1. Motion & Forces Action and Reaction  Newton’s Third Law  Momentum  Conservation of Momentum

2. Dynamic Carts

3. With carts only  Describe the motion of the carts  What evidence do you have that there is a force exerted on each of the carts?  Are the accelerations of the carts equal? Is the same amount of force exerted on each cart?

4. Dynamic Carts With additional weights on 1 cart  Describe the motion of the carts  What evidence do you have that there is a force exerted on each of the carts?  Are the accelerations of the carts equal? Is the same amount of force exerted on each cart?

5. Demo with Balloon Helicopter http://videos.howstuffworks.com/nasa/2139-how- helicopters-work-video.htm

6. A. Newton’s Third Law Newton’s Third Law of Motion When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first.

7. A. Newton’s Third Law “For every action there is an equal and opposite reaction.”

8. A. Newton’s Third Law Action-Reaction Pair

9. A. Newton’s Third Law Action-Reaction Pair

10. A. Newton’s Third Law  Action-Reaction Pairs

11. A. Newton’s Third Law Action-Reaction Pairs The rocket exerts a downward force on the exhaust gases. The gases exert an equal but opposite upward force on the rocket. FGFG FRFR

12. A. Newton’s Third Law Physics of walking

13. A. Newton’s Third Law Propulsion of fish through water  A fish uses its fins to push water backwards. In turn, the water reacts by pushing the fish forwards.

14. A. Newton’s Third Law Action-Reaction Pairs The hammer exerts a force on the nail to the right. The nail exerts an equal but opposite force on the hammer to the left.

15. A. Newton’s Third Law Consider the interaction between a baseball bat and baseball.  Action: the baseball forces the bat to the right.  Reaction: ?

16. A. Newton’s Third Law Using a diving board to spring into the air before a dive is a good example of Newton’s third law of motion. Explain.

17. Newton’s Third Law Newton vs. Elephant Who will move fastest?

18. A. Newton’s Third Law Action-Reaction Pairs Both objects accelerate. The amount of acceleration depends on the mass of the object. F m Small mass  more acceleration Large mass  less acceleration

19. A. Newton’s Third Law http://esamultimedia.esa.int/docs/ issedukit/en/activities/flash/start_t oolbar.html#ex03_gm01.swf http://esamultimedia.esa.int/docs/ issedukit/en/activities/flash/start_t oolbar.html#ex03_gm01.swf

20. A.Newton’s Third Law Collision between club head and ball  Experience equal but opposite forces  Ball experiences greater acceleration due to its smaller mass.

21. A. Newton’s Third Law Collision between moving 7-ball and an 8- ball at rest  Each ball experiences an equal force directed in opposite directions.  Each ball experiences the same acceleration.

22. A. Newton’s Third Law Pair figure skaters – what happens when he gracefully throws her through the air onto the ice?  She experiences a forward force which causes her to accelerate (speed up)  He experiences a backward force which causes him to decelerate (slow down)  Her acceleration is greater than his deceleration due to her smaller mass.

23. A. Newton’s Third Law While driving, you observe a bug striking the windshield of your car. Obviously, a case of Newton’s third law! The bug hits the windshield and the windshield hits the bug. Is the force on the bug or the force on the windshield greater?

24. A. Newtons’s Third Law A gun recoils when fired. The recoil is the result of action-reaction force pairs. As the gases from the gunpowder explosion expand, the gun pushes the bullet forwards and the bullet pushes the gun backwards. The acceleration of the recoiling gun is… a.Greater than the acceleration of the bullet. b.Smaller than the acceleration of the bullet. c.The same size as the acceleration of the bullet.

25. B. Momentum Momentum  “mass in motion”  Depends on object’s mass and velocity.  The more momentum an object has, the harder it is to stop.  It would require a greater amount of force or a longer amount of time or both to bring an object with more momentum to a stop.

26. B. Momentum Momentum p = mv p:momentum (kg ·m/s) m:mass (kg) v:velocity (m/s) m p v

27. B. Momentum and Impulse Newton’s 2 nd Law Impulse = Change of momentum

28. B. Momentum and Impulse

29. B. Momentum Find the momentum of a bumper car if it has a total mass of 280 kg and a velocity of 3.2 m/s. GIVEN: m = 280 kg v = 3.2 m/s p = ? WORK : p = mv p = (280 kg)(3.2 m/s) p = 896 kg·m/s m p v

30. B. Momentum The momentum of a second bumper car is 675 kg·m/s. What is its velocity if its total mass is 300 kg? GIVEN: p = 675 kg·m/s m = 300 kg v = ? WORK : v = p ÷ m v = (675 kg·m/s)÷(300 kg) v = 2.25 m/s m p v

31. C. Conservation of Momentum Law of Conservation of Momentum  The total momentum in a group of objects doesn’t change unless outside forces act on the objects. p before = p after

32. C. Conservation of Momentum If momentum is lost by one object, it must be gained by another object in the system so that the total momentum of the system is constant.

33. C. Conservation of Momentum Collision between 1-kg cart and 2-kg dropped brick Momentum of the loaded cart- dropped brick system is conserved

34. C. Conservation of Momentum Big fish in motion catches little fish

35. C. Conservation of Motion Little fish in motion is caught by big fish

36. C. Conservation of Momentum Elastic Collision (KE conserved) Inelastic Collision (KE not conserved)

37. C. Conservation of Momentum A 5-kg cart traveling at 4.2 m/s strikes a stationary 2-kg cart and they connect. Find their speed after the collision. BEFORE Cart 1: m = 5 kg v = 4.2 m/s Cart 2 : m = 2 kg v = 0 m/s AFTER Cart 1 + 2: m = 7 kg v = ? p = 21 kg·m/s p = 0 p before = 21 kg·m/sp after = 21 kg·m/s m p v v = p ÷ m v = (21 kg·m/s) ÷ (7 kg) v = 3 m/s

38. C. Conservation of Momentum A 50-kg clown is shot out of a 250-kg cannon at a speed of 20 m/s. What is the recoil speed of the cannon? BEFORE Clown: m = 50 kg v = 0 m/s Cannon: m = 250 kg v = 0 m/s AFTER Clown: m = 50 kg v = 20 m/s Cannon: m = 250 kg v = ? m/s p = 0 p before = 0 p = 1000 kg·m/s p after = 0 p = -1000 kg·m/s

39. C. Conservation of Momentum So…now we can solve for velocity. GIVEN: p = -1000 kg·m/s m = 250 kg v = ? WORK : v = p ÷ m v = (-1000 kg·m/s)÷(250 kg) v = - 4 m/s (4 m/s backwards) m p v