2. Units Motion Scalar / vector quantities Displacement / Velocity / Acceleration Vector components Forces 1 st & 2 nd law ∑ F = 0 & ∑ F = ma Force diagrams Momentum Change in Momentum Conservation of Momentum (1 & 2D collisions) Impulse / Newton’s 3 rd Law Energy Work Kinetic & Potential Energies (gravitational & spring) Conservation of Energy Level 2 Translational Motion

3. Success criterion:- merit YOU are able to rearrange an equation and determine the units for the new subject of the formulae. Success criterion:- merit YOU are able to rearrange an equation and determine the units for the new subject of the formulae. Make G the subject of the formulae and determine the units for G. The unit of Force (F) is the Newton (N = kg m s -2 ) The unit of the radius squared (r 2 ) is m 2 The units of mass squared (m 2 ) is kg 2 Show all working. Make G the subject of the formulae and determine the units for G. The unit of Force (F) is the Newton (N = kg m s -2 ) The unit of the radius squared (r 2 ) is m 2 The units of mass squared (m 2 ) is kg 2 Show all working. Writing the correct units.

4. Hence the units of G are m 3 kg -1 s -2 Writing the correct units. Make G the subject of the formulae and determine the units for G. The unit of Force (F) is the Newton (N = kg m s -2 ) The unit of the radius squared (r 2 ) is m 2 The units of mass squared (m 2 ) is kg 2 Show all working. Make G the subject of the formulae and determine the units for G. The unit of Force (F) is the Newton (N = kg m s -2 ) The unit of the radius squared (r 2 ) is m 2 The units of mass squared (m 2 ) is kg 2 Show all working. Success criterion:- merit YOU are able to rearrange an equation and determine the units for the new subject of the formulae. Success criterion:- merit YOU are able to rearrange an equation and determine the units for the new subject of the formulae.

5. Vectors Vector Addition (Graphical)

6. Vectors Vector Subtraction (Graphical)

7. x y Vector Components

8. The twin farmers are trying to drag a stubborn donkey into new pasture. The maximum force the donkey can resist the twins is 820 N. 1.Use vector calculations to determine the resistive force of the donkey; and 2.Explain why the farmers are unable to drag her into the next field of pasture. 3.Use vectors to determine how the farmers could drag the donkey into the next field of pasture without having to exert any more force than they already are. 1.Use vector calculations to determine the resistive force of the donkey; and 2.Explain why the farmers are unable to drag her into the next field of pasture. 3.Use vectors to determine how the farmers could drag the donkey into the next field of pasture without having to exert any more force than they already are. A free body sketch of the situation. F 1 = 450 N F 2 = 375 N 15 0 30 0 Resisting Force Translational MotionVectors

9. 2.Explain why the farmers are unable to drag her into the next field of pasture. F 1 = 450 N F 2 = 375 N Resisting Force By measurement the resisting force of the donkey = 760N The farmer boys are unable to drag the donkey into greener pastures By measurement the resisting force of the donkey = 760N The farmer boys are unable to drag the donkey into greener pastures Vector diagram 30º 15º Translational MotionVectors A free body sketch of the situation. F 1 = 450 N F 2 = 375 N 15 0 30 0 Resisting Force 1.Use vectors to determine the resistive force of the donkey The twin farmers are trying to drag a stubborn donkey into new pasture. The maximum force the donkey can resist the twins is 820 N. 780 N

10. The twin farmers are trying to drag a stubborn donkey into new pasture. The maximum force the donkey can resist the twins is 820 N. Formative practice enables students to gain frequent feedback and feedforward on the progress of their education Click  for answers A free body sketch of the situation. F 1 = 450 N F 2 = 375 N 15 0 30 0 Resisting Force Translational MotionVectors

11. F 1 + F 2 = 450 + 375 = 825 N The resistive force of the donkey has maxed out at 820 N The farmer twins combined forces are 825 N. The twins can drag the donkey into greener pastures because the donkey's maximum resisting force is only 820 N The farmer twins combined forces are 825 N. The twins can drag the donkey into greener pastures because the donkey's maximum resisting force is only 820 N Vector diagram By decreasing the 15 0 + 30 O = 45 0 angle between the farmers the twins are able to increase the force that they apply to the donkey as shown below The twin farmers are trying to drag a stubborn donkey into new pasture. The maximum force the donkey can resist the twins is 820 N. A free body sketch of the situation. F 1 = 450 N F 2 = 375 N 15 0 30 0 Resisting Force 3. Use vectors to determine how the farmers could drag the donkey into the next field of pasture without having to exert any more force than they already are.

12. Kinematic Equations Assume a is constant!

13. If no external force acts, an object maintains a constant velocity, or is at rest. 1) Newton’s First Law (Law of Inertia) 2) Newton’s Second Law Units of Force System Mass Acceleration Force SI kg ms -2 N = kg ms -2

14. Applications of Newton’s Laws F v i = 20 m/s m v = 0 F = -10 N m = 5 kg d Find distance block moves

15. Translational MotionVectors Success criteria: You are able to draw neat, accurate and correctly labelled free body diagrams. Success criteria: You are able to draw neat, accurate and correctly labelled free body diagrams. 1.Draw the free body diagram of the space shuttle sitting on the launch pad. 2.Determine the unbalanced force ( Δ F) required to lift off the space shuttle. 3.Draw the free body diagram at this instant. 1.Draw the free body diagram of the space shuttle sitting on the launch pad. 2.Determine the unbalanced force ( Δ F) required to lift off the space shuttle. 3.Draw the free body diagram at this instant.

16. 1.Draw the free body diagram of the space shuttle sitting on the launch pad. Support force = 24 MN Weight force = 24 MN free body diagram Scale 1cm : 2 MN Translational MotionVectors

17. Lift force = 4.1 MN 24 MN + −  Sketch your solution before you draw your answer Unbalanced force required to lift the space shuttle off the pad: Δ F = 4.1 + 24 = 28 MN Unbalanced force required to lift the space shuttle off the pad: Δ F = 4.1 + 24 = 28 MN Weight F W = 9.8 x 2.4 x 10 6 = 23.52 x 10 6 N (24 MN) Initial acceleration = 1.7 m s -2 now F = ma F = 4.1 MN Weight F W = 9.8 x 2.4 x 10 6 = 23.52 x 10 6 N (24 MN) Initial acceleration = 1.7 m s -2 now F = ma F = 4.1 MN Translational MotionVectors Scale 1 cm : 5 MN

18. Lift force = 28 MN F w = 24 MN free body diagram Scale 1 cm : 5 MN 3.Draw the free body diagram at this instance. Translational MotionVectors

19. Applications of Newton’s Laws m = 20 kg  = 60 o  m T 1 sin(60) T 1 cos(60) T1T1 T2T2 mg Find Tensions T 1 and T 2

20. Applications of Newton’s Laws m = 20 kg  = 60 o  m Find Tensions T 1 and T 2 T1T1 T2T2 mg

21. T1T1 T2T2  m T2T2 T1T1  Another method

22. Hooray for Algebra....

23.   N mg cos(  ) mg sin(  ) y x mg Force Components ~ Frictionless incline

24.   N mg sin(  ) mg fkfk mg cos(  ) y x Acceleration on a rough incline a

25. d  Frictionless  mg N mg sin  mg cos  A block is released from rest at the top of an incline. Find the final speed and the time to slide to the bottom.

26. d  Frictionless  mg N A block is released from rest at the top of an incline. Find the final speed and the time to slide to the bottom.

27. d  Now add Friction  mg N mg sin  mg cos  A block is released from rest at the top of an incline. Find the final speed and the time to slide to the bottom.

28. F m 3m2m Three mass system - find acceleration

29. m1m1 m2m2 N m1gm1g T T m2gm2g Forces on m 1 Forces on m 2

30. m1m1 m2m2 T T m1gm1g m2gm2g Mass 1 Mass 2

31. m1m1 m2m2 T T m1gm1g m2gm2g Mass 1 Mass 2 Forces Investigation ~ Add friction into the mix How much is it? f f Is f related to mass weight?