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Higher Physics Unit 1 – Our Dynamic Universe Printed Notes

Higher Physics Unit 1 – Our Dynamic Universe Printed Notes. Section 1 – Motion Vectors and Scalars. A scalar quantity is completely defined by stating its magnitude. A vector quantity is completely defined by stating its magnitude and direction. Distance and Displacement.

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Higher Physics Unit 1 – Our Dynamic Universe Printed Notes

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  1. Higher PhysicsUnit 1 – Our Dynamic UniversePrinted Notes

  2. Section 1 – MotionVectors and Scalars

  3. A scalar quantity is completely defined by stating its magnitude. A vector quantity is completely defined by stating its magnitude and direction.

  4. Distance and Displacement Distance is the total path length. It is described by magnitude (size) alone. Displacement is the direct length from a starting point to a finishing point. To describe displacement both magnitude and direction must be given.

  5. Speed and Velocity Average Speed = distance time Speed has magnitude but no direction. Average Velocity = displacement time

  6. Velocity has magnitude AND direction. Direction of velocity is same as displacement. Units for both is ms-1 (metres per second).

  7. Example A runner sprints 100 m along a straight track in 12 s and then takes a further 13 s to jog 20 m back towards her starting point. • What distance does she run during the 25 s? • What is her displacement from her starting point after the 25 s? • What is her average speed during the 25 s? • What is her average velocity during the 25 s?

  8. Solution • Distance = 100 m + 20 m = 120 m • Displacement = 100 m – 20 m = 80 m in original direction • Av. Speed = 120 / 25 = 4.8 ms-1 • Av. Velocity = 80 / 25 = 3.2 ms-1 in original direction Answers to (b) and (d) have direction as well as magnitude.

  9. Addition of Vectors

  10. Two or more vectors can be added (combined) to get a resultant vector. Magnitude and direction must be taken into account. We use a scale diagram for this, add the vectors ‘nose to tail’, then join the start and finish point.

  11. Vector: Resultant Vector: Always drawn with two arrows to represent a resultant. tail nose

  12. Example – Scale Diagram A girl walks 30 m North, then 40 m East. What is her resultant displacement? To solve: • Choose a suitable scale, e.g. 1 cm : 10 m. • Using a ruler and protractor, arrange arrows ‘nose to tail’. • Draw in resultant vector, measuring its length and direction.

  13. Solution N 4 cm (40 m) Her resultant displacement, s, is 50 m at an angle x of 53º East of North. W E 3 cm (30 m) s S x

  14. Example 2 A plane flies North at 120 ms-1 in a wind blowing East at 50 ms-1. What is the plane’s resultant velocity? Hint: Use a scale of 1cm : 10 ms-1

  15. Solution N 5 cm (50 ms-1) The plane’s resultant velocity, v, is 130 ms-1 at an angle y of 23º East of North. W E 12 cm (120 ms-1) S v y

  16. Resultant of a Number of FORCES This is the single force which has the same effect, in both magnitude and direction, as the sum of the individual forces. We can also use scale diagrams to find the magnitude and direction of the resultant of a number of forces.

  17. Resolution (Rectangular Components) of a Vector

  18. Any vector, x, can be resolved into two components at right angles to each other. The horizontal component xh The vertical component xv x xv is equivalent to θ xh

  19. sin θ = xv / x xv = x sin θ cos θ = xh / x xh = x cos θ x xv θ xh

  20. Velocity The vertical and horizontal components of a velocity vector, v, are, respectively: vv = v sin θ vh = v cos θ

  21. Force The vertical and horizontal components of a Force vector, F, are, respectively: Fv = F sin θ Fh = F cos θ

  22. Acceleration Acceleration = change in velocity time taken = final velocity – initial velocity time taken a = v – u t a: acceleration v: final velocity (ms-1) u: initial velocity (ms-1) t : time taken (s)

  23. What does this mean? ACCELERATION IS THE CHANGE IN VELOCITY PER UNIT TIME. An acceleration of 2 ms-2 means the velocity of the body changes by 2 ms-1 every second. Units are metres per second per second or ms-2. Acceleration is a VECTOR.

  24. Experiment to Measure Acceleration Light gates Stopclock

  25. Points to Note In Method • t1 = time for card to pass light gate 1 • t2 = time for card to pass light gate 2 • t3 = time for card to go between light gates • Card length = s • u = s / t1 v = s / t2 a = (v - u) / t3

  26. Graphs of Motion

  27. Note that in the following graphs, a = acceleration v = velocity s = displacement Just as the area under a speed-time graph gives the distance travelled, The area under a velocity-time graph gives the displacement.

  28. Graphs Showing Constant Acceleration a v s t t t 0 0 0

  29. Graphs Showing Constant Velocity a v s t t t 0 0 0

  30. Graphs Showing Constant Deceleration a v s t 0 t t 0 0

  31. The Equations of Motion

  32. 1) v = u + at 2) s = ut + ½at² 3) v² = u² + 2as u – initial velocity at time t = 0 v – final velocity at time t a – acceleration of object t – time to accelerate from u to v s – displacement of object in time t

  33. Example 1 Q A space rocket travelling at 20 ms-1 accelerates at 5 ms-2 for 2 s. How far does the rocket travel during the 2 s? A u = 20 ms-1; a = 5 ms-2; t = 2 s; s = ? Use s = ut + ½at² s = 40 + 10 s = 50 m

  34. Example 2 Q A train travelling at 45 ms-1 decelerates to 15 ms-1 at 2 ms-2. How far does the train travel while it is decelerating? A u = 45 ms-1; v = 15 ms-1; a = -2 ms-2; s = ? (note –ve sign) Use v² = u² + 2as 15² = 45² - 4s 4s = 45² - 15² s = 450 m

  35. Example 3 Q A ball is thrown vertically into the air with an initial velocity of 20 ms-1. What will its velocity be after 3 s? A u = 20 ms-1; t = 3 s; a = -10ms-2; v = ? Use v = u + at v = 20 – 30 v = -10 ms-1 (i.e. the ball is on the way down)

  36. Derivations The three equations of motion can be derived as described below.

  37. 1) v = u + at By definition, acceleration is given by: a = v – u t v – u = at v = u + at

  38. 2) s = ut + ½at² s = displacement = area under a v/t graph. s = = ut + ½(v – u)t = ut + ½(at)t i.e. s = ut + ½at² Velocity v v - u u u t time 0 +

  39. 3) v² = u² + 2as v = u + at (eqn 1) Squaring both sides, we get: v² = (u + at)² = u² + 2uat + a²t² = u ² + 2a(ut + ½at²) i.e. v² = u² + 2as

  40. These equations only apply to uniform acceleration in a straight line. The vector quantities displacement, velocity and acceleration have direction associated with them, and so they will have a positive or negative sign depending on their direction.

  41. Ex 4 A ball is launched vertically into the air with an initial velocity of 35 ms-1. What is the maximum height it will reach? Ex 5 A rock is thrown upwards with an initial velocity of 30 ms-1. After how long will it be 10 metres off the ground?

  42. Method for Tackling Problems Write down all the symbols like this: u = v = s = a = t = Fill in all numbers & values given in question. If there are two directions, use this diagram for + and – values: Choose the best equation to suit the problem. Up + Left - Right + Down -

  43. Section 2 – Forces, Energy and Power

  44. Newton’s First Law • If there are NO forces acting on a moving object, it will continue to move at a constant velocity. • If BALANCED FORCES act on a moving object, it will continue to move at a constant velocity.

  45. Newton’s Second Law Motion can exist without forces, but for change of motion to occur, forces must be involved. i.e. Unbalanced forces cause acceleration.

  46. Fun = ma Fun = Unbalanced Force in N m = mass in kg a = acceleration in ms-2 Definition of the Newton: One Newton is the (unbalanced) force that gives a 1 kg mass an acceleration of 1 ms-2. Force is a vector.

  47. An accelerating object has an unbalanced (resultant) force, F, acting on it in the same direction as the acceleration. If an object is not accelerating then the unbalanced force F = 0.

  48. A man pushes a trolley of mass 20 kg along a flat surface of 40 N. If the effects of friction can be ignored, what is the acceleration of the shopping trolley? a = F / m = 40 / 20 = 2 ms-2 Example – Single Force, Single Mass a 40 N 20 kg

  49. A rocket of mass 1000 kg is fired vertically into the air. The rocket motors provide a thrust of 20 000 N, and there is a drag force of 2000 N. What is the acceleration of the missile? Fun = 20000 - 9800 - 2000 = 8200 N a = F / m = 8200 / 1000 = 8.2 ms-2 Example 1 – Multiple Force, Single Mass Thrust = 20000 N N A S A a Drag = 2000 N W = mg = 1000 x 9.8 = 9800 N

  50. A ski-tow pulls two skiers, who are connected by a thin nylon rope, along a frictionless surface. The tow uses a force of 70 N and the skiers have masses of 60 kg and 80 kg. a) What is the acceleration of the system? b) What is the tension in the rope? a) a = F / m = 70 / (80 + 60) = 70 / 140 = 0.5 ms-2 b) Tension T is a Force. It is caused by 80 kg person. T = ma = 80 x 0.5 = 40 N Example 2 – Single (External) Force, Multiple Mass 70 N T 80 kg 60 kg

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