1. Momentum

2. Essential idea: Conservation of momentum is an example of a law that is never violated. Nature of science: The concept of momentum and the principle of momentum conservation can be used to analyse and predict the outcome of a wide range of physical interactions, from macroscopic motion to microscopic collisions. Topic 2: Mechanics 2.4 – Momentum and impulse Understandings: Newton’s second law expressed in terms of rate of change of momentum Impulse and force – time graphs Conservation of linear momentum Elastic collisions, inelastic collisions and explosions

3. Applications and skills: Applying conservation of momentum in simple isolated systems including (but not limited to) collisions, explosions, or water jets Using Newton’s second law quantitatively and qualitatively in cases where mass is not constant Sketching and interpreting force – time graphs Determining impulse in various contexts including (but not limited to) car safety and sports Qualitatively and quantitatively comparing situations involving elastic collisions, inelastic collisions and explosions Topic 2: Mechanics 2.4 – Momentum and impulse

4. Guidance: Students should be aware that F = ma is the equivalent of F =  p /  t only when mass is constant Solving simultaneous equations involving conservation of momentum and energy in collisions will not be required Calculations relating to collisions and explosions will be restricted to one-dimensional situations A comparison between energy involved in inelastic collisions (in which kinetic energy is not conserved) and the conservation of (total) energy should be made Topic 2: Mechanics 2.4 – Momentum and impulse

5. Data booklet reference: p = mv F =  p /  t E K = p 2 / (2m) Impulse = F  t =  p Topic 2: Mechanics 2.4 – Momentum and impulse

6. International-mindedness: Automobile passive safety standards have been adopted across the globe based on research conducted in many countries Theory of knowledge: Do conservation laws restrict or enable further development in physics? Utilization: Jet engines and rockets Martial arts Particle theory and collisions (see Physics sub-topic 3.1) Topic 2: Mechanics 2.4 – Momentum and impulse

7. Aims: Aim 3: conservation laws in science disciplines have played a major role in outlining the limits within which scientific theories are developed Aim 6: experiments could include (but are not limited to): analysis of collisions with respect to energy transfer; impulse investigations to determine velocity, force, time, or mass; determination of amount of transformed energy in inelastic collisions Aim 7: technology has allowed for more accurate and precise measurements of force and momentum, including video analysis of real-life collisions and modelling/simulations of molecular collisions Topic 2: Mechanics 2.4 – Momentum and impulse

8. Momentum The linear momentum of an object of mass m moving with a velocity v is defined as the product of the mass and the velocity The linear momentum of an object of mass m moving with a velocity v is defined as the product of the mass and the velocity The symbol for momentum is the letter p The symbol for momentum is the letter p p = m v p = m v SI units for momentum are kg m/s or N s SI units for momentum are kg m/s or N s Momentum is a vector quantity Momentum is a vector quantity The direction of the momentum is the same as the direction of the velocity The direction of the momentum is the same as the direction of the velocity

9. Momentum So let’s say you have a 10 kg model car, and it is going 5 m/s. So let’s say you have a 10 kg model car, and it is going 5 m/s. What is it’s momentum? What is it’s momentum? p = mv = (10 kg )(5 m/s) = 50 kg m/sec p = mv = (10 kg )(5 m/s) = 50 kg m/sec

10. Impulse

11. Impulse cont. When a single, constant force acts on the object When a single, constant force acts on the object Impulse = Δp = mΔv = FΔt Impulse = Δp = mΔv = FΔt FΔt is defined as the impulse (units N sec) FΔt is defined as the impulse (units N sec) Δp then is also defined as the impulse Δp then is also defined as the impulse The impulse is the average force x the time interval over which it acts The impulse is the average force x the time interval over which it acts Impulse is a vector quantity, the direction is the same as the direction of the force Impulse is a vector quantity, the direction is the same as the direction of the force

12. This side mv the same Impulse = Δp = mΔv = FΔt

13. Impulse-Momentum Theorem

14. Average Force in Impulse The average force can be thought of as the constant force that would give the same impulse to the object in the time interval as the actual time-varying force gives in the interval The average force can be thought of as the constant force that would give the same impulse to the object in the time interval as the actual time-varying force gives in the interval

15. Average Force cont. The impulse imparted by a force during the time interval Δt is equal to the area under the force-time graph from the beginning to the end of the time interval The impulse imparted by a force during the time interval Δt is equal to the area under the force-time graph from the beginning to the end of the time interval Or, the average force multiplied by the time interval Or, the average force multiplied by the time interval

16. Impulse example The force shown in the force-time diagram to the right acts on a 1.5 kg object. Find a)The impulse of the force. Area under curve = (2)(3) + ½(2)(2) = 8 N-s b)The final velocity of the object if it is initially at rest. FΔt = mΔv 8 N-s = 1.5 (v final – v initial ) = 1.5 v final V final = 8 N-s / 1.5 kg = 5.3 m/s

17. Impulse Applied to Auto Collisions

18. Air Bags The air bag increases the time of the collision The air bag increases the time of the collision It will also absorb some of the energy from the body It will also absorb some of the energy from the body It will spread out the area of contact It will spread out the area of contact decreases the pressure decreases the pressure helps prevent penetration wounds helps prevent penetration wounds

19. Example Prob. 1 2200 kg vehicle with v i = 26 m/s 2200 kg vehicle with v i = 26 m/s Can be stopped in 21 sec by gently applying brakes. Can be stopped in 21 sec by gently applying brakes. Can be stopped in 3.8 sec if slam on brakes. Can be stopped in 3.8 sec if slam on brakes. Can be stopped in 0.22 sec if hits brick wall. Can be stopped in 0.22 sec if hits brick wall. What average force is exerted on the vehicle each time? What average force is exerted on the vehicle each time? FΔt = Δp = p f – p i =(2200)(0) – (2200)(26) = FΔt = Δp = p f – p i =(2200)(0) – (2200)(26) = -5.7 x10 4 kg m/sec So F = -5.7 x 10 4 / Δt So F = -5.7 x 10 4 / Δt Case 1 21 sec F = -2.7 x 10 3 N braking force Case 1 21 sec F = -2.7 x 10 3 N braking force Case 2 3.8 sec F = - 1.5 x 10 4 N Case 2 3.8 sec F = - 1.5 x 10 4 N Case 3 0.22 sec F = -2.6 x 10 5 N Case 3 0.22 sec F = -2.6 x 10 5 N

20. Example Problem 2

21. Example Prob. 2 (continued) The driver of the 725 kg car suddenly applies the brakes hard for 2 sec. As a result, an average force of 5000 N is exerted on the car to slow it down. The driver of the 725 kg car suddenly applies the brakes hard for 2 sec. As a result, an average force of 5000 N is exerted on the car to slow it down. What is the change in momentum? (in other words, what is the direction and magnitude of the impulse?) What is the change in momentum? (in other words, what is the direction and magnitude of the impulse?) FΔt = Δp so FΔt = (5000 N) (2 sec) = 10,000 N s to the west (he was driving east).

22. Example Prob. 2 (continued) Make before and after sketches and find final momentum and velocity. Make before and after sketches and find final momentum and velocity. FΔt = Δp = -10,000 N s braking FΔt = Δp = -10,000 N s braking V i = 31.9 m/s v f = 13160/725 = 18 m/s p i = 23160 N s p f = 13160 N s

23. Example prob. 3 A driver of a 240 kg snowmobile accelerates, which results in a force being exerted that speeds up the snowmobile from 6 to 28 m/s over a time interval of 60 seconds. A driver of a 240 kg snowmobile accelerates, which results in a force being exerted that speeds up the snowmobile from 6 to 28 m/s over a time interval of 60 seconds. Give the initial and final situations Give the initial and final situations Initial p = mv i = (240)(6) = 1440 kg m / sec Initial p = mv i = (240)(6) = 1440 kg m / sec Final p = mv f = (240)(28) = 6720 kg m / sec Final p = mv f = (240)(28) = 6720 kg m / sec What is the change in momentum? What is the change in momentum? Δp = 6720-1440 = 5280 kg m /sec (this is the impulse) Δp = 6720-1440 = 5280 kg m /sec (this is the impulse) What is the magnitude of the average force that is exerted on the snowmobile by the engine? What is the magnitude of the average force that is exerted on the snowmobile by the engine? F = Δp/Δt = 5280 / 60 sec = 88 N F = Δp/Δt = 5280 / 60 sec = 88 N

24. Conservation of Momentum Momentum in an isolated system in which a collision occurs is conserved Momentum in an isolated system in which a collision occurs is conserved A collision may be the result of physical contact between two objects A collision may be the result of physical contact between two objects “Contact” may also arise from the electrostatic interactions of two charged particles (more on that in January). “Contact” may also arise from the electrostatic interactions of two charged particles (more on that in January). An isolated system will have no external forces An isolated system will have no external forces

25. Conservation of Momentum The principle of conservation of momentum states when no external forces act on a system consisting of two objects that collide with each other, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. The principle of conservation of momentum states when no external forces act on a system consisting of two objects that collide with each other, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. The total momentum of an isolated system of objects is conserved regardless of the nature of the forces between the objects The total momentum of an isolated system of objects is conserved regardless of the nature of the forces between the objects

26. Conservation of Momentum, cont. Mathematically if m 1 and m 2 collide: Mathematically if m 1 and m 2 collide: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Momentum is conserved for the system of objects, not individual objects Momentum is conserved for the system of objects, not individual objects The system includes all the objects interacting with each other The system includes all the objects interacting with each other Assumes only internal forces are acting during the collision Assumes only internal forces are acting during the collision Can be generalized to any number of objects Can be generalized to any number of objects

27. Types of Collisions Momentum is conserved in any collision Momentum is conserved in any collision Inelastic collisions Inelastic collisions Perfectly inelastic collisions occur when the objects stick together after they collide. Perfectly inelastic collisions occur when the objects stick together after they collide. Elastic collisions Elastic collisions Example: perfectly elastic balls that bounce off each other without deforming. Example: perfectly elastic balls that bounce off each other without deforming. Actual collisions Actual collisions Most collisions fall between elastic and perfectly inelastic collisions Most collisions fall between elastic and perfectly inelastic collisions

28. More About Perfectly Inelastic Collisions When two objects stick together after the collision, they have undergone a perfectly inelastic collision When two objects stick together after the collision, they have undergone a perfectly inelastic collision Conservation of momentum becomes Conservation of momentum becomes m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f total

30. Great video on momentum from space Richard Garriott Space Video Blog Richard Garriott Space Video Blog Richard Garriott Space Video Blog Richard Garriott Space Video Blog

31. Some General Notes About Collisions Momentum is a vector quantity Momentum is a vector quantity Direction is important Direction is important Be sure to have the correct signs Be sure to have the correct signs In order for conservation of momentum to occur it needs to be a closed system that does not gain or lose mass. In order for conservation of momentum to occur it needs to be a closed system that does not gain or lose mass. Net external force on the system must be zero for cons. of momentum to hold. Net external force on the system must be zero for cons. of momentum to hold.

32. Elastic Collisions Momentum is conserved Momentum is conserved m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f

33. Problem Solving for One - Dimensional Collisions Set up a coordinate axis and define the velocities with respect to this axis Set up a coordinate axis and define the velocities with respect to this axis It is convenient to make your axis coincide with one of the initial velocities It is convenient to make your axis coincide with one of the initial velocities In your sketch, draw all the velocity vectors with labels including all the given information In your sketch, draw all the velocity vectors with labels including all the given information

34. Sketches for Elastic Collision Problems Draw “before” and “after” sketches Draw “before” and “after” sketches Label each object Label each object include the direction of velocity include the direction of velocity keep track of subscripts keep track of subscripts

35. Sketches for Perfectly Inelastic Collisions The objects stick together The objects stick together Include all the velocity directions Include all the velocity directions Don’t forget: The “after” collision combines the masses!! Don’t forget: The “after” collision combines the masses!! m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f total

36. Types of collisions

37. Problem Solving for One- Dimensional Collisions, cont. Write the expressions for the momentum of each object before and after the collision Write the expressions for the momentum of each object before and after the collision Remember to include the appropriate signs Remember to include the appropriate signs Write an expression for the total momentum before and after the collision Write an expression for the total momentum before and after the collision Remember the momentum of the system is what is conserved (not the velocity…) Remember the momentum of the system is what is conserved (not the velocity…)

38. Perfectly Inelastic Collison Example on next page Note that once the cars collide, they stick together so the total mass afterwards is the combined mass of the two cars.

39. Rear-End Example

40. Head-on Example

41. Recoil Example Initially two skaters are at rest, and then they push off each other and go in opposite directions on the frictionless ice. p 1i + p 2i = p 1f + p 2f 0 + 0 = p 1f + p 2f so - p 1f = p 2f The backward motion of the skater on the left is an example of recoil. Another example is how a gun kicks backwards when the bullet is shot out in the forward direction.

42. Recoil example – lab carts Two lab carts on a track are pushed together with a spring mechanism compressed between them. Upon release, the 6 kg cart recoils one way with a velocity of 0.1 m/s while the 3 kg cart goes the opposite direction. What is the velocity of the 3 kg cart? Two lab carts on a track are pushed together with a spring mechanism compressed between them. Upon release, the 6 kg cart recoils one way with a velocity of 0.1 m/s while the 3 kg cart goes the opposite direction. What is the velocity of the 3 kg cart? m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 0 + 0 = (6)(-0.1) + (3)(v 2f ) 0 + 0 = (6)(-0.1) + (3)(v 2f ) 0.6 = 3v 2f 0.6 = 3v 2f v 2f = 0.2 m/s

43. Propulsion in space gas astronaut An astronaut at rest in space fires a thruster pistol that expels 35g of hot gas at 875 m/s. The combined mass of astronaut and pistol is 84 kg. How fast and in what direction is the astronaut moving after firing the pistol? An astronaut at rest in space fires a thruster pistol that expels 35g of hot gas at 875 m/s. The combined mass of astronaut and pistol is 84 kg. How fast and in what direction is the astronaut moving after firing the pistol? p i = p cf + p df 0 = m cf v cf + m df v df v cf = - (m df v df ) / m c Vcf = -(0.035)(-875)/84 v cf = 0.36 m/s

44. Uranium decay – 1D

45. Rocket Propulsion The operation of a rocket depends on the law of conservation of momentum as applied to a system, where the system is the rocket plus its ejected fuel The operation of a rocket depends on the law of conservation of momentum as applied to a system, where the system is the rocket plus its ejected fuel This is different than propulsion on the earth where two objects exert forces on each other This is different than propulsion on the earth where two objects exert forces on each other road on car road on car train on track train on track

46. Rocket Propulsion, cont. The rocket is accelerated as a result of the thrust of the exhaust gases The rocket is accelerated as a result of the thrust of the exhaust gases This represents the inverse of an inelastic collision This represents the inverse of an inelastic collision Before: rocket + fuel Before: rocket + fuel After: rocket in one direction, fuel in the other After: rocket in one direction, fuel in the other Momentum is conserved Momentum is conserved Kinetic Energy is increased (at the expense of the stored energy of the rocket fuel) Kinetic Energy is increased (at the expense of the stored energy of the rocket fuel)

47. Two Dimensional Momentum Problems The law of conservation of momentum holds for all closed systems in 1-D, 2-D and 3-D. The law of conservation of momentum holds for all closed systems in 1-D, 2-D and 3-D. For a general collision of two objects in three-dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved For a general collision of two objects in three-dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved m 1 v 1ix + m 2 v 2ix = m 1 v 1fx + m 2 v 2fx m 1 v 1ix + m 2 v 2ix = m 1 v 1fx + m 2 v 2fx m 1 v 1iy + m 2 v 2iy = m 1 v 1fy + m 2 v 2fy m 1 v 1iy + m 2 v 2iy = m 1 v 1fy + m 2 v 2fy Use subscripts for identifying the object, initial and final, and x,y components. Use subscripts for identifying the object, initial and final, and x,y components.

48. Glancing Collisions Now since ball C and D are not moving in a simple x or y direction following the collision, we need to find the x-component and y-component of their velocities.

49. Glancing Collisions The “after” velocities have x and y components Momentum is conserved both in the x direction and in the y direction Apply separately to each direction

50. Problem Solving for Two- Dimensional Collisions Set up coordinate axes and define your velocities with respect to these axes Set up coordinate axes and define your velocities with respect to these axes It is convenient to choose the x axis to coincide with one of the initial velocities It is convenient to choose the x axis to coincide with one of the initial velocities In your sketch, draw and label all the velocities and include all the given information In your sketch, draw and label all the velocities and include all the given information Write expressions for the x and y components of the momentum of each object before and after the collision Write expressions for the x and y components of the momentum of each object before and after the collision Write expressions for the total momentum before and after the collision Write expressions for the total momentum before and after the collision in the x-direction, then repeat for the y-direction in the x-direction, then repeat for the y-direction

51. To solve 2-D momentum problems, you may have up to three equations, conservation of momentum in x-direction, conservation of momentum in y-direction, and conservation of energy (if elastic).

52. Example Car C collides with moving car D. The two cars stick together. In what direction and with what speed do they move after the collision? Given: Unknown: m C = 1325 kg v fx m D = 2165 kg v fy v Ciy = 27.0 m/s θ v Dix = 11.0 m/s

53. Example (continued)

54. Uranium decay 1 2D example 3 An unstable radioactive nucleus with a total mass of 15 x 10 -27 kg initially at rest experiences a radioactive decay and splits into 3 particles. An unstable radioactive nucleus with a total mass of 15 x 10 -27 kg initially at rest experiences a radioactive decay and splits into 3 particles. m 1 = 5 x 10 -27 kg moves along y-axis at 6x10 6 m/s m 1 = 5 x 10 -27 kg moves along y-axis at 6x10 6 m/s m 2 = 7 x 10 -27 kg moves along the –x-axis at 3x10 6 m/s m 2 = 7 x 10 -27 kg moves along the –x-axis at 3x10 6 m/s Therefore m 3 = 15x10 -27 - 5x10 -27 - 7x10 -27 = 3x10 -27 kg Therefore m 3 = 15x10 -27 - 5x10 -27 - 7x10 -27 = 3x10 -27 kg Determine the angle and velocity of the third mass x-dir: 0 = 0 – m 2 v 2f + m 3 v 3fx y-dir: 0 = m 1 v 1f + 0 + m 3 v 3fy So v 3fx = m 2 v 2f / m 3 = (7E-27)(3E6)/(3E-27) = 7E6 m/s So v 3fx = m 2 v 2f / m 3 = (7E-27)(3E6)/(3E-27) = 7E6 m/s v 3fy = -m 1 v 1f / m 3 = (5E-27)(6E6)/(3E-27) = 10E6 or 1E7 m/s v 3fy = -m 1 v 1f / m 3 = (5E-27)(6E6)/(3E-27) = 10E6 or 1E7 m/s Pythagorean v 3fx and v 3fy to get v 3f = 1.22E7 m/s And θ = tan -1 ( 10E6/7E6) = 55 o 2 θ

55. Example – car and truck collision A 985 kg car traveling south at 20 m/s hits a ?? kg truck traveling west at 18 m/s. After the collision they stick together and travel with a final momentum of 4E4 kg m/sec at an angle of 45 degrees. What is the mass of the truck? A 985 kg car traveling south at 20 m/s hits a ?? kg truck traveling west at 18 m/s. After the collision they stick together and travel with a final momentum of 4E4 kg m/sec at an angle of 45 degrees. What is the mass of the truck? X- direction: m c v cix + m t v tix = (m c + m t )v f = p f X- direction: m c v cix + m t v tix = (m c + m t )v f = p f 0 + 18m t = 4x10 4 cos 45 0 + 18m t = 4x10 4 cos 45 So m t = 4x10 4 cos 45 / 18 = 1571 kg or about 1600 kg So m t = 4x10 4 cos 45 / 18 = 1571 kg or about 1600 kg 45 car truck

56. Example – 2 cars, 2D θ m 1 =1383 kg, 11.2 m/s m 2 =1732 kg, 31.3 m/s

57. Angular Momentum  Momentum resulting from an object moving in linear motion is called linear momentum.  Momentum resulting from the rotation (or spin) of an object is called angular momentum.

58. Angular momentum Angular momentum is important because it obeys a conservation law, as does linear momentum. Angular momentum is important because it obeys a conservation law, as does linear momentum. The total angular momentum of a closed system stays the same. The total angular momentum of a closed system stays the same.

59. Calculating Angular Momentum Angular momentum is calculated in a similar way to linear momentum, except the mass m and velocity v are replaced by the moment of inertia I and angular velocity ω Angular velocity (rad/sec) Angular momentum (kg m/sec 2 ) L = I  Moment of inertia (kg m 2 ) SI Unit of Angular Momentum: kg·m 2 /s Requirement: The angular speed must be expressed in rad/s.

60. Moments of Inertia ~0

61. 1.You are asked for angular momentum. 2.You are given mass, shape, and angular velocity.  Hint: both rotate about y axis. 3.Use L= I , I hoop = mr 2, I bar = 1 / 12 ml 2 (where l is the length of the bar) An artist is making a moving metal sculpture. She takes two identical 1 kg metal bars and bends one into a hoop with a radius of 0.16 m. The hoop spins like a wheel. The other bar is left straight with a length of 1 meter. The straight bar spins around its center. Both have an angular velocity of 2 rad/sec. Calculate the angular momentum of each and decide which would be harder to stop.

62. Angular Momentum example 3.For hoop: I hoop = (1 kg) (0.16 m) 2 = 0.026 kg m 2 L hoop = I ω = (0.026 kg m 2 ) (2 rad/s) = 0.052 kg m 2 /s 4.Solve bar: I bar = ( 1 / 12 )(1 kg) (1 m) 2 = 0.083 kg m 2 L bar = I ω = (0.083 kg m 2 ) (2 rad/s) = 0.166 kg m 2 /s 5.The bar has more than 3x the angular momentum of the hoop, so it is harder to stop.

63. Example A Satellite in an Elliptical Orbit An artificial satellite is placed in an elliptical orbit about the earth. Its point of closest approach is 8.37x10 6 m from the center of the earth, and its point of greatest distance is 25.1x10 6 m from the center of the earth. The speed of the satellite at the perigee is 8450 m/s. Find the speed at the apogee.

64. Angular Momentum Example (continued)