Physical Sciences momentum and impulse

Physical Sciences GRAD E 12 SELF STUDY GUIDE MOMENTUM AND IMPULSE

TABLE OF CONTENTS Page 1. Introduction 2 2. How to use this self-study guide 3 3. Momentum and Impulse 3.1 Summary sheets 4–8 3.2 Study and examination tips 9 – 12 3.3 Answering questions in the examination 13 – 28 4. Exercises 4.1 Multiple choice questions 29 - 34 4.2 Structured questions 35 - 41 5. Solutions to exercises 42 – 53 6. Message to grade 12 learners 54 7. Aclnowledgements 54 8. References 54 1

1. Introduction The declaration of COVID-19 as a global pandemic by the World Health Organisation led to the disruption of effective teaching and learning in many schools in South Africa. The majority of learners in various grades spent less time in class due to the phased-in approach and rotational/ alternate attendance system that was implemented by various provinces. Consequently, the majority of schools were not able to complete all the relevant content designed for specific grades in accordance with the Curriculum and Assessment Policy Statements in most subjects. As part of mitigating against the impact of COVID-19 on the current Grade 12, the Department of Basic Education (DBE) worked in collaboration with subject specialists from various Provincial Education Departments (PEDs) developed this Self-Study Guide. The Study Guide covers those topics, skills and concepts that are located in Grade 12, that are critical to lay the foundation for Grade 12. The main aim is to close the pre-existing content gaps in order to strengthen the mastery of subject knowledge in Grade 12. More importantly, the Study Guide will engender the attitudes in the learners to learning independently while mastering the core cross-cutting concepts 2

2. How to use this Self Study Guide? The intention of this booklet is to guide you through the topic of momentum and impulse. It should be used in conjunction with the CAPS document as well as the examination guidelines. Other resources should be used to reinforce content knowledge and understanding, e.g. textbooks, previous question papers and other study materials. The intent of this booklet is to make sure you understand the basics and fundamentals of the topic momentum and impulse. Read through the glossary of terms, make sure you understand these terms. Know your definitions. Work through the worked examples to get comfortable with same type of questions. Work through the exercises given on your own, before looking at the answers. Only when you are done, then compare your answer to the memorandum. If your answer is incorrect study the memorandum to see where you made a mistake. Be in possession of the formula sheet, there is no need to study the formulae off by heart. – Formulae for momentum and impulse will be provided in this guide. Extracts from the examination guidelines will be included before exercises are given, use these guidelines to make sure you know what is expected of you to study for the examination. 3

Definition: The product of an object’s mass and its velocity. Momentum is a vector Formula: ࢖ ൌ ࢓࢜ Vector: A vector is a physical quantity with Momentum is directly ି૚ ࢖ ൌ ࢓࢕࢓ࢋ࢔࢚࢛࢓ሺ࢑ࢍǤ ࢓Ǥ ࢙ ሻ proportional to both the mass magnitude and direction. and velocity of an object. Units for momentum: ࢓ ൌ ࢓ࢇ࢙࢙ሺ࢑ࢍሻ If the mass is constant, then ࢖ ൌ ࢓࢜, in the momentum formula velocity is ିଵ the greater the velocity the ݇݃Ǥ ݉Ǥ ‫ݏ‬ a vector. ࢜ ൌ ࢜ࢋ࢒࢕ࢉ࢏࢚࢟ሺ࢓Ǥ ࢙ି૚ ሻ greater the momentum. x We have to consider direction when substituting velocity or momentum. x Linear momentum is a vector with EXAMPLES: CALCULATIONS: the same direction as the velocity of 1. A dog with mass 30 kg runs at a velocity of x Always convert mass to kg and the object. ି૚ Ͷͷ݉Ǥ ‫ି ݏ‬ଵ . Calculate the momentum of the dog. velocity to ࢓Ǥ ࢙ Ǥ x Always choose a direction as positive ‫ ݌‬ൌ ݉‫ݒ‬ x Remember to indicate direction in your questions. ‫ ݌‬ൌ ͵Ͳ ൈ Ͷͷ using “+” and “-“ signs. x All movements in the opposite ିଵ x Provide your answer with a unit direction may then be substituted as ‫ ݌‬ൌ ͳ͵ͷͲ݇݃Ǥ ݉Ǥ ‫ݏ‬ in the direction of motion. and direction. negative (–) for vectors! Magnitude units Direction x If you obtain a negative answer for x Remember momentum will always 2. A car is travelling at 60݇݉Ǥ ݄ିଵ , if the mass of the a vector it only means the have magnitude and direction car is 250 kg, calculate the momentum of the car. direction is opposite to what you because it is a vector. ‫ ݌‬ൌ ݉‫ݒ‬ chose as positive. x Vector diagrams can be used to ‫ ݌‬ൌ ʹͷͲ ൈ ͳ͸ǡ͸͸͸͸͸ǤǤ represent the magnitude and MOMENTUM direction of a vector, e.g. ‫ ݌‬ൌ Ͷͳ͸͸Ǥ͸͹݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ in the direction of motion. Formula: ‫ ݌‬ൌ ‫ ݌‬ൌ ͵݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ East ݉‫ݒ‬ ଺଴ 4 Hint: to convert ݇݉Ǥ ݄ିଵ to݉Ǥ ‫ି ݏ‬ଵ ; ଷǤ଺ Units: ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ

Remember the initial momentum, Newton’s second law states that when a net / resultant force CHANGE IN MOMENTUM: acts on an object, the object will accelerate in the direction of final momentum and change in the net / resultant force, at an acceleration directly proportional momentum are vector quantities. Formula: ο‫ ݌‬ൌ ‫݌‬௙ െ ‫݌‬௜ ൌ ݉‫ ݂ݒ‬െ ݉‫݅ݒ‬ to the force and inversely proportional to the mass of the Direction is very important and object. Units: ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ should be included in your So whenever a net / resultant force acts on an object the object substitutions. Hence choose a will move faster or slower. direction as “+” e.g. take to the Formula: ο࢖ ൌ ࢖ࢌ െ ࢖࢏ - This means the velocity will change. right as “+”, the opposite direction - When the velocity changes the momentum will will then be “-“. ο࢖ ൌ ࢉࢎࢇ࢔ࢍࢋ࢏࢔࢓࢕࢓ࢋ࢔࢚࢛࢓ሺ࢑ࢍǤ ࢓Ǥ ࢙ି૚ ሻ also change since it is dependent on velocity. ࢖ࢌ ൌ ࢌ࢏࢔ࢇ࢒࢓࢕࢓ࢋ࢔࢚࢛࢓ ሺ࢑ࢍǤ ࢓Ǥ ࢙ି૚ ሻ ο‫ ݌‬ൌ ݉ο‫ݒ‬ Can momentum be zero? ࢖࢏ ൌ ࢏࢔࢏࢚࢏ࢇ࢒࢓࢕࢓ࢋ࢔࢚࢛࢓ ሺ࢑ࢍǤ ࢓Ǥ ࢙ି૚ ሻ ο‫ ݌‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ Yes, the object will have zero ο‫ ݌‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ momentum when the velocity of ο࢖ ൌ ࢉࢎࢇ࢔ࢍࢋ࢏࢔࢓࢕࢓ࢋ࢔࢚࢛࢓ሺ࢑ࢍǤ ࢓Ǥ ࢙ି૚ ሻ the object is zero. The velocity of an object will be zero when the Take note of the following notations: ࢓ ൌ ࢓ࢇ࢙࢙ሺ࢑ࢍሻ object is at rest. x Momentum changes by ‫݃݇ݔ‬Ǥ ݉Ǥ ‫ି ݏ‬ଵ relates to ο࢖ ࢜ࢌ ൌ ࢌ࢏࢔ࢇ࢒࢜ࢋ࢒࢕ࢉ࢏࢚࢟ሺ࢓Ǥ ࢙ି૚ ሻ An object that is not moving is x Momentum changes to ‫݃݇ݔ‬Ǥ ݉Ǥ ‫ି ݏ‬ଵ relates to ࢖ࢌ said to be: at rest or stationery. ࢜࢏ ൌ ࢏࢔࢏࢚࢏ࢇ࢒࢜ࢋ࢒࢕ࢉ࢏࢚࢟ሺ࢓Ǥ ࢙ି૚ ሻ EXAMPLE: EXAMPLE: 1. A tennis player serves a tennis ball towards his opponent, the ball 2. An arrow is shot with an initial momentum of ʹͲ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ towards a moves through the air with momentum of ͳͳ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ east. The target. It penetrates the target and its momentum changes by opponent hits the ball back in the opposite direction and the Ͷ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ in the opposite direction. Calculate the final momentum momentum of the ball changes to ͺ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ west. of the arrow. Take towards the a) Calculate the change in momentum of the tennis ball. a) ο‫ ݌‬ൌ ‫݌‬௙ െ ‫݌‬௜ target as “+” ο‫ ݌‬ൌ ‫݌‬௙ െ ‫݌‬௜ െͶ ൌ ‫݌‬௙ െ ʹͲ Take east as “+” 5 ‫݌‬௙ ൌ ͳ͸݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ towards the target. ο‫ ݌‬ൌ ሺെͺሻ െ ሺͳͳሻ ο‫ ݌‬ൌ െͳͻ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ  ‫݃݇ͻͳ ׵‬Ǥ ݉Ǥ ‫ି ݏ‬ଵ ࢃࢋ࢙࢚

Law of Conservation of Momentum: The total linear momentum in an isolated system remains constant (in magnitude and direction.) POSSIBILITY 1: Formula: σ࢖࢏ ൌ σ࢖ࢌ x Two separate objects move towards each other and ࢓૚ ࢜૚࢏ ൅ ࢓૛ ࢜૛࢏ ൌ ࢓૚ ࢜૚ࢌ ൅ ࢓૛ ࢜૛ࢌ collide, after the collision they bounce off each other ࢓૚ ൌ ࢓ࢇ࢙࢙࢕ࢌ࢕࢈࢐ࢋࢉ࢚૚ሺ࢑ࢍሻ and move away. x Think of snooker balls colliding. ࢓૛ ൌ ࢓ࢇ࢙࢙࢕ࢌ࢕࢈࢐ࢋࢉ࢚૛ሺ࢑ࢍሻ ࢓૚ ࢜૚࢏ ൅ ࢓૛ ࢜૛࢏ ൌ ࢓૚ ࢜૚ࢌ ൅ ࢓૛ ࢜૛ࢌ ࢜૚࢏ ൌ ࢏࢔࢏࢚࢏ࢇ࢒࢜ࢋ࢒࢕ࢉ࢏࢚࢟࢕ࢌ࢕࢈࢐ࢋࢉ࢚૚ሺ࢓Ǥ ࢙ି૚ ሻ ࢜૛࢏ ൌ ࢏࢔࢏࢚࢏ࢇ࢒࢜ࢋ࢒࢕ࢉ࢏࢚࢟࢕ࢌ࢕࢈࢐ࢋࢉ࢚૛ሺ࢓Ǥ ࢙ି૚ ሻ POSSIBILITY 2: Conservation of linear momentum x Two separate objects move towards each other ࢜૚ࢌ ൌ ࢌ࢏࢔ࢇ࢒࢜ࢋ࢒࢕ࢉ࢏࢚࢟࢕ࢌ࢕࢈࢐ࢋࢉ࢚૚ሺ࢓Ǥ ࢙ି૚ ሻ and collide, after the collision they combine and ି૚ x Should be an Isolated system ࢜૛ࢌ ൌ ࢌ࢏࢔ࢇ࢒࢜ࢋ࢒࢕ࢉ࢏࢚࢟࢕ࢌ࢕࢈࢐ࢋࢉ࢚૛ሺ࢓Ǥ ࢙ ሻ move as one system. x Isolated system is when the resultant (or net) external forces x Think of cars colliding and sticking together. A girl with mass 30 kg runs at 5 ݉Ǥ ‫ି ݏ‬ଵ and acting on the system is zero. x There can only be one velocity for objects that jumps on a 2 kg skateboard that is at rest. σ࢖࢏ ൌ σ࢖ࢌ are joined. For this scenario it will be࢜ࢌ . Calculate the velocity of the skateboard and the ࢓૚ ࢜૚࢏ ൅ ࢓૛ ࢜૛࢏ ൌ ሺ࢓૚ ൅ ࢓૛ ሻ࢜ࢌ girl after she jumps on it. POSSIBILITY 3: Solution: Take direction of motion as (+) σ࢖࢏ ൌ σ࢖ࢌ x Where there are 2 objects that start as one (connected) system, after an explosion they ࢓૚ ࢜૚࢏ ൅ ࢓૛ ࢜૛࢏ ൌ ሺ࢓૚ ൅ ࢓૛ ሻ࢜ࢌ move apart and are now two separate objects. (30)(5) + (2)(0) = (30 + 2) ‫ݒ‬௙ x Think of a trailer connected to a car. 150 + 0 = (32)‫ݒ‬௙ x There can only be one velocity for objects that 6 ͳͷͲ are joined. For this scenario it will be࢜࢏ . ‫ݒ‬௙ ൌ ͵ʹ ሺ࢓૚ ൅ ࢓૛ ሻ࢜࢏ ൌ ࢓૚ ࢜૚ࢌ ൅ ࢓૛ ࢜૛ࢌ ‫ݒ‬௙ = 4,69 ݉Ǥ ‫ି ݏ‬ଵ in the direction of motion.

Impulse: The product of the resultant / net force acting on an object and the time Impulse: ࡲࡺࡱࢀ ο࢚ ൌ ο࢖ the resultant / net force acts on the object. Newton’s second law expressed in terms Like momentum, impulse is a of momentum. Formula: ࡲࡺࡱࢀ ο࢚ ൌ ο࢖ vector quantity, and the direction of the impulse is the ࡲࡺࡱࢀ ൌ ࡾࢋ࢙࢛࢒࢚ࢇ࢔࢚ࢌ࢕࢘ࢉࢋሺࡺሻ Units for impulse: Definition: The resultant / net force acting on an same as the direction of the net force that causes it. object is equal to the rate of change of ο࢚ ൌ ࢚࢏࢓ࢋሺ࢙ሻ (Contact time) ܰǤ ‫݃݇ݎ݋ݏ‬Ǥ ݉Ǥ ‫ି ݏ‬ଵ momentum of the object in the direction of the ି૚ resultant / net force. ο࢖ ൌ ࢉࢎࢇ࢔ࢍࢋ࢏࢔࢓࢕࢓ࢋ࢔࢚࢛࢓ሺ࢑ࢍǤ ࢓Ǥ ࢙ ሻ x Newton’s second law states that when a net force acts on an object, it EXAMPLES: IMPULSE: will cause the object to accelerate in 1. A 300 g soccer ball is kicked against the goal post. Formula: ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ݊݁‫ ݁ܿݎ݋݂ݐ‬ൈ ‫݁݉݅ݐ‬ the direction of the net force at an ିଵ It hits the post with a velocity of 15݉Ǥ ‫ݏ‬ , and acceleration directly proportional to ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬ bounces off the post after 0.2 seconds with a the net force and inversely ିଵ Units: ܰǤ ‫݃݇࢘࢕ݏ‬Ǥ ݉Ǥ ‫ି ݏ‬ଵ velocity of 10݉Ǥ ‫ݏ‬ . proportional to the mass of the object. a) Calculate the impulse that the post exerts on the GRAPH: Net force vs time x We can now conclude that the net ball. force acting on an object is changing Direction towards goal post is + Any collision that takes place between two objects ୒୉୘ ο– ൌ ο’ will cause the net force to change from 0 N to a the velocity of the object and hence ’—Ž•‡ ൌ ͲǤ͵ሺെͳͲሻ െ ͲǤ͵ሺͳͷሻ maximum value and then return back to 0 N after the the momentum will change. ’—Ž•‡ ൌ െ͹ǤͷǤ • (Remember ୒୉୘ ο– ൌ ’—Ž•‡ሻ collision. The contact time between the two objects is ‫ݒ‬௙ െ ‫ݒ‬௜ ο‫ݒ‬ ‫ܨ‬ோ் ൌ ݉ܽǢ ܽ ൌ ൌ ‫ ׵‬͹ǤͷǤ •‫ܜܛܗܘܔ܉ܗ܏܍ܐܜܕܗܚ܎ܡ܉ܟ܉‬Ǥ referred to asο࢚. ο‫ݐ‬ ο‫ݐ‬ ο‫ݒ‬ Direction Maximum force ‫ܨ‬ோ் ൌ ݉ ο‫ݐ‬ b) Calculate the average force acting on the ball during collision. ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ during the collision with the post. ‫ܨ‬ோ் ൌ Ǣο‫ ݌‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ Area under the ο‫ݐ‬ ο࢚ሺ࢙ሻ ୒୉୘ ο– ൌ ο’ ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ graph which is shaded represents ο‫݌‬ 7 ୒୉୘ ሺͲǡʹሻ ൌ Ͳǡ͵ሺെͳͲ െ ͳͷሻ the impulse of the ‫ܨ‬ோ் ൌ  ο‫ݐ‬ Net force (N) Ͳǡ͵ሺെͳͲ െ ͳͷሻ object. ୒୉୘ ൌ ൌ  െ͵͹ǡͷܰ ሺͲǡʹሻ ‫ܨ ׵‬ோ் ο‫ ݐ‬ൌ ο‫݌‬ ‫ܨ ׵‬ோ் ൌ ͵͹ǡͷܰ away from the goal post.

ELASTIC AND INELASTIC COLLISIONS: Kinetic energyሺࡱ࢑ ሻȀࡷ: Elastic collision: A collision in which the total linear momentum and kinetic energy is Kinetic energy is the energy that an object possess because it is in conserved. motion. Total kinetic energy before the collision is EQUAL to the total kinetic energy after the collision. ଵ Formula:‫ܧ‬௞ ൌ ݉‫ ݒ‬ଶ ; Units: ‫( ܬ‬Joules) ଶ Inelastic collision: A collision during which the total linear momentum is conserved, but ࢓ ൌ ࢓ࢇ࢙࢙ሺ࢑ࢍሻ kinetic energy is NOT conserved. ࢜ ൌ ࢜ࢋ࢒࢕ࢉ࢏࢚࢟ሺ࢓Ǥ ࢙ି૚ ሻ Total kinetic energy before the collision is NOT EQUAL to the total kinetic energy after the collision. Elastic collisions: Inelastic collisions: x Objects can combine after collisions, and be considered as x Objects cannot combine together after collisions. one system. x We accept no energy is lost to the environment or transferred to other x Kinetic Energy is not conserved and may be lost to other forms e.g. heat or sound. forms of energy e.g. energy transferred to heat or sound during a collision. ͳ ଶ ͳ ଶ The answers for these 2 x σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  ݉ଵ ‫ݒ‬௜ଵ ൅ ݉ଶ ‫ݒ‬௜ଶ Kinetic energy ሺ‫ܧ‬௞ ሻ of an isolated system is conserved: ʹ ʹ calculations will NOT be ͳ ͳ ͳ ͳ equal!! ଶ ଶ ଶ ଶ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  ݉ଵ ‫ݒ‬௜ଵ ൅ ݉ଶ ‫ݒ‬௜ଶ The answers for these 2 σ‫ܧ‬௞௙௜௡௔௟ ൌ ݉ଵ ‫ݒ‬௙ଵ ൅ ݉ଶ ‫ݒ‬௙ଶ ʹ ʹ ʹ ʹ calculations will be equal!! ͳ ͳ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  σ‫ܧ‬௞௙௜௡௔௟ ଶ ଶ σ‫ܧ‬௞௙௜௡௔௟ ൌ ݉ଵ ‫ݒ‬௙ଵ ൅ ݉ଶ ‫ݒ‬௙ଶ ʹ ʹ x Think of two cars that are part of a collision and they x A ball that bounces to the same height from which it was dropped has an combine after they collided. x NOTE: Always do the calculation for ‫ܧ‬௞௜ separate from the almost elastic collision with the ground. one for ‫ܧ‬௞௙ . It does not matter whether EXAMPLE: ANSWER: it is an elastic or inelastic ଵ ଶ ଵ ଶ ଵ ଵ A truck travelling at 40 ݇݉Ǥ ݄ିଵ east, with a mass σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  ݉ଵ ‫ݒ‬௜ଵ ൅ ݉ଶ ‫ݒ‬௜ଶ ൌ ሺͻͷͲሻሺͳͳǡͳͳǤ Ǥ ሻଶ ൅ ሺ͵ͲͲሻሺͳ͸ǡ͸͸Ǥ Ǥ ሻଶ ൌ ଶ ଶ ଶ ଶ collision, the total linear of 950 kg crashes into a car with a mass of 300 ૚૙૙૜૙ૡǡ ૟૝ࡶ momentum remains kg that is travelling at 60݇݉Ǥ ݄ିଵ west. ଵ ଵ ଵ ଵ ଶ ଶ σ‫ܧ‬௞௙௜௡௔௟ ൌ ݉ଵ ‫ݒ‬௙ଵ ൅ ݉ଶ ‫ݒ‬௙ଶ ൌ ሺͻͷͲሻሺʹǡ͹͹Ǥ Ǥ ሻଶ ൅ ሺ͵ͲͲሻሺͻǡ͹ʹǤ Ǥ ሻଶ ൌ conserved in an isolated Immediately after the collision, the truck travels at ଶ ଶ ଶ ଶ system. 10 ݇݉Ǥ ݄ିଵ in its original direction of motion, and ૚ૠૡ૝૜ǡ ૜૟ࡶ the car travels at 35݇݉Ǥ ݄ିଵ in the direction east. 8 (Ignore all effects of friction) σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  σ‫ܧ‬௞௙௜௡௔௟ Was this an elastic or inelastic collision? ‫ܿ݅ݐݏ݈ܽ݁݊݅ݏ݅݊݋݅ݏ݈݈݅݋ܿ ׵‬Ǥ

3.2. Study and examination tipsǤ 3.2.1. Format of the question paper: x Momentum and impulse are part of paper 1. x It falls under the topic mechanics. x It can be examined in the multiple choice questions, as well as the structured questions. x Remember momentum can be integrated with topics such as vertical projectile motion and work, energy and power. 3.2.2. Resources in the examination: x Every question paper will be accompanied by a formula and data sheet at the end of the paper. x It is of utmost importance that you are comfortable with the formula sheet, and that you know which formulae are applicable to momentum and impulse. x Formulae labeled under FORCE in the formula sheet bears reference to momentum and impulse. x The kinetic energy formula can be found under work, energy and power. 3.2.3. Table 2: Formulae FORCE    ‫ ݌‬ൌ ݉‫ݒ‬ ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬ Momentum and Impulse ο‫ ݌‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ WORK, ENERGY AND POWER Kinetic energy, used in ૚ ૚ ࡷ ൌ ૛ ࢓࢜૛ OR ࡱ࢑ ൌ ૛ ࢓࢜૛ calculations for elastic and inelastic collisions. 3.2.4. Laws, definitions and principals. x Study the definitions from the examination guidelines. x Pay careful attention to keywords. – Definitions used in this booklet are taken from the examination guidelines. x Try to memorise the definitions with understanding. x Most of the times the definition can be remembered by looking at the formula. e.g. ‫ ݌‬ൌ ݉ ൈ ‫ ; ݒ‬Linear momentum is the product of an object’s mass and its velocity. 9

3.2.5. Glossary of terms: Momentum The product of an object’s mass and its velocity. Linear momentum Linear momentum is the product of an object’s mass and its velocity. A vector quantity with the same direction as the velocity of the object. Contact forces Contact forces arise from the physical contact between two objects (e.g. a soccer player kicking a ball.) or e.g. Tension. Non-contact forces Non-contact forces arise even if two objects do not touch each other (e.g. the force of attraction of the earth on a parachutist even when the earth is not in direct contact with the parachutist.) e.g. ‫ܨ‬௚ Newton’s second law of The resultant/net force acting on an object is equal to the rate of motion in terms of momentum change of momentum of the object in the direction of the resultant/net force. Principle of conservation of The TOTAL linear momentum of an isolated system remains linear momentum constant (is conserved). Closed system / Isolated A system in which the resultant/net external force acting on the system (in Physics) system is zero. Impulse The product of the resultant/net force acting on an object and the time the resultant/net force acts on the object. Elastic collision A collision in which both total linear momentum and total kinetic energy are conserved. Inelastic collision A collision during which total linear momentum is conserved but total kinetic energy is not conserved. System A system is a part of space that has been chosen for studying the changes that takes place within it. Environment Everything outside the system is called the environment. Internal forces Forces that objects exert on each other within a system, e.g. contact forces. External forces Forces that are outside the system, but acts on the objects within the system, e.g. frictional force. 10

3.2.6. Extract from examination guidelines: (Examiners will follow these guidelines when the NSC exam is set). Momentum and Impulse (This section must be read in conjunction with the CAPS, p. 99–101.) Momentum • Define momentum as the product of an object's mass and its velocity. • Describe the linear momentum of an object as a vector quantity with the same direction as the velocity of the object. • Calculate the momentum of a moving object using p = mv. • Describe the vector nature of momentum and illustrate it with some simple examples. • Draw vector diagrams to illustrate the relationship between the initial momentum, the final momentum and the change in momentum for each of the above examples. Newton's second law of motion in terms of momentum • State Newton's second law of motion in terms of momentum: The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force. ο௣ • Express Newton's second law of motion in symbols: ‫ܨ‬ோ் ൌ ο௧ • Calculate the change in momentum when a resultant/net force acts on an object and its velocity: nd o Increases in the direction of motion, e.g. 2 stage rocket engine fires. o Decreases, e.g. brakes are applied. o Reverses its direction of motion, e.g. a soccer ball kicked back in the direction it came from. Impulse • Define impulse as the product of the resultant/net force acting on an object and the time the resultant/net force acts on the object. • Deduce the impulse-momentum theorem: Fnet Δt = mΔv. • Use the impulse-momentum theorem to calculate the force exerted, the time for which the force is applied and the change in momentum for a variety of situations involving the motion of an object in one dimension. • Explain how the concept of impulse applies to safety considerations in everyday life, e.g. airbags, seatbelts and arrestor beds. 11

Conservation of momentum and elastic and inelastic collisions • Explain what is meant by a closed/an isolated system (in Physics), i.e. a system on which the resultant/net external force is zero. A closed/an isolated system excludes external forces that originate outside the colliding bodies, e.g. friction. Only internal forces, e.g. contact forces between the colliding objects, are considered. • State the principle of conservation of linear momentum: The total linear momentum of a closed system remains constant (is conserved). • Apply the conservation of momentum to the collision of two objects moving in one dimension (along a straight line) with the aid of an appropriate sign convention. • Distinguish between elastic collisions and inelastic collisions by calculation. 12

3.3. Answering questions in the examination. 3.3.1. Momentum: x Remember momentum is a vector quantity. x Make use of sign conventions, preferably choose one direction of motion to be positive. x Indicate this direction at the top, next to your solution so that you will remember. x All motions for vectors that occur in the opposite direction to what you chose as positive will be substituted as negative. x Make use of this sign convention throughout the question, do not change it at sub- questions, i.e., the sign convention chosen at 2.1 should be used in 2.2 etc. x Make sure you copy formulae from the formula sheet, do not write it from memory. x Write down the formula as it is from the formula sheet. Do NOT manipulate the formula. x Always attempt substitution, you will not be awarded a mark for the formula if substitution is not attempted. x Substitute into the original formula, then solve or calculate the unknown value. x Remember direction is very important when vectors are being substituted. x We use “+” and “-“ sign conventions to indicate direction. x SI-units for mass and velocity are kg and ࢓Ǥ ࢙ି૚ respectively. x SI-units should always be used when substituting, meaning when substituting mass it must be measured in kg, and when substituting velocity it must be measured in݉Ǥ ‫ݏ‬െͳ . x If the SI-units in the question differs, make use of conversions to obtain the correct SI-unit. ݃ x To convert grams to kg:ͳͲͲͲ. ௞௠Ǥ௛షభ x To convert ݇݉Ǥ ݄ିଵ to ݉Ǥ ‫ି ݏ‬ଵ : ଷǤ଺ x Your final answers should consist of magnitude, units and direction. Remember momentum is a vector, however if the magnitude of a vector is only required then you don’t have to give direction. x Try to use un-rounded off values in your question – it will assist in the final answer being more accurate, e.g. if given 0,1768 do not substitute 0,18. Use it as it is. 13

x Round off your final answer to a minimum of two decimal places. Worked examples: a) What will the momentum of an object moving at 15 ݉Ǥ ‫ି ݏ‬ଵ to the east be, if the object has a mass of 2 kg? Solution: Step 1: With the use of sign conventions, choose a positive direction. East = Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI- units. ‫ ݌‬ൌǫ ‫ ݒ‬ൌ15 ݉Ǥ ‫ି ݏ‬ଵ ݉ ൌ ʹ݇݃ Step 3: Choose a suitable formula from the formula sheet. ‫ ݌‬ൌ ݉‫ݒ‬ Step 4: Substitute the values from step 2 into the formula and do the calculations. Remember to write down the units and direction for the answer. ‫ ݌‬ൌ ݉‫ݒ‬ ‫ ݌‬ൌ ሺʹሻሺͳͷሻ ‫ ݌‬ൌ ͵Ͳ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ ‫ݐݏܽܧ‬ b) The momentum of a truck with mass 1.5 ton is 19 590݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ . At what velocity is the truck travelling? Step 1: With the use of sign conventions, choose a positive direction. Motion of truck = “+” Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI-units. ‫ ݌‬ൌ ͳͻͷͻͲ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ. ‫ ݒ‬ൌǫ݉Ǥ ‫ି ݏ‬ଵ ݉ ൌ ͳǤͷ‫݊݋ݐ‬ Convert to kg: ૚Ǥ ૞ ൈ ૚૙૙૙ ൌ ૚૞૙૙࢑ࢍ Step 3: Choose a suitable formula from the formula sheet. ‫ ݌‬ൌ ݉‫ݒ‬ 14

Step 4: Substitute the values from step 2 into the formula and do the calculations. Remember to write down the units and direction for the answer. ‫ ݌‬ൌ ݉‫ݒ‬ ͳͻͷͻͲ ൌ ሺͳͷͲͲሻሺ˜ሻ ‫ ݒ‬ൌ ͳ͵ǡͲ͸݉Ǥ ‫ି ݏ‬ଵ ݅݊‫݊݋݅ݐ݋݂݉݋݊݋݅ݐܿ݁ݎ݄݅݀݁ݐ‬Ǥ c) In a cricket match the bowler bowls a 150 g cricket ball towards the batsman. The velocity of the ball is 135݇݉Ǥ ݄ିଵ . Calculate the momentum of the ball. Step 1: With the use of sign conventions, choose a positive direction. Towards the batsman is taken as “+” (Siyavula textbook) Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI-units. ‫ ݌‬ൌǫ ૚૜૞ ‫ ݒ‬ൌ135 ݇݉Ǥ ݄ିଵ Convert to࢓Ǥ ࢙ି૚ : ൌ ૜ૠǡ ૞࢓Ǥ ࢙ି૚  ૜Ǥ૟ ૚૞૙ ݉ ൌ ͳͷͲ݃ Convert to kg: ൌ ૙ǡ ૚૞࢑ࢍ ૚૙૙૙ Step 3: Choose a suitable formula from the formula sheet. ‫ ݌‬ൌ ݉‫ݒ‬ Step 4: Substitute the values from step 2 into the formula and do the calculations. Remember to write down the units and direction for the answer. ‫ ݌‬ൌ ݉‫ݒ‬ ‫ ݌‬ൌ ሺͲǤͳͷሻሺ͵͹ǡͷሻ ‫ ݌‬ൌ ͷǡ͸͵݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ ܶ‫݊ܽ݉ݏݐܾ݄ܽ݁ݐݏ݀ݎܽݓ݋‬Ǥ 15

3.3.2. Change in momentum. x Momentum and velocity are vectors. x Make use of sign conventions, meaning always choose one direction of motion to be positive. x All motions for vectors that occur in the opposite direction to what you chose as positive will be substituted as negative. x SI-units should always be used when substituting, meaning when substituting mass it must be measured in kg, and when substituting velocity it must be measured in࢓Ǥ ࢙െ૚ . x Change in momentum: ο࢖ ൌ ࢖ࢌ െ ࢖࢏ ο࢖ ൌ ࢓࢜ࢌ െ ࢓࢜࢏ ο࢖ ൌ ࢓ሺ࢜ࢌ െ ࢜࢏ ሻ Worked examples: a) At the US open a tennis player serves a tennis ball with mass of 100 g towards his opponent with a velocity ofʹʹͲ݇݉Ǥ ݄ିଵ east. The serve is returned by the opponent atͳͷͶǡͺ݇݉Ǥ ݄ିଵ in the opposite direction. Calculate the change in momentum of the ball. Step 1: With the use of sign conventions, choose a positive direction. East = “+” Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI-units. ο‫ ݌‬ൌǫ ૛૛૙ ‫ݒ‬௜ ൌ220 ݇݉Ǥ ݄ିଵ Convert to࢓Ǥ ࢙ି૚ : ൌ ૟૚ǡ ૚૚૚૚Ǥ Ǥ ࢓Ǥ ࢙ି૚  ૜Ǥ૟ ି૚૞૝ǡૡ Negative sign only indicates that ‫ݒ‬௙ ൌ െͳͷͶǡͺ݇݉Ǥ ݄ିଵ Convert to࢓Ǥ ࢙ି૚ : ൌ െ૝૜࢓Ǥ ࢙ି૚  ૜Ǥ૟ the direction of the ball is now ૚૙૙ West. ݉ ൌ ͳͲͲ݃ Convert to kg: ൌ ૙ǡ ૚࢑ࢍ ૚૙૙૙ Step 3: Choose a suitable formula from the formula sheet. ο‫ ݌‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ 16

Step 4: Substitute the values from step 2 into the formula and do the calculations. Remember to write down the units and direction for the answer. Note: ‫݌‬௜ ൌ ݉‫ݒ‬௜ Velocity is substituted as a positive value because it is in the same ‫݌‬௜ ൌ ሺͲǡͳሻሺ͸ͳǡͳͳͳͳሻ direction of motion chosen as “+” ‫݌‬௜ ൌ ͸ǡͳͳ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ  East. (East) ‫݌‬௙ ൌ ݉‫ݒ‬௙ Velocity is substituted as a negative ‫݌‬௙ ൌ ሺͲǡͳሻሺെͶ͵ሻ value because it is in the opposite direction of motion chosen as “+”, ‫݌‬௙ ൌ െͶǤ͵݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ (West) ‫݌‬௙ ൌ ͶǤ͵݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ West. NOTE: ο‫ ݌‬ൌ ‫݌‬௙ െ ‫݌‬௜ You don’t have to use separate calculations in the examination like we did in this example. ο‫ ݌‬ൌ െͶǡ͵ െ ͸ǡͳͳ You can do it as one calculation as shown in option 2. ο‫ ݌‬ൌ െͳͲǡͶͳ݇݃Ǥ ݉Ǥ ‫ݏ‬െͳ We only wanted to emphasize the importance of direction and how it will ο‫ ݌‬ൌ ͳͲǡͶͳ݇݃Ǥ ݉Ǥ ‫ݏ‬െͳ West. influence the initial momentumሺ࢖࢏ ሻ, final momentum (࢖ࢌ ሻ and change in momentum (ο࢖ሻ. Graphical representation: Graphical representation for an object where the momentum changes to the opposite direction. ‫݌‬௜ =͸ǡͳͳ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ ‫݌‬௙ ൌ െͶǤ͵ ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ Negative values only indicates that the direction of motion is in the opposite direction. ο‫=݌‬െͳͲǡͶͳ݇݃Ǥ ݉Ǥ ‫ݏ‬െͳ Option 2: ο‫ ݌‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ This is the preferred ο‫ ݌‬ൌ ͲǡͳሺെͶ͵ െ ͸ͳǡͳͳͳͳሻ method to use in the examination. ο‫ ݌‬ൌ െͳͲǡͶͳ݇݃Ǥ ݉Ǥ ‫ݏ‬െͳ ο‫ ݌‬ൌ ͳͲǡͶͳ݇݃Ǥ ݉Ǥ ‫ݏ‬െͳ West. 17

b) Graphical representation for an object where the momentum decrease in the same direction. At the US open a tennis player serves a tennis ball with mass 100 g towards his opponent with a velocity ofʹʹͲ݇݉Ǥ ݄ିଵ east. The tennis ball hits the net and continues to move towards the opponent with a velocity of 120݇݉Ǥ ݄ିଵ ‫ݒ‬௙ ܽ݊݀‫ݒ‬௜ must be converted ο‫ ݌‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ to ݉Ǥ ‫ି ݏ‬ଵ You will notice that even though the ο‫ ݌‬ൌ Ͳǡͳሺ͵͵ǡ͵͵͵͵ െ ͸ͳǡͳͳͳͳሻ initial momentum and final െͳ ο‫ ݌‬ൌ െʹǡ͹ͺ݇݃Ǥ ݉Ǥ ‫ݏ‬ momentum of the tennis ball remains in the same direction, the change in ο‫ ݌‬ൌ ʹǡ͹ͺ݇݃Ǥ ݉Ǥ ‫ݏ‬െͳ ࢃࢋ࢙࢚Ǥ momentum is in the opposite direction because the magnitude of ‫݌‬௜ ൌ ͸ǡͳͳ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ the final momentum is less than the ିଵ ‫݌‬௙ ൌ ͵Ǥ͵͵ ݇݃Ǥ ݉Ǥ ‫ݏ‬ ο‫ ݌‬െ ʹǡ͹ͺ݇݃Ǥ ݉Ǥ ‫ݏ‬െͳ magnitude of the initial momentum. 3.3.3. Conservation of linear momentum. x Linear Momentum will always be conserved when the collision occurs in a closed or isolated system. x Linear Momentum is conserved when the resultant or net external forces acting on the system is zero. x Always start the questions with the following equation.σ࢖࢏ ൌ σ࢖ࢌ x Remember the above formula is not given on the formula sheet. x All the information substituted before the “=” sign refers to ‘before the collision’ (initial), all the information substituted after the “=” sign refers to ‘after the collision’ (final). The following three scenarios are most frequently asked. Remember conservation of momentum only occurs in closed or isolated systems. 1. Two separate objects which are initially not in contact, collide and bounce of each other remain two separate objects, each with its own velocity. 18

2. Two separate objects which are initially not in contact, collide and are now joined together moving with only one final velocity. 3. Two objects that are joined, with one initial velocity experience a collision and separate into two different objects, each with its own velocity. x Remember direction is important when substituting velocity in the above equations. – make use of sign conventions. x You must be able to manipulate the equations to solve any of the variables in that equation when a final answer is calculated. Do not manipulate the formula before substitution. 19

Worked examples: a) The diagram below shows two trolleys, P and Q, held together by means of a compressed spring on a flat, frictionless horizontal track. The masses of P and Q are 400 g and 600 g respectively. Ignore the effects of friction. When the spring unwinds Trolley Q moves to the right at 4 ݉Ǥ ‫ି ݏ‬ଵ Ǥ Calculate the Velocity of trolley P after the trolleys are released. Step 1: With the use of sign conventions, choose a positive direction. Motion towards the right = “+” Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI-units. ସ଴଴ ݉௉ ൌ ͶͲͲ݃ Convert to kg: ൌ ͲǡͶ݇݃ ଵ଴଴଴ ଺଴଴ ݉ொ ൌ ͸ͲͲ݃ Convert to kg: ൌ Ͳǡ͸݇݃ ଵ଴଴଴ ‫ݒ‬௣௜ ൌ Ͳ݉Ǥ ‫ି ݏ‬ଵ ‫ݒ‬௤௜ ൌ Ͳ݉Ǥ ‫ି ݏ‬ଵ Because P and Q are initially combined they will have the same ‫ݒ‬ொ௙ ൌ Ͷ݉Ǥ ‫ି ݏ‬ଵ initial velocity! ‫ݒ‬௉௙ ൌǫ ݉Ǥ ‫ି ݏ‬ଵ Step 3: Choose a suitable formula. Conservation of momentum should always be started with the following formula:σ࢖࢏ ൌ σ࢖ࢌ , it can then be expanded to 1 of the 3 scenarios. σ୮୧ ൌ σ୮୤ ሺ୔ ൅ ୕ ሻ˜୧ ൌ ୔ ˜୔୤ ൅ ୕ ˜୕୤ (Scenario 3 will work best- 2 objects that starts of as one system, and after the spring is released are now two separate objects.) 20

Step 4: Substitute the values from step 2 into the formula and do the calculations. Remember to write down the units and direction of the answer. σ୮୧ ൌ σ୮୤ We have changed the subscript to suit our ሺ୔ ൅ ୕ ሻ˜୧ ൌ ୔ ˜୔୤ ൅ ୕ ˜୕୤ scenario. Using P and Q as stated in the question. This will help us to easily relate ሺͲǡͶ ൅ Ͳǡ͸ሻͲ ൌ ሺͲǡͶሻ˜୔୤ ൅ ሺͲǡ͸ሻሺͶሻ to our information. Ͳ ൌ ሺͲǡͶሻ˜୔୤ ൅ ʹǡͶ െʹǡͶ ൌ ˜୔୤ Remember velocity is a vector, the ͲǡͶ negative answer for ˜୔୤ indicates that the motion of direction will be opposite to what െ͸Ǥ • ିଵ ൌ ˜୔୤ we chose as positive. Hence the direction of P is towards the left because we chose ‫୔˜ ׵‬୤ ൌ ͸݉Ǥ ‫ି ݏ‬ଵ ܶ‫ݐ݂݈݄݁݁ݐݏ݀ݎܽݓ݋‬Ǥ towards the right to be positive. We always write our final answer as a positive value. If the answer is negative like in the example above, we re-write it as positive with the correct indication of direction represented by the negative sign. a) In the illustration below, ball A and ball B both with the SAME mass are moving towards each other at 15 ݉Ǥ ‫ି ݏ‬ଵ and 8 ݉Ǥ ‫ି ݏ‬ଵ respectively. They collide and after the collision ball A is moving at 5 ݉Ǥ ‫ି ݏ‬ଵ towards the right. Calculate the velocity of ball B after the collision. Ignore the effects of friction. Step 1: With the use of sign conventions, choose a positive direction. Motion towards the right = “+” 21

Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI-units. ݉஺ ൌ ݉஻ ‫ݒ‬஺௜ ൌ ͳͷ݉Ǥ ‫ି ݏ‬ଵ ‫ݒ‬஻௜ ൌ െͺ݉Ǥ ‫ି ݏ‬ଵ (The negative sign indicates that ball B is travelling towards the left) ‫ݒ‬஺௙ ൌ ͷ݉Ǥ ‫ି ݏ‬ଵ ‫ݒ‬஻௙ ൌǫ ݉Ǥ ‫ି ݏ‬ଵ Step 3: Choose a suitable formula. Conservation of momentum can always be started with the following formula: σ࢖࢏ ൌ σ࢖ࢌ, it can then be expanded to 1 of the 3 scenarios. Exam tip: 1 mark is awarded for the formula, but only if substitution is attempted. Marks are awarded for correct substitutions. A mark is awarded for the answer, but only if it has units. If the answer is a vector quantity direction should be included in the answer to obtain the mark, unless magnitude only is required. σ௣௜ ൌ σ௣௙ ࢓૚ ࢜૚࢏ ൅ ࢓૛ ࢜૛࢏ ൌ ࢓૚ ࢜૚ࢌ ൅ ࢓૛ ࢜૛ࢌ (Scenario 1 will work best- 2 separate objects colliding and remain two separate objects after the collision) Step 4: Substitute the values from step 2 into the formula and do the calculations. Remember to write down the units and direction of the answer. ݉஺ ‫ݒ‬஺௜ ൅ ݉஻ ‫ݒ‬஻௜ ൌ ݉஺ ‫ݒ‬஺௙ ൅ ݉஻ ‫ݒ‬஻௙ We have changed the subscript to suit our scenario. Using A and B as stated in the ݉஺ ሺͳͷሻ ൅ ݉஻ ሺെͺሻ ൌ ݉஺ ሺͷሻ ൅ ݉஻ ‫ݒ‬஻௙ question. This will help us to easily relate to our information. ݉ሺͳͷ െ ͺሻ ൌ ݉ሺͷ ൅ ‫ݒ‬஻௙ ሻ ሺͳͷ െ ͺሻ ൌ ሺͷ ൅ ‫ݒ‬஻௙ ሻ Always remember to make use of your sign conventions. The velocity of B is substituted ͹ െ ͷ ൌ ‫ݒ‬஻௙  as a negative “-“ value because it is moving towards the left. ‫ݒ‬஻௙ ൌ ʹ݉Ǥ ‫ି ݏ‬ଵ A and B have the same mass as stated in the question, we can use a maths technique and factorise mass on both sides. Our direction is towards the This will allow us to eliminate mass, hence we right because our answer is did not need a value for mass to solve the positive. And we chose in question. step 1, that motion towards the right is positive. 22

3.3.4. Impulse and momentum theorem. x Formula: ࡲࡺࡱࢀ ο࢚ ൌ ο࢖ x The SI-unit for impulse is ࡺǤ ࢙‫ࢍ࢑ݎ݋‬Ǥ ࢓Ǥ ࢙ି૚ x Impulse = change in momentum. x Impulse is a vector quantity, hence direction is important. x When writing the answer always include the units and direction. x If the question only asks for the magnitude of the impulse, you don’t have to include direction in the answer. x ο࢚, is the time interval that the net force is acting on the object. The focus will be during the collision. x We can accept that the net force will remain constant for this time interval. x The change in momentum will always be in the direction of the net force. x When using the formula for impulse, only apply it to one object because both objects will experience equal forces (magnitude) but in opposite directions. (Newton’s 3rd law). The contact time for both objects will always be the same. Therefore the impulse experienced by both objects will be the same, but in opposite directions. ሺࡲࡺࡱࢀ ο࢚ሻ࡭࢕࢔࡮ ൌ  െሺࡲࡺࡱࢀ ο࢚ሻ࡮࢕࢔࡭ x Impulse can be used to explain why safety belts, airbags, arrestor beds, crumple zones etc. are so vital in ensuring our safety. x The net force is directly proportional to the change in momentum, and ο࢖ inversely proportional to the contact time. ࡲࡺࡱࢀ ൌ , hence when the change ο࢚ in momentum remains constant and the contact time increases the net force experienced will decrease. Worked examples: b) At the US open a tennis player serves a tennis ball with mass 100 g towards his opponent with a velocity of 220 km.h-1 west. The serve is returned by the opponent at 154,8 km.h-1 . Calculate the impulse of the ball. 23

Step 1: With the use of sign conventions, choose a positive direction. West = “+” Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI-units. ૚૙૙ ݉஻௔௟௟ ൌ ͳͲͲ݃ Convert to kg: ൌ ૙ǡ ૚࢑ࢍ ૚૙૙૙ ૛૛૙ ‫ݒ‬஻௔௟௟௜ ൌ ʹʹͲ݇݉Ǥ ݄ିଵ Convert to࢓Ǥ ࢙ି૚: ൌ ૟૚ǡ ૚૚૚૚૚ ǥ࢓Ǥ ࢙ି૚ ૜ǡ૟ ૚૞૝ǡૡ ‫ݒ‬஻௔௟௟௙ ൌ ͳͷͶǡͺ݇݉Ǥ ݄ିଵ Convert to࢓Ǥ ࢙ି૚: ൌ ૝૜࢓Ǥ ࢙ି૚ ૜ǡ૟ Step 3: Choose a suitable formula from the formula sheet. HINT: If asked to calculate impulse, ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬ substitute “‫ܨ‬ோ் ο‫ ”ݐ‬with the word impulse. This will prevent you from using time and thus calculating the net force instead of the impulse. Step 4: Substitute the values from step 2 into the formula and do the calculations. Remember to write down the units and direction of the answer. ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬ We also know that: ο‫ ݌‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ The final velocity of the ball is substituted as a negative value because it is travelling in the ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ሺͲǡͳሻሺെͶ͵ሻ െ ሺͲǡͳሻሺ͸ͳǡͳͳͳͳǤ Ǥ ሻ opposite direction. ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ െͶǡ͵ െ ͸ǡͳͳͳͳ Always make sure to substitute SI-units. Mass should be in kg and velocity in ࢓Ǥ ࢙ି૚ ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ െͳͲǡͶͳ Re-write your answer as a positive value, the ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ͳͲǡͶͳܰǤ ‫ݐݏܽܧݏ‬ negative value indicates direction. We chose west = “+”, therefore a negative value will indicate that the motion is in the direction east. c) A cricket ball of mass 156 g is thrown horizontally towards a player at 37݉Ǥ ‫ି ݏ‬ଵ . It is hit back in the opposite direction with a velocity of 45݉Ǥ ‫ି ݏ‬ଵ. The ball is in contact with the bat for a period ofͲǡͲͶ‫ݏ‬. Calculate the average force exerted on the ball by the bat. 24

Step 1: With the use of sign conventions, choose a positive direction. Initial motion of ball = “+” Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI-units. ૚૞૟ ݉஻௔௟௟ ൌ ͳͷ͸݃ Convert to kg: ൌ ૙ǡ ૚૞૟࢑ࢍ ૚૙૙૙ ‫ݒ‬஻௔௟௟௜ ൌ ͵͹݉Ǥ ‫ି ݏ‬ଵ ‫ݒ‬஻௔௟௟௙ ൌ െͶͷ݉Ǥ ‫ି ݏ‬ଵ (The negative sign indicates that the ball is travelling in the opposite direction) ο‫ ݐ‬ൌ ͲǡͲͶ‫( ݏ‬Contact time) Step 3: Choose a suitable formula from the formula sheet. ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬ Step 4: Substitute the values from step 2 into the formula and do the calculations. Remember to write down the units and direction of the answer. ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬ The final velocity of the ball is substituted as a ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ negative value because it is travelling in the opposite direction. ‫ܨ‬ோ் ሺͲǡͲͶሻ ൌ ሺͲǡͳͷ͸ሻሺെͶͷሻ െ ሺͲǡͳͷ͸ሻሺ͵͹ሻ െ͹ǡͲʹ െ ͷǡ͹͹ʹ Always make sure to substitute SI-units. Mass ‫ܨ‬ோ் ൌ ͲǡͲͶ should be in kg and velocity in ࢓Ǥ ࢙ି૚ ‫ܨ‬ோ் ൌ െ͵ͳͻǡͺͲܰ Re-write your answer as a positive value, the ‫ܨ‬ோ் ൌ ͵ͳͻǡͺͲܰin the opposite direction. negative value indicates direction. We chose initial direction of motion = “+”, therefore a negative value will indicate that the motion is in the opposite direction. 25

What will the average force exerted by the ball on the bat be? ‫ܨ‬ோ் ൌ ͵ͳͻǡͺͲܰin the direction of motion. Newton’s 3rd law Only one object should be used when calculating the impulse, do not mix the information of both objects forο‫݌‬. When calculating impulse always use the information for the object ON which you want the impulse. The impulse will be the same for both objects, only in opposite directions. ሺࡲࡺࡱࢀ ο࢚ሻ࡭࢕࢔࡮ ൌ  െሺࡲࡺࡱࢀ ο࢚ሻ࡮࢕࢔࡭ 3.3.5. Elastic and inelastic collisions: x Remember to calculate the kinetic energy of every object before the collision and add the energies together. x Remember to calculate the kinetic energy of every object after the collision and add the energies together. x Remember to do these calculations separately ! x If the answers are equal for the above bullets it indicates an elastic collision. x If there is a difference in the answers for the above bullets it indicates an inelastic collision. a) In the illustration below, a car of mass 1000 kg travelling atͶͲ݉Ǥ ‫ି ݏ‬ଵ , collides head-on with a truck of mass 5000 kg moving atʹͲ݉Ǥ ‫ି ݏ‬ଵ . After the collision, the car and truck move together atͳͲ݉Ǥ ‫ି ݏ‬ଵ towards the left. Ignore the effects of friction. The car experiences a force of 100 000 N towards the left. Calculate the contact time of the collision. Step 1: With the use of sign conventions, choose a positive direction. Motion towards the right = “+” 26

Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI-units. ݉஼ ൌ ͳͲͲͲ݇݃ ݉ ் ൌ ͷͲͲͲ݇݃ ‫ݒ‬஼௜ ൌ ͶͲ݉Ǥ ‫ି ݏ‬ଵ ‫்ݒ‬௜ ൌ െʹͲ݉Ǥ ‫ି ݏ‬ଵ (The negative sign indicates that the truck is travelling towards the left) ‫ݒ‬௙ ൌ െͳͲ݉Ǥ ‫ି ݏ‬ଵ (The negative sign indicates that the truck and the car are travelling as one system towards the left after the collision) ‫ܨ‬ோ்௖௔௥ ൌ െͳͲͲͲͲͲܰ (Remember force is a vector. The negative sign indicates that the force experienced by the car is towards the left) Step 3: Choose a suitable formula from the formula sheet. ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬ Step 4: Substitute the values from step 2 into the formula and do the calculations. Remember to write down the units and direction of your answer. Remember when working with impulse you Remember when working with impulse you should only use the data of one object. In should only use the data of one object. In this example we are using the data given for this example we are using the data given for the CAR. the TRUCK. ࡲࡺࡱࢀ ο࢚ ൌ ο࢖ ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬ ࡲࡺࡱࢀ ο࢚ ൌ ࢓࢜࡯ࢌ െ ࢓࢜࡯࢏ ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉‫்ݒ‬௙ െ ݉‫்ݒ‬௜ ࡲࡺࡱࢀ ο࢚ ൌ ࢓ሺ࢜࡯ࢌ െ ࢜࡯࢏ ሻ ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉ሺ‫்ݒ‬௙ െ ‫்ݒ‬௜ ሻ ሺെ૚૙૙૙૙૙ሻο࢚ ൌ ૚૙૙૙ሺെ૚૙ െ ૝૙ሻ ሺͳͲͲͲͲͲሻο‫ ݐ‬ൌ ͷͲͲͲሺെͳͲ െ ሺെʹͲሻሻ െ૞૙૙૙૙ ͷͲͲͲͲ ο࢚ ൌ ο‫ ݐ‬ൌ െ૚૙૙૙૙૙ ͳͲͲͲͲͲ ο࢚ ൌ ૙ǡ ૞࢙ ο‫ ݐ‬ൌ Ͳǡͷ‫ݏ‬ x Note that the average force x Note that the average force experienced experienced by the car is to the left, by the truck is to the right, hence a hence a negative “-“ value for ࡲࡺࡱࢀ . positive “+“ value for ‫ܨ‬ோ் . FNET car on truck = - FNET truck on car ο࢚ሺࢉࢇ࢘ሻ ൌ  ο࢚ሺ࢚࢛࢘ࢉ࢑ሻ 27

Is the above collision elastic or inelastic? Show all your calculations . Step 1: Remember energy is a scalar quantity, hence direction is not needed. Step 2: Determine what information is given, and write it down. Always make sure you are working with the correct SI-units. ݉஼ ൌ ͳͲͲͲ݇݃ ݉ ் ൌ ͷͲͲͲ݇݃ This will be the final velocity for both the car and truck, because ‫ݒ‬஼௜ ൌ ͶͲ݉Ǥ ‫ି ݏ‬ଵ they are combined after the collision and moving as one system. ‫்ݒ‬௜ ൌ ʹͲ݉Ǥ ‫ି ݏ‬ଵ ‫ݒ‬௙ ൌ ͳͲ݉Ǥ ‫ି ݏ‬ଵ Step 3: Choose a suitable formula from the formula sheet. To determine if the collision was elastic or inelastic we have to calculate the kinetic energy of all objects before the collision, and the kinetic energy of all objects after the collision. That is why we use the phrase “the sum of” kinetic energy. It means we add all the kinetic energy of the objects before the collision, and we add all the kinetic energy of the objects after the collision. If these answers are equal it indicates an elastic collision, if the answers differ it will indicate an inelastic collision. ͳ ଶ ͳ ଶ σ‫ܧ‬௞௜௡௜௧௜௔௟  ൌ  ݉஼ ‫ݒ‬஼  ൅  ݉ ் ‫்ݒ‬ ʹ ʹ Sum of the initial kinetic Initial kinetic energy Initial kinetic energy energy of the car and of the car before the of the truck before the truck before the collision. the collision. collision. ͳ ଶ ͳ ଶ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  ݉஼ ‫ݒ‬஼  ൅  ݉ ் ‫்ݒ‬ ʹ ʹ ͳ ͳ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  ሺͳͲͲͲሻሺͶͲሻଶ  ൅  ሺͷͲͲͲሻሺʹͲሻଶ ʹ ʹ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ ͺͲͲͲͲͲ ൅ ͳͲͲͲͲͲͲ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ ͳǡͺ ൈ ͳͲ଺ ‫ ܬ‬- This is the total kinetic energy of the car and truck before the collision. ͳ σ‫ܧ‬௞௙௜௡௔௟  ൌ  ሺ݉஼ ൅ ݉ ் ሻ‫ݒ‬௙ଶ ʹ Final kinetic energy of the car and truck after the collision. Sum of the final kinetic energy of the car and We wrote it as one term because they are moving together the truck after the as one system. – They are having the same final velocity. collision. 28

ͳ σ‫ܧ‬௞௙௜௡௔௟ ൌ  ݉஼௔௡ௗ் ‫ ݒ‬ଶ ʹ ͳ σ‫ܧ‬௞௙௜௡௔௟ ൌ  ሺͳͲͲͲ ൅ ͷͲͲͲሻͳͲଶ ʹ σ‫ܧ‬௞௙௜௡௔௟ ൌ ͵ǡͲ ൈ ͳͲହ ‫ – ܬ‬This is the total kinetic energy of the car and truck after the collision. ͳǡͺ ൈ ͳͲ଺ ‫͵ > ܬ‬ǡͲ ൈ ͳͲହ ‫ܬ‬ ‫ ׵‬σ‫ ݈ܽ݅ݐ݅݊݅݇ܧ‬ൌ  σ‫݈݂ܽ݊݅݇ܧ‬ ‫ࢉ࢏࢚࢙ࢇ࢒ࢋ࢔࢏࢙࢏࢔࢕࢏࢙࢏࢒࢒࢕ࢉ ׵‬Ǥ 4. Exercises 4.1. Multiple-choice questions. x These questions will have 4 possible answers. x Only one answer will be correct, the other three answers are only distractors. x It is advisable to answer these questions at the end. If the distractors confuse you, it will show in the content questions as well. x Read the question and cover the answers, decide for yourself what you think the correct answer is. x Now uncover the options and see if one corresponds to your answer. x Choose the option that corresponds best to your answer. x Read all the options to make sure there is not a better answer than the one you have selected. x Elimination can also be used as a method – out of the four answers, eliminate those that you know cannot be correct. Then by using laws and principles of the relevant topic reason out which one is the correct answer. Multiple-choice questions: Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. 29

1. Two trolleys, X and Y, of masses m and 2m respectively, are held together by a compressed spring between them. Initially they are stationary on a horizontal floor, as shown below. Ignore the effects of friction The spring is now released and falls to the floor while the trolleys move apart. The magnitude of the MOMENTUM of trolley X while it moves away is… A zero. B half the magnitude of the momentum of trolley Y. C twice the magnitude of the momentum of trolley Y. D the same as the magnitude of the momentum of trolley Y. (2) 2. An object is dropped from rest and after falling a distance x, its momentum is p. Ignore the effects of air friction. The momentum of the object, after it has fallen a distance 2x, is… A p. B ඥʹ‫݌‬ C ‫݌‬ ʹ D 2p (2) 3. When the velocity of a moving object is doubled, and mass is kept constant, the A Net work done by the object is doubled. B Kinetic energy of the object is doubled. C Potential energy of the object is doubled. D Linear momentum of the object is doubled. (2) 4. A ball, moving horizontally, hits a wall with speed 2v. The ball then bounces back horizontally with a speed v, as shown in the diagram below. Ignore all forms of friction. 30

Which ONE of the following combinations regarding the linear momentum and the total kinetic energy of the ball for the collision above is CORRECT? LINEAR MOMENTUM TOTAL KINETIC ENERGY A Conserved Not conserved B Conserved Conserved C Not conserved Not conserved (2) D Not conserved Conserved 31

5 A person drops a glass bottle onto a concrete floor from a certain height and the bottle breaks. The person then drops a second, identical glass bottle from the same height onto a thick, woolen carpet, but the bottle does not break. Which ONE of the following is CORRECT for the second bottle compared to the first bottle for the same momentum change? AVERAGE FORCE ON SECOND TIME OF CONTACT WITH BOTTLE CARPET A Larger Smaller B Smaller Smaller C Larger Larger D Smaller Larger (2) 6 Which of the following physical quantities is equal to the product of net force and change in time? A Power B Impulse C Energy D Work (2) 7 Two bodies undergo an INELASTIC collision in the absence of friction. Which ONE of the following combinations of momentum and kinetic energy of the system is CORRECT? MOMENTUM KINETIC ENERGY A Not conserved Conserved B Conserved Not conserved C Not conserved Not conserved D Conserved Conserved (2) 8 Airbags in modern cars provide more safety during an accident. The statements below are made by a learner to explain how airbags can ensure better safety in a collision. (i) The time of impact increases (ii) The impact force decreases (iii) The impulse increases 32

Which of the statements above are CORRECT? A (i) only B (ii) only C (ii) and (iii) only D (i) and (ii) only (2) 9 The net (resultant) force acting on an object is equal to the… of the object in the direction of the net force. A change in momentum B change in kinetic energy C rate of change of momentum D rate of change of kinetic energy (2) 10 If the total momentum of a system is changing... A particles of the system must be exerting forces on each other. B the system must be under the influence of gravity. C the system must move at constant velocity. D a net external force must be acting on the system. (2) 11 The linear momentum of an object is a…. A vector quantity with the same direction as the velocity of the object. B scalar quantity with the same direction as the velocity of the object. C vector quantity with direction opposite to that of the velocity of the object. D scalar quantity with direction opposite to that of the velocity of the object. (2) 12 Ball M, moving at speed v to the right, collides with a stationary ball N on a smooth horizontal surface. Immediately after the collision, ball M comes to rest and ball N moves to the right with speed v. Which ONE of the following statements about the collision of the balls is CORRECT? A Total momentum is conserved and the masses of the balls are unequal. B Total kinetic energy is conserved and the masses of the balls are unequal. C Total momentum and total kinetic energy are conserved and the masses of the balls are equal. D Total momentum is conserved but total kinetic energy is not conserved and the masses of the balls are equal. (2) 33

13 Learners perform an experiment using identical trolleys, each of mass m. The trolleys are arranged, as shown in the diagram below. They are initially at rest on a frictionless surface and are connected with a compressed, massless spring. When the spring is released it falls vertically down and the single trolley moves with momentum p to the left. The magnitude of the momentum of the two trolleys moving to the right will be: A ʹ‫݌‬ B ‫݌‬ C ͳ ‫݌‬ ʹ D ͳ ‫݌‬ Ͷ (2) 14 An object X, mass m, is moving at constant velocity v along a smooth, horizontal surface. An identical object Y is dropped onto object X and stays there. Which one of the following represents the velocity of the combination X + Y? A Zero B ͳ ‫ݒ‬ ʹ C ‫ݒ‬ D ʹ‫ݒ‬ (2) 15 Three separate, identical blocks are in contact with each other in a straight line. They are at rest on a smooth, horizontal surface. Each of these blocks has a mass m. Another block with the same dimensions, but with mass 2m, moving at a velocity v, collides ELASTICALLY and in the same straight line with the three stationary blocks. Which one of the following diagrams represents the situation immediately after the collision? 34

At rest 2v 2m m m At rest v 2m m m At rest 2v 2m m m At rest Vv 2m m m (2) 35

4.2. Structured questions. Question 1 A 400 g ball is moving horizontally towards a cricketer at a velocity ofͶͲ݉Ǥ ‫ି ݏ‬ଵ . Then it is hit away by the cricketer in the opposite direction it had come from with a net force of 400 N. The bat makes contact with the ball forͲǡͳ‫ݏ‬. Ignore the vertical motion of the ball. ͶͲ݉Ǥ ‫ି ݏ‬ଵ 1.1 Define impulse. (2) 1.2 Calculate the initial momentum of the ball before it hits the bat. (3) 1.3 Calculate the magnitude of the final velocity of the cricket ball. (4) 1.4 When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, car A with mass 1 750 kg, traveling to the right atͳǡͷ݉Ǥ ‫ି ݏ‬ଵ , collides with car B of mass 1 450 kg going to the left atͳǡͳ݉Ǥ ‫ି ݏ‬ଵ . Measurements show that the speed of car A just after the collision has been Ͳǡʹͷ݉Ǥ ‫ି ݏ‬ଵ in its original direction. Ignore the effects of friction during the collision. 1.4.1 Calculate the velocity of car B immediately after the collision. (5) 1.4.2 “When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage.” Explain how flexible bumpers reduce damages during low speed collisions. (3) [17] Question 2 36

In a Physics laboratory, a trolley of unknown mass, m, moves along a track at a constant velocity ofͲǡͶ݉Ǥ ‫ି ݏ‬ଵ . A wooden block, mass 500 g, is released directly above and it lands on top of the trolley. The trolley, along with the block, continues moving atͲǡͳͷ݉Ǥ ‫ି ݏ‬ଵ in the same direction, as the diagram below illustrates. 2.1 Define, in words, the term momentum as used in Physics. (2) 2.2 Is the collision between the wooden block and the trolley ELASTIC or INELASTIC? Use a suitable calculation to justify the answer. (7) 2.3 Draw a velocity versus time sketch graph to illustrate the motion of the trolley BEFORE and AFTER the block lands on it. (No values required). (2) [11] Question 3 The graph below represents the NET FORCE F (versus time) on a 58 g ball that collides with a wall. The initial speed of the ball is͵Ͷ݉Ǥ ‫ି ݏ‬ଵ , perpendicular to the wall. The ball re-bounds at the same speed, perpendicular to the wall. Disregard the effect of friction. 3.1 Define the term impulse in words. (2) 3.2 For how long was the ball in contact with the wall? (1) 3.3 Calculate the magnitude of the maximum net force,‫ܨ‬ோ்ሺெ஺௑ሻ , on the ball during the collision. (4) 37

3.4 A Ͷǡʹ݃bullet, moving horizontally to the right at a velocity of͵͸Ͳ݉Ǥ ‫ି ݏ‬ଵ , strikes a stationary block with mass ݉ଵ of 1150 g. The velocity of the block immediately after the bullet has passed through it is Ͳǡͷͷ݉Ǥ ‫ି ݏ‬ଵ to the right. The bullet proceeds and strikes another stationary block with mass ݉ଶ of 1 530 g. THE BULLET REMAINS EMBEDDED IN THE SECOND BLOCK. Ignore the effects of friction. 3.4.1 State the principal of the conservation of linear momentum in words. (2) 3.4.2 Calculate the magnitude of the velocity of the second block after the bullet has been embedded in the block. (6) [15] Question 4 A bullet is fired from a stationary gun. Assume that the bullet moves horizontally. Immediately after firing, the gun recoils (moves back) with a velocity ofͳǡͶ݉Ǥ ‫ି ݏ‬ଵ . The momentum of the gun after firing the bullet is observed to beͶǡʹ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ . It is calculated that the gun is 150 times heavier than the bullet. Ignore all effects of friction. 4.1 Would you consider this to be a closed or isolated system? Give a reason for your answer. (2) 4.2 State Newton’s second law expressed in terms of momentum in words. (2) 4.3 How does the force that the bullet experience compares to the force that the gun experience? Choose from: GREATER; SMALLER OR THE SAME AS. (1) 4.4 A learner observes that the speed of the bullet is much greater than the speed at which the gun recoils. Give a reason to explain this observation. (2) 4.5 Calculate the velocity of the bullet just after it has been fired. (7) [14] 38

Question 5 A sample of xenon gas (Xe) is sealed in a closed container. The xenon atoms are all moving at high velocities and constantly colliding with each other as well as the container walls. A single xenon atom has a mass of ʹǡʹ ൈ ͳͲିଶହ ݇݃Ǥ Consider a single collision between two of the atoms. Just before the collision, Atom 1 is moving with a velocity of ʹͲͺ݉Ǥ ‫ି ݏ‬ଵ to the left and Atom 2 is moving with a velocity of ʹ͹ʹ݉Ǥ ‫ି ݏ‬ଵ to the left. 1 2 BEFORE Immediately after the two atoms collide, Atom 1 has a velocity of ʹ͹ʹ݉Ǥ ‫ି ݏ‬ଵ to the left. ? 1 2 IMMEDIATELY AFTER 5.1 State the principle of conservation of linear momentum in words. (2) 5.2 Determine the momentum of Atom 2 immediately after the collision. (5) 5.3 According to kinetic molecular theory, this collision should be elastic. Perform a suitable calculation to show that the collision between the two atoms is perfectly elastic. (4) 5.4 Determine the magnitude of the impulse experienced by Atom 1 during the collision. (3) [13] 39

Question 6 A boy on roller blades with his hands on a fully loaded trolley, mass 18 kg, moves west at ͷ݉Ǥ ‫ି ݏ‬ଵ over a frictionless surface as shown in the sketch. The boy now pushes the trolley so that he moves at ͳ݉Ǥ ‫ି ݏ‬ଵ east after this push. The mass of the boy and his roller blades is 45 kg. 6.1 State the conservation principle that is applicable during the interaction between the boy and the trolley. (2) 6.2 Calculate the velocity of the trolley directly after the boy pushed it. (4) 6.3 During the pushing motion of the boy on the trolley, the trolley experiences an impulse. How does the magnitude of the impulse that the boy experiences compare to that of the trolley? Write down INCREASES, DECREASES or REMAINS THE SAME and explain your answer. (3) 6.4 If the force exerted on the trolley lastsͲǡͶ‫ݏ‬, calculate the force that the boy exerts on the trolley. (4) [13] 40

Question 7 A 2 kg block is sliding to the right on a frictionless horizontal surface atͶ݉Ǥ ‫ି ݏ‬ଵ. A force of 2 500 N is now exerted on the block for a short period of time as indicated in the graph below. 7.1 Define the term impulse. (2) 7.2 Calculate the magnitude of the impulse on the block. (3) 7.3 Calculate the velocity of the block immediately after the force stops acting on the block if the force was exerted to: 7.3.1 The right (4) 7.3.2 The left (3) [12] Question 8 Percy, mass 75 kg, rides at ʹͲ݉Ǥ ‫ି ݏ‬ଵ on a quad bike (motorcycle with four wheels) with a mass of 100 kg. He suddenly applies the brakes when he approaches a red traffic light on a wet and slippery road. The wheels of the quad bike lock and the bike slides forward in a straight line. The force of friction causes the bike to stop in 8 s. 8.1 Define the concept momentum in words. (2) 8.2 Calculate the change in momentum of Percy and the bike, from the moment the brakes lock until the bike comes to a stop. (4) 8.3 Calculate the average frictional force exerted by the road on the wheels to stop the bike. (4) [10] 41

Question 9 The graph below shows how the momentum of car A changes with time just before and just after a head-on collision with car B. Car A has a mass of 1 500 kg, while the mass of car B is 900 kg. Car B was travelling at a constant velocity of ͳͷ݉Ǥ ‫ି ݏ‬ଵ west before the collision. Take east as positive and consider the system as isolated. 9.1 What do you understand by the term isolated system as used in physics? (1) Use the information in the graph to answer the following questions. 9.2 Calculate the: 9.2.1 Magnitude of the velocity of car A just before the collision. (3) 9.2.2 Velocity of car B just after the collision. (5) 9.2.3 Magnitude of the net average force acting on car A during the collision. (4) [13] 42

5. Answers 5.1. Answers to multiple-choice questions. 1 D 6 B 11 A 2 B 7 B 12 C 3 D 8 D 13 B 4 A 9 C 14 B 5 D 10 D 15 D 5.2. Answers to structured questions. Question 1 1.1 The product of the resultant/net force 9acting on an object and the time the resultant/net force acts on the object.9 (2) Option 1 Option 2 Towards the cricketer = “+” Away from cricketer = “+” 1.2 ‫ ݌‬ൌ ݉‫ ݒ‬9 ‫ ݌‬ൌ ݉‫ ݒ‬9 ‫ ݌‬ൌ ሺͲǡͶሻሺͶͲሻ 9 ‫ ݌‬ൌ ሺͲǡͶሻሺെͶͲሻ 9 ‫ ݌‬ൌ ͳ͸݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ Towards the ‫ ݌‬ൌ െͳ͸݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ cricketer 9 ‫ ݌ ׵‬ൌ ͳ͸݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ Towards the cricketer 9 (3) 1.3 Option 1 Option 2 Towards the cricketer = “+” Away from cricketer = “+” ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫ ݌‬9 ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫ ݌‬9 ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ ሺെͶͲͲሻሺͲǡͳሻ9 ൌ ͲǡͶሺ‫ݒ‬௙ െ ͶͲሻ 9 ሺͶͲͲሻሺͲǡͳሻ9 ൌ ͲǡͶሺ‫ݒ‬௙ െ ሺെͶͲሻሻ 9 െͳͲͲ ൌ ‫ݒ‬௙ െ ͶͲ ͳͲͲ ൌ ‫ݒ‬௙ ൅ ͶͲ ‫ݒ‬௙ ൌ െ͸Ͳ݉Ǥ ‫ି ݏ‬ଵ ‫ݒ‬௙ ൌ ͸Ͳ݉Ǥ ‫ି ݏ‬ଵ ‫ݒ ׵‬௙ ൌ ͸Ͳ݉Ǥ ‫ି ݏ‬ଵ 9 ‫ݒ ׵‬௙ ൌ ͸Ͳ݉Ǥ ‫ି ݏ‬ଵ 9 (4) We don’t have to indicate direction in the answer for 1.3 because the question only asked for MAGNITUDE. 43

1.4.1 Option 1: Towards the right as positive σ୮୧ ൌ σ୮୤ 9 ݉஺ ‫ݒ‬஺௜ ൅ ݉஻ ‫ݒ‬஻௜ ൌ ݉஺ ‫ݒ‬஺௙ ൅ ݉஻ ‫ݒ‬஻௙ ሺͳ͹ͷͲሻሺͳǡͷሻ ൅ ሺͳͶͷͲሻሺെͳǡͳሻ9 ൌ ሺͳ͹ͷͲሻሺͲǡʹͷሻ ൅ ሺͳͶͷͲሻ‫ݒ‬஻௙ 9 ͷͻʹǤͷ ൌ ሺͳͶͷͲሻ‫ݒ‬஻௙ (5) ‫ݒ‬஻௙ ൌ ͲǡͶͳ݉Ǥ ‫ି ݏ‬ଵ 9 to the right. 9 Option 2: Towards the left as positive σ୮୧ ൌ σ୮୤ ݉஺ ‫ݒ‬஺௜ ൅ ݉஻ ‫ݒ‬஻௜ ൌ ݉஺ ‫ݒ‬஺௙ ൅ ݉஻ ‫ݒ‬஻௙ ሺͳ͹ͷͲሻሺെͳǡͷሻ ൅ ሺͳͶͷͲሻሺͳǡͳሻ9 ൌ ሺͳ͹ͷͲሻሺെͲǡʹͷሻ ൅ ሺͳͶͷͲሻ‫ݒ‬஻௙ 9 െͷͻʹǤͷ ൌ ሺͳͶͷͲሻ‫ݒ‬஻௙ (5) ‫ݒ‬஻௙ ൌ െͲǡͶͳ݉Ǥ ‫ି ݏ‬ଵ ‫ݒ ׵‬஻௙ ൌ ͲǡͶͳ݉Ǥ ‫ି ݏ‬ଵ 9 to the right. 9 1.4.2 Flexible bumpers reduce the net force 9by increasing the time 9required to bring about the same change in momentum. 9 OR ο௣ According to ‫ܨ‬ோ் ൌ 9; For the same change in momentum9 ο௧ (3) increasing ο‫ ݐ‬decreases the net force. 9 [17] 44

Question 2 2.1 The product of an object’s mass and its velocity. 99 (2) 2.2 We have to calculate the mass of the trolley first. – We need this value to calculate the kinetic energy, which will tell us if the collision was elastic or inelastic. Hints: σ୮୧ ൌ σ୮୤ The motion of the trolley is taken ݉ ் ‫்ݒ‬௜ ൅ ݉஻ ‫ݒ‬஻௜ ൌ ሺ݉ ் ൅ ݉஻ ሻ‫ݒ‬௙ as the positive direction. The mass of the block must be ݉ ் ሺͲǡͶሻ ൅ ሺͲǡͷሻሺͲሻ ൌ ሺ݉ ் ൅ ͲǡͷሻሺͲǡͳͷሻ 9 converted to kg. ͲǡͶ݉ ் ൌ Ͳǡͳͷ݉ ் ൅ ͲǡͲ͹ͷ The block has a velocity of 0 ݉Ǥ ‫ି ݏ‬ଵ before it is dropped. ݉ ் ൌ Ͳǡ͵݇݃. 9 ଵ ଶ ଵ ଶ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  ଶ ݉ ் ‫்ݒ‬  ൅  ଶ ݉஻ ‫ݒ‬஻ 9 (Before the block is dropped.) ͳ ͳ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  ሺͲǡ͵ሻሺͲǡͶሻଶ  ൅  ሺͲǡͷሻሺͲሻଶ ʹ ʹ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ ͲǡͲʹͶ‫ ܬ‬9 ଵ σ‫ܧ‬௞௙௜௡௔௟ ൌ  ଶ ሺ݉ ் ൅ ݉஻ ሻ‫ݒ‬௙ଶ (After the block is dropped on the trolley) ͳ σ‫ܧ‬௞௙௜௡௔௟ ൌ  ሺͲǡ͵ ൅ ͲǡͷሻሺͲǡͳͷሻଶ ʹ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ ͲǡͲͲͻ‫ ܬ‬9 ‫ ׵‬σ‫ ݈ܽ݅ݐ݅݊݅݇ܧ‬ൌ  σ‫ ݈݂ܽ݊݅݇ܧ‬9 OR σ‫ܧ‬௞௜௡௜௧௜௔௟ is greater than σ‫ܧ‬௞௙௜௡௔௟ ‫ܿ݅ݐݏ݈ܽ݁݊݅ݏ݅݊݋݅ݏ݈݈݅݋ܿ ׵‬9ǡ ݇݅݊݁‫݀݁ݒݎ݁ݏ݊݋ܿݐ݋݊ݏ݅ݕ݃ݎ݁݊݁ܿ݅ݐ‬Ǥ (7) 2.3 GRAPH OF VELOCITY VS TIME 9 9 (2) [11] 45

Question 3 3.1 The product of the resultant/net force 9acting on an object and the time the resultant/net force acts on the object.9 (2) 3.2 ିଷ ͸ ൈ ͳͲ ‫ ݏ‬9 OR ͲǡͲͲ͸‫ݏ‬ (1) 3.3 Hint: Remember the question asks to calculate the maximum force. If you only calculate ‫ܨ‬ோ் it will represent the average force between the ball and the wall during the contact time. Option 1: ‫ܨ‬ோ்ሺெ஺௑ሻ ο‫ ݐ‬ൌ ο‫݌‬ 9 ‫ܨ‬ோ்ሺெ஺௑ሻ ο‫ ݐ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ ଵ ሺʹ ൈ ͳͲିଷ ൅ ͸ ൈ ͳͲିଷ ሻ‫ܨ‬ோ்ሺெ஺௑ሻ ൌ ͲǡͲͷͺሺ͵Ͷ 9 െ ሺെ͵Ͷሻሻ9 9 ଶ Area of Trapezium ͳ ‫ܣ‬ൌ ሺܽ ൅ ܾሻ݄ ʹ ‫ܶܧܰܨ‬ሺ‫ܺܣܯ‬ሻ ൌ ͻͺ͸ܰ 9 Option 2: Impulse = area under the graph Impulse = ο‫݌‬ ‫ ׵‬ο‫ ݌‬ൌ ܽ‫݄݌ܽݎ݄݃݁ݐݎ݁݀݊ݑܽ݁ݎ‬ ଵ ο‫ ݌‬ൌ ሺܽ ൅ ܾሻ݄ 9 ଶ ͳ ݉൫‫ݒ‬௙ െ ‫ݒ‬௜ ൯ ൌ ሺܽ ൅ ܾሻ݄ ʹ ଵ ሺͲǡͲͷͺሻሺ͵Ͷሻ െ ሺͲǡͲͷͺሻሺെ͵Ͷሻ9 ൌ ሺʹ ൈ ͳͲെ͵ ൅ ͸ ൈ ͳͲെ͵ ሻ‫ܶܧܰܨ‬ሺ‫ܺܣܯ‬ሻ9 ଶ ‫ܶܧܰܨ‬ሺ‫ܺܣܯ‬ሻ ൌ ͻͺ͸ܰ 9 Option 3: If you don’t feel comfortable working with the area of a trapezium, divide the graph up into easier shapes. For this graph we will use 2 triangles and 1 rectangle. Impulse = area under the graph Impulse = ο‫݌‬ ‫ ׵‬ο‫ ݌‬ൌ ܽ‫݄݌ܽݎ݄݃݁ݐݎ݁݀݊ݑܽ݁ݎ‬ ଵ ଵ ο‫ ݌‬ൌ ܾ݄ ൅ ሺ݈ ൈ ܾሻ ൅ ܾ݄ 9 ଶ ଶ ͳ ͳ ݉൫‫ݒ‬௙ െ ‫ݒ‬௜ ൯ ൌ ܾ݄ ൅ ሺ݈ ൈ ܾሻ ൅ ܾ݄ ʹ ʹ 46

1 1 ሺ0,058ሻሺ34ሻ-ሺ0,058ሻሺ-34ሻ 9= ൫2×10-3 ൯FNET ሺMAXሻ +൫2×10-3 ൯൫FNET ሺMAXሻ ൯+ ൫2×10-3 ൯FNET ሺMAXሻ 9 2 2 ‫ܶܧܰܨ‬ሺ‫ܺܣܯ‬ሻ ൌ ͻͺ͸ܰ 9 (4) 3.4.1 The TOTAL linear momentum in an isolated system remains constant (is conserved). 9 9 (2) 3.4.2 Option 1: (Right positive) + σ୮୧ ൌ σ୮୤ 9 ݉஻௨௟௟௘௧ ‫ݒ‬஻௨௟௟௘௧௜ ൅ ݉஻௟௢௖௞ଵ ‫ݒ‬஻௟௢௖௞ଵ௜ ൌ ݉஻௨௟௟௘௧ ‫ݒ‬஻௨௟௟௘௧௙ ൅ ݉஻௟௢௖௞ଵ ‫ݒ‬஻௟௢௖௞ଵ௙ ሺͲǡͲͲͶʹሻሺ͵͸Ͳሻ ൅ ሺͳǡͳͷሻሺͲሻ9 ൌ ሺͲǡͲͲͶʹሻ‫ݒ‬஻௨௟௟௘௧௙ ൅ ሺͳǡͳͷሻሺͲǡͷͷሻ 9 ͳǡͷͳʹ െ Ͳǡ͸͵ʹͷ ൌ ͲǡͲͲͶʹ‫ݒ‬஻௨௟௟௘௧௙ ‫ݒ‬஻௨௟௟௘௧௙ ൌ ʹͲͻǡͶͲ݉Ǥ ‫ି ݏ‬ଵ (This will be the velocity of the bullet after it passes through block 1). Hint: The question states we should ignore friction, hence we can accept that the velocity with which the bullet leaves block 1 will be the same velocity with which the bullet strikes block 2. σ୮୧ ൌ σ୮୤ ݉஻௨௟௟௘௧ ‫ݒ‬஻௨௟௟௘௧௜ ൅ ݉஻௟௢௖௞ଶ ‫ݒ‬஻௟௢௖௞ଶ௜ ൌ ሺ݉஻௨௟௟௘௧ ൅ ݉஻௟௢௖௞ଶሻ ‫ݒ‬௙ ሺͲǡͲͲͶʹሻሺʹͲͻǡͶͲሻ ൅ ሺͳǡͷ͵ሻሺͲሻ9 ൌ ሺͲǡͲͲͶʹ ൅ ͳǡͷ͵ሻ‫ݒ‬௙ 9 Ͳǡͺ͹ͻͶͺ ൌ ሺͳǡͷ͵Ͷʹሻ‫ݒ‬௙ No direction is needed with the answer since the question only asked for magnitude. ‫ݒ‬௙ ൌ Ͳǡͷ͹݉Ǥ ‫ି ݏ‬ଵ 9 47

Option 2: Left positive + σ୮୧ ൌ σ୮୤ 9 ݉஻௨௟௟௘௧ ‫ݒ‬஻௨௟௟௘௧௜ ൅ ݉஻௟௢௖௞ଵ ‫ݒ‬஻௟௢௖௞ଵ௜ ൌ ݉஻௨௟௟௘௧ ‫ݒ‬஻௨௟௟௘௧௙ ൅ ݉஻௟௢௖௞ଵ ‫ݒ‬஻௟௢௖௞ଵ௙ ሺͲǡͲͲͶʹሻሺെ͵͸Ͳሻ ൅ ሺͳǡͳͷሻሺͲሻ9 ൌ ሺͲǡͲͲͶʹሻ‫ݒ‬஻௨௟௟௘௧௙ ൅ ሺͳǡͳͷሻሺെͲǡͷͷሻ 9 െͳǡͷͳʹ ൅ Ͳǡ͸͵ʹͷ ൌ ͲǡͲͲͶʹ‫ݒ‬஻௨௟௟௘௧௙ ‫ݒ‬஻௨௟௟௘௧௙ ൌ െʹͲͻǡͶͲ݉Ǥ ‫ି ݏ‬ଵ (This will be the velocity of the bullet after it passes through block 1). σ୮୧ ൌ σ୮୤ ݉஻௨௟௟௘௧ ‫ݒ‬஻௨௟௟௘௧௜ ൅ ݉஻௟௢௖௞ଶ ‫ݒ‬஻௟௢௖௞ଶ௜ ൌ ሺ݉஻௨௟௟௘௧ ൅ ݉஻௟௢௖௞ଶሻ ‫ݒ‬௙ ሺͲǡͲͲͶʹሻሺെʹͲͻǡͶͲሻ ൅ ሺͳǡͷ͵ሻሺͲሻ9 ൌ ሺͲǡͲͲͶʹ ൅ ͳǡͷ͵ሻ‫ݒ‬௙ 9 െͲǡͺ͹ͻͶͺ ൌ ሺͳǡͷ͵Ͷʹሻ‫ݒ‬௙ ‫ݒ‬௙ ൌ െͲǡͷ͹݉Ǥ ‫ି ݏ‬ଵ No direction is needed with the answer since the question only asked for magnitude. ‫ݒ ׵‬௙ ൌ Ͳǡͷ͹݉Ǥ ‫ି ݏ‬ଵ 9 (6) [15] Question 4 4.1 Yes9, all effects of friction can be ignored. 9 (No external forces acting on the system) (2) OR The net external force acting on the system is equal to zero. 4.2 The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force. 99 (2) 4.3 THE SAME AS 9 (1) 4.4 The mass of the bullet is much smaller9 than the mass of the gun, both the gun and bullet will experience the same change in momentum9, hence the velocity of the bullet should be greater than the recoil speed of the gun. (2) 48

4.5 Consider the motion towards the right as “+” ‫ீ݌‬௙ ൌ ݉‫ݒ‬ െͶǡʹ ൌ ݉ሺെͳǡͶሻ ݉ ൌ ͵݇݃9 (This is the mass of the Gun) We know the ratio of bullet mass to gun mass: 1 : 150 9 ‫׷ ܾ݈݈݁݅ݓݐ݈݈݁ݑܾ݄݁ݐ݂݋ݏݏ݄ܽ݉݁ݐ݃݇͵ݏ݅݊ݑ݄݃݁ݐ݂݋ݏݏ݄ܽ݉݁ݐܨܫ ׵‬ ଷ ൌ ͲǡͲʹ݇݃ (using the ratio) ଵହ଴ σ୮୧ ൌ σ୮୤ 9 ሺ݉ீ ൅ ݉஻ ሻ‫ݒ‬௜ ൌ ݉ீ ‫ீݒ‬௙ ൅ ݉஻ ‫ݒ‬஻௙ ሺ݉ீ ൅ ݉஻ ሻͲ9 ൌ െͶǡʹ ൅ ሺͲǡͲʹሻ‫ݒ‬஻௙ 9 (7) Ͷǡʹ ‫ݒ‬஻௙ ൌ ͲǡͲʹ ‫ݒ‬஻௙ ൌ ʹͳͲ݉Ǥ ‫ି ݏ‬ଵ Ǥ 9 Towards the right. 9 [14] Question 5 5.1 The TOTAL linear momentum in an isolated system remains constant (is conserved). 99 (2) 5.2 Take right as positive. + σ୮୧ ൌ σ୮୤ 9 ݉ଵ ‫ݒ‬ଵ௜ ൅ ݉ଶ ‫ݒ‬ଶ௜ ൌ ݉ଵ ‫ݒ‬ଵ௙ ൅ ݉ଶ ‫ݒ‬ଶ௙ 9 ሺʹǡʹ ൈ ͳͲିଶହ ሻሺെʹͲͺሻ ൅ ሺʹǡʹ ൈ ͳͲିଶହ ሻሺെʹ͹ʹሻ ൌ ሺʹǡʹ ൈ ͳͲିଶହ ሻሺെʹ͹ʹሻ ൅ ‫݌‬௙ െͳǡͲͷ͸ ൈ ͳͲିଶଶ ൌ െͷǡͻͺͶ ൈ ͳͲିଶଷ ൅ ‫݌‬௙ ‫݌‬௙ ൌ െͶǡͷͺ ൈ ͳͲିଶଷ ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ  ‫݌‬௙ ൌ Ͷǡͷͺ ൈ ͳͲିଶଷ ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ 9 ‫ݐ݂݈݄݁݁ݐݏ݀ݎܽݓ݋ݐ‬Ǥ 9 49

OR Take left as positive. + σ୮୧ ൌ σ୮୤ 99 ݉ଵ ‫ݒ‬ଵ௜ ൅ ݉ଶ ‫ݒ‬ଶ௜ ൌ ݉ଵ ‫ݒ‬ଵ௙ ൅ ݉ଶ ‫ݒ‬ଶ௙ ሺʹǡʹ ൈ ͳͲିଶହ ሻሺʹͲͺሻ ൅ ሺʹǡʹ ൈ ͳͲିଶହ ሻሺʹ͹ʹሻ ൌ ሺʹǡʹ ൈ ͳͲିଶହ ሻሺʹ͹ʹሻ ൅ ‫݌‬௙ ͳǡͲͷ͸ ൈ ͳͲିଶଶ ൌ ͷǡͻͺͶ ൈ ͳͲିଶଷ ൅ ‫݌‬௙ ‫݌‬௙ ൌ Ͷǡͷͺ ൈ ͳͲିଶଷ ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ 9 ‫ݐ݂݈݄݁݁ݐݏ݀ݎܽݓ݋ݐ‬Ǥ 9 (5) 5.3 ଵ ଵ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  ଶ ݉ଵ ‫ݒ‬ଵଶ  ൅  ଶ ݉ଶ ‫ݒ‬ଶ ଶ 9 (Before the collision.) ଵ ଵ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ  ሺʹǡʹ ൈ ͳͲିଶହ ሻሺʹͲͺሻଶ ൅ ሺʹǡʹ ൈ ͳͲିଶହ ሻሺʹ͹ʹሻଶ 9 ଶ ଶ σ‫ܧ‬௞௜௡௜௧௜௔௟ ൌ ͳǡʹͻ ൈ ͳͲିଶ଴ ‫ܬ‬ ଵ ଵ σ‫ܧ‬௞௙௜௡௔௟ ൌ  ଶ ݉ଵ ‫ݒ‬ଵଶ  ൅  ଶ ݉ଶ ‫ݒ‬ଶ ଶ (After the collision.) ଵ ଵ σ‫ܧ‬௞௙௜௡௔௟ ൌ  ሺʹǡʹ ൈ ͳͲିଶହ ሻሺʹ͹ʹሻଶ ൅  ሺʹǡʹ ൈ ͳͲିଶହ ሻሺʹͲͺሻଶ 9 ଶ ଶ σ‫ܧ‬௞௙௜௡௔௟ ൌ ͳǡʹͻ ൈ ͳͲିଶ଴ ‫ܬ‬ ‫ ׵‬σ‫ ݈ܽ݅ݐ݅݊݅݇ܧ‬ൌ  σ‫ ݈݂ܽ݊݅݇ܧ‬9 ‫ࢉ࢏࢚࢙ࢇ࢒ࢋ࢙࢏࢔࢕࢏࢙࢏࢒࢒࢕ࢉࢋࢎࢀ ׵‬Ǥ (4) Remember atoms are 5.4 ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫ ݌‬9 moving towards the left, ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ because we chose right as positive in this option, we ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ should substitute the ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ሺʹǡʹ ൈ ͳͲିଶହ ሻሺെʹ͹ʹ െ ሺെʹͲͺሻሻ 9 velocities as negative. ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ െͳǡͶͳ ൈ ͳͲିଶଷ ܰǤ ‫ݏ‬ (3) ିଶଷ ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ͳǡͶͳ ൈ ͳͲ ܰǤ ‫ ݏ‬9 We can determine the velocity It is allowed to use of Atom 2 after the collision in [13] the unit ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ the following manner. ‫݌‬௙ ൌ ݉‫ݒ‬௙ We do not need to include direction, the Ͷǡͷͺ ൈ ͳͲିଶଷ ൌ ሺʹǡʹ ൈ ͳͲିଶହ ሻ‫ݒ‬௙ question only asked for magnitude. ‫ݒ‬௙ ൌ ʹͲͺ݉Ǥ ‫ି ݏ‬ଵ 50

Question 6 6.1 The TOTAL linear momentum in an isolated system remains constant (is conserved). 99 You may note how many times in questions this definition is asked, just an indication of its importance – make sure to study it for the exam. (2) 6.2 Option 1: West taken as “+” σ୮୧ ൌ σ୮୤ 9 ݉ଵ ‫ݒ‬ଵ௜ ൅ ݉ଶ ‫ݒ‬ଶ௜ ൌ ݉ଵ ‫ݒ‬ଵ௙ ൅ ݉ଶ ‫ݒ‬ଶ௙ ሺͳͺ ൅ Ͷͷሻሺͷሻ9 ൌ Ͷͷሺെͳሻ ൅ ሺͳͺሻ‫ݒ‬௙ 9 ‫ݒ‬௙ ൌ ʹͲ݉Ǥ ‫ି ݏ‬ଵ ܹ݁‫ݐݏ‬Ǥ 9 Option 2: East taken as “+” σ୮୧ ൌ σ୮୤ 9 ݉ଵ ‫ݒ‬ଵ௜ ൅ ݉ଶ ‫ݒ‬ଶ௜ ൌ ݉ଵ ‫ݒ‬ଵ௙ ൅ ݉ଶ ‫ݒ‬ଶ௙ ሺͳͺ ൅ Ͷͷሻሺെͷሻ9 ൌ Ͷͷሺͳሻ ൅ ሺͳͺሻ‫ݒ‬௙ 9 (4) ‫ݒ‬௙ ൌ െʹͲ݉Ǥ ‫ି ݏ‬ଵ  ‫ݒ ׵‬௙ ൌ ʹͲ݉Ǥ ‫ି ݏ‬ଵ ܹ݁‫ݐݏ‬Ǥ 9 6.3 Remains the same, 9 The force experienced by both the boy and the trolley is the same 9– Newton’s 3rd law. The change in momentum ሺο‫݌‬ሻ remains constant. 9 OR ο‫݌‬௧௥௢௟௟௘௬ ൌ െο‫݌‬஻௢௬ 9; ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ο‫݌‬9 (3) (NOTE: Always answer the question first!) 6.4 Option 1: Information of trolley is used. Take West as “+” ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫ ݌‬9 ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ ‫்ܨ‬௥௢௟௟௘௬ ሺͲǡͶሻ9 ൌ ͳͺሺʹͲ െ ͷሻ 9 OR ‫்ܨ‬௥௢௟௟௘௬ ሺͲǡͶሻ9 ൌ ͳͺሺʹͲሻ െ ͳͺሺͷሻ 9 ‫்ܨ‬௥௢௟௟௘௬ ൌ ͸͹ͷܹܰ݁‫ݐݏ‬Ǥ 9 Option 2: Information of trolley is used. Take East as “+” ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫ ݌‬9 ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ ‫்ܨ‬௥௢௟௟௘௬ ሺͲǡͶሻ9 ൌ ͳͺሺെʹͲ െ ሺെͷሻሻ 9ࡻࡾ‫்ܨ‬௥௢௟௟௘௬ ሺͲǡͶሻ ൌ ͳͺሺെʹͲሻ െ ͳͺሺെͷሻ ‫்ܨ‬௥௢௟௟௘௬ ൌ െ͸͹ͷܰ ‫்ܨ‬௥௢௟௟௘௬ ൌ ͸͹ͷܹܰ݁‫ݐݏ‬Ǥ 9 51

Option 3: Information of boy is used. Take West as “+” ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫ ݌‬9 Remember the question asked the force ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ exerted on the trolley. Because of ‫ܨ‬஻௢௬ ሺͲǡͶሻ9 ൌ Ͷͷሺെͳ െ ͷሻ 9 Newton’s 3rd law we know that the force experienced by the boy will be the same ‫ܨ‬஻௢௬ ൌ െ͸͹ͷܰ as the force experienced by the trolley but in the opposite direction. For this option ‫ܨ‬஻௢௬ ൌ ͸͹ͷܰ‫ݐݏܽܧ‬Ǥ we will have to write a conclusion answer stating the force of the trolley. (4) ‫ ݕ݈݈݁݋ݎܶܨ  ׵‬ൌ ͸͹ͷܹܰ݁‫ݐݏ‬Ǥ 9 [13] Question 7 7.1 The product of the resultant/net force 9acting on an object and the time the resultant/net force acts on the object.9 (2) 7.2 Option 1: ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ‫ܨ‬ோ் ο‫ ݐ‬9 Hint: The difference in time ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ሺʹͷͲͲሻሺͳ ൈ ͳͲ ሻ 9 ିଷ from 10 to 11 is ૚ ൈ ૚૙ି૜ s. (x-axis of graph) ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ʹǡͷܰǤ ‫ ݏ‬9 Option 2: ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ܽ‫݄݌ܽݎ݄݃݁ݐݎ݁݀݊ݑܽ݁ݎ‬ We don’t need to indicate ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ݈ ൈ ܾ9 the direction, the question only asks for the magnitude. ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ሺʹͷͲͲሻሺͳ ൈ ͳͲିଷ ሻ 9 ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ʹǡͷܰǤ ‫ ݏ‬9 (3) 7.3.1 Option 1: Right as positive ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ 9 Remember!! ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ ‫ ݁ݏ݈ݑ݌݉ܫ‬ൌ ‫ܨ‬ோ் ο‫ݐ‬ ሺʹǡͷሻ9 ൌ ʹሺ‫ݒ‬௙ െ Ͷሻ 9 You calculated the impulse in 7.2 ‫ݒ‬௙ ൌ ͷǡʹͷ݉Ǥ ‫ି ݏ‬ଵ ‫ݐ݄݃݅ݎ‬Ǥ 9 Option 2: Left as positive ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ 9 ሺെʹǡͷሻ9 ൌ ʹሺ‫ݒ‬௙ െ ሺെͶሻሻ 9 ‫ݒ‬௙ ൌ െͷǡʹͷ݉Ǥ ‫ି ݏ‬ଵ  ‫ݒ‬௙ ൌ ͷǡʹͷ݉Ǥ ‫ି ݏ‬ଵ ‫ݐ݄݃݅ݎ‬Ǥ 9 (4) 7.3.2 Option 1: Right as positive ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ ሺെʹǡͷሻ9 ൌ ʹሺ‫ݒ‬௙ െ Ͷሻ 9 ‫ݒ‬௙ ൌ ʹǡ͹ͷ݉Ǥ ‫ି ݏ‬ଵ ‫ݐ݄݃݅ݎ‬Ǥ 9 52

Option 2: Left as positive ‫ܨ‬ோ் ο‫ ݐ‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ ሺെʹǡͷሻ9 ൌ ʹሺ‫ݒ‬௙ െ ሺെͶሻሻ 9 ‫ݒ‬௙ ൌ െʹǡ͹ͷ݉Ǥ ‫ି ݏ‬ଵ  ‫ݒ‬௙ ൌ ʹǡ͹ͷ݉Ǥ ‫ି ݏ‬ଵ ‫ݐ݄݃݅ݎ‬Ǥ 9 (3) [12] Question 8 8.1 Momentum is the product of an object’s mass and its velocity. 99 (2) 8.2 Option 1: Direction of motion is taken as positive. ο‫ ݌‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ 9 Always write your final answer as a positive value, the negative sign only indicates that ο‫ ݌‬ൌ ሺͳ͹ͷሻሺͲ െ ʹͲሻ 9 the motion was in the opposite direction. ο‫ ݌‬ൌ  െ͵ͷͲͲ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ ο‫ ݌‬ൌ ͵ͷͲͲ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ 9‘’’‘•‹–‡–Š‡†‹”‡ –‹‘‘ˆ‘–‹‘Ǥ 9 Option 2: Direction of motion is taken as negative. ο‫ ݌‬ൌ ݉ሺ‫ݒ‬௙ െ ‫ݒ‬௜ ሻ 9 ο‫ ݌‬ൌ ሺͳ͹ͷሻሺͲ െ ሺെʹͲሻሻ 9 ο‫ ݌‬ൌ ͵ͷͲͲ݇݃Ǥ ݉Ǥ ‫ି ݏ‬ଵ 9‘’’‘•‹–‡–Š‡†‹”‡ –‹‘‘ˆ‘–‹‘Ǥ9 (4) 8.3 Option 1: Direction of motion is taken as positive. ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬9 ݂ሺͺሻ ൌ െ͵ͷͲͲ9 ݂ ൌ െͶ͵͹ǡͷܰ ݂ ൌ Ͷ͵͹ǡͷܰ9‫݊݋݅ݐ݋݂݉݋݊݋݅ݐܿ݁ݎ݅݀݋ݐ݁ݐ݅ݏ݋݌݌݋‬Ǥ 9 Option 2: Direction of motion is taken as negative. ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬9 ݂ሺͺሻ ൌ ͵ͷͲͲ9 ݂ ൌ Ͷ͵͹ǡͷܰ9‫݊݋݅ݐ݋݂݉݋݊݋݅ݐܿ݁ݎ݅݀݋ݐ݁ݐ݅ݏ݋݌݌݋‬Ǥ 9 (4) [10] Question 9 9.1 A system on which the resultant/net external force is zero9 (1) 9.2.1 Option 1: Option 2: ‫ ݌‬ൌ ݉‫ݒ‬9 ο‫ ݌‬ൌ ݉‫ݒ‬௙ െ ݉‫ݒ‬௜ 9 ͵ͲͲͲͲ ൌ ሺͳͷͲͲሻ‫ݒ‬9 Ͳ ൌ ሺͳͷͲͲሻ‫ݒ‬௙ െ ͵ͲͲͲͲ9 ‫ ݒ‬ൌ ʹͲ݉Ǥ ‫ି ݏ‬ଵ 9 ‫ ݒ‬ൌ ʹͲ݉Ǥ ‫ି ݏ‬ଵ 9 (3) 53

9.2.2 Option 1: Remember the instruction was that east must be taken as positive. σ‫݌‬௜ ൌ σ‫݌‬௙ 9 ݉஺ ‫ݒ‬஺௜ ൅ ݉஻ ‫ݒ‬஻௜ ൌ ݉஺ ‫ݒ‬஺௙ ൅ ݉஻ ‫ݒ‬஻௙ ͵ͲͲͲͲ ൅ ሺͻͲͲሻሺെͳͷሻ9 ൌ ͳͶͲͲͲ ൅ ͻͲͲ‫ݒ‬஻ 9 ‫ݒ‬஻ ൌ ʹǡ͹ͺ݉Ǥ ‫ି ݏ‬ଵ East.9 Option 2: Remember the instruction was that east must be taken as positive. ο‫݌‬஺ ൌ െο‫݌‬஻ 9 ‫݌‬஺௙ െ ‫݌‬஺௜ ൌ െሺ݉஻ ‫ݒ‬஻௙ െ ݉஻ ‫ݒ‬஻௜ ሻ ͳͶͲͲͲ െ ͵ͲͲͲͲ9 ൌ ͻͲͲ‫ݒ‬஻௙ െ ͻͲͲሺെͳͷሻ9 ‫ݒ‬஻௙ ൌ ʹǡ͹ͺ݉Ǥ ‫ି ݏ‬ଵ East.9 (4) 9.2.3 Option 1: ο‫݌‬ ݈ܵ‫ ݁݌݋‬ൌ ൌ ‫ܨ‬ோ் 9 ο‫ݐ‬ ሺͳͶͲͲͲ െ ͵ͲͲͲͲሻ9 ݈ܵ‫ ݁݌݋‬ൌ ሺʹͲǡʹ െ ʹͲǡͳሻ9 ݈ܵ‫ ݁݌݋‬ൌ െͳ͸ͲͲͲͲܰ ‫ܨ ׵‬ோ் ൌ ͳ͸ͲͲͲͲܰ9 Option 2: ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬9 Remember: There is no need to show direction ‫ܨ‬ோ் ሺͲǡͳሻ9 ൌ ͳͶͲͲͲ െ ͵ͲͲͲͲ9 because the question only ‫ܨ‬ோ் ൌ െͳ͸ͲͲͲͲܰ asked for the magnitude. ‫ܨ ׵‬ோ் ൌ ͳ͸ͲͲͲͲܰ9 Option 3: ‫ܨ‬ோ் ο‫ ݐ‬ൌ ο‫݌‬9 ‫ܨ‬ோ் ሺͲǡͳሻ9 ൌ ͻͲͲሾሺʹǡ͹ͺሻ െ ሺെͳͷሻሿ ‫ܨ‬ோ் ൌ െͳ͸ͲͲʹͲܰ ‫ܨ‬஺ ൌ െ‫ܨ‬஻ ‫ܨ‬ோ் ൌ ͳ͸ͲͲʹͲܰ 54

6. MESSAGE TO GRADE 12 LEARNERS Allow me this opportunity to wish you the best of luck during the upcoming exams. There will be times where you are stressed and confused, it may feel like a hopeless situation and that you will not be able to get the job done – we understand that, but keep going. Nothing in life that is worth having will come easy. We are behind you, cheering you on to the finish line. Remember the following quotes from two brilliant scientists. Nothing in life is to be feared, it is only to be understood. Now is the time to understand more, so that we may fear less. ― Marie Curie Life is like riding a bicycle. To keep your balance you must keep moving. – Albert Einstein We know you can do this, good luck! 7. ACKNOWLEDGEMENTS The Department of Basic Education (DBE) gratefully acknowledges the following officials for giving up their valuable time and families and for contributing their knowledge and expertise to develop this resource booklet for the children of our country, under very stringent conditions of COVID-19: Writer: Mr Danie du Toit (Subject Specialist, PED: Limpopo). Reviewers: Mr SN Nkosi, Ms N September, Ms Z Letile, Mr SJ Sibanyoni and Ms A Smit DBE Subject Specialist: Ms JSK Maharaj (Veena) The development of the Study Guide was managed and coordinated by Ms Cheryl Weston and Dr Sandy Malapile 8. Reference Examination guidelines 2014 Siyavula Gr.12 textbook Mind the gap Gr. 12 textbook CAPS document Previous question papers from DBE Previous question papers from Provincial departments Phet simulations. Science clinic DBE v3.1 55

Uy, U; / Mi Mi Ui ISBN: 978-1-4315-3525-5 SSSA S So ae High Enrolment Self Study Guide Series basic education = —_ This publication is not for sale. « bene eueston © Copyright Department of Basic Education REPUBLIC OF SOUTH AFRICA www.education.gov.za | Call Centre 0800 202 993 |