ge Vt = OY) = 324 000 a = 3 hu. a 7. How long will it take for a body moving with a uniform speed of 7 ms“! to cover 98 m? be 705" deme ts 4. B= 4s. £22 ~ 8. A cricket ball is thrown a distance of 90 m from the boundary to the wicket at an average horizontal speed of 30 m s“1, How long will it take for the ball to reach the wicket? detox 4. 2 ®. 205 Ve pomst Vv 30 = t.% 9. Determine the time taken for a horse running with an average speed of 14 m st to complete ‘one circuit of a 840 m track. d . 3 _ _— 40 vs ous t- 4= <2 @s dz H0u tag = €= 7? 10. How long will it take to complete a trip from Perth to Kalgoorlie, a distance of 595 km, travelling at an average speed of 70 km h-1? SIL dese e- > So ash - =< Ve VOlunh” t:7 14. Atrain moving with an average speed of 80 km hv! takes 0.25 hours to travel from one railway siding to the next. What is the distance between the two sidings? Ve Bo buh" d es o-2sh -vt 1a = (wos) 26 ku ‘Answers: 1) 6.9s; 2) 5.5h; 3) 60s; 4) 150m; 5) 30m; 6) 324km; 7) 14s; 8) 3s; 9) 60s; 10) 8.5h; 11) 20km 26Year 10 Physics Booklet: Forces and Motion 2023 Lesson 8: VELOCITY ‘The average velocity is the rate of change of displacement of a body. Velocity is a vector quantity; therefore you MUST include magnitude and direction. The formula is similar to that of average speed. The average velocity of a body can be found by: Velocityiav) = displacement/time we th where vv = average velocity, displacement, time interval The unit of velocity is the same as for speed but a direction must be given with the answer. @ Rearrange the formula for average velocity below. Ss ZS vow = tz te Vz Converting speed (velocity) units In the SI system the unit for speed (and velocity) is metres per second (m s“*). However, in everyday travel, we usually measure speed in kilometres per hour (km h-). Note that: akm h? = 1x 1000M/ 36995 = (1/3.6) m/sHence 1ms?=3.6kmh* + so to convert m s* into km h~, multiply by 3.6 + so to convert km h- into m s*, divide by 3.6 ‘Converting km h™ into m s* Example: convert 50.0 km h™ to ms?50.0kmh*= 50.0/3.6 =13.9mst . 50 km h-1is the same speed as 13-9 ms Converting ms* into kmh Example: convert 10.0 ms“ to km h> 10.0ms* =10.0 x 3.6 = 36.0kmh* + 10.0 ms*is the same speed as 36.0 km h™ 27Year 10 Physics Booklet: Forces and Motion 2023 WS4: QUESTIONS 4.1 Converting speed units |, Did you know: the first person to measure speed was the famous astronomer and physicist Galileo Galile LY 1. Convert 30 km h~* into m s* 20 x O+3hO = 207 2. Express a speed of 45 m s“as kilometres per hour. 445! 2 As simystoo = Ys XC = [fo huh” — 3. Convert 110 km ht into ms 0436 = 30:56 a5 4, Convert 110 m s* into km h Wo «3-6 = 346kuh 5. Convert a walking pace of 4.00 km h~ into m s* ~ 4-0xdb = (4-4 us = 6. Convert 110 km min“ into ms x (lO x (ooo go = | 833-3 ms 7. It is more helpful to know the velocity of a plane rather than just its speed. Why is this? volodity theo dncleds its direction (Wscastig) qluidh & wore vafl Lhcwn just its seed, 8.A person may travel many kilometres during a day and yet have a final displacement of zero. Explain how this is oon 6 trol yor ond yp (A bee aol of has pene “slit chet , Law Vis lr derplecbui® 6s Ze". 9. Acheetah can run with a speed of 90 km h-! a. How fast is 90 km hin ms? ot36 =< 2Sms! b, How far could the cheetah run at this speed in 1.0 minute? t= lms bos de vt = JsxW Vz es mst = (SoD m 28 EeYear 10 Physics Booklet: Forces and Motion 2023 Calculating Velocity Example 1: Calculate the average velocity of a car that travels 300 m east in 10 seconds.s = 300m — t = 10s vv = Sfp = 300M/yq, = 30m/sEast Example 2: Determine the displacement of a body that moves with a uniform velocity of 2.4 m/s for 50 ‘S.Vav = 24m/s t 50s Vev = Sfp sos = vavxt = 24m/s x 50s =120m (in DOM WS5 Calculating Velocity - QUESTIONS 4.2 Draw diagrams where necessary to help complete these exercises. 4 1. A girl takes 100 seconds to walk 90 m in an easterly direction. What is her average velocity? £ ot Ve g = #2. O4ui 700 = 2. The girl in the above question runs back along the same path a distance of 50.m. It takes 40 seconds to travel this distance. (a) Whatis the total distance travelled and what is her displacement now? d= 9O+s0 =/40u S=< 40-50 = You: EAST- (b) What is her average speed and her average velocity for the entire trip? ~ 4 2 te = |.Quz! (a tee ooo 740 eke A= 0.24m5' Ease € > M0 3. A car travels 150 km south east in 2 hours and then 30 km north west in 15 mins. (a) Whatis the total distance travelled and what is the car's displacement now? der = (04+ 30~ [gO ku. 4,5 2x30 = 7200 +(sxlo)= a $ (so~30 = ole SE. = Fils (b) What isthe average speed and the average velocity for the entire tip? = (30 we at Soe - VE gee 3 eT ve ogg? Mibu! SE 29 == eeYear 10 Physics Booklet: Forces and Motion pies 4. A ball is dropped from a cliff, 20 m above the ground and bounces and is caught at a height of 1 m after 0.5 seconds. (a) Whatis the total distance travelled and what is the ball's displacement now? = 2o+l > 2lu the ground 2 seconds later. It 20-1 = 14d woun —— (b) What is the average speed for the flight of the ball? de es Kens! veg? oe * Ete (c). Whatis the average velocity for the flight of the ball? = T.bms' Down 5. A person walks for 10 s at 1 m/s then for another 10 sin the same direction at 2 m/s. Find the: (a) total displacement. 3, = Ve= 1x0 su VE= 2K Sp = (04 = ZOU Saune Diceok (b) average velocity. =— “ (Ou ton. “ = Uitte LR = LS wei! saseat Div Mes Hite Aes feat se 6. A body travels 200 m at a uniform velocity of 4 m s* and a further 150 m in the same direction at a uniform velocity of 7.5 ms. Determine the: r (a) total time taken 6 = £ = 22 =s0 tp = bitte vo 4 = & = Iso <~ 5O te => =z = + ws =< Sot2o VO 9 = = (es — (b) average velocity for the whole journey = 2004180 _ 350 Viv. 0 To = sus! sie a 30Year 10 Physics Booklet: Forces and Motion Ey 7. Acyclist travels at 6 ms“ south for 10 minutes and then returns 600 m along the same path. What is the final displacement of the cyclist? Sz VE = 0)LioK60) = 360m. Sb = 3600-600 = 3000m S — 8. Arunner runs a full lap of a 400 m oval and finishes back where she started. What is her displacement and average velocity? ee Vz Ooms’ ‘Answers: 1) 0.9m/sE; 2) 140m; 40m east; 1m/s; 0.29m/s E; 3)180km;120kmSE;80km/h; 53.33km/hSE; 4) 21m;19m down; 8.4m/s; 7.6m/sdown; 5) 30m in the direction of motion; 1.5m/s in the direction of motion; 6) 705; Sm/s in the direction of motion; 7)3000mS; 8) 0; m/s indirection of motion. 31Year 10 Physics Booklet: Forces and Motion 2023 Education Perfect Module: Forces and Motion Acceleration; 5. Using the Acceleration Formula to Calculate Final Velocity Use the space below to take notes. Did you know: Gravitational attraction ofthe proton and the electron in hyérogen atom is 3.6*10~"N. The gravitational free between the ‘earth and the Sun, on the other hand, is 3.5% 10 N 32Year 10 Physics Booklet: Forces and Motion eal Lesson 9: ACCELERATION Acceleration occurs whenever the velocity of a body changes. e.g. moving faster or slower, or changing direction. For the moment we will only discuss how to calculate acceleration due to change in speed. Acceleration is a vector quantity. Acceleration is defined as the rate of change of velocity. The average acceleration of a body can be found by: Average acceleration = change in velocity = final velocity - initial velocity time time where; v= final velocity and jal velocity Rearrange the formula for average acceleration below. The units for acceleration are ms~ and can be said as: + metres per second per second, or * metres per second squared. You may use either expression when speaking or thinking of acceleration. However, ‘per secondper second’ is more precise and helps you to remember what acceleration actually means. As acceleration is the rate of change of velocity then its units are those of velocity divided by time: m ms-2 (or m/s2).s NOTE: For a body starting from rest, the initial velocity u is zero, u = 0. Assume any stated acceleration to be in the direction of the initial motion, 33Year 10 Physics Booklet: Forces and Motion eal Example 1 Find the acceleration of a body that is accelerated from 2 ms“ to 10 ms1 in 2.5 s. initial velocity, u=2ms final velocity, v= 10 m st =2.5s a=(v-uyt =(10-2)/25 =8/25=3.2ms~ The acceleration is 3.2 ms” (in the direction of the initial motion) Example 2: What is the final velocity of a car starting from rest after 5 s if it is accelerated at 2s“? and has an initial velocity, u = 0 (from rest) a=2ms2 v-0=(2x5) t=5s =10mst ‘The velocity is 10 m/s (in the direction of the acceleration Example 3 What is the initial velocity of a body that reaches a velocity of 16 ms~1 when it is accelerated at 1.5 ms? for 8s? 6mst v-usat final velocity, v 1sms2 16-u=(1.5x8) 8s 16-u=12 initial velocity, u =? u = 4ms-2 The initial velocity is 4 m/s (in the direction of the final velocity} 34Year 10 Physics Booklet: Forces and Motion pas WSS: Accelerated Motion - QUESTIONS a What is the increase in velocity each second of a body accelerated at 1.8 ms? Pract Bas! cach Hive 2 Acar is accelerated from 10 ms? to 25 ms* in 2.5 s. What is the average acceleration? ot U> (ums = Vee - a= 2s ~10 Vea asus a Ss 15,222 a 3. A body starts from rest and is accelerated at 2.7 ms®, What is the final velocity after 12 s? uz Oust = ea ee Ve atot = 012M) = BYES t-ts 4, A ball rolling down a slope accelerates at 3.5 ms” from rest. What is its velocity after 6 s? a= asus or = Omit ve asad = 0+ BsXb0) = 21-0mS t=6s 5. Find the acceleration of a plane that increases its velocity from 100 ms* to 200 ms? in 40s. 1 V= 20oms = V-4 ~ 100-100 _ gp. st U= [00s ase = SG 2.5m t= 40s 6. _ Whats the average acceleration of a body that is accelerated from rest to 15 ms? in 5 s? uous a Ve isms! az Ve 2 IO 2 gous t Ss 8. Avehicle starting from rest is accelerated at 2.5 ms®, Determine the time taken to reach avelocity of 5 ms". t= UUM. 50-0 ~ 05 a ra = Find the time taken for a body travelling with a velocity of 15 ms* to reach a velocity of, 29 ms*when accelerated at 0.7 ms®. of un Is a5 ~ Yu, US. Ws vy =2ques! tal Ye Ones a 07 10. Calculate the final velocity of a car initially travelling at 10 ms* that is accelerated at 1.4 ms*for a period of 10 s. : ye at ol 2 ory) uz (Oust, = S de tapas = 24.0u5¢ ‘Answers: 1) 1.8m/s 2) 6m/s? 3)32.4m/s 4) 2im/s 5)2.5m/s2_6)3m/s2_7)3m/s? 8)2s 9)20s 10) 24m/s 35Year 10 Physics Booklet: Forces and Motion eral @ Lesson 10: What is “negative acceleration?” 7.0 DECELERATION (Negative Acceleration) IF the speed of an object is increasing, we say the acceleration is positive. If an object is slowing down it is decelerating. Its speed is decreasing, We say its acceleration is negative. For example, -3 ms“refers to a decrease in speed of 3 ms“ every second. Ifa body travelling in a positive direction slows down uniformly, the final velocity is going to be lessthan the initial velocity. will give a negative value for acceleration because the acceleration is in the opposite direction tothe velocity. Negative acceleration is also known as deceleration or retardation. SIGN CONVENTION FOR ACCELERATION IN ONE STRAIGHT LINE 1, Take the direction of initial motion to be positive. All vectors in this direction will be given apositive value. 2. If, in a calculation, an acceleration is found to be: (a) positive, then itis in the direction of initial motion and velocity increases unit Lm (b) negative, then itis in the opposite direction to the i n and velocity decreasesuniformly. Example 1: Acar is uniformly slowed down from 26 m st to 6 ms“ in a period of 8 s. Calculate theacceleration of the car. initial velocity, u=26 mst a=(v-u)/t final velocity, v= 6 m s t=8s = (6-26) a 8 = 220 8 25ms2 The acceleration is -2.5 m s~? or the retardation is +2.5 m s~, and is in the opposite direction to the initial motion. 36Yoar 10 Physics Booklet: Forces and Motion 2023 Q WSS: Accelerated Motion - QUESTIONS j 1. A train approaching a railway station at 30 ms” is brought to rest in 10 s. Determinethe 1 retardation. U> 2m vV-4 = 9-30 a vc ome oe a cake ees! t= (0s 10 = 2. Aplane on landing touches down at a speed of 80 ms*, Determine the acceleration if the plane comes to rest in a time of 8 s, ue sous’ : Lu be -2 Vz ous" a= Ve = O-P_ _igus t= Bs c z = 3. Atruck travelling at 36 ms" is decelerated uniformly at 1.5 ms. Calculate the: (a) velocity after 6 s. (b) time taken to come to rest. =V-4 _ 9-36 Ve at at a, oe = 36 + Cls)b%) = 2409 —e= 4 = ui" —_ 4, Aplane fiying at 170 ms* slows down steadily at 2 ms What is its velocity after 30 s? Us ows! a= 20 ws* t = 30s 5. How long will it take a vehicle travelling at 30 ms* to reach a velocity of 12 ms* if the acceleration is -3 ms*? eT Ve [tums = Vu _ 12-30 Us 30us t i Qa sous ee 1 6. A bus travelling at 25 ms* is decelerated at 2.5 ms®. Find the time taken for the bus to cometo rest. ~ U= us ve 0-2 arn 2st to Wt = CE « 9s Va Oui! ~ tz! 7. Amotor bike brakes suddenly with a uniform acceleration of -4 ms? and comes to rest in 2.5 s. Determine the initial velocity of the motor bike, ot. ae- 4us ~ yeat = 0- Cred er uz v-ot 7 o~6¢ a t<2rs = (0-0ums Answers: 1) 3m/s?; 2) -10m/s?; 3) 27m/s, 245; 4)110m/s; 5) 6s; 6)10s, 7) 10.0m/s 37Year 10 Physics Booklet: Forces and Motion eal Lesson 13 - 14: DISPLACEMENT OF UNIFORMLY ACCELERATED BODIES We can find the displacement (straight line distance from start to finish) of a moving body by using: Equations of Motion Average speed: Yavg = = Average Velocity vat Accelerating velocity, veutat a Displacement: at? tut Velocity square uw + 2as The GUESS Problem Solving Strategy ‘When solving a physics problem, it is essential that a clear, full and logical working out is necessary in order to gain the full mark, ‘The GUESS Method/Strategy is quite a simple yet powerful way to solve a written question. GUESS stands for: Given, Unknown, Equation, Substitute and Solve. Here is a break-down of the GUESS problem solving strategy: = Given: This is the information or data you are provided/given. List this information/data, including correct units. You may need to convert the information/data to SI units. = Unknown: This is what you are instructed to calculate/find out. Write down this unknown, = Equation: you need to identify what is the most suitable equation to use. Select the appropriate equation which has BOTH what is Given and the Unknown. = Substitute: in this step, you may need to re-arrange the equation to make the Unknown the subject. Once this is done, substitute the Given information/data in the Equation = Solve: solve the equation to find the Unknown (answer). Do not forget to include the correct units in your final answer. Finally ~ but not always required - write a short sentence to express your final answer. 38Year 10 Physics Booklet: Forces and Motion 2023 Here are some worked examples. Carefully study the examples and then use the problem solving strategy to complete the questions that follow. Problem 1: A year 10 student performs a sprint. He accelerates from rest to 10 m/s in 3 seconds. a. Calculate his acceleration. [3] b. What is the distance travelled by the student? [3] Solution a. Calculate his acceleration. Given: initial velocity _u = 0 m/s {he was at rest: not moving), © final velocity v= 10.m/s (he now moved at 10m/s after the time of 3 seconds) ©. time taken t=3s = Unknown: Acceleration a = ? (we need to calculate this unknown!) ~ Equation ae e ta] ~ substitute a=(10-0)/3 rey Solve: 3 m/s (1) b. What is the distance travelled by the student? © u=0 © v=10m/s © te3s . 3.3ms? 2 ss? © s=fatt tut Pn 3.3 x3? + Ox3 (1) 149m (0) ‘The student travels a distance of 14.9 m in the time of 3 seconds 39Year 10, Physics Booklet: Forces and Motion 2003 Problem 2: A car accelerates from 10 m/s to 20 m/s ina time of 5.5 seconds. a. Determine the acceleration of the car during this time. b. Calculate the distance travelled by the car. Solution a. Determine the acceleration of the car during this time. v-u)/t (20-10) / 5.5 &ms* The acceleration of the car is 1.8 ms? b. Calculate the distance travelled by the car. 10 m/s v=20m/s t=55s 1.8ms? 1 sat? +ut s= zat? + Y%18x5.5'+10x5.5 =82.2m The car has travelled 82.2 m after the time of 5.5 s, Did you know: We all now the Earth is revolving around the Sun, And, the average orbital speed of Earth is about 30 km/s (108000 km/h). |And, at the equator, the surface of the Earth moves ata speed of about 460 meters/second, That translates into about 1656 km per hour. 40Year 10 Physics Booklet: Forces and Motion aa Problem 3: a cheetah chased her prey. She accelerated very quickly from rest to her top speed with an acceleration of 22.5 ms®, covered a distance of 25 m when the top speed was reached. a. Calculate her top speed. b. How much time did it need to cover the distance of 25 m? c. She continued with this speed for a distance of 200 m. What was the time required to cover the 200 m? d. The cheetah caught her prey at the end of the chase. She slowed down to a complete stop with a deceleration of 40 ms®. Calculate the distance covered by the cheetah while he was slowing down to a stop. Solution a. Calculate her top speed. u=0 s=25m a= 22,5 ms? ve? v?= uw +2as =042x22.5x25 = 1125 (1) Hence v = SQRT(1125) = 33.5 m/s (1) The cheetah’s top speed was 33.5 m/s b. How much time did it need to cover the distance of 25 m? 1p s= 5at* + ut D t= 29/a > t=SQRT(2s / a) = SQRT(2x25 /22.5) 2 215s. ‘The cheetah took 1.5 s to reach her top speed and cover the 25 m distance. She continued with this speed for a distance of 200 m. What was the time required to cover the 200 m? v= 33.5 m/s s= 200m te? tes/v = 200 / 33.5, = 6 seconds ‘The cheetah covered the distance of 200 m in 6 seconds, a1Year 10 Physics Booklet: Forces and Motion eal WS10: Equations of Motion Questions Apply appropriate equations to answer the following questions. Show FULL working. During a Gymkhana session, Ken Block accelerates his car from rest to 65 m/s in 10 seconds. What is its acceleration? i= ows! Ve bus dv Ve, t= as t BO Lbs us o = b. How far does the car travel in the 10 seconds? ge ot thal’ <0 + tGAel > ssw Aan arrow is shot from rest to 120 m/s in 0.02 seconds a. What is its acceleration? ae eer oo Ve 2 en haces © e booo us — b. The arrow hits the target located at a distance of 55 m from the arrow. How long does it take the arrow to strike the target? - tatt Sa tor = 2s ~ # 2x CAE + JZ - 0.135. 3. Arocket is travelling at a steady speed of 76m/s. Its velocity then changes to 98 m/s in 1.5 s. en een U= bus! ~ ve Y= 4305 € t= bss = Qytb Te = YT us oe! b. What is the distance travelled by the rocket in this time? s-u + tod = (6lis) + E04 Uy + 6S 42Year 10 4, Physics Booklet: Forces and Motion a 2023 A cyclist accelerates from 0 m/s to & m/s in 3 seconds. What is his acceleration? ue Oms" Vek = BO > 267 a5” eae Bee 3 = Calculate the distance covered by the cyclist after the time. s- a+ tat’ - ot £067%3)" = (202M Find the car’s average acceleration. U=0 us! 2 V4 Ve roluk'= (4.4 a5" eae te t= 7s Find the minimum distance the car needs to travel to achieve this speed. e se ats ted 2 0 + ¢@r7lz)” = 67-4us~ Acar advertisement states that a certain car can accelerate from rest to 70 km/h in 7 seconds. = Some species of lizards can run on water. A lizard accelerates on land from 2 m/s to 10 m/s in 4 seconds before entering the water. ‘What is the lizard’s average acceleration on land? as Ue por - 7 2 = 20s b. Determine the distance the lizard has travelled before entering the water. 5 = ut +40 © Qfa+ 5eLe =< 8416 = 240m = ‘The water spans a distance of 20.2 m. How much time does the lizard need to run on the water before reaching the other end? S= 202m » V4ous! a 2-028 Vv 2) 43Year 10 Physics Booklet: Forces and Motion sl 7. Aracing car travels from 10 m/s to 65 m/s in a distance of 120m. a, What sits acceleration? ie sas" Vi ud ras = nou (68= (io) + (dolleo) Aus -too+ efor -1 A= (lus b. How much time does the car need to accelerate to 65 m/s? a (Ww a< VE t= V4_ 65-0 € ~ t fF =" £ §-2S. a 1F2 2 8. Your mum is taking you to school in her trusted Toyota Yaris. Before entering the school zone, she slows down from 60km/h to 40 km/h in 2.5 seconds a. Whatisthe deceleration ofhercar? = Gokul’ = 67m 5 Ve 4Ohaht 2 (fas tress Ae Vow 2 WAM? € os =) as © Decobmbtn = Layo s* b. How far does the car travel in the 2.5 seconds? = iz (—2. sz ute tat? « (6ned+ teres) = 4s +E20) = 34 75m c. After the school zone, your mum presses the gas pedal to the max, the car gives out a mighty roar and propels the car and its passengers from 40 km/h to 90 km/h in a respectable time of 20 seconds. i, Calculate the car's acceleration, and hence Me folk (|! te dws Vz Ghul s-0us 4 Be eee a “ E- 25 = 0.695 a= ii. Calculate the distance covered by the car during this time. 2 sz th + het . (ied + £ C-ABxz2) ~ 2 + IM = 26l uw = 44Year 10 Physics Booklet: Forces and Motion 2023 9. Aplane is travelling at a steady speed of 200 m/s and its velocity changes to 398m/s with an acceleration of 2.5 ms? a. How long does it take the plane to reach its final speed? : us 2ems’ = ats ees isu ve. = we a 4 < bs Siow Oo Vw eae e ae b. What is the distance travelled by the car in this time? Ss ut +4 a” ae = a4?) + £@sdo44) +8 = 23600-3u 23-7 hu 1s840 + 7€40-8 = 236WBu & 10. A cyclist accelerates from 0 m/s with an acceleration of 2.9 ms® for 5 seconds. a. What is his final speed? b. Calculate the time taken for him to cover the distance of 25 m, 62 2 = Piz 14s — 11. Acar advertisement states that a certain car can accelerate from rest to 100 km/h in a distance of 50 m. a. Find tha car's average acceleration. Ve uct 2as we Our! - = Ve (ookul! = 27-8 ws <= s0u b. Find the minimum time the car needs to achieve this speed. Vou _ 21-8-0 t= = = 2 BGs CS TB = 45Yeer 10 Physics Booklet: Forces and Motion 2023 12. Achameleon propelled his tongue to catch an insect fram a distance. Slow action camera showed that his tongue accelerated from zero to 95 km/h in 0.01 second. ‘What was the average acceleration of the tongue? ows u = Ve qshak' = 16-4m5" x oble b. Determine the distance from the chameleon to the insect. se ts dtl’ = 0, 4euxooi = 0-(32m Did you know: They found thatthe animals’ tongues are capable of impressive acceleration, doing 0 kilometers to 100 kilometers per hour in one-hundredth of a second, twice as fast asthe fastest car. And, lke sports cars, the smallest chameleons are the most powerful 46Year 10 Physios Booklet: Forces and Motion Newton’s Laws of Motion ‘was old that Sirleaae Newton decovered gravity when he saw an apple fling fom the ree 2028 a7Year 10 Physics Booklet: Forces and Motion 2023 Education Perfect Module: Forces and Motion 3. Force: 4. Newton's Second Law Use the space below to take notes. NEWTON'S 1" LAW (sey ABODY AT REST WANTS TO STAY AT REST. 48Year 10 Physics Booklet: Forces and Motion 2023 Recap on MOTION and FORCES Everything in the universe is in a state of motion. The earth itself is filled with moving things, from the microscopic to the macroscopic. Electrons move around the nucleus. A person cycles to school. A car drives down the street. The earth spins on its axis. The planets of our solar system orbit the sun. Our solar system moves through space as part of the Milky Way Galaxy. The stars and galaxies move away from each other. It might seem impossible to find a simple way to describe and understand the motions of all these objects. However, this is just what physicists have done. & HOW MUCH DO YOU REMEMBER ABOUT FORCE? a, What are some words we use to describe forces? fusk, Gull, crush, strobe, fe. Pressure , ae. b. List as many examples of forces as you can. oh Ke ei giat, gravity, contri. , HAC c. Can the forces you named above be grouped into 2 specific groups of forces? What are thetwo groups of forces? tet bun Vi; Non corckedt d. What are the characteristics of a force? How could we represent forces diagrammatically? vecrbr quot y . 30 Ey 49Year 10 Physics Booklet: Forces and Motion So you remember that forces are a push, pull, twist or stretch which can produce a change in theshape, speed or direction of an object. BUT did you know that objects move because they have unbalanced forces on them? Usually there is more than one force acting on an object, so the movement is a result of the combined effect of those forces. Forces have both size (magnitude) and direction, so when we add forces together, we have to separate them into horizontal forces and vertical forces. Any time you must provide a size and direction for a quantity, it is referred to as a vector quantity. Vector and Scalar Quantities ‘A quantity is something you can measure, directly or indirectly. A vector quantity has both sizeand direction, whereas a scalar quantity has size only. LIST as many different quantities that you have measured (directly) or calculated (indirectly) peed veloicty , dis touce Mam, Volume, weiglt, Time, disploomuut 2023 Which of these quantities are scalar? Which of these are vector? Fill in the table below. SCALAR QUANTITY VECTOR QUANTITY Mass ergot Yee Vokou ty _ Volune at - D ‘spladuaut 50Year 10 Physics Booklet: Forces and Motion oes Combining Forces Arrows are often used to represent forces with the point of the arrow showing which way the force is acting. The length of the arrow is used to show the size of the force i.e. the longer the arrow the bigger the force. The arrows describing the vertical forces on the kayaker are even in length, which means that the forces are the same magnitude. These forces are acting in opposite direction, which means the force of the kayakers mass (called the weight force) is pushing against the force of the water {called the reaction force) which is pushing up EQUALLY (i.e. he doesn’t sink or float up!). oS longeron the diagram, than the arrow pointing up (reaction force due to the water)? What would it mean if the arrow pointing down (weight force) was shown as being 51Year 10 Physics Booklet: Forces and Motion pal z Look at the picture above. Explain the motion of the kayaker using the idea of unbalanced forces causing motion. (HINT: label each arrow with a name and draw a diagram using the arrows). Did you know? Every day you experience forces on your body which push and pull you in a range of directions. These forces include that of the air pushing on you; which is surprisingly high; some 100 000 N for every square metre of body size (the average human body is approx. 0.8 m’). We call this air pressure and it is a measure of the force exerted per square metre of body area. As the force of gravity is pulling the air particles towards the earth’s surface then the air pressure is greatest closest to the surface and decreases as you rise through the atmosphere. Air pressure at the surface varies depending upon the air temperature and movements. ‘The forces involved in skydiving 1. Skydivers can jump from various Ieights ut most Ueginners jmp from abut four Blametes above the ground. When the plane is over th jump site, the skydver leaps from the plane, Beginners ave taught to release their parachute as wom as they ere clear ofthe plane, More experienced divers foe fll for some time fore opening their parachute. 2. When skyivers jump from a plane, the Earth's pul of gravity causes thera to fal towards the ground. Near the start Of the jump, a diver does not all very quickly At this point in (he jump, the diver does mot experince much air resistance. But asthe diver’ sped increases during the fll 0 does the sse of the air resistance pushing against him or her een tacvacanne es tases ses ar es a en 4. During np yon fel fst and fase ‘The air ressiance pushing against a diver got bigger and bigger asthe 5. When the pevachute opens, ge airresitance iorce pushes againat it, When the parachute fst opens, (he air resistance is Digger than the gravity force pulling the sksdiver dwn, £0 the diver slows den ‘The shytiver reaches seed increases, Eventually the uproard push ofthe air resistance and the cass seer ownseard pull of gravity balance out. There is no overall foree acting on the diver eer ale ae ‘anymore. When this happens, the diver falls at steady sped. The steady speed is fer the parachute Callad terminal velocity. The lerminel velocity ofa skydiver without a parachutes very fast. A diver could not land safely at this sper, 0 Zz, opens ‘3 parachute is mended. ‘The shydiver lands Epo safely atthe drop zone 52Yoar 10 Physics Booklet: Forces and Motion paral ilarly, water particles push against your body exerting a considerable force when you are immersed underwater. The deeper you go the water particles push down more strongly on your body so the greater the force. In fact, for every 10 m of depth there is another 100 000 N pushing inon your body. There are other forces acting on your body under the water too including the force of the diver’smass (weight) and the force of the water pushing up on the diver (buoyancy). The following diagram shows some of these forces. When a diver has these forces, all perfectly balanced it is called neutral buoyancy. | Wecait Morwall Vesstign force | Ate cestefuce (ticki ), Clad toe outide- e « . ii, Which forces are unbatenced?__Ewd motion gud _ dr Mesis Ipuce brcher ) iv. Isthe rider's speed increasing, decreasing or constant?__.cv£a fag 53Year 10 Physics Booklet: Forces and Motion os Applying Newton’s Laws of Motion WS12 - Activity: Finding Balance 1 In each of the images below add arrows and a description of forces acting on abjects. Are forces balanced or unbalanced? 1. What forces are acting on the basketball as it approaches the hoop? Crvitetionsl force ¥ Alv vasistouce 2. What forces are acting onthe hammer when the athlete lets go ofthe wire? eee | Grovitettoual force Aiv resistence 3. What forces are acting on the ball at the moment the boot makes contact? leick (Pus roewaeo) AR fests pwce jravide tonal larchYear 10 Physics Booklet: Forces and Motion 2023 4. What forces are acting on this windsurfer’s board? Friction wilh watir wind foree Lorward Groviferhigndl lye orwuel vain. fre Buoyouny) Flud bre (bs) im Lave from pin Friction ee wet fre Mojouk ashen bre Equel ¢ opostte boom our Plasys. Gauity F friclter 2 55Year 10 7. What forces are acting on the white ball as the cue strikes it? 8 What forces are acting on the tennis ball as it’s hit? fora frou (ue | Physics Booklet: Forces and Motion 2003 vation fore from bill fra di iw ee (Afr dial ' “saath, he | Ale O-1Sbus : aw 4. Acceleration = 7.9m/s2 ~£- Abe sh Force = 91N Ms a a4 2S Mass =? 60Year 10 1. 12. 13. 14. Physics Booklet: Forces and Motion 2023, Acceleration = 2.4m/s2 F f me £ 2. spark Force = 75N a Tt Shee Mass =? v = Acceleration = 0.6m/s2 Force = 5,000N - F = sow to ao a ape BBh Force = 13kN E- Ize _ ae Mass = 58kg Le oa cosa eeu —— Acceleration =? Force = 960N K gE qe. 4ST us Mass = 21kg . _ Tas Acceleration = ? o~ ar eee Force = 0.4N Mass = 76kg a~ & OF = 06-0083 mS Acceleration =? aA 6 — What force is needed to accelerate a mass of 10kg at a rate of 2m/s2? fz me = (>) = 2 ‘An unknown force causes a mass of 180kg to accelerate at a rate of 90cm/s2. = 0-45 Calculate the magnitude of the force. = Fe mas (I8)(08) = eV Nasa wish to accelerate a satellite of mass 460kg at a rate of 98m/s2. What force will they need? es mo. = (460\47) = as 090 a“ 4s pekW/ An unknown mass is accelerated at 43m/s2 when a force of 700N is applied to it. Find the mass. — Fs WO. y. me £ w= 6.3% (on 61Year 10 15. 16. 17. 19. Physics Booklet: Forces and Motion pial What is the mass of an abject if a force of 57N is needed to accelerate it by 0.6m/s"? S1 Me Ge yi 9s ky ‘A force of 0.5N causes an object to accelerate at 24cm/s2. Find the mass of the object. = oryus™ ee US 2 ita a qe 200k A force of 430N is applied to a mass of 2.8kg. Find the acceleration. Ep 4S = 153-6 us e iy ee What acceleration will an object of 5kg undergo if a force of 90N is applied? = 4 -2 a -& = (eu mM S$ Find the acceleration of a 360kg object when a 0.2N force is applied. = 0-000S6 us~ THE FREQUENCY OF BAD PHYSICS JOKES HERTZ 62Year 10 Physics Booklet: Forces and Motion oma Lesson 16: Applying Newton’s Laws of Motion WS 13 Activity: Finding Balance 2 Key concepts: ‘* Forces have strength and direction. © The forces acting on an object can be balanced or unbalanced. © Forces are balanced when they are equal in strength but opposite in direction, © Balanced forces don’t change the motion of objects. © Unbalanced forces cause objects to accelerate. Part 1 ‘Answer questions 1-4 and draw labelled arrows on the diagrams to show forces acting on the object (gravity, normal force, applied force and friction). The length of the arrows should indicate the relative strength of the force. 1. Acar is parked on a level driveway. Are the forces on it balanced or unbalanced? 2. rock is dropped from the top of a cliff. Are the forces on it balanced or unbalanced? AB ALAMCE® 0. te, Aes... weet Lovee a3. gutter ‘sbonste.: 63Year 10 Physics Booklet: Forces and Motion rea 3. A skydiver is descending on a parachute at constant speed. Are the forces on him balanced or unbalanced? MB AAME?.....09, los.erv4 Part2 4. Why does the boy have to push against the ground to start the skateboard moving? 5. This skateboard is m forces acting on it? Thay, LH, Lolaseel. to:...wleat= Mocmak, auhion ore jing at constant speed. What does this tell us about the vertical and horizontal 6. What happens to the skateboard when it starts rolling down the slope? What forces make this happen? wslheas,.s4f fi. a hed auitehtent. duce. on eentic. a. vahion SE. 64Year 10 Physics Booklet: Forces and Motion eal 7, a) b) I 4) e) f) On a level path, the skateboard eventually slows down and stops. What makes it stop? TAX... ALY Complete the following statements: Forces are balanced when they are equal in AeA debut opposite in_ HC! . bros “ is a force that acts on all objects on Earth. When balanced forces act on a stationary object, the object will__ udu _S Yoh onavy { / When balanced forces act on a moving object, the object will coukiaue bo uous! When an unbalanced force acts on a stationary object, the object will__ ove, /¥ Le ‘ abled Lyre tar querpue str Goce ! When an unbalanced force acts on a moving object, the object will__s/*e2d_ yr o- / Slow elawn Motion and Force Workbook 65