Momentum. Slide 1 / 47. Slide 2 / 47. Slide 3 / 47. Conservation of Momentum. Conservation of Momentum

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1 Slide 1 / 47 Momentum 2009 by Goodman & Zavorotniy onservation of Momentum Slide 2 / 47 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. nergy was the first conserved property that we studied. Momentum is the second. onservation of Momentum Slide 3 / 47 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.

2 Momentum is a Vector Quantity Slide 4 / 47 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. The Momentum of a Single Object Slide 5 / 47 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 The Momentum of a System of Objects Slide 6 / 47 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 1v 1 + m 2v 2 + m 3v It's critically important to note that momenta add as vectors, not as scalars.

3 The Momentum of a System of Objects Slide 7 / 47 p system = m 1v 1 + m 2v 2 + m 3v In order to determine the total momentum of a system you need to first: etermine 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. The Momentum of a System of Objects etermine 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 Slide 8 / 47 p system = p 1 + p 2 p system = m 1v 1 + m 2v 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 The onservation of Momentum Slide 9 / 47 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

4 The onservation of Momentum & Impulse Slide 10 / 47 oth the onservation of Momentum and the concept of Impulse follow directly from Newton's Second Law: SF = ma We can derive a mathematical expression for Impulse. The onservation of Momentum & Impulse Slide 11 / 47 SF = ma ssume one force and substitute a = v/t F = m(v/t) multiply both sides by t Ft = mv move the outside the parenthesis Ft = (mv) Ft = I; where I stands for Impulse I = p mv = p; where p stands for "momentum" use the definition of I = p f - p 0 Or, like our energy equation " 0 + W = f" p 0 + I = p f where p = mv and I = Ft The onservation of Momentum & Impulse Slide 12 / 47 p 0 + I = p f where p = mv and I = Ft These equations tell us that if no external force acts on system, it's momentum will not change. If F = 0, then Ft = 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.

5 onservation of nergy and Momentum in ollisions Slide 13 / 47 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, lastic ollisions, Mechanical nergy is conserved. In the other two cases, Inelastic ollisions, Mechanical nergy is not conserved...some of the energy is transformed into heat, bonding, etc. onservation of Momentum Slide 14 / 47 uring 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 onservation of nergy and Momentum in ollisions Slide 15 / 47 V V pproaching Momentum is conserved in all collisions. ollision V' If elastic V' ollisions in which kinetic energy is conserved as well are called elastic collisions. V' V' If inelastic Those in which it is not are called inelastic.

6 onservation of Momentum Slide 16 / 47 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 Inelastic ollisions Slide 17 / 47 m v l M vm = 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 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 Sp = Sp' m 1v 1 + m 2v 2 = m 1v 1' + m 2v 2' m 1v 1 = m 1v' + m 2v' m 1v 1 = (m 1+ m 2)v' v' = m 1v 1 / (m 1+ m 2) p system = Sp; p' system = Sp' v 2 = 0 amd v 1' = v 2' = v' 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 Slide 18 / 47

7 onservation of Momentum Slide 19 / 47 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 y lastic ollisions in One imension Here we have two objects colliding elastically. Slide 20 / 47 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 lastic ollisions Slide 21 / 47 m 1v 1 + m 2v 2 = m 1v 1' +m 2v 2' m 1v 1 - m 1v 1' = m 2v 2' - m 2v 2 m 1(v 1 - v 1') = m 2(v 2' - v 2) 1/2m 1v /2m 2v 2 2 = 1/2m 1v 1' 2 +1/2m 2v 2' 2 m 1v m 2v 2 2 = m 1v 1' 2 +m 2v 2' 2 m 1v m 1v 1' 2 = m 2v 2' m 2v 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 '

8 lastic ollisions Slide 22 / 47 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. ollisions Slide 23 / 47 Type of collision Is Momentum onserved? Is nergy onserved? Ine las tic Ye s No No Perfectly Inelastic Yes No Yes lastic Yes Yes No oes it stick? 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? Slide 24 / 47 v 1/2 v 3/4 v 1.5 v 2v

9 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. Slide 25 / 47.5v v 2v 3v 4v lastic ollisions with Identical Masses Slide 26 / 47 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! Impulse Slide 27 / 47 The formula for Impulse is I = Ft The units for Impulse are either N-s or kg-m/s which are equivalent. nd since I = p The Impulse delivered to an object, or system, is exactly equal to the change in its momentum.

10 Graphical Interpretation of Impulse Slide 28 / 47 The area under the Force vs. time graph is: F (N) rea = length x width rea = F x t Since the formula for Impulse is I = Ft t (s) rea = Impulse The impulse applied on an object is equal to an objects change in momentum so rea = p 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 16 N-s 18 N-s 20 N-s F (N) t (s) Slide 29 / 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? Slide 30 / 47 4 N-s 6 N-s 8 N-s 10 N-s F (N) N-s t (s)

11 5 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 velocity of this object after 6s in m/s? 10 m/s 8 m/s 6 m/s 5 m/s 4 m/s F (N) t (s) Slide 31 / 47 Implications of Impulse Slide 32 / 47 uring 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 avidhazy Implications of Impulse Slide 33 / 47 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.

12 onservation of Momentum in 2 or 3 imensions Slide 34 / 47 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' onservation of Momentum in 2 or 3 imensions Slide 35 / 47 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. onservation of Momentum in 2 or 3 imensions Note that the component parallel to the wall does not change at all. Slide 36 / 47 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#

13 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? Slide 37 / 47 mv 2mv 2mvcosθ 2mvsinθ zero θ θ 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? Slide 38 / 47 mvsinθ 2mvsinθ mvcosθ 1.5mvcosθ 2mvcosθ v/2 v θ θ ollisions in 2 imensions Slide 39 / 47 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 ollision all fter ollision all fter ollision?

14 onserve momentum in the x-dimension: ollisions in 2 imensions efore ollision Slide 40 / 47 onserve momentum in the y-dimension: all fter ollision dd the components all fter ollision? 8 fter the collision shown below, what is a possible velocity vector for the second ball? Slide 41 / 47 efore ollision Red all fter ollision Inelastic ollisions in 2 imensions Slide 42 / 47 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 ollision fter ollision

15 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? Slide 43 / m/s m/s 30 m/s m/s m/s 10 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.)? Slide 44 / o o o o o xplosions in 2 imensions Slide 45 / 47 The green object explodes into three equal mass pieces (blue and red), determine the velocity of the missing piece. uring the explosion of an object its momentum is unchanged, since no XTRNL 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.

16 11 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? Slide 46 / 47 Slide 47 / 47