Higher Physics Our Dynamic Universe Notes

Transcription

1 Higher Physics Our Dynamic Universe Notes Teachers Booklet

2 Previous knowledge This section builds on the knowledge from the following key areas from Dynamics and Space Booklet 1 - Dynamics Velocity and Displacement vectors and scalars Velocity-time graphs Acceleration Learning Outcomes At the end of this section you should be able to: o Carry out calculations using the following kinematic relationships s = vt v = u + at s = ut + ½ at 2 v 2 = u 2 + 2as s = ½ (u + v) t o Draw displacement-time graphs o Draw velocity-time graphs o Draw acceleration-time graphs given a velocity-time graph o Draw the velocity time graph for objects thrown vertically upwards o Draw the velocity time graph for bouncing objects All graphs are restricted to constant acceleration in one dimension, inclusive of change of direction o State that the gradient of a displacement-time graph gives velocity o State that the gradient of a velocity-time graph gives acceleration o State that the area under a velocity-time graph is equal to the displacement

3 Equations of Motion The equations of motion are 1. v = u + at 2. s = ut + ½ at 2 3. v 2 = u 2 + 2as where s = displacement (m) u = initial velocity (ms -1 ) v = final velocity (ms -1 ) a = acceleration (ms -2 ) t = time (s) If you have any three pieces of information from the list suvat you are able to find the other two using the appropriate equation. Deriving the Equations (for information) 1. Acceleration is defined as the rate of change of velocity so a = v-u t ð v u = at ð v = u + at 2. velocity v u 2 Displacement = area under Velocity-time graph s = Area 1 + Area 2 = ut + ½ (v-u) t 1 But v-u = at (from equation 1) 0 t (time) ð s = ut + ½ at x t ð s = ut + ½ at 2 3. v = u + at v 2 = (u + at) 2 square both sides of the equation v 2 = u 2 + 2uat + a 2 t 2 But s = ut + ½ at 2 v 2 = u 2 + 2a (ut + ½ at 2 ) v 2 = u 2 + 2as Motion Equations and Graphs 3

4 Equations of Motion Example 1 A car accelerates from rest. Its velocity after 8 seconds is 26ms -1. Calculate: (a) the acceleration (b) how far it travels in 15 seconds. s =? u = 0 ms -1 v = 26 ms -1 a =? t = 8s (a) a = v-u = 26 0 = 3.25ms -2 t 8 (b) Using the value for a calculated in part (a) s = ut + ½ at 2 ð s = (0 x 15) + (½x 3.25 x 15 2 ) ð s = 365.6m Example 2 SQA 2008 Q21 adapted A car is driven at a constant speed of 30ms -1 until it reaches the start of the braking zone at P. The brakes are then applied. Calculate the length of the braking zone if the car slows down at a constant rate of 9 ms -2 and the speed at Q is recorded as 12 ms -1. s =? u = 30 ms -1 v = 12 ms -1 a = -9ms -2 t =? v 2 = u 2 + 2as 12 2 = x (-9) x s 144 = (-18s) 18s = s = 756 = 42m 18 Motion Equations and Graphs 4 ODU Problem booklet 1 P7 Q1-10

5 Graphs Displacement, velocity and acceleration are all interrelated. The gradient of a displacement-time graph gives the velocity. The gradient of a velocity-time graph gives the acceleration. The area under a velocity-time graph gives the displacement. When drawing graphs o Use a ruler o o Label the axes Put zero at the origin Constant Increasing Decreasing Velocity velocity velocity Velocity Displacement s 0 t v 0 t s 0 t v 0 t s 0 t v 0 t Acceleration a 0 t a 0 t a 0 t Motion Equations and Graphs 5

6 Comparison between v-t and a-t graphs for an object moving in a straight line Velocity (ms -1 ) A B C D Time (s) E F 4 A B E F 3 Acceleration (ms - 2 ) C D Time (s) From A to B a = v-u = 6 0 = 3 ms -2 t 2 From B to C a = v-u = 6 6 = 0 ms -2 t 2 From C to E a = v-u = -6 6 = -3 ms -2 t 4 From E to F a = v-u = 0- (-6) = 3 ms -2 t 2 Motion Equations and Graphs 6 ODU Problem booklet 1 P9 Q1-7

7 Object thrown upwards B v A Displacement upwards 0 B t A C Displacement downwards C Gravity acts downwards at a constant value of 9.8 ms -2. At A Ball has velocity, which decreases as it goes up. At B Ball has zero velocity (at top of flight) At C Ball increases in velocity as it falls (opposite direction so negative). Velocity has the same magnitude at A but is in the opposite direction. Total displacement = zero speed Speed-time Graph for comparison A C 0 t Distance up Distance down B Motion Equations and Graphs 7

8 Velocity-time graph for bouncing ball A D Velocity ms -1 B E F Time (s) C AB B BC CD DE E EF ball leaves hand and decelerates upwards - top of flight ball accelerates downwards ball touches ground with a large force causing it to change direction in a very short time period. ball decelerates upwards top of flight - ball falls again Each time this repeats some energy is lost as heat (and sound), so the ball does not rise to the same height as before. The gradient of the graph is negative and constant which shows that acceleration due to gravity is constant at -9.8 ms -2. The area under the graph represents the displacement during each part of the motion. Velocity time graph Displacementtime graphs Displacement Velocity Acceleration Area under the graph Read straight from line of best fit Read straight from line of best fit v = Δs Δt = Δy Δx = m v = gradient of graph a = Δv Δt = v u = Δy t Δx = m a = gradient of graph Motion Equations and Graphs 8 ODU Problem booklet 1 P15 Q8-10

9 Previous knowledge This section builds on the knowledge from the following key area from Dynamics and Space Booklet Projectiles Learning Outcomes At the end of this section you should be able to o Resolve initial horizontal and vertical velocity from velocity acting at an angle to the horlzontal o State that horizontal motion for projectiles is constant o State that vertical motion for projectiles is affected by the force due to gravity o Compare projectiles to objects in freefall Gravitation 9

10 1. No horizontal motion Direction of motion Vertical Forces Velocity Acceleration Weight Changes with time -9.8 ms -2 No horizontal motion Initial vertical motion = 0 ms -1 Final vertical motion => v = u + at = at (since u = 0) Acceleration = -9.8 ms -2 (Remember if upwards is positive, acceleration due to gravity is negative because it is a vector quantity) Example 3 A ball is dropped off a bridge. It takes 4 seconds to land in the water. How high is the bridge? s =? u = 0 ms -1 v =? a = 9.8 ms -2 t = 4s s = ut + ½ at 2 ð s = (0 x 4) + (½x 9.8 x 4 2 ) ð s = ð s = 78.4m Example 4 A ball is dropped 1.6 m onto a force sensor. Show that the final velocity is 5.6m. SQA 2006 Q22 adapted s = 1.6m u = 0 ms -1 v =? a = 9.8 ms -2 t =? v 2 = u 2 + 2as v 2 = x 9.8 x 1.6 v = v = 5.6 ms -1 as required. Gravitation 10

11 1. No horizontal motion B A At A - the object has maximum upwards velocity - u v At B the object has reached its maximum height and vertical velocity is zero. At C the objects velocity has the same magnitude as at A but is in the opposite direction. A C Example 5 A ball is thrown into the air with a velocity of 7ms -1. How long does it take to get to the maximum height? What is that height? s =? u = 7 ms -1 v = 0 ms -1 a = -9.8ms -2 t =? Time t = v-u = 0 7 a -9.8 = 0.71 s Height v 2 = u 2 + 2as 0 2 = (2 x -9.8 x s) 0 = s 19.6s = 49 s = 49 = 2.5 m 19.6 Example 6 A person on the end of a bungee cord is fired upwards to a height of 29.9m.. They take 2.47s to reach their maximum height. What was their initial velocity? s = 29.9m u =? ms -1 v = 0 ms -1 a = -9.8ms -2 t = 2.47s u = v at = 0 (- 9.8 x 2.47) = 24.2 ms -1 Gravitation 11

12 2. An object thrown horizontally height range Look at horizontal and vertical velocity separately. Horizontal velocity is constant. The higher the horizontal velocity the further the object travels as it falls. Initial vertical velocity = zero - use v= u + at to calculate final velocity. Calculate displacement using area under a graph OR one of the equations of motion with s in it. Direction of Forces Velocity Acceleration motion Horizontal None Constant None Vertical Weight Changing with time -9.8 ms -2 Example 7 A ball rolls off a table with a velocity of 2.3 ms -1. What is the height of the table if it lands 1.2 m away from the table? Horizontal s h = 1.2 m u h = 2.3 ms -1 v h = 2.3 ms -1 a h = 0 ms -2 t h =? t = s h u h = = 0.52s Vertical s v =? u v = 0 ms -1 v v =? a v = 9.8 ms -2 t v = 0.52 s s = ut + ½ at 2 ð s = (0 x 0.52) + (½x 9.8 x ) ð s = 1.3 m Gravitation 12

13 2. An object thrown horizontally Example 8 A stuntman on a motorcycle jumps a river, which is 5.1 m wide. He lands on the edge of the far bank, which is 2.0 m lower than the bank from which he takes off. Calculate his minimum take off speed to ensure a safe landing. Vertical s v = 2.0m u v = 0 ms -1 v v =? ms -1 a v = 9.8 ms -2 t v =? (SQA 2004 Q2 adapted) s = ut + ½ at 2 ð 2 = (0 x t) + (½x 9.8 x t 2 ) ð 2 = 4.9t 2 ð t = (2/4.9) ð t = 0.64s Horizontal s h = 2.0m u h =? ms -1 v v = u v ms -1 a h = 0 ms -2 t h = 0.64s u h = s h t = = = 8.0 ms -1 Gravitation 13

14 Graphs for an object thrown horizontally The information about the horizontal and vertical velocity for an object can be combined to calculate the resultant velocity. V h (ms -1 ) 10 V v (ms -1 ) time (s) From the two graphs we can tell that at 4s the horizontal velocity is 10ms -1 and the vertical velocity is 40ms -1. To find the resultant velocity either A draw a vector diagram. Or B use Pythagoras and sohcahtoa 10ms time (s) Length of X = 10! + 40! = 41.2 ms -1 X 40ms -1 Tan θ = opp = 40 adj 10 => θ = 76 θ Gravitation 14 ODU Problem booklet 1 P17 Q1-3

15 3. An object thrown upwards at an angle θ Resolve the velocity into its horizontal and vertical components (this is possible because any vector can be split up into two other vectors at right angles to one another) V h = cosθ = adj = V h hyp v V v = sinθ = opp = V v hyp v => V h = v cosθ => V v = v sinθ height u v v v range The path of a projectile is symmetrical about the highest point in the horizontal plane. Initial vertical velocity = -final vertical velocity U v = -V v The time of flight = twice the time to the highest point V vertical at highest point = zero Direction of motion Horizontal Vertical Forces Velocity Acceleration Air resistance is negligible so no forces Air resistance is negligible so only the force of gravity Constant Changing at a constant rate Zero Constant or uniform acceleration of 9.8 ms -2 Gravitation 15

16 3. An object thrown upwards at an angle Example 9 A gun is fired with a velocity of 16 ms -1 at an angle of 40 above the horizontal. Find (a) the horizontal and vertical components of the balls velocity (b) The maximum height reached by the ball (c) The horizontal distance the ball travels. a) 16 ms -1 u v 40 U h = 16 cos 40 = 12.3 ms -1 U v = 16 sin 40 = 10.3 ms -1 u h b) Vertical s =? v 2 = u 2 + 2as u v = 10.3 ms = (2 x -9.8 x s) v v = 0 ms s = a = -9.8 ms -2 s = 5.4 m t =? c) Vertical v v = u v + at Total time in air = 2 x 1.05 = 2.1s 0 = (-9.8 x t) t = = 1.05 s Horizontal -9.8 s = vt = 12.3 x 2.1 = 25.8m Gravitation 16

17 3. An object thrown upwards at an angle Example 10 A basketball player throws a ball with initial velocity of 6.5ms -1 at an angle of 50 to the horizontal. The ball is 2.3 m above the ground when released. The ball travels a horizontal distance of 2.9m to reach the top of the basket. The effects of air resistance can be ignored. (SQA H 2009 adapted). (a) Calculate: (i) the horizontal component of the initial velocity of the ball (ii) the vertical component of the initial velocity of the ball. (b) If the time taken for the ball to reach the basket is 0.69s calculate the height h of the top of the basket. (a) (i) u h = v cos θ = 6.5 cos 50 = 4.18 ms -1 (ii) u V = v sin θ = 6.5 sin 50 = 4.98 ms -1 (b) s =? u v = 4.98 ms -1 v v =? ms -1 a = - a = -9.8 ms -2 t = 0.69s s = ut + ½ at 2 ð s = (4.98 x 0.69) + (½x -9.8 x ) ð s = (-2.33) ð s = 1.11m Actual height to top of basket = = 3.41m Gravitation 17

18 3. An object thrown upwards at an angle Example 11 A golfer on an elevated tee hits a golf ball with an initial velocity of 35.0 ms -1 at an angle of 40 to the horizontal. The ball travels through the air and hits the ground at point R. Point R is 12 m below the height of the tee, as shown. The effects of air resistance can be ignored. SQA 2003 Q21 adapted a) Calculate: i) The horizontal component of the initial velocity of the ball ii) The vertical component of the initial velocity of the ball. iii) The time taken for the ball to reach its maximum height at point P. b) From its maximum height at point P, the ball falls to point Q, which is at the same height as the tee. It then takes a further 0.48s to travel from Q until it hits the ground at R. Calculate the total horizontal distance d travelled by the ball. (a) (i) u h = v cos θ = 35 cos 40 = 26.8 ms -1 (ii) u V = v sin θ = 35 sin 40 = 22.5 ms -1 (iii) s =? u v = 22.5 ms -1 v v = 0 ms -1 a = - a = -9.8 ms -2 t =? t = v-u = = 2.3s a -9.8 (b) t total = = 5.08s s h = u h t total = 26.8 x 5.08 = 136.1m Gravitation 18 ODU Problem booklet 1 P17 Q4-12

19 Previous Knowledge This section builds on the knowledge from the following key area from Dynamics and Space Booklet 2 - Forces Newton s Laws It also builds on the knowledge from the following key area from Electricity and Energy Booklet 1 - Conservation of Energy Learning Outcomes At the end of this section you should be able to o Analyse motion using Newton s first and second laws (for forces acting in one plane only) o Recognise balanced and unbalanced forces o Describe friction as a force acting in a direction to oppose motion o Describe tension as a pulling force exerted by a string or cable on another object o Draw the velocity-time graph of a falling object when air resistance is taken into account, including changing the surface area of the falling object o Analyse the motion of a rocket which may involve a constant force on a changing mass as fuel is used up. o Resolve forces acting at an angle into two perpendicular forces o Calculate forces acting at an angle to the direction of movement. o Resolve the weight of an object on a slope into a component acting down the slope and a component acting normal to the slope. o Combine systems of two or more forces acting in two dimensions to obtain a resultant force. o Carry out calculations using the following equations o E w = Fd E k =½mv 2 E p =mgh E= Pt o Apply conservation of energy where appropriate Forces, Energy and Power 19

20 Newton s Laws 1. An object will remain at rest or move with constant velocity (constant speed in a straight line) unless acted on by an external force. F m F 2. An object will accelerate when acted upon by an unbalanced force. m F Acceleration => F = ma F = Force (newtons N) m = mass (kilogramme kg) a = acceleration (metres per second squared ms -2 ) Definition of the Newton One newton is the force required to accelerate a mass of 1kg at a rate of 1ms -2 An acceleration of 1ms -2 means that the velocity changes by 1ms -1 each second. Force is a vector quantity. If several forces act on an object they can be added together. This allows us to calculate whether they are balanced or unbalanced. Balanced forces 4N 4N Forces are balanced so object will remain at rest (or move with constant velocity if it is moving) Forces, Energy and Power 20

21 Analysing Forces Unbalanced forces 4N 2kg 6N Original diagram 2kg 2N Equivalent diagram If forces are positioned directly opposite one another then the resultant force is the difference between the forces. If the forces are not directly opposite you need to draw a vector diagram. 6N 4N X 4N 6N 2kg 9N 2kg 3N 2kg θ 3N 2N To calculate the resultant force either a) Do a scale drawing and measure the resultant vector and angle. b) Use Pythagoras to calculate the magnitude of the resultant vector and sohcahtoa to calculate the angle. To calculate the length of side X X 2 = = 25 X= 5 To calculate the angle Tan θ = opp = 4 Adj 3 θ = 53.1 Forces, Energy and Power 21

22 Analysing Forces Example 12 What mass is required to lift a mass of 600 kg upwards with an acceleration of 2 ms -2. Air resistance is 800 N. F W = mg = 600 x 9.8 = 5880 N F = ma = 600 x 2 = 1200 N W = mg Air resistance Total force required = = 7880 N upwards Example 13 Calculate the minimum force required to accelerate a mass of 400 kg upwards with an initial acceleration of 6 ms -2, when air resistance is 250 N. a = 6 ms -2 W = mg = 400 x 9.8 = 3920 N F = ma = 400 x 6 = 2400 N W = mg 400 kg 250 N Total force required = = 6570 N upwards Forces, Energy and Power 22 ODU Problem booklet 1 P21 Q1 8, 11

23 Tension Tension is a pulling force applied by a string or cable on another object. a) Frictionless surface 1kg T 2kg F Example 14 Calculate the tension in the rope at T if a force of 6N is applied to the rope. Total mass = = 3kg If a force of 6N is applied then the acceleration will be a = F = 6 = 2ms -2 m 3 The tension in the rope at T = 2 x 1 = 2N Example 15 m 1 = 2000kg T m 2 = 3000kg An articulated lorry is returning from dropping off its load. The mass of the lorry and the trailer are shown in the diagram. Calculate: a) The acceleration of the system b) The tension in connection at T a) a = F = 9000 = 9000 = 1.8 ms -2 m ( ) 5000 b) F = ma = 2000 x 1.8 = 3000N Forces, Energy and Power 23

24 Tension b) Surface including friction SQA 2011 Q3 adapted Example 16 A car of mass 1200kg pulls a horsebox of mass 700kg along a straight, horizontal road. They have an acceleration of 2ms -2. If the friction between the car tyres and the road is 70 N and the horsebox and the road is 50N, what engine force is required to maintain that acceleration? Total mass = = 1900 kg F = ma = 1900 x 2 = 3800N Total force required = = 3920N m 1 = 6000kg T m 2 = 20,000kg 200N 600N Example 17 A train pulls a carriage. The train has a mass of 20,000 kg and the carriage has a mass of 6000 kg. The force of friction on the train is 600 N and on the carriage is 200 N. The whole system accelerates at 4 ms -2. Calculate: a) the force of the engine b) the force that the coupling exerts on the carriage a) F = ma = (20, ) x 4 = 2-4,000 N Total force required = 104, = 104,800 N b) F = ma = 6000 x 4 = 24,000N Total force = 224, = 24,200 N Forces, Energy and Power 24 ODU Problem booklet 1 P24 Q 16-18

25 Tension c) Mass over a pulley Tension in rope at T = ma = 1 x 2 = 2N m 1 m 2 T Both blocks accelerate. Total mass = m 1 + m 2 Force = weight of m 2 = m 2 g a = F = F = m 2 g m m 1 +m 2 m 1 +m 2 Tension in string at T = pulling force on m 1 => T = m 1 a Example 18 If m 1 = 2 kg and m 2 = 5 kg, calculate the tension at T, assuming the surface is frictionless. Total mass = = 7kg W = m 2 g = 5 x 9.8 = 49N a = F = F = 49 = 7ms -2 m m 1 +m 2 7 Tension in string at T T = m 1 a = 2 x 7 = 14N d) Masses suspended over a pulley m 1 m 2 1. Which is the greater mass? The system will rotate in that direction. e.g. if m 2 > m 1 it will rotate clockwise 2. W = mg = m 2 x The whole system moves, so a = W m 1 + m 2 Example 19 If m 1 = 5kg and m 2 = 3kg state the direction of rotation and find the tension in the string. m 1 > m 2 so system will rotate anticlockwise. W = mg = 5 x 9.8 = 49N a = F = 49 = ms -2 m Tension in string = pulling force on m 2 F = 3 x = 18.4N Forces, Energy and Power 25 ODU Problem booklet 1 P25 Q 20

26 Blocks Calculate: a) The force between the 6kg block and the 4 kg block b) The force between the 4 kg block and the 2 kg block. a) Frictionless surface 24N 6kg 4kg 2kg Example 20 a = F = 24 = 24 = 2ms -2 m Force between 6kg block and 4 kg block F= ma = (4 +2) x 2 = 12 N Force between 4kg block and 2 kg block F = ma = 2 x 2 = 4 N b) Friction between the surface and the 6kg, 4kg and 2kg blocks is 3N, 2N and 1N respectively. 30N 6kg 4kg 2kg 3N 2N 1N Example 21 a = F = 24 = 24 = 2ms -2 m Force between 6kg block and 4 kg block F= ma = (4 +2) x 2 = 12 N Force between 4kg block and 2 kg block F = ma = 2 x 2 = 4 N Forces, Energy and Power 26 ODU Problem booklet 1 P24 Q 19

27 Weight and Lifts A person of mass 60kg stands on a set of scales in a lift. The reading on the scales shows the upwards force on the person in newtons. This is equal to their weight (Newton s third law) W = mg = 60 x 9.8 = 588N a) Lift standing still OR lift travelling at constant speed. There is no unbalanced force so the reading is the persons weight. 588N b) Acceleration upwards has the same effect as deceleration downwards. You feel as if your weight has increased. If the rate of acceleration is 2.5 ms -2 F = ma = 60 x 2.5 Reading on scales = W + F = 150N = = 738N c) Acceleration downwards has the same effect as deceleration upwards. You feel as if your weight has decreased. If the rate of acceleration is 1.5 ms -2 F = ma = 60 x 1.5 Reading on scales = W - F = 90N = = 498N d) Freefall if the lift cable breaks If the lift cable breaks both the lift and the person will fall at the same rate. The person does not exert a force on the scales, so the reading will be 0N. Forces, Energy and Power 27

28 Weight and Lifts Example 22 A person of mass 85kg stands on a set of bathroom scales in a lift. What would be the reading, in newtons, if the lift was (a) stationary (b) accelerating upwards at 2ms -2 (c) decelerating upwards at 3ms -2 (d) accelerating downwards at 1.5ms -2 (e) moving at constant speed. (a) W = mg = 85 x 9.8 = 833N (b) F = ma = 85 x 2 = 170 N F tot = W + F = = 1003 N (c) F= ma = 85 x (-3) = -255 N F tot = W + F = (-255) = 578 N W = mg (d) F = ma = 85 x -1.5 = N F tot = W + F = (-127.5) = N (e) W = mg = 85 x 9.8 = 833N Example 23 A fully loaded lift has a total mass of 300kg. For safety reasons the tension in the pulling cable must never exceed 4440N. What is the maximum upwards acceleration of the lift? W = mg = 300 x 9.8 = 2940N F un = = 1500N a = F = 1500 = 5 ms -2 m 300 Forces, Energy and Power 28 ODU Problem booklet 1 P22 Q12-15

29 Rockets 1. At the instant of lift-off F un = Thrust (weight + air resistance) If F un > 0 then the rocket can take off. a = F un m 2. As lift off continues 1. Fuel is used up => mass decreases, so weight decreases. F un increases so a increases. 2. Air resistance decreases (as air gets thinner) 3. Gravitational field strength decreases (further from planet). Example 24 A rocket produces a thrust of 400,000 N it has a mass of 60,000 kg. Calculate the acceleration on take off. W = mg = 6000 x 9.8 = 58,800 N F un = 400,000 58,800 = 341,200 N a = F = 341,200 = 56.9 ms -2 m 6000 Forces, Energy and Power 29 ODU Problem booklet 1 P22 Q9-10

30 Resolving Forces θ F F h cos θ = adj = F h hyp F Fv Force is a vector quantity. This means a force can be resolved into two component forces, which are perpendicular to one another. sin θ = opp = F V hyp F F h = F cos θ F V = F sin θ Example 25 A box of weight 120N is placed on a smooth horizontal surface. A force of 20 N is applied to the box as shown in the diagram. Calculate the work done in pulling the box a distance of 50m along the surface. Component of weight acting parallel to the surface F parallel = F cos 30 Work done = Force x distance = 20 cos 30 = 17.3 x 50 = 17.3 N = 866J SQA 2009 Q3 adapted Forces, Energy and Power 30 ODU Problem booklet 1 P26 Q1-4

31 Forces in Two Dimensions Example N Find the acceleration of the mass shown in the diagram on the right. 60 kg Starting point 110N You have two choices 1. Draw a vector diagram to scale and measure the resultant force OR 2. The forces are symmetrical so you can work out the horizontal component for one force, then double it. Horizontal component of 110N at 25 = 110 cos 25 = 99.7 N Total force = 2 x 99.7 = N a = F = = 3.3 ms -2 m 60 Example 27 a) Calculate the vertical component of the force in each cord if the tension in the elastic cord is 4.5 x 10 3 N. b) If the mass of the capsule and occupants is 236 kg, calculate the initial acceleration a) F = 4.5 x 10 3 x cos 21 = 4.2 x 10 3 N b) a = F tot = 2 x 4.2 x 10 3 = 35.6 ms -2 m 236 SQA 2005 Q22 adapted Forces, Energy and Power 31 ODU Problem booklet 1 P27 Q5

32 Forces on a Slope F parallel θ θ W = mg F perpendicular Example 28 4kg 20º The trolley moves down the slope with constant speed. Calculate the component of the weight acting parallel with the slope. F = mg sin θ = 9.8 x 4 x sin 20 = 13.4 N Forces, Energy and Power 32

33 Forces on a Slope Example kg Friction = 250N 25º A car of mass 800 kg rolls down a hill. Assuming friction of 250 N calculate the acceleration of the car down the slope. F = mg sin θ =800 x 9.8 x sin 25 = N Unbalanced force down slope = = N a = F = = 3.8 ms -2 m 800 Example 30 The force of friction acting on the crate is 44N. The rope snaps and the acceleration of the crate is -6ms -2. It comes to a halt, then accelerates down the slope. Why is the acceleration down the slope less than the acceleration up the slope? Going up the slope the component of weight and friction act in the same direction. Going down the slope the component of weight and friction act in opposite directions, so the unbalanced force is less. Forces, Energy and Power 33 ODU Problem booklet 1 P27 Q6-9

34 Conservation of Energy We can use the idea of conservation of energy to allow us to predict what happens in certain situations. The pendulum swings between A and C, passing through point B. At points A and C it has potential energy. At point B it has kinetic energy. Using conservation of energy we can predict that all the potential energy at A is converted to kinetic energy at B. E p = mgh E k = ½ mv 2 E p = E k mgh = ½ mv 2 mgh = ½ mv 2 v =!2gh A c c B C Example 31 A pendulum of mass 0.6kg is released from a height of 0.2m. What is the maximum velocity of the pendulum? v =!2gh ð v = 2x 9.8 x 0.2 ð v = 1.98ms -1 Example 32 A pendulum travels at 4.5 ms -1 when it is at the bottom of its swing. a) What is the maximum height gain? v =!2gh ð h = v 2 = = 1.03m 2g 2 x 9.8 b) If the mass is 0.1kg what is the potential energy at this point? Ep = mgh = 0.1 x 9.8 x 1.03 = 1.01J Forces, Energy and Power 34

35 Conservation of Energy Example 33 SQA 2014 Q23 adapted A block of wood of mass 0.20kg is suspended from the ceiling by thin cords of negligible mass. A dart of mass 0.050g is thrown at the stationary block of wood. Just before the dart hits the block it is travelling horizontally at a velocity v. The dart sticks into the block. The dart and block then swing to a maximum height of 0.15m as shown. Use conservation of energy to show that the velocity of the dart and block just after the collision is 1.7 ms -1. E p = E k mgh = ½ mv 2 v = 2gh ð v = 2x 9.8 x 0.15 ð v = 2.94 ð v = 1.7 ms -1 as required Forces, Energy and Power 35 ODU Problem booklet 1 P29 Q1-6

36 Momentum Learning Outcomes At the end of this section you should be able to o State that momentum can be calculated using p=mv o State the law of conservation of momentum When two objects collide (or explode) the total momentum before the collision (or explosion) is equal to the total momentum after the collision (or explosion) in the absence of net external forces. o State that momentum is always conserved in inelastic collisions, elastic collisions and explosions o State that total energy is always conserved in inelastic collisions, elastic collisions and explosions o State that kinetic energy is conserved in elastic collisions but not in inelastic collisions. o Use the law of conservation of momentum to calculate unknown quantities in collisions or explosions in one dimension and in cases where the objects may move in opposite directions. o State that impulse= change in momentum o State that impulse = area under a force-time graph o Use change in momentum = area under force-time graph to calculate unknown quantities o Explain crumple zones, air bags, crash helmets etc. In terms of impulse and change in momentum. o Describe the relationship between explosions and Newton s third law Collisions, Explosions and Impulse 36

37 Momentum = mass x velocity Momentum p = mv p = momentum m = mass v = velocity ** CARE = don t mix up p for momentum with P for pressure or ρ for density Law of conservation of momentum The total momentum before a collision or explosion is equal to the total momentum after a collision or explosion in the absence of net external forces. Any momentum question should always start off with a statement of the law of conservation of momentum it qualifies what you do next. Remember that velocity is a vector quantity so direction is important. Collisions, Explosions and Impulse 37

38 Collisions where the objects stick together Before collision After collision 11ms -1 Oms -1 V =? m 1 = 1000kg m 2 = 1200kg m 1 = 1000kg m 2 = 1200kg Always start of with a statement of the law of conservation of momentum Label the objects and velocities from left to right. o Use u for velocities before collision o Use v for velocities after the collision Substitute in the numbers and work out the answer. Total momentum before collision = total momentum after collision m 1 u 1 + m 2 u 2 = ( m 1 +m 2 ) v (11 x 1000) + (0 x 1200) = ( ) v 11, = 2200 v v = 11,000 = 5ms -1 to the right 2200 Example ms -1 v 0 ms kg kg 0.65 kg 0.9 kg The two vehicles collide and stick together. What will the velocity of the vehicles be after the collision? Total momentum before collision = total momentum after collision m 1 u 1 + m 2 u 2 = ( m 1 +m 2 ) v (0.65 x 1.2) + (0 x 0.9) = ( ) v = 1.55v v = 0.78 = 0.5 ms -1 to the right 1.55 Collisions, Explosions and Impulse 38 ODU Problem booklet 1 P31 Q1-4, 6, 9

39 Collisions where the objects stick together Example ms ms ms -1 2 kg? kg 2 kg? kg After the collision the trolleys stick together and travel to the left at 0.7 ms -1. What is the mass of the second trolley? Total momentum before collision = total momentum after collision m 1 u 1 + m 2 u 2 = ( m 1 +m 2 ) v (2.0 x 1.2) + (-3.0 x m 2 ) = (2.0 + m 2 ) x (-3 m 2 ) = (-0.7 m 2 ) = (-0.7 m 2 ) + 3 m = 2.3 m 2 m 2 = 3.8 = 1.65 kg 2.3 Prove that the collision is inelastic. If the collision is inelastic then kinetic energy will not be conserved. E k before E k after E k before E k after 2 2 ½ m 1 u 1 + ½ m 2 u 2 ½ (m 1 +m 2 ) v 2 [½ x 2 x ] + [½ x 1.65 x 3 2 ] [½ x (2+1.65) x ] Total E k after = 0.89J Total E k before = 8.87J Since total E k before total E k after this is an inelastic collision Collisions, Explosions and Impulse 39 ODU Problem booklet 1 P31 Q1-4, 6, 9

40 Collisions where the objects do not stick together Before collision After collision 10ms -1 1ms -1 5 ms -1 V 2 m 1 = 2kg m 2 = 10kg m 1 = 2kg m 2 = 10kg Always start of with a statement of the law of conservation of momentum Label the objects and velocities from left to right. o Use u for velocities before collision o Use v for velocities after the collision Substitute in the numbers and work out the answer. Total momentum before collision = total momentum after collision m 1 u 1 + m 2 u 2 = m 1 v 1 +m 2 v 2 (2 x 10) + (1 x 10) = (-5 x 2) + 10v = v 2 40 = 10 v 2 Example 36 6 ms -1 0 ms -1 4 ms -1? ms -1 5kg 2 kg 5kg 2 kg Calculate the velocity of the right hand vehicle after the collision. Total momentum before collision = total momentum after collision m 1 u 1 + m 2 u 2 = m 1 v 1 +m 2 v 2 (6 x 5) + (2 x 0) = (4 x 5) + 2v = v = 2v 2 v 2 = 10 = 5ms -1 to the right 2 Collisions, Explosions and Impulse 40

41 Collisions where the objects do not stick together Example 37 Before collision After collision 20 ms -1 2 ms ms -1 V kg 2 kg 0.4 kg 2 kg Trolley A hits trolley B and bounces back with a velocity of 10ms -1 to the left. Find the velocity of trolley B after the collision and prove that the collision was elastic. Total momentum before collision = total momentum after collision m 1 u 1 + m 2 u 2 = m 1 v 1 +m 2 v 2 (0.4 x 20) + (2 x 2) = (0.4 x -10) + 2v = v 2 16 = 2v 2 v 2 = 16 = 8ms -1 to the right 2 If the collision is elastic then kinetic energy will be conserved. E k before = E k after E k before E k after ½ m 1 u 1 + ½ m 2 u 2 ½ m 1 v 1 + ½ m 2 v 2 [½ x 0.4 x 20 2 ] + [½ x 2 x 2 2 ] [½ x 0.4 x 10 2 ] + [½ x 2 x 8 2 ] Total E k before = 84J Total E k after = 84J Since total E k before = total E k after this is an elastic collision Collisions, Explosions and Impulse 41 ODU Problem booklet 1 P31 Q 5,7,8

42 Types of Collision There are two types of collision In both types of collision the total energy and total momentum are always conserved. 1. Elastic collisions In elastic collisions the total kinetic energy is conserved. Practical collisions are never 100% efficient, but some collisions can appear to be very close to 100% efficient such as the collision between two snooker balls. 2. Inelastic collisions In inelastic collisions the total kinetic energy is not conserved. The only way to be sure which type of collision has taken place is to check the values for the kinetic energy before and after the collision and see whether or not it has been conserved. Summary of momentum Inelastic Collision Elastic Collision Explosion Total Energy Kinetic Energy Total momentum Conserved Not Conserved Conserved Conserved Conserved Conserved Conserved Not Conserved Conserved Collisions, Explosions and Impulse 42

43 Explosions Before explosion After explosion The total momentum before an explosion is zero. In these questions there will only be two objects moving in opposite directions since the force exerted by object A on object B is equal but opposite to the force exerted by object B on object A. (Newton s Third Law) Total momentum before explosion = total momentum after explosion 0 = m 1 v 1 + m 2 v 2 Example 38 A cannon of mass 1200 kg fires a cannonball of mass 15 kg at a velocity of 60 ms -1 East. Assuming the force of friction is negligible, calculate the velocity of the cannon just after firing. Total momentum before explosion = total momentum after explosion 0 = m 1 v 1 + m 2 v 2 0 = (1200 x v 1 ) + (15 x 60) v 1 = 900 v 1 = v 1 = 0.75 ms -1 West SQA Q4 Example 39 Two people stand together on an ice rink. One has a mass of 55kg and the other has a mass of 45 kg. They push apart and the lighter one moves off with a velocity of 0.7 ms -1. Calculate the velocity of the other skater. Total momentum before explosion = total momentum after explosion 0 = m 1 v 1 + m 2 v 2 0 = (45 x -0.7) + (55 x v 2 ) 0 = v 2 55 v 2 = 31.5 v 2 = 0.6 ms -1 to the right Collisions, Explosions and Impulse 43 ODU Problem booklet 1 P32 Q10-13

44 Momentum and Impulse Starting from Newton s second law F = ma F = m! v u t! Not examinable Ft = mv - mu Impulse = force x time Change in momentum = mv mu During a collision there is a change in momentum caused by the impulse. If the time of contact is short the applied force must be large. The force can be decreased if the contact time is increased. This is the science behind crumple zones, air bags and soft surfaces in playgrounds. We can illustrate the impulse by drawing a graph. F F 0 t 0 t Impulse = area under a force-time graph. If the graph is curved you would be given the area. If the graph is triangular you could be expected to calculate the area. If you are showing the comparison between two surfaces you must make the differences clear use the same graph, but label it clearly. Example 40 Two samples of foam for the inside of crash helmets are tested. The force time graph for sample 1 is given. If the time of impact is doubled using sample 2, show how this alters the applied force if the change in momentum is the same. F Sample 1 Sample 2 0 t Collisions, Explosions and Impulse 44

45 Momentum and Impulse Example 41 A golf club hits a stationary ball of mass 0.4 kg. The ball moves off at 4.2 ms -1 and the time of contact is 0.025s. Calculate the average force exerted on the ball by the club. Ft = mv mu F x = 0.4 (4.2 0) F = 1.68 = 67.2N Example 42 A cue exerts a force of 40N on a stationary ball of mass 0.3 kg. If the impact lasts for 60 ms find the speed the ball leaves the cue. Ft = mv mu 40 x 60 x 10-3 = 0.3 (v 0) 2.4 = 0.3v v = 8 ms -1 Example 43 SQA 2005 Q5 A car is designed with a crumple-zone so that the front of the car collapses during impact. The purpose of the crumple-zone is to A decrease the driver s change in momentum per second B increase the driver s change in momentum per second C decrease the driver s final velocity D increase the driver s total change in momentum E decrease the driver s total change in momentum Collisions, Explosions and Impulse 45

46 Momentum and Impulse Example 44 A gymnast of mass 40kg is practising on a trampoline SQA 2010 Q 23 adapted The speed of the gymnast, as she lands on the trampoline, is 6.3 ms -1. She rebounds with a speed of 5.7 ms -1. a) Calculate the change in momentum of the gymnast. b) The gymnast was in contact with the trampoline for 0.50s. Calculate the average force exerted by the trampoline on the gymnast. a) Change in momentum = m(v-u) = 40 x ((-5.7) -6.3) = 40 x -12 = -480 kg ms -1 b) Ft = change in momentum F x 0.5 = -480 F = -960N Collisions, Explosions and Impulse 46 ODU Problem booklet 1 P33 Q14-26