1. Momentum

2. Wednesday November 10 2010 Momentum and Momentum Change

3. Announcements Quiz today – final 10 minutes Test – Energy - Friday Egg drop on Thursday next week Wednesday sessions – Energy Test Review

4. Momentum Momentum is a measure of how hard it is to stop or turn a moving object. Momentum is related to both mass and velocity. Momentum is possessed by all moving objects.

5. Calculating Momentum For one particle p = mv For a system of multiple particles P =  p i =  m i v i Momentum is a vector with the same direction as the velocity vector. The unit of momentum is… kg m/s or Ns

6. Which has the most momentum?

7. Sample Problem Calculate the momentum of a 65-kg sprinter running east at 10 m/s.

8. Sample Problem Calculate the momentum of a system composed of a 65-kg sprinter running east at 10 m/s and a 75-kg sprinter running north at 9.5 m/s. North p=mv=(75kg)*(9.5m/s)=712.5 Ns East p=mv=(65kg)*(10m/s)=650 Ns Total p=(712.5^2+650^2)^.5=964 Ns Angle tan -1 (712.5/650)= 48 degrees north of east

9. Change in momentum Like any change, change in momentum is calculated by looking at final and initial momentums.  p = p f – p i –  p: change in momentum – p f : final momentum – p i : initial momentum

10. Momentum change demonstration Using only a meter stick, find the momentum change of each sphere when it strikes the desk from a height of exactly one meter. Which sphere, Bouncy or Lazy, has the greatest change in momentum?

11. Wording dilemma In which case is the magnitude of the momentum change greatest? In which case is the change in the magnitude of the momentum greatest?

12. Impulse (J) Impulse is the product of an external force and time, which results in a change in momentum of a particle or system. J = F t J =  P Units: N s or kg m/s (same as momentum)

13. Impulsive Forces Usually high magnitude, short duration. Suppose the ball hits the bat at 90 mph and leaves the bat at 90 mph, what is the magnitude of the momentum change? What is the change in the magnitude of the momentum?

14. Impulse (J) on a graph F(N) t (ms) 01234 0 1000 2000 3000 area under curve

15. Sample Problem Suppose a 1.5-kg brick is dropped on a glass table top from a height of 20 cm. a)What is the magnitude and direction of the impulse necessary to stop the brick? b)If the table top doesn’t shatter, and stops the brick in 0.01 s, what is the average force it exerts on the brick? c)What is the average force that the brick exerts on the table top during this period?

16. Solution a) Find the velocity of the brick when it strikes the table using conservation of energy. mgh = ½ mv 2 v = (2gh) 1/2 = (2*9.8 m/s 2 *0.20 m) 1/2 = 2.0 m/s Calculate the brick’s momentum when it strikes the table. p = mv = (1.5 kg)(2.0 m/s) = 3.0 kg m/s (down) The impulse necessary to stop the brick is the impulse necessary to change to momentum to zero. J =  p = p f – p i = 0 – 3.0 kg m/s = -3.0 kg m/s or 3.0 kg m/s (up)

17. Solution b) and c) b) Find the force using the other equation for impulse. J = Ft 3.0 N s = F (0.01 s) F = 300 N (upward in the same direction as impulse) c) According the Newton’s 3 rd law, the brick exerts an average force of 300 N downward on the table.

18. Sample Problem This force acts on a 1.2 kg object moving at 120.0 m/s. The direction of the force is aligned with the velocity. What is the new velocity of the object? 0.200.400.600.80t(s) F(N) 1,000 2,000

19. Solution Find the impulse from the area under the curve. A = ½ base * height = ½ (.1 s)(2500 N) = 125 Ns J = 125 N s Since impulse is equal to change in momentum and it is in the same direction as the existing momentum, the momentum increases by 125 kg m/s.  p = 125 kg m/s  p = p f - p i = mv f - mv i mv f = mv i +  p = (1.2 kg)(120 m/s) + 125 kg m/s = 269 kg m/s v f = (269 kg m /s) / (1.2 kg) = 224 m/s

20. Law of Conservation of Momentum If the resultant external force on a system is zero, then the vector sum of the momentums of the objects will remain constant.  P before =  P after

21. Sample problem A 75-kg man sits in the back of a 120-kg canoe that is at rest in a still pond. If the man begins to move forward in the canoe at 0.50 m/s relative to the shore, what happens to the canoe?

22. Solution The momentum before the man moves is equal to the momentum after the man moves.  p b =  p a 0 = m m v m + m c v c 0 = (75 kg)(0.50 m/s) + (120 kg)v v = - (75 kg)(0.50 m/s)/(120 kg) v = -0.31 m/s The canoe slips backward in the water at -0.31 m/s

23. Tuesday, November 16, 2010 Collisions

24. Announcements Wednesday Review canceled Online HW are required – 50% of points possible = full credit. Momentums 1 and 2 are due at 2 PM on ll/30/2010 (Tuesday after break) EGG drop – Thursday! Dress warm. Container must be packed and ready at the beginning of class. LARGE eggs will be used.

25. External versus internal forces External forces: forces coming from outside the system of particles whose momentum is being considered. –External forces change the momentum of the system. Internal forces: forces arising from interaction of particles within a system. –Internal forces cannot change momentum of the system.

26. An external force in golf Consider the collision between the club head and the golf ball in the sport of golf. The club head exerts an external impulsive force on the ball and changes its momentum. The acceleration of the ball is greater because its mass is smaller.

27. An internal force in pool Consider the collision between two balls in pool. The forces they exert on each other are internal and do not change the momentum of the system. Since the balls have equal masses, the magnitude of their accelerations is equal.

28. Explosions When an object separates suddenly, as in an explosion, all forces are internal. Momentum is therefore conserved in an explosion. There is also an increase in kinetic energy in an explosion. This comes from a potential energy decrease due to chemical combustion.

29. Recoil Guns and cannons “recoil” when fired. This means the gun or cannon must move backward as it propels the projectile forward. The recoil is the result of action-reaction force pairs, and is entirely due to internal forces. As the gases from the gunpowder explosion expand, they push the projectile forwards and the gun or cannon backwards.

30. Sample problem Suppose a 5.0-kg projectile launcher shoots a 209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher?

31. Solution Momentum conservation is used to calculate recoil speed.  p b =  p a 0 = m p v p + m l v l 0 = (0.209 kg)(350 m/s) + (5.0 kg)v v = - (0.209 kg)(350 m/s)/(5.0 kg) v = - 14.6 m/s

32. Sample Problem An exploding object breaks into three fragments. A 2.0 kg fragment travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s. The third fragment has mass 3.0 kg. What is the magnitude and direction of its velocity?

33. Solution The momentum before is zero, so the momentum after is zero. This is a vector addition problem. Each fragment has a momentum magnitude of 400 kg m/s according to the formula p = mv. 400 kg m/s (400 2 + 400 2 ) 1/2 566 kg m/s due southwest v = p/m = 566/3 = 189 m/s due SW

34. Sample problem An exploding object breaks into three fragments. A 2.0 kg fragment travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s. The third fragment has mass 3.0 kg. What is the magnitude and direction of its velocity?

35. Today Bumper cars demo lab. You will need to be clever and quick as you attempt to simulate various types of collisions using the carts and cart track. For each collision type, you will be given a few minutes to create the simulation and answer the associated questions. You are not to harm the carts or each other in your simulations.

36. Simulation #1 Your mission: Create a situation in which an impulse changes the momentum of a moving cart. You express (a) the magnitude of the momentum change, and (b) the change in the magnitude of the momentum in terms of m and v. You must identify the force causing the impulse. Rule 1: Use just one cart. Rule 2: No kinetic energy may be lost or gained by the cart in this collision.

37. Simulation #2 Your mission: Create a situation in which an impulse changes the momentum of a moving cart. You express (a) the magnitude of the momentum change, and (b) the change in the magnitude of the momentum in terms of m and v. You must identify the force causing the impulse. Rule 1: Use just one cart. Rule 2: All kinetic energy must be lost by the cart in this collision.

38. Simulation #3 Your mission: Create a collision between two carts in which momentum and kinetic energy (of the system of two carts) are both conserved. You must be able to express momentum and kinetic energy of each cart before and after the collision. Rule 1: One cart must lose ALL of its momentum and kinetic energy in this simulation.

39. Simulation #4 Your mission: Create a collision between two carts in which the following happens: Rule 1: The two carts become one cart. Rule 2: The system of two carts loses all of its kinetic energy. Be prepared to identify the initial and final momentum and kinetic energy of the carts in terms of m and v.

40. Simulation #5 Your mission: Create simulation of an explosion creating two equal mass fragments. You must express (a) the initial and final momentum of the system and (b) the final momentum of each of the fragments. Rule 1: Kinetic energy must be zero initially. Rule 2: You must identify the source of the kinetic energy using Conservation of Energy thinking.

41. Collisions When two moving objects make contact with each other, they undergo a collision. Conservation of momentum is used to analyze all collisions. Newton’s Third Law is also useful. It tells us that the force exerted by body A on body B in a collision is equal and opposite to the force exerted on body B by body A.

42. Collisions During a collision, external forces are ignored. The time frame of the collision is very short. The forces are impulsive forces (high force, short duration).

43. Collision Types Elastic collisions –Also called “hard” collisions –No deformation occurs, no kinetic energy lost Inelastic collisions –Deformation occurs, kinetic energy is lost Perfectly Inelastic (stick together) –Objects stick together and become one object –Deformation occurs, kinetic energy is lost

44. (Perfectly) Inelastic Collisions Simplest type of collisions. After the collision, there is only one velocity, since there is only one object. Kinetic energy is lost. Explosions are the reverse of perfectly inelastic collisions in which kinetic energy is gained!

45. Sample Problem An 80-kg roller skating grandma collides inelastically with a 40- kg kid. What is their velocity after the collision?

46. Sample Problem A train of mass 4m moving 5 km/hr couples with a flatcar of mass m at rest. What is the velocity of the cars after they couple?

47. Sample Problem A fish moving at 2 m/s swallows a stationary fish which is 1/3 its mass. What is the velocity of the big fish after dinner?

48. Elastic Collision After the collision, there are still two objects, with two separate velocities Kinetic energy remains constant before and after the collision. Therefore, two basic equations must hold for all elastic collisions  p b =  p a (momentum conservation)  K b =  K a (kinetic energy conservation)

49. Sample Problem A pool ball traveling at speed v strikes a second pool ball at rest such that the first pool ball stops completely. Show that the second pool ball now must have speed v. Assume that the collision is elastic.

50. Sample Problem A 500-g cart on an air track strikes a 1,000-g cart at rest. What are the resulting velocities of the two carts? (Assume the collision is elastic, and the first cart is moving at 2.0 m/s when the collision occurs.)

51. Solution before after m 1 v 1 = m 1 v 1 + m 2 v 2 1.0 = 0.50v 1 + v 2 ½ m 1 v 1 2 = ½ m 1 v 1 2 + ½ m 2 v 2 2 2.0 = 0.50v 1 2 + v 2 2 Solve simultaneously v 1 = -0.67 m/s v 2 = 1.33 m/s

52. Sample Problem Suppose three equally strong, equally massive astronauts decide to play a game as follows: The first astronaut throws the second astronaut towards the third astronaut and the game begins. Describe the motion of the astronauts as the game proceeds. Assume each toss results from the same-sized "push." How long will the game last?

53. Monday, November 29 Lab

54. Announcements Test Wednesday Clicker Quiz today Online HW due tomorrow at 2 PM – 50% mastery required Review for test – Tuesday 7:20 AM

55. Lab – turn in data, calculations, and results only Analyze collisions of the carts in terms of momentum and energy conservation. 3 or 4 Trials –Perfectly inelastic collision: equal masses –Perfectly inelastic collision: unequal masses –Elastic collision: equal masses –Elastic collision: unequal masses (BONUS!!) Clearly show all data collected (mass, width, time) for each trial Clearly show a comparison of the momentum before and after collision. Clearly show a comparison of kinetic energy before and after collision.

56. 2D-Collisions Momentum in the x-direction is conserved. –  P x (before) =  P x (after) Momentum in the y-direction is conserved. –  P y (before) =  P y (after) Treat x and y coordinates independently. –Ignore x when calculating y –Ignore y when calculating x Let’s look at a simulation: –http://surendranath.tripod.com/Applets.htmlhttp://surendranath.tripod.com/Applets.html

57. Sample problem Calculate velocity of 8-kg ball after the collision. 3 m/s 2 kg 8 kg 0 m/s Before 2 m/s 2 kg 8 kg v After 50 o x y x y

58. Solution x-direction (2kg)(3m/s)=(2kg)(2m/s*cos(50))+(8k g)v x (8kg)v x = (2kg)(3m/s)- (2kg)(2m/s*cos(50))=3.43kg*m/s V x =.429m/s Y-direction 0=(2kg)(2m/s*sin(50))-(8kg)v y v y =.383m/s Magnitude V= Direction  =tan -1 (.383/.429)=41.8 degrees below the positive x-axis