Slide 1 / 47. Momentum by Goodman & Zavorotniy

Transcription

1 Slide 1 / 47 Momentum 2009 by Goodman & Zavorotniy

2 Slide 2 / 47 Conservation of Momentum s we pointed out with energy, the most powerful concepts in science are called "conservation principles". These principles allow us to solve problems without worrying about the details of a process. We just have to take a snapshot of a system initially and finally; by comparing those two snapshots we can learn a lot. Energy was the first conserved property that we studied. Momentum is the second.

3 Slide 3 / 47 Conservation of Momentum Like energy, momentum is a conserved property of nature. It is not created or destroyed; so in a closed system we will always have the same amount of momentum. The only way the momentum of a system can change is if momentum is added or taken away by an outside force.

4 Slide 4 / 47 Momentum is a Vector Quantity key difference between momentum and energy is that while energy is a scalar (there is no direction associated with it), momentum is a vector (there is a direction associated with it). When there is more than one object in a system, the total momentum of the system is found by the vector addition of the momenta of each object. nother key difference is that there is only one formula for momentum.

5 Slide 5 / 47 The Momentum of a Single Object The momentum (p) of a single object is the product of its mass and its velocity: p = mv The bolded symbols for p and v are to remind you that they are vectors. Mass (m) is a scalar. There is no special unit for momentum. It's just the product of the units in the formula: kg-m/s

6 Slide 6 / 47 The Momentum of a System of Objects If a system contains more than one object, it's total momentum is just the vector sum of the momenta of those objects. p system = Sp p system = p 1 + p 2 + p p system = m 1 v 1 + m 2 v 2 + m 3 v It's critically important to note that momenta add as vectors, not as scalars.

7 Slide 7 / 47 The Momentum of a System of Objects p system = m 1 v 1 + m 2 v 2 + m 3 v In order to determine the total momentum of a system you need to first: Determine a direction to be considered positive ssign positive values to momenta in that direction ssign negative values to momenta in the opposite direction. Then you can add the momenta to get a total.

8 Slide 8 / 47 The Momentum of a System of Objects Determine the momentum of a system of two objects: m 1, has a mass of 6 kg and a velocity of 13 m/s towards the east and m 2, has a mass of 14 kg and a velocity of 7 m/s towards the west. p system = Sp p system = p 1 + p 2 p system = m 1 v 1 + m 2 v 2 p system = (6 kg)(+13m/s) + (14kg)(-7m/s) p = mv Let's choose east as positive p system = (78 kg-m/s) + (-98kg-m/s) p system = (-20 kg-m/s) p system = (20 kg-m/s) to the west

9 Slide 9 / 47 The Conservation of Momentum If we call the amount of momentum that we start with "p o " and the amount we end up with as "p f " then if no momentum is added to or taken away from a system then p o = p f There is only one way to change the momentum of a system. That's by an outside force delivering an Impulse (I) to the system. So if an outside force acts on a system, the general equation for the momentum of a system becomes: p o + I = p f

10 Slide 10 / 47 The Conservation of Momentum & Impulse oth the Conservation of Momentum and the concept of Impulse follow directly from Newton's Second Law: SF = ma We can derive a mathematical expression for Impulse.

11 Slide 11 / 47 The Conservation of Momentum & Impulse SF = ma ssume one force and substitute a = Dv/Dt F = m(dv/dt) multiply both sides by Dt FDt = mdv move the D outside the parenthesis FDt = D(mv) FDt = I; where I stands for Impulse I = Dp mv = p; where p stands for "momentum" use the definition of D I = p f - p 0 Or, like our energy equation "E 0 + W = E f " p 0 + I = p f where p = mv and I = FDt

12 Slide 12 / 47 The Conservation of Momentum & Impulse p 0 + I = p f where p = mv and I = FDt These equations tell us that if no external force acts on system, it's momentum will not change. If F = 0, then FDt = 0, and I = 0 If I = 0, then p 0 + I = p f becomes p 0 = p f ; when I = 0 We'll talk more about I later, now let's look at cases where I = 0.

13 Slide 13 / 47 Conservation of Energy and Momentum in Collisions These equations tell us that if no external force acts on system, it's momentum will not change. We're going to look at three types of collisions. In all cases, momentum is conserved. In one case, Elastic Collisions, Mechanical Energy is conserved. In the other two cases, Inelastic Collisions, Mechanical Energy is not conserved...some of the energy is transformed into heat, bonding, etc.

14 Slide 14 / 47 Conservation of Momentum During a collision, measurements show that the total momentum does not change: m V m V m v + m v = m v ' + m v ' m V ' m V ' x

15 Slide 15 / 47 Conservation of Energy and Momentum in Collisions V V pproaching Momentum is conserved in all collisions. Collision V ' If elastic V ' Collisions in which kinetic energy is conserved as well are called elastic collisions. V ' If inelastic V ' Those in which it is not are called inelastic.

16 Slide 16 / 47 Conservation of Momentum More formally, the law of conservation of momentum states: The total momentum of an isolated system of objects remains constant. V = 4.5 m/s V = 0 m/s efore collision V' =? fter collision

17 Slide 17 / 47 Inelastic Collisions m v l M v M = 0 With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. It may also be gained during explosions, as there is the addition of chemical or nuclear energy. l perfectly inelastic collision is one where the objects stick together afterwards, so there is only one final velocity. M+m v' h

18 Slide 18 / 47 13,500 kg railroad freight car travels on a level track at a speed of 4.5 m/s. It collides and couples with a 25,000 kg second car, initially at rest and with brakes released. No external force acts on the system. What is the speed of the two cars after colliding? Let's choose the first car's direction as positive. p 0 + I = p f F external = 0; so I = 0 p 0 = p f p system = Sp; p' system = Sp' Sp = Sp' m 1 v 1 + m 2 v 2 = m 1 v 1 ' + m 2 v 2 ' v 2 = 0 amd v 1 ' = v 2 ' = v' m 1 v 1 = m 1 v' + m 2 v' m 1 v 1 = (m 1 + m 2 )v' v' = m 1 v 1 / (m 1 + m 2 ) v' = (13500 kg)(4.5 m/s)/ (13500 kg kg) v' = (60750 kg-m/s)/ (38500 kg) v' = 1.6 m/s in the direction of the original car's motion

19 Slide 19 / 47 Conservation of Momentum Momentum conservation works for a rocket as long as we consider the rocket and its fuel to be one system, and account for the mass loss of the rocket. p = 0 p gas p rocket

20 Slide 20 / 47 y Elastic Collisions in One Dimension Here we have two objects colliding elastically. m m y v v x We know the masses and the initial speeds. Since both momentum and kinetic energy are conserved, we can write two equations. m m This allows us to solve for the two unknown final speeds. v ' v ' x

21 Slide 21 / 47 Elastic Collisions m 1 v 1 + m 2 v 2 = m 1 v 1 ' +m 2 v 2 ' m 1 v 1 - m 1 v 1 ' = m 2 v 2 ' - m 2 v 2 m 1 (v 1 - v 1 ') = m 2 (v 2 ' - v 2 ) 1/2m 1 v /2m 2 v 2 2 = 1/2m 1 v 1 ' 2 +1/2m 2 v 2 ' 2 m 1 v m 2 v 2 2 = m 1 v 1 ' 2 +m 2 v 2 ' 2 m 1 v m 1 v 1 ' 2 = m 2 v 2 ' m 2 v 2 m 1 (v v 1 ' 2 ) = m 2 (v 2 ' 2 - v 2 2 ) m 1 (v 1 + v 1 ')(v 1 - v 1 ') = m 2 (v 2 ' + v 2 )(v 2 ' - v 2 ) m 1 (v 1 - v 1 ') = m 2 (v 2 ' - v 2 ) v 1 + v 1 ' = v 2 ' + v 2 v 1 - v 2 = v 2 ' - v 1 '

22 Slide 22 / 47 Elastic Collisions v 1 - v 2 = v 2 ' - v 1 ' For all elastic collisions, the magnitude of the difference of the velocities is the same before and after a collision. Physically, v 1 - v 2, is the relative velocity that the two objects collide with...it's how fast they're approaching each other. nd, v 2 ' - v 1 ', is the relative velocity of their separation, it's how fast they are going apart after the collision. For all elastic collisions, regardless of the masses of the objects, the objects separate, after the collision, with the same velocity that they collided with.

23 Slide 23 / 47 Type of collis ion Is Momentum Conserved? Collisions Is Energy Conserved? Ine las tic Ye s No No Perfectly Inelastic Yes No Yes Elastic Yes Yes No Does it stick?

24 Slide 24 / 47 1 bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest. The velocity of the bowling ball is virtually unaffected by the collision. What will be the speed of the ping pong ball? C D E v 1/2 v 3/4 v 1.5 v 2v

25 Slide 25 / 47 2 baseball bat has a velocity of +v when it collides with a baseball that has a velocity of -2v. The bat barely changes velocity during the collision. The speed of the baseball will be v..5v C D E v 2v 3v 4v

26 Slide 26 / 47 Elastic Collisions with Identical Masses v 1 - v 2 = v 2 ' - v 1 ' special case is when two objects with the same mass have an elastic collision. In that case, they exchange velocities: v 1 ' = v 2 and v 2 ' = v 1 This is only true if they have identical masses!

27 Slide 27 / 47 Impulse The formula for Impulse is I = FDt The units for Impulse are either N-s or kg-m/s which are equivalent. nd since I = Dp The Impulse delivered to an object, or system, is exactly equal to the change in its momentum.

28 Slide 28 / 47 Graphical Interpretation of Impulse The area under the Force vs. time graph is: F (N) rea = length x width rea = F x Dt Since the formula for Impulse is I = FDt t (s) rea = Impulse The impulse applied on an object is equal to an objects change in momentum so rea = Dp

29 Slide 29 / 47 3 n object starts from rest and moves along the x- axis. net horizontal force is applied to the object in +x direction. The force vs time graph is presented below. What is the net impulse delivered by this force in Newtons? 12 N-s 14 N-s F (N) 6 4 C D 16 N-s 18 N-s 2 E 20 N-s t (s)

30 Slide 30 / 47 4 n 2 kg object starts from rest and moves along the x-axis. net horizontal force is applied to the object in +x direction. The force vs time graph is presented below. What is the net impulse delivered by this force in Newtons? 4 N-s 6 N-s F (N) 6 4 C D 8 N-s 10 N-s 2 E 12 N-s t (s)

31 Slide 31 / kg object starts from rest and moves along the x-axis. net horizontal force is applied to the object in +x direction. The force vs time graph is presented below. What is the velocity of this object after 6s in m/s? 10 m/s 8 m/s F (N) 6 4 C D 6 m/s 5 m/s 2 E 4 m/s t (s)

32 Slide 32 / 47 Implications of Impulse During a collision, objects are deformed due to the large forces involved. We can determine the relationship between the force, the time it acts and the change of momentum (often velocity) of the object by using our definition of impulse: Photograph by ndrew Davidhazy

33 Slide 33 / 47 Implications of Impulse The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. Image taken from: This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete.

34 Slide 34 / 47 Conservation of Momentum in 2 or 3 Dimensions Momentum is conserved along every axis independently. The vector expression p + I = p' Is true for each component of momentum: p x + I x = p x ' p y + I y = p y ' p z + I z = p z '

35 Slide 35 / 47 Conservation of Momentum in 2 or 3 Dimensions Let's determine the Impulse delivered to the ball bouncing off this wall. We know that the I = #p. So what is the change in momentum due to this collision? First, we'll break the velocities into perpendicular components, such that all the change in velocity is found along one axis.

36 Slide 36 / 47 Conservation of Momentum in 2 or 3 Dimensions Note that the component parallel to the wall does not change at all. On the other hand, the component perpendicular to the wall, is reversed. #v perp = v' perp - v perp #v perp = -v perp -v perp #v perp = -2v perp #v perp = -2vcos#

37 Slide 37 / 47 6 tennis ball of mass m rebounds from a vertical wall with the same speed v as it had initially. What is the change in momentum of the ball? C D E mv 2mv 2mvcosθ 2mvsinθ zero θ θ

38 Slide 38 / 47 7 tennis ball of mass m and velocity v rebounds from a vertical wall with half its initial speed. What is the change in the ball's component of momentum perpendicular to the wall? C D E mvsinθ 2mvsinθ mvcosθ 1.5mvcosθ 2mvcosθ v/2 v θ θ

39 Slide 39 / 47 Collisions in 2 Dimensions Since momentum is independently conserved along each orthogonal axis, picking one axis to line up with the momentum of one object is a first step. If only one object is in motion, the perpendicular momentum must remain zero. For instance, in this case, what must be the direction of motion of the second ball after the collision? efore Collision all fter Collision all fter Collision?

40 Conserve momentum in the x-dimension: Slide 40 / 47 Collisions in 2 Dimensions efore Collision Conserve momentum in the y-dimension: all fter Collision dd the components all fter Collision?

41 Slide 41 / 47 8 fter the collision shown below, what is a possible velocity vector for the second ball? C D E efore Collision Red all fter Collision

42 Slide 42 / 47 Inelastic Collisions in 2 Dimensions The vector sum of the momenta is shown below. However, this will only work for velocities if the masses are equal. Otherwise, the velocities will have to be calculated. efore Collision fter Collision

43 Slide 43 / 47 9 Object with mass 20 kg travels to the east at 10 m/s and object with mass 5 kg travels south at 20 m/s. What is the magnitude of the velocity they have after the collision? C D E 8.9 m/s m/s 30 m/s m/s m/s

44 Slide 44 / Object with mass 20 kg travels to the east at 10 m/s and object with mass 5 kg travels south at 20 m/s. What is the direction of the velocity they have after the collision (use east as 0 o, south as 90 o, etc.)? C D E o o o o o

45 Slide 45 / 47 Explosions in 2 Dimensions The green object explodes into three equal mass pieces (blue and red), determine the velocity of the missing piece. During the explosion of an object its momentum is unchanged, since no EXTERNL force acts on it. So if it's original momentum is zero, so is it's final momentum. The third piece must have equal and opposite momentum to the sum of the other two.

46 Slide 46 / stationary cannon ball explodes in three pieces. The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece? C D E

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