Newton's Laws of Motion - Intermediate Level Physics PDF

Intermediate Level Physics 1st Year: Mechanics B. M. Valletta Page 1 of 86  IM Syllabus: Physics MECHANICS 2.1.5 Newton’s first law of motion Candidates should be able to: a. State and apply Newton’s f...

Intermediate Level Physics 1st Year: Mechanics B. M. Valletta Page 1 of 86  IM Syllabus: Physics MECHANICS 2.1.5 Newton’s first law of motion Candidates should be able to: a. State and apply Newton’s first law of motion. b. Link inertial mass with Newton’s first law of motion. c. Define density. d. State the equation for density and use it to solve problems. e. Name the two fundamental forces that can act outside the nucleus. f. Draw free body diagrams. The object can be represented using a sketch or a simple shape such as a box. g. Explain what is meant by centre of mass. h. Distinguish between smooth and rough surfaces. 2.1.6 Newton’s second law of motion Candidates should be able to: a. Define linear momentum. b. State Newton’s second law of motion in terms of the rate of change of momentum. !" !(%&) c. Express Newton’s second law of motion as: F = !# = where F is the resultant force, !# p is momentum, m is mass, v is velocity and t is time. d. Derive the equation F = ma, where m is the mass taken to be constant and a is the acceleration. e. Define the newton using Newton’s second law of motion. f. Write down the equation representing Newton’s second law of motion and use it to solve problems where only the mass is changing or only the velocity is changing, but not both changing at the same time. 2.1.7 Time of impact and impulse The candidate should be able to: a. Explain what is meant by time of impact. b. Discuss how changing the time of impact affects the force of impact. 2.1.8 Common forces The candidate should be able to: a. Determine the weight of an object given its mass and the acceleration due to gravity. b. Solve simple problems involving frictional force between two surfaces that do not include the coefficient of friction. Page 2 of 86 2.1.9 Drag and terminal velocity Candidates should be able to: a. Explain qualitatively what drag and viscosity are. b. Explain qualitatively how the drag depends on area, viscosity and velocity. Knowledge of the equations governing drag is not examinable. The candidate is only expected to be able to state if the drag decreases or increases with these quantities. c. Draw the velocity (speed)-time graph of a body falling in a viscous medium. d. State what is meant by terminal velocity (speed) and explain how this can be attained by a body falling in a viscous fluid. 2.1.10 Pressure Candidates should be able to: a. Define pressure in terms of the force and the area on which it acts. ( b. Write down the equation 𝑃 = ), where P is pressure and A is area and use it to solve problems. c. State the equation for hydrostatic pressure and use it to solve problems. 2.1.11 Newton’s third law of motion Candidates should be able to: a. State and use Newton’s third law of motion. b. Identify and give examples of Newton’s third law pairs of forces, i.e. the action and the reaction. 2.1.12 Conservation of linear momentum Candidates should be able to: a. State the principle of conservation of linear momentum. b. Describe in detail an experiment that verifies the principle of conservation of momentum. This included providing a diagram, the procedure, and adequate precautions and calculations to be made. c. Solve problems using the principle of conservation of momentum in one dimension that may not involve quadratic equations. d. Distinguish between perfectly elastic and inelastic collisions and determine if a given collision is perfectly elastic or inelastic. Page 3 of 86  In this topic the ways in which motion can be explained will be discussed1. Forces Ü A force is a push or pull exerted by one object on another object. When the interaction between the two objects stops, the two objects no longer experience the force. ® Forces exist only as a result of an interaction. Ü Forces can be broadly classified as either Contact forces or Action-at-a- distance forces. Contact forces are those forces that result when the two interacting objects are in direct contact with each other. Examples of contact forces: Frictional forces tension reaction 1 Dynamics is a subdivision of mechanics that is concerned with the motion of material objects in relation to the physical factors that affect them: force, mass, momentum, energy. It is concerned with the effect of forces and torques on the motion of bodies having mass. The ways (words, graphs, numbers etc.…) by which motion can be described were discussed in the topic: Linear Motion. Page 4 of 86  Action-at-a-distance forces are forces that result even when the two interacting objects are not in physical contact with each other. They are able to exert a push or a pull even though they are physically separated. Examples of action-at-a-distance forces: gravitational forces magnetic forces electric forces Ü All forces in nature may be classified into three fundamental types: § Gravitational forces § Electromagnetic forces These two forces act outside the nucleus ® Nuclear force – These forces are only experienced by sub-atomic particles. Therefore, it can only act within the nucleus. There are two types: © The strong nuclear force © The weak nuclear force Ü Free-body diagrams A free body diagram is a diagram showing only all the forces acting on \ the object. Page 5 of 86 Situation 1: Flower pot on a table Force of table Force of ground on flower pot on table Weight of Force of flower Weight of flower pot pot on table table Free body diagram of flower pot Free body diagram of table Situation 2: Box at rest on an inclined plane Normal reaction force (Force of the plane on box) Frictional force Weight of the box Free body diagram of the box Page 6 of 86 Situation 3: Rock climber using a rope to hang against a cliff Pull of rope on rock climber Frictional force Normal force of rock on rock climber. Weight of rock climber Free body diagram of the rock climber When drawing free body diagrams, the following points should be taken into consideration: © Only one body is drawn. © Although a whole diagram of the body may be drawn, it is customary in a free body diagram to represent the object by a box or a circle. © All forces acting on the body in the given situation must be identified. © Force arrows are drawn in the directions in which the forces are acting. The size of the arrow reflects the magnitude of the force. © Force arrows are drawn from the centre of the box or circle representing the object. © If a number of forces act in the same line of action, then the force arrows are drawn side by side close to the correct spot. © Each force arrow is labelled according to its type. Page 7 of 86  Sir Isaac Newton proposed three laws that explain why objects move (or don't move) as they do. These three laws are known as Newton's three laws of motion. Newton’s 1st Law of Motion Newton’s 1st Law of Motion – The Law of Inertia: A body continues in its state of rest or of uniform unaccelerated motion in a straight line, if no external resultant force acts on it. Ü The two natural states of motion The Law establishes two natural states of motion: Rest Motion with constant velocity. The law predicts (i) the behaviour of stationary objects In the absence of a resultant force an object at rest, will remain at rest. (ii) the behaviour of moving objects In the absence of a resultant force an object moving with constant velocity will remain moving with constant velocity. Note: The state of motion of an object is maintained as long as the object is not acted upon by an unbalanced force. Forces on object are balanced Resultant force = 0 N Object at rest Object in motion constant velocity (v = 0 m s ) -1 (v ¹ 0 m s-1) a = 0 m s-2 a = 0 m s-2 stays at rest remains in motion same speed and direction Page 8 of 86  Note: An unbalanced force changes an object’s natural state of motion. © An unbalanced force sets an object in motion if it is at rest © An unbalanced force changes the object’s velocity if it is moving. An object does not need a resultant force to keep moving. Ü Inertia and Inertial mass Newton’s 1st Law implies that matter has an inbuilt opposition to change its natural state of motion. This is known as inertia. Inertia is the reluctance of an object to move if at rest or to change its direction or speed when moving in a straight line. Example 1: An object at rest begins to move only when a force acts on it. Example 2: An object which is moving with constant speed in a straight line changes direction only when a force acts on the object Example 3: When a car starts moving forward, the passengers would stay at rest, unless the contact force of the seats would make them move forward. Page 9 of 86 Example 4: When a car brakes suddenly, the passengers would continue to move forward in a straight line unless the force of the seat belt stops them. Inertia depends on mass. The greater the mass, the greater the force needed to change the object’s natural state of motion. Objects with a greater mass have more inertia. Mass is a measure of how difficult it is to accelerate a body. It is referred to as the inertia of a body or its reluctance to accelerate the body. ® The larger the mass, the bigger the inertia i.e., more force is needed to produce a particular acceleration. ® The smaller the mass, the smaller the inertia i.e., less force is needed to produce a particular acceleration. If an object has a mass of 100 kg, it will be 100 times more difficult to accelerate it than the standard 1kg mass. Inertial mass measures how difficult it is to change the velocity of an object. The higher the inertial mass, the more difficult it is to change the velocity of an object. It is a mass parameter giving the inertial resistance to the acceleration of the body when responding to all types of forces. Inertial mass is defined by Newton’s 2nd law: F= ma. When a force F is applied to an object, it will accelerate proportionally. The constant of proportionality is the inertial mass. It is the ratio of force over acceleration. Page 10 of 86 To determine the inertial mass: § Apply a force F / N to the object. § Measure the acceleration a in m s-2. § !" gives the inertial mass in kg. Difference between mass and weight Mass Weight § is a measure of its resistance to § is a measure of the acceleration gravitational force on the § is a measure of the inertia of a object body § is the force of attraction of the § is the amount of matter in a planet / moon / star on the body object § is measured by means of a § is measured by means of a balance spring balance § Scalar § Vector § measured in kg § measured in N § constant throughout the § Varies from place to place Universe. When a person kicks an object An object of weight of 50 N on in order to accelerate it, he Earth would have a weight of feels the force of the kick on his 8.33 N on the moon. It would toes. The same feeling on the be easier to lift the object on toes would be obtained if the the moon. person were to kick the same object on the moon. Page 11 of 86 Ü Density Which is the heavier 2 kg of lead or 2 kg of feathers? The lead and the feathers both have the same mass. (Obviously, the volume of 2kg of lead would be smaller than the volume of 2 kg of feathers). In order to compare, the same volume must be considered. The amount of matter in a particular volume is the density of the material. Density is defined as the mass per unit volume. The density of a substance is given by the following equation: 𝑚𝑎𝑠𝑠 𝑚 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑜𝑟 𝜌= 𝑣𝑜𝑙𝑢𝑚𝑒 𝑉 𝜌 = density in kg m-3, m = mass in kg, and V = volume in m3 SI unit of density: kg m-3 Other unit of density: g cm-3 Note: To convert from g cm-3 to kg m-3 multiply by 1000 To convert from kg m-3 to g cm-3 divide by 1000 # The equation 𝜌= $ can be rearranged to give: # 𝑉= % and 𝑚 = 𝜌𝑉 Useful consideration: Sometimes the mass of the object is not given. The mass can always be expressed in terms of density. 𝑚 = 𝜌𝑉 In general, the volume of an object having a uniform cross-sectional area A and length L is given by: V = AL Therefore, the mass of the object can be found from: 𝑚 = 𝜌𝐴𝐿 Page 12 of 86 Ü Solving Newton’s 1st law of motion Problems: According to Newton’s 1st law of motion, whenever an object is at rest or moving with constant velocity, the resultant external force acting on the object must be zero. Thus, The forces acting on the object must be balanced. The forces acting on the object must be in equilibrium. When coplanar forces acting on a system are in equilibrium, the horizontal and vertical components of these forces are also in equilibrium. This means that: the resultant force in the vertical direction is zero, and the resultant force in the horizontal direction is zero. Therefore: ↑ forces = ↓ forces → forces = ← forces This method can be used to find unknown forces. Steps involved in answering questions: Draw a free body diagram. Resolve forces in two perpendicular directions. Use above equations to solve for any unknown. Page 13 of 86 Problem 1: 1. A toolbox of weight 400 N is on a slope which is inclined at 30o to the horizontal. (a) Draw a free body diagram of the toolbox, naming all the forces acting on the toolbox. (b) Find the values of the unknown forces. (a) Force of slope on toolbox (Normal Reaction R) Frictional force F Weight W (400 N) Page 14 of 86 (b) R F P 30° Q 30° W =400 N Resolve the weight into 2 components: A component parallel to the slope, P A component perpendicular to the slope, Q To find P: &'' * Sin 30 = ()' = +,, P = 400 Sin 30 = 200 N To find Q: "-. / Cos 30 = ()' = +,, Q = 400 Cos 30 = 346.4 N The car is stationary. By Newton’s 1st Law the resultant force acting on the car is zero. (i) Parallel to the slope: Forces up the slope = Forces down the slope F = P = 200 N (ii) Perpendicular to the slope: Forces out of the slope = Forces into the slope R = Q = 346.4 N Page 15 of 86 Problem 2: A boat was pulled forward by two 400N forces applied symmetrically to the front of the boat. The forces act at an angle of 50° to each other. i. What is the resultant force pulling the boat forward? ii. Find the drag force acting on the boat if the boat moves at constant velocity. i. N 400 25° 25° 400 N Since the two forces are equal and are applied symmetrically to 0, the boat, then the resultant forward force is at an angle of 1 = 25° to the original force. adj X N Cos 25 = = 400 hyp 400 X = 400 Cos 25 = 362.5 N 25° X P adj P 25° Cos 25 = = hyp 400 400 N P = 400 Cos 25 = 362.5 N Resultant forward force = X + P = 362.5 + 362.5 = 725 N The resultant forward force is 725 N acting at an angle of 25° to the original 400 N force. Page 16 of 86 ii. The boat was moving with constant velocity. Thus, according to Newton’s 1st law, there is no resultant force acting in the direction of motion. Resultant forward force, R Drag force, D Drag force – Resultant forward force = 0 D + 725 = 0 D = - 725 N The drag force has a magnitude of 725 N and acts in the opposite direction to the motion of the boat. Page 17 of 86  Newton’s 2nd Law of Motion According to Newton’s 1st law: if a resultant force acts on an object, the object’s velocity would change. This means that a force causes a change in momentum2. Newton’s 2nd law of motion summarises this. Newton’s 2nd law of Motion states that the rate of change of momentum of an object is directly proportional to the applied force and the change in momentum takes place in the direction of that force. Note: This is Newton’s 2nd law of motion expressed in terms of the rate of change of momentum. Ü Derivation of the equation F = ma: The momentum p of an object of constant mass m moving with velocity u is, by definition mu. That is momentum = mass x velocity Suppose a force F acts on the object for a time t and changes its velocity from u to v, then change of momentum = mv – mu #23#4 #(234) The object’s rate of change of momentum is 5 = 5 2 Momentum is the property of an object which makes it easy or difficult to start or stop motion. Momentum is defined as the product of mass and velocity. Momentum = mass x velocity p = mv Momentum is a vector quantity. Its direction is the same as the direction of the velocity. Unit of momentum: kgms-1 Page 18 of 86 Thus, by Newton’s 2nd law: Force µ rate of change of momentum 𝑚(𝑣 − 𝑢) 𝐹 ∝ 𝑡 If a is the acceleration of the object, then 𝑣−𝑢 𝑎= 𝑡 ∴ 𝐹 ∝ 𝑚𝑎 or F = kma k is the constant of proportionality. One newton is defined as the resultant force which gives a mass of 1 kg an acceleration of 1 ms-2. Thus, if m = 1 kg and a = 1 ms-2 then F = 1 N Substituting these values in F = kma: 1 = k (1) (1) k = 1. So, as k = 1 F = ma Note: Ü F is the resultant (or unbalanced) force acting in the direction of motion of the object. Ü The acceleration produced would be in the same direction as the unbalanced force. Ü When using the formula, the quantities should be in SI units: F ® in Newtons m ® in kilograms a ® in ms-2 Page 19 of 86  If the force is in the same direction as the velocity, acceleration occurs. If the force acts in the opposite direction to the velocity, deceleration occurs. Note: The equation F = ma shows that the force is directly proportional to the acceleration if the mass of the object remains constant. There is no acceleration, if the resultant force which acts on the object is zero. With no resultant force, the object must continue at the same velocity or remain at rest as stipulated by Newton’s 1st law. The equation from Newton’s 2nd law of motion can also be stated as: mv − mu F = t Ü When using the formula, the quantities should be in SI units: F ® in Newtons m ® in kilograms u ® in ms-1 v ® in ms-1 t ® in seconds Page 20 of 86  In mathematical notation, the above equation can be written as: d(mv) F= dt dp F= dt Ü Solving Newton’s 2nd law of motion Problems: Step 1: Draw a diagram. Include all the forces. Step 2: Resolve the forces of the system in two perpendicular directions: § a direction parallel to the direction of motion § a direction perpendicular to the direction of motion Step 3: Apply Newton’s 2nd law in the direction of motion. Use the #23#4 formula F = ma or 𝐹 = 5. It is important to use the same sign convention for all vectors (forces, velocities and accelerations). Step 4: Generally, forces in the direction perpendicular to motion would be in equilibrium. In this case, Newton’s 1st law applies. Thus, forces in this direction balance: ↑ forces = ↓ forces. Steps involved in answering questions: Draw a free body diagram Resolve forces in two perpendicular directions Use F = ma in the direction of motion. Solve for any unknown. Use ↑ forces = ↓ forces in the direction perpendicular to motion. Solve for any unknown. Page 21 of 86 Free Body diagrams – Newton’s 2nd law Example 1: Accelerating car acceleration, a Free body diagram of the car: Normal Reaction, R Drag and Frictional forces, D Engine force, E Weight, W Vertically: There is no motion in the vertical direction. Thus, forces in the vertical direction balance in accordance with Newton’s 1st law. R=W Horizontally: The car is accelerating in the horizontal direction. The resultant force in the horizontal direction (E – D) causes the acceleration. Thus, by Newton’s 2nd law: E – D = ma Page 22 of 86 Example 2: A car accelerating up an inclined plane Free body diagram of car n, a el e ratio acc Normal Reaction, R (force of slope on car) Engine force, E W Sin q Air drag and Frictional forces, D W Cos q Weight, W Perpendicular to slope: The car is not moving perpendicular to the slope. Thus, forces perpendicular to the slope balance in accordance with Newton’s 1st law. R = W Cos q Parallel to the slope: The car is accelerating up the slope. The resultant force parallel to the slope (E – D – W Sin q) causes the acceleration. Thus, by Newton’s 2nd law: E – D – W Sin q = ma Page 23 of 86 Example 3: A sledge accelerating down an inclined plane Free body diagram of sledge n, a el e ratio acc Normal Reaction, R (force of slope on sledge) Air drag and Frictional forces, D W Sin q W Cos q Weight, W Perpendicular to slope: The sledge is not moving perpendicular to the slope. Thus, forces perpendicular to the slope balance in accordance with Newton’s 1st law. R = W Cos q Parallel to the slope: The sledge is accelerating down the slope. The resultant force parallel to the slope (W Sin q - D) causes the acceleration. Thus, by Newton’s 2nd law: W Sin q - D = ma Page 24 of 86 Newton’s 2nd Law of Motion Problems: (Where necessary take g to be equal to 9.81ms-2) 1. A box of mass 8kg is pulled along a rough String horizontal surface by a string inclined at 30° to the horizontal. If the tension in the string is 10N and the frictional force is 7.5N, find (a) the acceleration of the object and (b) the normal reaction of the surface on the object Page 25 of 86 2. An object of mass 10kg is pulled up a rough slope inclined at 60° to the horizontal by a string parallel to the slope. If the acceleration of the object is 5 ms-2 and the frictional force is 0.5N, find the tension in the string. Page 26 of 86 3a) A block of mass 15kg rests on the floor of a lift. Find the reaction between the block and the floor of the lift if: i) the lift is accelerating down at 4ms-2 ii) the lift is moving down with a uniform speed of 4ms-1 iii) the lift is moving down with a deceleration of 4ms-2 iv) the lift is accelerating up at 4ms-2 v) the lift is moving up with a uniform speed of 4ms-1 vi) the lift is moving up with a deceleration of 4ms-2 Page 27 of 86 b) A person stands on a balance inside the lift. Comment on the readings given by the balance when the lift goes through stages (i) to (vi) above. b) Note: By Newton’s 3rd law of motion: Force of the balance on person R is equal and opposite to the force of person on the balance R’. Page 28 of 86 If a person (or an object) inside a lift is on a compression balance, the reading on the balance would vary during motion. The compression balance would give a reading greater than the weight if the lift is accelerating upwards or decelerating downwards; it would give a reading less than the weight if the lift accelerates downwards or decelerates upwards. This reading is the apparent weight of the person or object. The compression balance would give a reading equal to the actual weight of the person or object only if the lift is moving upwards or downwards with a constant speed. Moreover, if the lift is in a state of freefall, the compression balance would register 0N. The person or object would appear to be weightless even though the actual weight is still there. Page 29 of 86 4. A car of mass 1000kg tows a caravan of mass 600kg. There are constant frictional resistances of 200N and 100N to the motion of the car and to the motion of the caravan respectively. The combination has an acceleration of 1.2ms-2 with the engine exerting a constant driving force. Find (a) the driving force (b) the tension in the tow bar. Page 30 of 86 5. A toy train consists of an engine and two trolley. The toy engine produces a forward thrust of 8N. The mass of each trolley is 0.4 kg. The friction at the wheels of each trolley amounts to 0.5 N while that of the engine is 0.55 N. If the toy train is moving forward at an acceleration of 1.5 m s-2, i) what is the tension in each of the towing bars? ii) what is the mass of the engine? Page 31 of 86 6. A jet engine on a test bed takes in 20.0kg of air per second at a velocity of 100 m/s and burns 0.80 kg of fuel per second. After compression and heating the exhaust gases are ejected at 500 m/s relative to the aircraft. Calculate the thrust of the engine. Page 32 of 86 7. A water pipe with a nozzle 80mm in diameter ejects a horizontal stream of water at a rate of 0.044m3/s. The density of water is 1000kgm- 3. With what velocity will the water leave the nozzle. What will be the force exerted on a vertical wall situated close to the nozzle and at right angles to the stream of water, if after hitting the wall, the water (a) falls vertically to the ground (b) rebounds horizontally Page 33 of 86 8. Sand is deposited at a uniform rate of 20 kilograms per second and with negligible kinetic energy on to an empty conveyor belt moving horizontally at a constant speed of 10 metres per minute. Find a) the force required to maintain constant velocity, b) the rate of change of kinetic energy of the moving sand. Page 34 of 86 9. A sailing boat of mass 250.0kg has a sail of area 22.0m2. The sail is unfurled and made to face the wind in a perpendicular direction. If the boat is at rest and the wind speed is 2.0 knots, what is the initial acceleration of the boat? Assume that the wind does not rebound from the sail. (One knot is equal to 0.51 ms-1 and the density of air is equal to 1.293 kgm-3) Page 35 of 86 10. A bird of mass 0.50kg hovers by beating its wings of effective area 0.30 m2. (i) What is the upward force of the air on the bird? (ii) What is the downward force of the bird on the air as it beats its wings? (iii) Estimate the velocity imparted to the air, which has a density of 1.3 kgm-3, by the beating of the wings. Page 36 of 86 11. A helicopter of mass 810kg supports itself in a stationary position by imparting a downward velocity v to all the air in a circle of area 30m2. Given that the density of air is 1.20kg/m3, calculate the value of the velocity v. Page 37 of 86 12. A girl of mass 50 kg jumps on to the ground from a 2m high wall. Calculate the force on her when she lands if she bends her knees and stops in 0.2s. Page 38 of 86 13. A tennis player plays a ball, which reaches her with a velocity of 20m/s. If the maximum force she can exert on the ball is 200 N, how long must the ball be in contact with the racquet in order for the player to return it to her opponent with a velocity of 30m/s in the opposite direction? (Mass of a tennis ball = 0.058 kg) Page 39 of 86  Impulse and Time of impact Newton’s 2nd law of motion states that: The rate of change of momentum of a body is proportional to the resultant force and occurs in the direction of the force. 8("9:; =9 #&#;954# 𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒 = 5=#; ∆' 𝐹= ∆5 Rearranging the equation: F Dt = Dp = change in momentum For uniform acceleration: F Dt = Dp = mv – mu. Force x time = change in momentum § If a force F acts for a short time t, then the quantity Force x time is known as the impulse. § Impulse measures the effect of the force. § Impulse is measured in Newton second (Ns) § 1 Ns = 1 kg m s-1 § Impulse is a vector quantity. Ü Definition: Impulse is the product of the resultant force on an object and the time of impact during which the resultant force was acting on the object. Impulse is described as the time effect of a force on an object. Page 40 of 86  Formulae: Impulse = Force x time Impulse = FDt F ® force Dt ® time of impact Impulse = change in momentum Impulse = Dp Dp ® change in momentum From the equation: Force x time = change in momentum one can say that: § the greater the force of impact and the greater the time of impact, the greater the change in the object’s momentum. Increase Increase Increase force of time of in the change impact impact of momentum Impulse = F Dt = Dp A tennis player or a cricketer ‘ follows through’ with the racket or bat when striking the ball. The force applied then acts for a longer time, the impulse is greater and so also is the gain of momentum (and velocity) of the ball. Ü The effect of changing the time of impact on the force of impact Definitions: Force of impact ® The force of impact is a large force applied for a short time when two or more bodies come in contact. Time of impact ® The time during which the force of impact is applied. Page 41 of 86 If the change in momentum is kept constant: the force of impact is inversely proportional to the time of impact. Impulse = Dp = F Dt ? If Dp is constant then, 𝐹 ∝ ∆5 § If the momentum is changed over a longer time, than the force of impact would be reduced. Time of For a given impact change in increases. momentum the impulse Impulse = F Dt = Dp stays constant. Force of impact decreases § if the momentum is changed over a shorter time, than the force of impact would be larger. Force of For a given impact change in increases momentum the impulse Impulse = F Dt = Dp stays constant. Time of impact decreases In a car crash, the car’s momentum is reduced to zero in a very short time. If the time of impact can be extended by using crumple zones and extensible seat belts, the average force needed to stop the car and passengers is reduced. So, the injury to passengers should also be less. The airbag deforms considerably when a force is applied. The person will thus take a longer time to come to rest. Consequently, according to Newton’s 2nd law, the force of impact is much smaller. Therefore, airbags can protect car passengers from hurting themselves during collision. Page 42 of 86  When we want to stop a moving ball, its momentum has to be reduced to zero. An impulse is then required in the form of an opposing force acting for a certain time. While any number of combinations of force and time will give a particular impulse, the sting can be removed from the catch by drawing back the hands as the ball is caught. A smaller force is then applied for a longer time. The use of sand gives a softer landing for long jumpers as a smaller stopping force is applied over a longer time. Fragile objects are packed in soft containers. Containers deform when a force is applied. This increases the time of impact; therefore, a smaller stopping force is applied. Ü Force – time graphs: A force – time graph shows how the force applied to an object varies with time. Time / s Area under the graph gives the total impulse acting. The area under the graph also represents the total change in momentum produced. Area under a force – time graph = total impulse acting = total change in momentum produced Page 43 of 86  Common Forces Ü Weight: Weight is the gravitational pull of the earth on an object. It is the force of gravity acting on an object. It gives the object an acceleration g when it is falling freely near the Earth’s surface. From Newton’s 2nd law: W = mg This force is always drawn pointing downwards. Ü Frictional Force: Friction is the resistance that one surface or object encounters when moving over another solid surface. This force is always drawn in the opposite direction of motion. Page 44 of 86 Friction and the Laws of motion: Consider a box which is placed on an inclined plane. Will the box move? Explain making appropriate reference to the relevant laws of motion. The normal reaction R T he f r (the force of the inclined plane ic (which tional force on the box) acts o F n the box) q The weight W (the force of the Earth on the box) W = mg The forces acting on the box are: § the weight W. This is the force exerted by the Earth on the box. The weight can be split into 2 components: W Sin q which is the component parallel to the inclined plane W Cos q which is the component perpendicular to the inclined plane § the normal reaction R. This is the force which the inclined plane applies on the box. By Newton’s 3rd law, this force is equal in magnitude and opposite in direction to the force that the box applies to the inclined plane. § the frictional force. This force acts on the box in the opposite direction to the box’s direction of motion (or the potential direction of motion if box is stationary). There is no motion in the direction perpendicular to the inclined plane. So, according to Newton’s first law, there is no resultant force in this direction. The forces which are perpendicular to the plane balance: R = W Cos q. In a direction parallel to the inclined plane, there are 3 possibilities: § W Sin q > F There is a resultant force W Sin q - F acting down the slope. So, by Newton’s 2nd law the box accelerates. Thus, the box moves down the inclined plane with increasing velocity. Page 45 of 86  § W Sin q = F There is no resultant force acting on the box. So, by Newton’s 1st law, the box is either at rest or moving with constant velocity. An object at rest would remain at rest, whereas an object in motion would remain moving with constant velocity. § W Sin q < F There is a resultant force F – W Sin q acting up the slope. In this case: © if the box is stationary it would remain at rest © if the box is in motion, then it would move down the inclined plane with a decreasing velocity. In other words, the box would decelerate down the inclined plane. Note: Þ When the frictional force between two surfaces is absent or negligible, the contact between the two surfaces is said to be smooth. Þ The contact between two surfaces in which the frictional force is present is said to be rough. Ü Drag: Drag is the resistance to motion that is experienced by an object that is moving through a fluid (a liquid or a gas). This force is always drawn in the opposite direction of motion. Page 46 of 86 The drag force on a moving object depends on § the density r of the fluid the greater the density, the greater the drag force the smaller the density, the smaller the drag force. § the surface area A of the object the greater the surface area, the greater the drag force the smaller the surface area, the smaller the drag force. § the speed v of the object, the greater the speed, the greater the drag force the smaller the speed, the smaller the drag force. Speed v Drag Force D 2v 4D 3v 9D 4v 16 D § the drag coefficient the greater the drag coefficient, the greater the drag force the smaller the drag coefficient, the smaller the drag force. The drag coefficient takes into account factors such as shape, compressibility, texture and viscosity. Drag increases with an increase in: s Density s Area s Velocity s Viscosity The most important factor in determining the drag force is the velocity. Page 47 of 86 Ü Viscosity: Viscosity is a kind of internal frictional force in fluids which opposes the motion when they flow. Viscosity depends on the molecular structure of the medium. This effects the ease with which the fluid flows. Syrup and motor oil pour slowly. They are more viscous than water. Fluids with a high viscosity are described as thick. Gases flow much more easily than liquids. They are less viscous than liquids. the greater the viscosity, the greater the drag force the smaller the viscosity, the smaller the drag force. Ü Viscous drag: Viscous drag is the result of friction between the body's surface and fluid particles. It depends on the velocity of the object. § the greater the velocity, the greater the viscous drag § the smaller the velocity, the smaller the viscous drag. Page 48 of 86  Terminal Velocity When an object moves in a fluid, it always experiences some resistance to its motion. This resistance to motion that is experienced by the object moving in a fluid is known as drag. Drag is zero when the object is stationary but increases as the object speeds up. A) A ball bearing which is moving in a viscous fluid (example: motor oil). When the ball bearing is placed on the motor oil, it accelerates downwards due to the force of gravity. There is no drag acting in the upward direction. The weight is the force acting downwards so the ball bearing accelerates towards the bottom of the cylinder. Resultant force: W (downwards) Motion ! By Newton’s 2nd Law: 𝑎= (downwards) of ball " bearing Weight W The object accelerates downwards. As the ball bearing gains speed, the weight stays the same, but the drag increases. The ball bearing still accelerates but its acceleration decreases as time goes by. Its speed still increases but by a smaller amount. The resultant force still acts downwards but is decreasing. This is because the drag acting against it is increasing as the speed increases; but is still less than the weight of the object. Drag D W>D Resultant force: W – D (downwards) !#$ By Newton’s 2nd law: 𝑎′ = (downwards) " The ball bearing moves downwards with a decreasing acceleration. Weight W § Eventually a speed is reached where the drag becomes equal to the weight. As there is no resultant force acting the ball bearing stops accelerating. Page 49 of 86 The ball bearing continues to move downwards with constant velocity. This velocity is known as the terminal velocity. Drag D W=D Resultant force: 0N By Newton’s 2nd law: 𝑎** = 0 𝑚 𝑠 +, Ball bearing moves downwards with constant velocity. Weight W Note: The ball bearing does not stop falling once its resultant force is zero, unless it reaches the bottom of the cylinder. The terminal velocity is the constant final velocity achieved when opposing forces become equal to the forward forces. The following graphs are the velocity - time and the acceleration – time graphs for the ball bearing’s motion Page 50 of 86 B) A skydiver O § Immediately on leaving the plane the skydiver accelerates downwards due to the force of gravity. There is no air drag (air resistance) acting in the upward direction and there is a Skydiver accelerates resultant force acting downwards. So, the downwards. skydiver accelerates towards the ground O to A § As the skydiver gains speed, the weight stays the same but the air drag increases. As the air drag increases, the acceleration of the skydiver decreases because the resultant force is Weight > Drag smaller. The skydiver’s speed increases but As skydiver speeds up the more slowly. The skydiver thus continues to air drag increases. Downward resultant force move downwards with a decreasing decreases. acceleration. Acceleration decreases. A to B § Eventually, the skydiver's weight is balanced by the air drag. There is no resultant force and the skydiver stops accelerating. At this point the skydiver continues to move downwards with Weight = Drag the maximum possible velocity. This velocity is Resultant force = 0N known as the terminal velocity. Acceleration = 0 m s-2. Skydiver travels downwards at constant velocity. Page 51 of 86 § When the parachute is first opened, the air B to C drag it creates at this high speed will be greater than the skydiver’s weight. This upward resultant force makes the skydiver slow down rapidly, reducing the air drag. Weight < Drag Resultant force will be in the upward direction. Skydiver decelerates. Velocity decreases § Eventually the air drag will balance the weight C to D again, at a new lower terminal velocity. The skydiver continues to move downwards with this terminal velocity until he lands safely. Weight = Drag Resultant force = 0N Acceleration = 0 m s-2. Skydiver travels downwards at constant velocity. The following graph is the velocity - time graph for the skydiver’s motion: Page 52 of 86 C) Car speeding on a motorway When a car is stationary, the reaction drag force acting on the car is zero. When the car starts thrust moving, the thrust force no drag force causes the car to accelerate forward. weight Forward resultant force - car accelerates reaction As speed increases, the drag force increases. But the thrust force would be greater than thrust the drag force. So, there drag force would be a forward resultant force. Thus, the car continues weight to accelerate with a smaller Forward resultant force decreases – acceleration acceleration. of car decreases - car continues to move with increasing velocity reaction As the car moves faster the drag force continues to increase. This reduces the thrust drag force forward resultant force. The car continues to move forward weight with decreasing acceleration. Drag force increases – forward resultant force decreases further – acceleration decreases more – but car’s velocity continues to increase Eventually a point is reached when the drag force becomes equal to the thrust force. At reaction this point, the resultant force in the forward direction is zero. The car cannot thrust accelerate further but moves drag force forward with the final, steady weight velocity attained. This Drag force balances thrust force – No forward velocity is the terminal resultant force – No acceleration – car attains its velocity. terminal velocity Page 53 of 86 The following graph is the velocity - time graph for the vehicle’s motion Page 54 of 86  Pressure Ü Pressure is defined as the force acting normally per unit area. Force Pressure = Area F P= A There are many units in which pressure is measured. Some of these are: atm, bar, psi, mmHg etc. The SI unit of pressure is the pascal (Pa). 1 Pa = 1 N m-2 If the force acting on the object remains constant, then pressure is inversely proportional to the area. Examples: © Skiers increase the area in contact with the snow by using skis. Their body weight is no longer concentrated on their feet but is spread out over the area of the skis. The lower pressure of the skis means the skier glides on the snow rather than sink ing into it. In comparison, the smaller area of the boot creates a greater pressure, and as a result a person just wearing boots would sink into the snow. © Different styles of shoe can cause different pressures due to their area. Flat shoes spread the force over a large area, reducing the pressure. High heeled shoes transfer the force through a much smaller area, causing a much greater pressure. It will hurt more if a person steps on someone’s foot in high heels than in flat shoes. © Footballers wear boots with studs as they need to get a good grip on the ground. The studs have a small area to make the pressure large enough to dig into the surface. In this way a better grip is obtained. and will not slip. Page 55 of 86  © When using a sharp knife, the small area of the blade creates a large pressure, making cutting easier. © Having caterpillar tracks on vehicles means their weight acts over a large area. This reduces the pressure they exert and are less likely to sink into wet ground. Ü Liquid pressure § Pressure in a liquid increases with depth and all points at the same depth are at the same pressure. § Pressure at one depth acts equally in all directions. § Pressure in a liquid does not depend on the shape or area of the container or on the total mass of the liquid. § Pressure depends on the density of the liquid. The denser the liquid the greater the pressure. Pressure in a liquid depends only on depth and density. Ü Hydrostatic pressure 3can be found using the equation: p=hrg where: p is the pressure h is the height or depth of the fluid r is the density of the fluid g is the acceleration due to gravity Note that this equation only applies to incompressible fluids4. The density of these fluids does not change with depth. 3 Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. 4 Liquids are incompressible fluids. Page 56 of 86 Ü Example: What is the total pressure exerted on a diver at a depth of 80.0 m under the sea? Density of sea water = 1030 kg m-3 Atmospheric pressure = 1.01 x 105 Pa Pressure due to sea water = h r g = 80 x 1030 x 9.81 = 808344 Pa This is the pressure on the diver caused by a column of water 80 m high. The weight of the air above the sea also adds pressure on the diver. The pressure caused by the weight of the air is known as the atmospheric pressure. Atmospheric pressure is caused by the bombardment of atmospheric molecules that are moving about in a fast and random fashion. Thus, Total pressure on the diver = pressure due to sea water + atmospheric pressure = 808344 + 1.01 x 105 = 909344 = 9.09 x 105 Pa Ü The U-tube manometer The U-tube manometer is a device which can be used to measure the pressure of a gas If one side of the u-tube manometer is not connected to a gas supply (for example) each surface of the liquid is acted on equally by atmospheric pressure and the levels are the same. If one side is connected to a gas supply, the gas exerts a pressure on surface A and level B rises until B Pressure of gas = atmospheric pressure + pressure due to liquid column BC. A C The pressure of the liquid column BC equals the amount by which the gas pressure exceeds atmospheric pressure. Page 57 of 86 If the gas pressure is 140 kPa and the atmospheric pressure is 100 kPa, determine the value of h given that the density of mercury is 13.6 g cm-3. Pressure of gas = atmospheric pressure + pressure due to liquid column BC. Pressure of gas = atmospheric pressure + h r g Pressure of gas = 140 kPa = 140 000 Pa Atmospheric pressure = 100 kPa = 100 000 Pa Density of mercury = 13.6 g cm-3 = 13 600 kg m-3 140 000 = 100 000 + h (13 600)(9.81) 140 000 – 100 000 = 133 416 h 40 000 = 133 416 h h = 0.30 m Page 58 of 86  Newton’s 3rd Law of Motion Newton’s 3rd Law states that: If a body A exerts a force on a body B, then B exerts an equal and oppositely directed force on A. In other words: For every action there is an equal but opposite reaction. Note: Newton’s 3rd Law pair of forces act on two different objects – never on the same object. Ü Examples of Newton’s 3rd Law pair of forces. (i) Book resting on table: Force of table on book The book exerts a downward force on the table. The table exerts an equal and opposite upward force Force of book on table on the book. These two forces act on different objects and are represented by the red arrow. Note: The blue arrow is the weight of the book and does not form a pair with the upward force on the book. These two forces act on the same object. Page 59 of 86 (ii) A person stepping from a rowing boat The person steps forward ® the person pushes backwards on the boat. The boat moves backwards ® the Force of boat Force of person on person on boat boat pushes the person forwards with an equal force. (iii) Runner Action: The runner pushes backwards on the ground. Reaction: The ground pushes forwards on the runner. The two forces are equal and oppositely directed. They act on different bodies. (iv) Moving bicycle or car Action:wheels push Reaction: road pushes backwards on road forward on wheels. The size of the force on the road equals the size of the force on the wheels of the car. The direction of the force on the road (backwards) is opposite the direction of the force on the wheels (forwards). Action-reaction force pairs make it possible for cars to move along a roadway surface. Page 60 of 86 (v) Skydiver Action: The Earth pulls the skydiver downwards. This force is the weight of the skydiver. Reaction: The skydiver pulls the earth upwards. The two forces are equal and oppositely directed. They act on different bodies. Note: The Earth is moved upwards, but the Earth is so massive that the upward force on it has far too small an effect for any movement to be detected. (The magnitude of this force is equal to the magnitude of the weight of the skydiver.) (vi) Rocket Engine When rocket fuel is burned, hot gases are produced. These gases expand rapidly and are forced out of the back of the rocket. Action: Force of rocket on exhaust gas ® Exhaust gases are pushed backward. Reaction: Force of exhaust gas on rocket engine® Rocket engine pushed forward. These two forces are equal in magnitude and opposite in direction. Page 61 of 86  (vii) Explosions A gunpowder explosion creates hot gases that expand outward allowing the rifle to push forward on the bullet. Action: Forward force of rifle on bullet: the bullet is ejected. Reaction: Backward force of bullet on rifle: the rifle recoils Ü Properties of Newton’s 3rd Law pair of forces: The forces (i) have the same magnitude, (ii) have the same nature – either both electromagnetic or both gravitational, (iii) have opposite directions or lines of action, (iv) act on different bodies. Page 62 of 86 Ü Momentum is the property of an object which makes it easy or difficult to start or stop motion. Momentum is defined as the product of mass and velocity. Momentum = mass x velocity p = mv Momentum is a vector quantity. Its direction is the same as the direction of the velocity. Unit of momentum: kgms-1 Ü Principle of conservation of linear momentum: If no external forces act on a system of colliding objects, the total momentum of the objects in a given direction before collision is equal to the total momentum in the same direction after collision. Examples: § An object A of mass mA and an object B of mass mB are moving in the same direction as shown. The speed of A is greater than the speed of B. A collides with B. Following the collision, the speed of A changes from uA to vA whereas that of B changes from uB to vB. Object A Object B Object A Object B uA uB vA vB mA mB mA mB Before Collision After collision Before collision: After collision: Total Momentum: mAuA + mBuB Total Momentum: mAvA + mBvB Principle of conservation of momentum: mAuA + mBuB = mAvA + mBvB Page 63 of 86  § An object A of mass mA and an object B of mass mB are moving in opposite directions as shown. A collides with B. Following the collision, the speed of A changes from uA to vA whereas that of B changes from uB to vB. Object A Object B Object A Object B uA uB vA vB mA mB mA mB Before Collision After collision Before collision: After collision: Total Momentum: mAuA + mB (-uB) Total Momentum: mA (-vA) + mBvB mAuA – mBuB – mAvA + mBvB Principle of conservation of momentum: mAuA – mBuB = - mAvA + mBvB § A home-made cannon of mass M is used to project a tennis ball of mass m. Before the explosion both cannon and tennis ball were at rest. After the explosion the tennis ball moved with speed vB in one direction; whereas the cannon recoiled with speed vc in the opposite direction. vB vC Mass of cannon: M Mass of tennis ball: m At rest Before Explosion After Explosion Before Explosion: After Explosion: Total Momentum: M x 0 + m x 0 Total Momentum: M (-vc) + m vB 0 kg m s-1 – M vc + m vB Principle of conservation of momentum: 0 = - M vc + m vB M vc = m vB The momentum of the cannon is equal and opposite to the momentum of the tennis ball. Principle of conservation of linear momentum: Whenever objects interact, their total momentum in any direction remains constant, provided that no external force acts on the objects in that direction. Page 64 of 86 Ü Experiment to verify the Principle of Conservation of Linear Momentum – Completely inelastic collision Diagram: Before Collision After collision Procedure: § A pin was attached to the front of vehicle A and a piece of cork was stuck to the rear of vehicle B. An opaque interrupter card was attached to vehicle A. § The mass of vehicle A and the mass of vehicle B were measured by means of an electronic balance. § The electronic timers were set to time the passage of the interrupter card on vehicle A through the light gates. Page 65 of 86 § The apparatus was set up as shown in the diagram. § Vehicle A was pushed towards vehicle B which was at rest. § As vehicle A passed through light gate 1, timer 1 was activated. The time taken for the interrupter card to pass through light gate 1 was recorded. § The velocity uA of vehicle A before collision was calculated by dividing the length of the ‘interrupter card’ by the time displayed on timer 1. § Vehicle A collided with vehicle B. The pin on the front of A stuck to the cork on B. Vehicles A and B continued to move together. § As the two vehicles passed through light gate 2, timer 2 was activated. The time taken for the interrupter card to pass through light gate 2 was recorded. § The velocity v of vehicles A and B after collision was calculated by dividing the length of the ‘interrupter card’ by the time displayed on timer 2. § The above steps were repeated several times using a different initial speed for vehicle A each time. Results: Mass of vehicle A, mA : / kg Mass of vehicle B, mB : / kg Length of the interrupter card, L : / m Time of timer 1, t1 : / s Time of timer 2, t2 : / s Velocity of vehicle B before collision : 0 m s-1 Calculations: Velocity of A before collision, uA: 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑟𝑢𝑝𝑡𝑒𝑟 𝑐𝑎𝑟𝑑 𝐿 𝑢! = 𝑠𝑝𝑒𝑒𝑑 = = = 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟 𝑡" Velocity of A and B after collision, v: 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑟𝑢𝑝𝑡𝑒𝑟 𝑐𝑎𝑟𝑑 𝐿 𝑣 = 𝑠𝑝𝑒𝑒𝑑 = = = 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟 𝑡# $ Momentum of A before collision: mA uA = mA %% Momentum of B before collision: mB 0 = 0 kg m s-1 Page 66 of 86  $ Momentum of A and B before collision: mA uA + 0 = mA uA = mA %% $ Momentum of A and B after collision: (mA + mB) v = (mA + mB) % & The experiment should confirm that the total momentum before collision is equivalent to that after collision. If a graph of the total momentum before against the total momentum after is drawn, a straight line graph passing through the origin having a gradient of 1 would be obtained showing that momentum is conserved. Precautions: § An air track was used to eliminate friction. § The air track was levelled to avoid experimental errors § The glider was stopped as soon as it reached the end of the track, to prevent it from bouncing back. § The electronic balance was reset to zero before taking any readings. § Repeated readings of mass and length are taken to minimize random errors. Ü Experiment to verify the Principle of Conservation of Linear Momentum – Elastic collision Diagram: Before collision Page 67 of 86 After collision Procedure: § A stretched rubber band is attached to the front of each vehicle. An opaque interrupter card was attached to each vehicle. § The mass of vehicle A and the mass of vehicle B were measured by means of an electronic balance. § The electronic timers were set to time the passage of the interrupter cards through the light gates. § The apparatus was set up as shown in the diagram. § The two vehicles were pushed towards each other. § Vehicle A passed through light gate 1, activating timer 1. The time taken for the interrupter card on vehicle A to pass through light gate 1 was recorded. Vehicle B passed through light gate 2, activating timer 2. The time taken for the interrupter card on vehicle B to pass through light gate 2 was also recorded. § The velocity uA of vehicle A before collision was calculated by dividing the length of its ‘interrupter card’ by the time displayed on timer 1 and the velocity uB of vehicle B before collision was calculated by dividing the length of the ‘interrupter card’ on B by the time displayed on timer 2. § Vehicle A collided with vehicle B. The rubber bands act as buffers so that the collision is almost perfectly elastic. Vehicles A and B reverse their direction on impact. § Vehicle A passes back through light gate 1 reactivating timer 1; whereas vehicle B passes back through light gate 2, reactivating timer 2. The time taken for each interrupter card to pass through the light gate was recorded. § The velocities vA and vB of vehicles A and B after collision were calculated. Page 68 of 86 § The above steps were repeated for a number of different values of uA, uB, mA and mB Results: Mass of vehicle A, mA : / kg Length of card A, LA : / m Time of timer 1 before collision, t1 : / s Time of timer 1, after collision, t3 : / s Mass of vehicle B, mB : / kg Length of card B, LB : / m Time of timer 2 before collision, t2 : / s Time of timer 2, after collision, t4 : / s Note: Timers were reset before the 2nd time was taken Calculations: Velocity of A before collision, uA: 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑎𝑞𝑢𝑒 𝑐𝑎𝑟𝑑 𝐿! 𝑢! = 𝑠𝑝𝑒𝑒𝑑 = = = 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟 𝑡" Velocity of A after collision, vA: 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑎𝑞𝑢𝑒 𝑐𝑎𝑟𝑑 𝐿! 𝑣! = 𝑠𝑝𝑒𝑒𝑑 = = = − 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟 𝑡# Velocity of B before collision, uB: 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑎𝑞𝑢𝑒 𝑐𝑎𝑟𝑑 𝐿$ 𝑢$ = 𝑠𝑝𝑒𝑒𝑑 = = = − 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟 𝑡% Velocity of B after collision, vB: 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑎𝑞𝑢𝑒 𝑐𝑎𝑟𝑑 𝐿$ 𝑣$ = 𝑠𝑝𝑒𝑒𝑑 = = = 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟 𝑡& '- Momentum of A before collision: mA uA = mA (. ' Momentum of B before collision: mB uB = mB e− (/f 0 Total Momentum of A and B before collision: '- '/ mA uA + mB uB = mA (. - mB (0 Page 69 of 86  ' Momentum of A after collision: mA vA = mA e− (- f 1 '/ Momentum of B after collision: mB vB = mB (2 Total Momentum of A and B after collision: '- '/ mA vA + mB vB = - mA (1 + mB (2 The experiment should confirm that the total momentum before collision is equivalent to that after collision. If a graph of the total momentum before against the total momentum after is drawn, a straight line graph passing through the origin having a gradient of 1 would be obtained showing that momentum is conserved. Precautions: § An air track was used to eliminate friction. § The air track was levelled to avoid experimental errors § The glider was stopped as soon as it reached the end of the track, to prevent it from bouncing back. § The electronic balance was reset to zero before taking any readings. § Repeated readings of mass and length are taken to minimize random errors. Ü Experiment to verify the Principle of Conservation of Linear Momentum – Explosion type collision Before Page 70 of 86 Procedure: § An opaque interrupter card was attached to each vehicle. § The mass of vehicle A and the mass of vehicle B were measured by means of an electronic balance. § The electronic timers were set to time the passage of the interrupter cards through the light gates. § The apparatus was set up as shown in the diagram with the two vehicles at rest between the two light gates. A compressed spring was placed between the two vehicles. The two vehicles were kept together with the compressed spring between them by means of a string which was attached to the vehicles. § The string was cut and the two vehicles moved in opposite directions. § Vehicle A passed through light gate 1, activating timer 1. The time taken for the interrupter card on vehicle A to pass through light gate 1 was recorded. Vehicle B passed through light gate 2, activating timer 2. The time taken for the interrupter card on vehicle B to pass through light gate 2 was also recorded. § The velocity vA of vehicle A after the explosion was calculated by dividing the length of its ‘interrupter card’ by the time displayed on timer 1 and the velocity vB of vehicle B after the explosion was calculated by dividing the length of the ‘interrupter card’ on B by the time displayed on timer 2. § The above steps were repeated for a number of different values of mA and mB or for different amounts of spring compression Results: Mass of vehicle A, mA : / kg Length of card A, LA : /m Time of timer 1 after explosion, t1 : /s Mass of vehicle B, mB : / kg Length of card B, LB : /m Time of timer 2 after explosion, t2 : /s Calculations: Velocity of A before explosion, uA: 0 m s-1 (vehicle A was initially at rest) Velocity of A after explosion, vA: 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑎𝑞𝑢𝑒 𝑐𝑎𝑟𝑑 𝐿! 𝑣! = 𝑠𝑝𝑒𝑒𝑑 = = = 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟 𝑡" Page 71 of 86 Velocity of B before explosion, uB: 0 m s-1 (vehicle B was initially at rest) Velocity of B after explosion, vB: 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑎𝑞𝑢𝑒 𝑐𝑎𝑟𝑑 𝐿$ 𝑣$ = 𝑠𝑝𝑒𝑒𝑑 = = = − 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟 𝑡% Momentum of A before explosion: mA uA = mA x 0 = 0 kg m s-1 Momentum of B before explosion: mB uB = mB x 0 = 0 kg m s-1 Total Momentum of A and B before explosion: mA uA + mB uB = 0 kg m s-1 ' Momentum of A after explosion: mA vA = mA e (-f. ' Momentum of B after explosion: mB vB = mB e− (/ f 0 Total Momentum of A and B after explosion: '- '/ mA vA + mB vB = mA (. - mB (0 The experiment should confirm that the total momentum before the explosion-type collision is equivalent to that after the explosion-type collision. Precautions: § An air track was used to eliminate friction. § The air track was levelled to avoid experimental errors § The glider was stopped as soon as it reached the end of the track, to prevent it from bouncing back. § The electronic balance was reset to zero before taking any readings. § Repeated readings of mass and length are taken to minimize random errors. Note: Some conservation of linear momentum experiments may be viewed at: https://www.youtube.com/watch?v=rG5qTzbTbk8 https://www.youtube.com/watch?v=lS9k8bM0YBs Page 72 of 86  Types of Collision There are 3 types of collisions: Elastic collisions Inelastic collisions Super-elastic or explosion-type collisions Ü Elastic collisions: A perfectly elastic collision is one in which there is no loss of kinetic energy in collision. K.E. before collision = K.E. after collision Example of a perfectly elastic collision: § All collisions between ideal gas atoms are perfectly elastic according to the Kinetic Theory of Gases. An elastic collision is one in which the kinetic energy before is equal to the kinetic energy after. The loss in kinetic energy which occurs is negligible. K.E. before collision = K.E. after collision Examples of an elastic collision: § Collision between smooth billiard balls. Page 73 of 86  § Collision between 2 vehicles which do not stick together on collision. § A bouncy ball is dropped, it bounces and returns to the initial height. In a perfectly elastic collision or in an elastic collision: § Linear momentum is conserved in accordance with the Principle of Conservation of linear momentum. ® Total momentum before collision = Total momentum after collision § Kinetic energy is conserved ® Kinetic Energy before collision = Kinetic Energy after collision § Total energy is conserved in accordance with the Principle of Conservation of Energy. ® Total energy before collision = Total energy after collision Page 74 of 86 Ü Inelastic collisions: A perfectly inelastic collision is one when the maximum amount of kinetic energy is lost. It is a collision in which the objects stick together after the collision. K.E. before collision > K.E. after collision Some of the K.E. which the objects possess before collision changes to other forms on collision. Example of a perfectly inelastic collision: § A railway engine collides with and couples to a truck § A ballistic pendulum § A piece of play dough which is dropped on the ground Page 75 of 86 An inelastic collision is one in which the kinetic energy before is greater than the kinetic energy after. Some kinetic energy changes into other forms on collision. Examples of an inelastic collision: § A cricket ball is dropped, it bounces but does not regain its original height. Some of the cricket ball’s kinetic energy changes to heat energy and sound energy on impact. In a perfectly inelastic collision or in an inelastic collision: § Linear momentum is conserved in accordance with the Principle of Conservation of linear momentum. ® Total momentum before collision = Total momentum after collision § Kinetic energy is not conserved ® Kinetic Energy before collision > Kinetic Energy after collision ® Some K.E. is converted into other forms of energy (example: heat energy and sound energy) as a result of the impact § Total energy is conserved in accordance with the Principle of Conservation of Energy. ® Total energy before collision = Total energy after collision Page 76 of 86 Ü Super-elastic or explosion-type collisions: A super-elastic collision or an explosion-type collision is one in which the kinetic energy before is smaller than the kinetic energy after. Some form of energy changes into kinetic energy on collision. K.E. before collision < K.E. after collision Some form of energy changes to K.E. Example of a super-elastic collision or an explosion-type collision: § Two spring loaded trolleys which are placed back-to-back. The spring in one of the trolleys is released. The two trolleys will explode and move apart. The elastic potential energy in the compressed spring changes to kinetic energy. § A bullet fired from a gun or a cannon ball fired from a cannon Chemical energy changes to kinetic energy § Bomb explosion Chemical energy changes to kinetic energy Page 77 of 86  § Person jumping from a boat Chemical energy changes to kinetic energy In a super-elastic collision or an explosion-type collision: § Linear momentum is conserved in accordance with the Principle of Conservation of linear momentum. In the explosion-type collision, both objects would be at rest before the collision occurs; thus the total momentum before the explosion-type collision is equal to zero. ® Total momentum before collision = Total momentum after collision 0 = Total momentum after collision In the explosion-type collision, the momentum of the two objects after the collision would be equal and opposite. § Kinetic energy is not conserved ® Kinetic Energy before < Kinetic Energy after ® Some form of energy is converted to K.E. on explosion § Total energy is conserved in accordance with the Principle of Conservation of Energy. ® Total energy before collision = Total energy after collision Page 78 of 86  Linear Momentum Problems: 1a) A car of mass 1000 kg moving at 20 m s-1 collides with a car of mass 1200 kg moving at 5.0 m s-1 in the same direction. If the second car is shunted forwards at 15 m s-1 by the impact, what is the velocity of the first car immediately after the crash? Is this collision elastic or inelastic? b) If the cars collide head on at the same speeds as in (a), what would their combined velocity be after the collision if they stick together on impact? Calculate the loss of kinetic energy as a result of the impact. Page 79 of 86 2a) A bullet of mass 10 g is fired at 400 m s-1 from a rifle of mass 4.0 kg. What is the recoil velocity of the rifle? b) The bullet is fired into a block of wood of mass 390 g resting on a smooth surface. If the bullet remains embedded in the wood, calculate the velocity that the block moves off at. Page 80 of 86 3) A car of mass 1500 kg travelling at 12 m s-1 collides head-on with a lorry moving at 20 m s-1. The lorry has a mass of 9000 kg. If the collision reduces the speed of the lorry to 15 m s-1, what is the car’s velocity after the impact? In which direction is this? 4) A rocket of total mass 3500 kg is moving at 250 m s-1 through space. When the booster rockets are fired, 1200 kg of burnt fuel is ejected from the back of the rocket at 20 m s-1. What is the new speed of the rocket? Page 81 of 86 5) A gas molecule of mass 2.0 x 10-26 kg moving at 600 m s-1 collides with a stationary molecule of mass 10 x 10-26 kg. The first molecule rebounds at 400 m s-1. Is this collision elastic or inelastic? 6) A block of wood of mass 1kg is freely suspended by a string. A bullet of mass 10g is fired horizontally at the block from a close range and becomes embedded in it. The block swings to one side rising a vertical distance of 50cm. With what speed did the bullet hit the block? Page 82 of 86 7) A toy car A of mass 0.1 kg travelling at a velocity u collides elastically with a stationary toy car B of mass 0.4 kg. By the impact, toy car B is shunted forward with a velocity of 2 m s-1, whereas toy car A moves with velocity v. Calculate the initial and final velocities of toy car A given that the collision is perfectly elastic. 8) A toy train moving eastwards with a velocity of 6 m s-1 collides head on with another toy train which was moving westwards with a velocity of 4 m s-1. After collision the two toy trains stick together and move with a common velocity of 3 m s-1 in the eastward direction. Given that 100 J of energy were transformed to sound and heat energy during collision, calculate the mass of each toy train. Page 83 of 86  Appendix: Ü The Air Track The air track is a hollow tube of triangular cross-section through which air is blown; the air emerges through holes in each side of the track. It has adjustable feet allowing it to be made accurately horizontal so that a plastic vehicle has no tendency to drift along it in either direction. The air track acts as a frictionless runway. Ü Finding velocity The plastic vehicle can carry an opaque card of known length (e.g. 10 cm) which is arranged to interrupt a beam of light falling on a photodiode. The circuitry is such that the millisecond timer is inoperative whilst light is falling on the photodiode to which it is connected. When the light beam is broken by the leading edge of the card, the timer switches on and remains operative for as long as the card is in the beam. The timer therefore records the time for the vehicle to travel a distance equal to the length of the card and so allows the speed to be found. 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑎𝑞𝑢𝑒 𝑐𝑎𝑟𝑑 𝑠𝑝𝑒𝑒𝑑 = = 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟. Page 84 of 86 Ü Finding acceleration The plastic vehicle can carry an opaque card of known length (e.g. 10 cm) which is arranged to interrupt a beam of light falling on a photodiode. The circuitry is such that each of the millisecond timers is inoperative whilst light is falling on the photodiode to which it is connected. When the light beam is broken by the leading edge of a card, the associated timer switches on and remains operative for as long as the card is in the beam. The timer therefore records the time for the vehicle to travel a distance equal to the length of the card and so allows the speed to be found. 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑎𝑞𝑢𝑒 𝑐𝑎𝑟𝑑 𝑠𝑝𝑒𝑒𝑑 = = 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑜𝑛 𝑡𝑖𝑚𝑒𝑟. The initial velocity is found from timer 1, whilst the final velo