# Summary Notes. to avoid confusion it is better to write this formula in words. time

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1 National 4/5 Physics Dynamics and Space Summary Notes The coloured boxes contain National 5 material. Section 1 Mechanics Average Speed Average speed is the distance travelled per unit time. distance (m) average speed (m/s) Note to avoid confusion it is better to write this formula in words. average speed = distance time SI unit of average speed is m/s (metres per second) since the SI unit for distance is metres (m) and SI unit for time is seconds (s). s of alternative units for average speed are kilometres per hour (km/h) or miles per hour (mph). Measurement of average speed To measure an average speed, you must: measure the distance travelled with a measuring tape, metre stick or trundle wheel measure the time taken with a stop clock calculate the average speed by dividing the distance by the time 1

2 Calculations involving distance, time and average speed Note: care must be taken to use the correct units for time and distance. Calculate the average speed in metres per second of a runner who runs 15 m in 5 minutes. Solution distance = 15 m t = 5 minutes = 5 6 seconds = 3 s v =? Instantaneous speed The instantaneous speed of a vehicle at a given point can be measured by finding the average speed during a very short time interval as the vehicle passes that point. Average speed and instantaneous speed are often very different e.g. the average speed for a car over an entire journey from Glasgow to Edinburgh could be 4 mph, but at any point in the journey, the instantaneous speed of the car could be 3 mph, 7 mph, 6 mph or mph if the car is stationary at traffic lights. Measuring Instantaneous Speed To measure instantaneous speeds, it is necessary to measure very short time intervals. With an ordinary stop clock, human reaction time introduces large uncertainties. These can be avoided by using electronic timers. The most usual is a light gate. A light gate consists of a light source aimed at a photocell. The photocell is connected to an electronic timer or computer. photocell light source electronic timer or computer 2

3 The timer measures how long an object takes to pass through the light beam. The distance travelled is the length of the object which passes through the beam. Often a card is attached so that the card passes through the beam. Procedure to measure the instantaneous speed of a trolley card photocell TIMER trolley light source Set up apparatus as shown Measure length of card Push trolley through light beam Timer measures time taken for card to pass through light gate Calculate instantaneous speed by In the above experiment, the following measurements are made: Length of card: 1 cm Time on timer:.25 s Calculate the instantaneous speed of the vehicle Solution Length of card = 1 cm =.1 m Time on timer =.25 s =.4 m/s 3

4 Vector and Scalar Quantities A scalar quantity is fully described by its magnitude (size) and unit, e.g. quantity time = 22 s unit magnitude A vector quantity is fully described by its magnitude, unit, and direction, e.g. unit quantity Force = 8 N upwards direction magnitude Distance and Displacement Distance is a scalar quantity. It measures the total distance travelled, no matter in which direction. Displacement is a vector quantity. It is the length measured from the starting point to the finishing point in a straight line. Its direction must be stated. A girl walks 3 km due north then turns and walks 4 km due east as shown in the diagram. 4 km FINISH km 18 START Calculate (a) The total distance travelled (b) The girl s final displacement relative to her starting position. 4

5 Solution (a) Total distance = 3 km + 4 km = 7 km (b) (to calculate displacement, we need to draw a vector diagram) (this solution involves using a scale diagram). 4 km FINISH Choose an appropriate scale Remember to join the vectors tip-to-tail 3 km θ Resultant displacement Measure the resultant displacement and the angle. Resultant displacement = 5 km at 53 START Speed and Velocity Speed is a scalar quantity. As discussed above, speed is the distance travelled per unit time. Velocity is a vector quantity (the vector equivalent of speed). Velocity is defined as the displacement per unit time. Since velocity is a vector, you must state its direction. The direction of velocity will be the same as the displacement. The girl s walk in the previous example took 2.5 hours. Calculate (a) The average speed (b) The average velocity for the walk, both in km/h 5

6 Solution (a) Distance = 7 km Time = 2.5 hours = 2.8 km/h (b) Displacement = 5 km (53 ) Time = 2.5 hours v = s t at 53 Acceleration Acceleration is defined as the change in velocity per unit time. The change in velocity can be given by the moving object s final velocity (v) initial velocity (u) Therefore we can write our equation for acceleration as final velocity (m/s) initial velocity (m/s) acceleration (m/s 2 ) metres per second per second Acceleration is a vector quantity. If the final speed is less than the initial speed, the acceleration is in the opposite direction to motion and will be negative. This indicates a deceleration. The equation for acceleration can be rearranged in terms of the final velocity, v 6

7 A car is travelling with an initial velocity of 15 m/s. The driver presses the accelerator pedal and the car accelerates at a rate of 2 m/s 2 for 4 s. Calculate the final velocity of the car. v =? u = 15 m/s a = 2 m/s 2 t = 4 s v = u + at = 15 + (2x4) = 23 m/s Measuring Acceleration mask photocells computer trolley.453 runway light sources Set up apparatus as shown Measure length of mask Let the trolley accelerate past the two photocells Computer measures time taken for card to pass through both light gates. Computer also measures the time between both light gates u is calculated using the formula for distance, instantaneous speed and time v is calculated using the formula for distance, instantaneous speed and time acceleration is then calculated using a = v - u t 7

8 speed (m/s) Speed Time Graphs Speed time graphs are useful to physicists because they can be used to: a) describe the motion of the object b) calculate the acceleration of the object c) calculate the distance travelled by the object 2 18 C A B O D a) OA: Constant acceleration from to 1 m/s in 5 seconds. b) OA: acceleration = Δv/t acceleration = (1 )/5 acceleration = 2m/s 2 c) To calculate the distance the area under the graph must be used C speed (m/s) A B III O I II IV D

9 velocity (m/s) velocity (m/s) velocity (m/s) velocity (m/s) Area I : Triangle area = ½ x b x h = ½ x 5 x 1 = 25m Area II : Rectangle area = l x b = 4 x 1 = 4m Area III : Triangle area = ½ x b x h = ½ x 2 x 8 = 8m Area IV : Triangle area = ½ x b x h = ½ x 3 x 18 = 27m Total Area Distance travelled = 1m = 1m Velocity-time graphs Velocity-time graphs provide a useful way of displaying the motion of an object. They can be used to: 1) Describe the motion of the object in detail. 2) Calculate values for accelerations and decelerations using gradients. 3) Calculate distances travelled and resultant displacements using area under graph. 4) Calculate the average velocity for a journey. constant velocity constant acceleration constant deceleration constant deceleration to rest,then constant acceleration in the opposite direction 9

10 velocity (m/s) The graph below shows the motion of a car over 35 seconds (a) Describe the motion of the car during the 35 seconds (b) Calculate the acceleration between and 1 seconds (c) Calculate the acceleration between 3 and 35 seconds (d) Calculate the final displacement of the car from the starting position (e) Calculate the average velocity during the 35 seconds. Solution (a) 1 seconds: constant acceleration from rest to 16 m/s 1 3 seconds: constant velocity of 16 m/s 3 35 seconds constant deceleration from 16 m/s to rest. (b) a =? v = 16 m/s u= t = 1 s a = v-u t a = 16-1 a = 1.6 m/s 2 (c) a =? v = m/s u= 16 t = 5 s a = v-u t 1

11 a = 16-5 a = 3.2 m/s 2 (d) displacement = area under v-t graph = (½ x 1 x 16) + (2 x 16) + (½ x 5 x 16) = = 44 m (e) average velocity =? displacement = 44 m time = 35 s v = s t = = 12.6 m/s 11

12 Newton s Laws Effects of Forces We can t see forces themselves, but we can observe the effects that they have. Forces can change: the shape of an object the speed of an object the direction of travel of an object Measuring Force The unit of force is the Newton. The size of a force can be measured using a device called a Newton Balance. scale A force is applied to the hook and the spring is stretched. The greater the force applied, the greater the extension of the spring. spring The size of the force is read from the scale. force to be measured is applied to the hook here Mass and Weight Mass and Weight mean different things! Mass is a measure of the amount of matter that an object is made of. The units of mass are kilograms (kg). Weight is the size of the force that gravity exerts on an object. The units of weight are Newtons (N) since weight is a force. 12

13 Gravitational field strength Gravitational field strength is the weight per unit mass, i.e. gravitational field strength is the amount of force that gravity exerts on every kilogram of an object. Therefore the units of gravitational field strength are Newtons per kilogram (N/kg). This can be rearranged to give: weight (N) mass (kg) gravitational field strength (N/kg) The gravitational field strength on the surface of Earth is 1 N/kg. The gravitational field strength on the surface of the Moon is 1.6 N/kg. Calculate the weight of an astronaut of mass 6 kg (a) on the surface of the Earth (b) on the surface of the Moon Solution (a) W=? m=6 kg g=1 N/kg W = m g = 6 x 1 = 6 N (b) W=? m=6 kg g=1.6 N/kg W = m g = 6 x 1.6 = 96 N 13

14 Friction Friction is a resistive force, which opposes the motion of an object. This means that it acts in the opposite direction to motion. Friction acts between any two surfaces in contact. When one surface moves over another, the force of friction acts between the surfaces and the size of the force depends on the surfaces, e.g. a rough surface will give a lot of friction. Friction caused by collisions with particles of air is usually called air resistance. It depends mainly on two factors: the shape and size of the object the speed of the moving object. Air resistance increases as the speed of movement increases. Increasing and Decreasing Friction Where friction is making movement difficult, friction should be reduced. This can be achieved by: lubricating the surfaces with oil or grease separating the surfaces with air, e.g. a hovercraft making the surfaces roll instead of slide, e.g. use ball bearings streamlining to reduce air friction. Where friction is used to slow an object down, it should be increased. This can be achieved by: choosing surfaces which cause high friction e.g. sections of road before traffic lights have higher friction than normal roads increasing surface area and choosing shape to increase air friction, e.g. parachute. 14

15 Force is a vector Force is a vector quantity. The direction that a force acts in will dictate the effect that it has. Balanced Forces Two forces acting in opposite directions are said to be balanced. The effect is equivalent to no force acting on the object at all. Newton s First Law An object will remain at rest or move at a constant velocity in a straight line unless acted on by an unbalanced force. This means that if the forces acting on an object are balanced, then the object will remain stationary if it was already stationary. For a moving object, if the forces acting on it are balanced, then it will continue to move in a straight line at a constant velocity. balanced forces Object at rest OR object moving at a constant velocity in a straight line Newton s Second Law If an overall resultant (or unbalanced) force is applied to an object, then the object will accelerate in the direction of the unbalanced force. The greater the resultant force, the greater the acceleration. The smaller the mass, the greater the acceleration. These two relationships are combined to give (resultant or unbalanced) force (N) acceleration (m/s 2 ) mass (kg) This equation shows that a force of 1 N is the force needed to accelerate a mass of 1 kg at a rate of 1 m/s 2 A trolley of mass 2 kg accelerates at a rate of 2.5 m/s 2. Calculate the resultant force acting on the trolley. 15

16 2.5x1 6 kg Solution F u =? m=2kg a=2.5 m/s 2 F u = m a F u = 2 x 2.5 F u = 5 N Free Body Diagrams and Newton s Second Law Free body diagrams are used to analyse the forces acting on an object. The names of the forces are labelled on the diagram. (It is also useful to mark the sizes of the forces on the diagram). This enables us to calculate the resultant force, and then apply F un =ma. A space rocket of mass 2.5x1 6 kg lifts off from the earth with a thrust of 3.5x1 7 N. (a) Draw a free body diagram of the rocket at lift-off (b) Calculate the resultant force acting on the rocket (c) Calculate the initial acceleration of the rocket. Solution (a) thrust weight (Initially, no air resistance will act on the rocket at the instant it lifts off. Air resistance will only begin to act once the rocket starts moving). (b) F un = thrust weight W = mg = 3.5x x1 7 = 2.5x1 6 x 1 = 2.5x1 7 N = 1. x 1 7 N 16

17 (c) F u = 1 x1 7 N m = 2.5x1 6 kg a =? 1 x1 7 = 2.5x1 6 x a F u = 4 m/s 2 Forces at Right Angles The resultant force arising from two forces acting at right angles can be found by (a) (b) Drawing the force vectors nose-to-tail Then find the magnitude using a ruler and your chosen scale, and the direction using a protractor. Two forces act on an object at right angles to each other as shown. 2 N 4 N Calculate (a) The magnitude of the resultant force (b) The direction of the resultant force, relative to the 4 N force. 17

18 Solution (arrange vectors nose-to-tail and draw resultant from tail of first to nose of last) Using a scale diagram, work out the size of the resultant force and measure the angle. Resultant Force θ 4 N 2 N Resultant force = 44.7 N at 26.6 to 4 N force (question asks for direction relative to 4 N force) Acceleration Due to Gravity and Gravitational Field Strength Weight is the force which causes an object to accelerate downwards and has the value mg, where g is the gravitational field strength. The value of the acceleration caused by weight can be calculated from Newton s second law, using F un = ma where F is now the weight W, and W = mg. a = F u m a = W m a = mg m a = g This shows that the acceleration due to gravity and the gravitational field strength are numerically equivalent. 18

19 Therefore the acceleration due to gravity on Earth is 1 m/s 2 since the gravitational field strength 1 N/kg. is The acceleration due to gravity on the moon is 1.6 m/s 2 since the gravitational field strength is 1.6 N/kg. An astronaut drops a hammer on the Moon where the gravitational field strength is 1.6 m/s 2. The hammer lands on the surface of the Moon 2.2 s after being dropped. Calculate the velocity of the hammer at the instant it strikes the surface of the Moon. Solution u= v=? a=1.6 m/s 2 t=2.2 s v = u + at v = + (1.6x2.2) v = 3.52 m/s Free-fall and Terminal Velocity An object falling from a height on Earth will accelerate due to its weight. This is called free-fall. As the velocity increases, the air resistance acting on the object will increase. This means that the unbalanced force on the object will decrease, producing a smaller acceleration. Eventually, the air resistance will balance the object s weight, meaning that the object will fall with a constant velocity. This final velocity is called terminal velocity. The velocity-time graph shown below is for a parachutist undergoing free-fall before opening her parachute. Constant velocity Constant velocity B C D A 19

20 Point A Point B Point C Point D parachutist jumps from plane and undergoes acceleration due to gravity (free-fall) air resistance has increased to balance weight constant velocity: terminal velocity 1 parachutist opens parachute increased air resistance causes deceleration weight and air resistance balanced again so new slower constant velocity reached: terminal velocity 2 Newton s Third Law For every action, there is an equal and opposite reaction. Newton noticed that forces occur in pairs. He called one force the action and the other the reaction. These two forces are always equal in size, but opposite in direction. They do not both act on the same object. If an object A exerts a force (the action) on object B, then object B will exert an equal, but opposite force (the reaction) on object A. s Skateboarder pushes wall to the left (action force) Water pushes swimmer to left (reaction force) Wall pushes skateboarder to right (reaction force) Swimmer pushes water to right (action force) Fuel forces rocket upwards (reaction force) Cannon fires cannonball to the right (action force) Rocket forces fuel downwards (action force) Cannonball forces cannon back to the left (reaction force) These action and reaction forces are also known as Newton Pairs. 2

21 Projectile Motion A projectile is an object which has been given a forward motion through the air, but which is also pulled downward by the force of gravity. This results in the path of the projectile being curved. A projectile has two separate motions at right angles to each other. Each motion is independent of the other. The horizontal motion is at a constant velocity since there are no forces acting horizontally (air resistance can be ignored). horizontal velocity (m/s) So horizontal distance (m) The vertical motion is one of constant acceleration, equal to g (1 m/s 2 on earth). For projectiles which are projected horizontally, the initial vertical velocity is zero. For vertical calculations, use final vertical velocity (m/s) v v = u v + at where u v = and a = g initial vertical velocity (m/s) acceleration due to gravity (m/s 2 ) 21

22 velocity (m/s) velocity (m/s) Velocity time graphs for horizontal and vertical motion Horizontal motion Vertical motion constant velocity constant acceleration from rest horizontal distance can be found using vertical height can be found from area area under graph (equivalent to d h =v h t) under graph (equivalent to h=½ v max t) An aircraft flying horizontally at 2 m/s drops a food parcel which lands on the ground 12 seconds later. Calculate: Solution a) the horizontal distance travelled by the food parcel after being dropped b) the vertical velocity of the food parcel just before it strikes the ground c) the height that the food parcel was dropped from. (a useful layout is to present the horizontal and vertical data in a table) (then select the data you need for the question, and clearly indicate whether you are using horizontal or vertical data). initial vertical velocity always for a projectile launched Horizontal Vertical horizontally d h =? u v = v h = 2 m/s v v =? t = 12 s a = 1 m/s 2 time of flight the same for both t = 12 s acceleration due to gravity 22

23 velocity (m/s) (a) Horizontal: d h =? v h = 2 m/s t = 12 s d h = v h t d h = 2 x 12 d h = 24 m (b) Vertical: u v = v v =? a = 1 m/s 2 t = 12 s v v = u v + at v v = + 1 x 12 v v = 12 m/s (c) (Draw a velocity time graph of vertical motion) 12 height = area under v-t graph = ½ x 12 x 12 = 72 m 12 23

24 Work Work is a measure of how much energy has been used or transferred. work done = force x distance E w = F x d distance (m) work done (J) force (N) An engine exerts a pull of 5 N to pull a train a distance of 4 m. How much work is done? Solution E w =? F = 5 N d = 4 m E w = F x d E w = 5 x 4 E w = 2 J Gravitational potential energy When an object is lifted it will gain energy. This energy is known as gravitational potential energy. gravitational potential energy = mass x gravitational field strength x height E P = m x g x h gravitational potential energy (J) E P = m x g x h mass (kg) height (m) gravitational field strength (N/kg) 24

25 A lift raises a person of mass 55 kg to a height of 6 m. How much potential energy does the person gain? Solution E P =? m = 55 kg g = 1 N/kg h = 6 m E P = m x g x h E P = 55 x 1 x 6 E P = 33 J Power Power is a measure of how quickly energy is converted. The more powerful an appliance, machine or person the greater the quantity of energy converted each second. Power = energy time power (W) P = E t energy (J) time (s) A man pushes a wheelbarrow for 6 m using a 5 N force. If he takes 1 s, what is his average power? Solution This must be answered in two steps. Calculate the work done, then the power. E w =? F = 5 N d = 6 m E w = F x d E w = 5 x 6 E w = 3 J 25

26 P =? E = 3 J t = 1 s P = E t P = 3 1 P = 3 W Kinetic energy Kinetic energy is the energy of movement. The kinetic energy of an object increases with its speed. Kinetic energy = ½ x mass x speed 2 Kinetic energy (J) E k = ½ x m x v 2 mass (kg) speed (m/s) Calculate the kinetic energy of a.2 kg bullet moving at 1 m/s. Solution E k =? m =.2 kg v = 1 m/s E k = ½ x m x v 2 E k = ½ x.2 x 1 2 E k = ½ x.2 x 1 E k = 1 J Conservation of Energy Energy cannot be created or destroyed but it can change from one form to another. Sometimes the energy produced is not useful; we say that this energy has been lost to the system. Normally this energy is lost as heat due to friction. 26

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