Momentum. Slide 2 / 69. Slide 1 / 69. Slide 4 / 69. Slide 3 / 69. Slide 5 / 69. Slide 6 / 69. Conservation of Momentum. Conservation of Momentum

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1 Slide 1 / 69 Momentum 2009 by Goodman & Zavorotniy Slide 2 / 69 onservation of Momentum The most powerful concepts in science are called "conservation principles". Without worrying about the details of a process, conservation principles can be used to solve problems. If we were to take a snapshot of the initial and final system, comparing the two would provide a lot of information. While energy is the first conserved property we studied, momentum is the second. Slide 3 / 69 onservation of Momentum Since momentum is a conserved property of nature, it is not created nor destroyed. For example, if we have a closed system we will have the same momentum. It is conserved. Momentum of a system can only change if momentum is added or taken away by an outside force. Slide 4 / 69 Momentum is a Vector Quantity key difference between momentum and energy is that energy is a scalar, while momentum is a vector. 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. Slide 5 / 69 The Momentum of a Single Object The momentum (p) of a single object is the product of its mass and its velocity: 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. Slide 6 / 69 1 What is the momentum of a 20kg object with a velocity of +5.0 m/s? 10 kg-m/s 50 kg-m/s 100 kg-m/s 150 kg-m/s 160 kg-m/s It's just the product of the units in the formula: kg-m/s

2 Slide 7 / 69 2 What is the momentum of a 20kg object with a velocity of -5.0 m/s? -150 kg-m/s -100 kg-m/s 100 kg-m/s 150 kg-m/s 200 kg-m/s Slide 8 / 69 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 1v 1 + m 2v 2 + m 3v It's critically important to note that momenta add as vectors, not as scalars. Slide 9 / 69 The Momentum of a System of Objects p system = m 1v 1 + m 2v 2 + m 3v In order to determine the total momentum of a system you need to first: Slide 10 / 69 The Momentum of a System of Objects etermine the momentum of a system of two objects: m 1, has a mass of 15 kg and a velocity of 16 m/s towards the east and m 2, has a mass of 42 kg and a velocity of 6 m/s towards the west. p system = Sp 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. p system = p 1 + p 2 p system = m 1v 1 + m 2v 2 p system = (15 kg)(+16m/s) + (42kg)(-6m/s) p system = (240 kg-m/s) + (-252kg-m/s) p system = (-12 kg-m/s) p system = (12 kg-m/s) to the west p = mv Let's choose east as positive Slide 11 / 69 3 etermine the magnitude of the momentum of a system of two objects: m 1, which has a mass of 6.0 kg and a velocity of 13 m/s north and m 2, which has a mass of 14 kg and a velocity 7.0 m/s south. 10 kg-m/s 15 kg-m/s 20 kg-m/s -15 kg-m/s -20 kg-m/s Slide 12 / 69 4 etermine the momentum of a system of 3 objects: m 1, which has a mass of 7.0 kg and a velocity 23 m/s north; m 2, which has a mass of 9.0 kg and a velocity 7 m/s north; and m 3, which has a mass of 5 kg and a velocity of 42 m/s south. -12 kg-m/s 12 kg-m/s 13 kg-m/s 14 kg-m/s 16 kg-m/s

3 Slide 13 / 69 The onservation 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: 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: Slide 14 / 69 The onservation of Momentum & Impulse oth the onservation of Momentum and the concept of Impulse follow directly from Newton's Second Law: We can derive a mathematical expression for Impulse by examining how momentum is related to acceleration. p 0 + I = p r Slide 15 / 69 The onservation of Momentum & Impulse 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 mv = p; where p stands for "momentum" I = p 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 Slide 16 / 69 Impulse The formula for Impulse is The units for Impulse are either N-s or kg-m/s which is equivalent to p, therefore I = p The Impulse delivered to an object, or system, is exactly equal to the change in its momentum. Slide 17 / 69 5 n external force of 25N acts on a system for 10s. What is the magnitude of the Impulse delivered to the system? 25 N-s 100 N-s 150 N-s 200 N-s 250 N-s Slide 18 / 69 6 In the previous problem, an external force of 25N acted on a system for 10s. We found that the Impuse delivered was 250 N-s. In that case, what is the magnitude of the change in momentum of the system? 200 N-s 250 N-s 300 N-s 350 N-s 400 N-s

4 Slide 19 / ,000 N force acts for s on a 2.5 kg object that was initially at rest. What is its final velocity? 800 m/s 700 m/s 600 m/s 500 m/s 400 m/s Slide 20 / 69 The onservation of Momentum & Impulse p 0 + I = p f where and 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 We'll talk more about I later, now let's look at cases where I = 0. F (N) t (s) Slide 21 / 69 Graphical Interpretation of Impulse The area under the Force vs. time graph is: rea = length x width rea = F x t Since the formula for Impulse is I = F t: rea = Impulse The impulse applied on an object is equal to an objects change in momentum, so rea = p Slide 22 / 69 8 n object starts from rest and moves along the x- axis. net horizontal force is applied to the object in the +x direction. The force vs time graph is presented below. What is the net impulse delivered by this force in Newtons per second? 12 N-s 14 N-s 16 N-s 18 N-s 20 N-s F (N) t (s) Slide 23 / 69 Slide 24 / 69 9 n 2 kg object starts from rest and moves along the x-axis. net horizontal force is applied to the object in the +x direction. The force vs time graph is presented below. What is the net impulse delivered by this force in Newtons per second? 10 2 kg object starts from rest and moves along the x-axis. net horizontal force is applied to the object in the +x direction. The force vs time graph is presented below. What is the velocity of this object after 6s in m/s? 4 N-s 6 N-s 8 N-s F (N) m/s 8 m/s 6 m/s F (N) N-s 2 5 m/s 2 12 N-s t (s) 4 m/s t (s)

5 Slide 25 / 69 Implications of Impulse 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: Slide 26 / 69 Implications of Impulse 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. This is why one should bend their knees when landing; why airbags work; and why landing on a pillow hurts less than landing on concrete. Slide 27 / 69 onservation of nergy and Momentum in ollisions 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. Slide 28 / 69 onservation of Momentum uring a collision, measurements show that the total momentum does not change m v m v In all cases, momentum is conserved. m v + m v = m v ' + m v ' In one case, lastic ollisions, Mechanical nergy is conserved. In the other two cases, of Inelastic ollisions, Mechanical nergy is not conserved...some of the energy is transformed into heat, bonding, etc. m v ' x m v ' Slide 29 / 69 onservation of nergy and Momentum in ollisions V pproaching V Momentum is conserved in all collisions. Slide 30 / 69 onservation 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 ollision V' If elastic V' ollisions in which kinetic energy is conserved are called elastic collisions. efore collision V' =? V' V' If inelastic ollisions in which kinetic energy is not conserved are called inelastic collisions. fter collision

6 Slide 31 / 69 Slide 32 / 69 xample 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 v l M Inelastic ollisions 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. 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) v 2 = 0 amd v 1' = v 2' = v' l perfectly inelastic collision is one where the objects stick together afterwards, so there is only one final velocity. v' = (13500 kg)(4.5 m/s)/ (13500 kg kg) M+m v' h v' = 1.6 m/s in the direction of the original car's motion Slide 33 / 69 Slide 34 / cannon ball with a mass of 100 kg flies in the horizontal direction with a speed of 800 m/s and strikes a ship initially at rest. The mass of the ship is 15,000 kg. Find the speed of the ship after the ball becomes embedded in it. 4.4 m/s 5.3 m/s 6.7 m/s 7.2 m/s 7.6 m/s kg girl skates at 5.5 m/s on ice toward her 70 kg friend who is standing still, with open arms. s they collide and hold each other, what is the speed of the couple? 1 m/s 2 m/s 3 m/s 4 m/s 5 m/s Slide 35 / g bullet is fired at 460 m/s into a stationary 1 kg block, and makes a completely inelastic collision with it. Find the velocity of the bullet-block system. Slide 36 / g bullet is fired at 460 m/s into a stationary 1 kg block that hangs from a 2 m long string, and makes a completely inelastic collision with it. fter impact, the block swings up to a maximum height. Find the maximum height of the bullet-block system. 3.2 m/s 4.7 m/s 5.0 m/s 5.9 m/s 6.3 m/s v 1.0 m 1.53 m 1.78 m 2.15 m 2.32 m efore v fter 2 m h

7 Slide 37 / g bullet is fired at 460 m/s into a stationary 1 kg block that hangs from a 2 m long string, and makes a completely inelastic collision with it. fter impact, the block swings up and forms an angle. Find formed by the bullet-block system. Slide 38 / 69 onservation 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 o o o 2 m p = o o h p gas p rocket y y Slide 39 / 69 lastic ollisions in One imension m v m v x Here we have two objects colliding elastically. We know the masses and the initial speeds. Since both momentum and kinetic energy are conserved, we can write two equations. Slide 40 / 69 lastic ollisions onservation of Momentum Kinetic nergy m 1v 1 + m 2v 2 = m 1v 1' +m 2v 2' 1/2m 1v /2m 2v 2 2 = 1/2m 1v 1' 2 +1/2m 2v 2' 2 m 1v 1 - m 1v 1' = m 2v 2' - m 2v 2 m 1v m 2v 2 2 = m 1v 1' 2 +m 2v 2' 2 m m 1(v 1 - v 1') = m 2(v 2' - v 1v m 1v 1' 2 = m 2v 2' m 2v 2 2) m 1(v v 1' 2 ) = m 2(v 2' 2 - v 2 2 ) m m This allows us to solve for the two unknown final speeds. v' v' v 1 + v 1' = v 2' + v 2 x v 1 - v 2 = v 2 ' - v 1 ' Slide 41 / 69 lastic ollisions 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 moving apart after the collision. Type of collision Is Momentum onserved? Slide 42 / 69 ollisions Is nergy onserved? Inelastic Yes No No Perfectly Inelastic Yes No Yes lastic Yes Yes No oes it stick? 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.

8 Slide 43 / Two object of equal mass have an elastic collision. efore they collide they are approaching each other with a velocity of 4.0 m/s relative to each other. With what velocity do they go apart from one another? 2 m/s 3 m/s 4 m/s 5 m/s 6 m/s Slide 44 / Two objects have an elastic collision. One object, m 1, has an initial velocity of +4.0 m/s and m 2 has a velocity of -3.0 m/s. fter the collision, m 1 has a velocity of 1.0 m/s. What is the velocity of m 2? 2 m/s 4 m/s 6 m/s 8 m/s 10 m/s Slide 45 / 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? 1/2 v 3/4 v v 1.5 v 2 v Slide 46 / 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. 1/2 v v 2v 3v 4v Slide 47 / 69 lastic ollisions 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! Slide 48 / Two objects with identical masses have an elastic collision: the initial velocity of m 1 is +6.0 m/s and of m 2 is -3.0 m/s. What is the velocity of m 1 after the collision? 3 m/s 6 m/s -3 m/s -6 m/s -2 m/s

9 Slide 49 / Two objects with identical masses have an elastic collision: the initial velocity of m 1 is +6.0 m/s and m 2 is -3.0 m/s. What is the velocity of m 2 after the collision? Slide 50 / Two objects with identical masses have an elastic collision: the initial velocity of m 1 is +3.0 m/s and m 2 is +2.0 m/s. What is the velocity of m 1 after the collision? -3 m/s 1 m/s 6 m/s 2 m/s 5 m/s 3 m/s 3 m/s 4 m/s 7 m/s 5 m/s Slide 51 / Two objects with identical masses have an elastic collision: the initial velocity of m 1 is +3.0 m/s and m 2 is +2.0 m/s. What is the velocity of m 2 after the collision? Slide 52 / 69 onservation of Momentum in 2 or 3 imensions Momentum is conserved along every axis independently. 1 m/s 2 m/s 3 m/s 4 m/s 5 m/s The vector expression: p + I = p' This 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' Slide 53 / 69 onservation of Momentum in 2 or 3 imensions Let's determine the Impulse delivered to the ball bouncing off this wall. We know that the I = Δp. 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. Slide 54 / 69 onservation of Momentum in 2 or 3 imensions 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. Δv perp = v' perp - v perp Δv perp = -v perp -v perp Δv perp = -2v perp Δv perp = -2vcosθ

10 Slide 55 / 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? mv 2mv 2mvcosθ 2mvsinθ Zero Slide 56 / 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? mvsinθ 2mvsinθ mvcosθ 1.5mvcosθ 2mvcosθ v/2 v θ θ Since momentum is independently conserved along each orthogonal axis, picking one axis to line up with the momentum of another object is a first step. Slide 57 / 69 ollisions in 2 imensions efore ollision Red all fter ollision onserve momentum in the x-dimension: onserve momentum in the y-dimension: Slide 58 / 69 ollisions in 2 imensions efore ollision Red all fter ollision If only one object is in motion, the perpendicular momentum must remain zero. lue all fter ollision? dd the components lue all fter ollision For instance, in this case, what must be the direction of motion of the blue ball after the collision? Slide 59 / 69 Inelastic ollisions in 2 imensions 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. Slide 60 / 69 xplosions in 2 imensions 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. efore ollision fter ollision 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.

11 Slide 61 / 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 62 / 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? m/s m/s m/s m/s m/s Slide 63 / 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 ast as 0 o, north as 90 o, etc.)? Slide 64 / 69 enter of Mass We can restate the conservation of momentum in a useful way by using the concept of center of mass. If we were to receive the coordinates of several masses we can define center of mass of the system o o o o o The position vector of the center of mass can be expressed in terms of the position vectors Slide 65 / 69 enter of Mass For solid bodies, the sums of the equations mentioned before have to be replaced with integrals. For example, bodies such as a basketball, a sugar cube, or a cell phone, all have their center of mass at its geometric center. Slide 66 / 69 Motion of the enter of Mass y taking the derivatives of the center of mass, we can find the velocity of the center of mass system. single vector equation can be made into: When bodies have an axis of symmetry, such as a pulley or wheel, the center of mass always lies along that axis. Then multiplying by the masses, we can find the momentum of the system. If the net external force is zero, the momentum is constant.

12 Slide 67 / 69 xternal Forces and enter-of-mass Motion However if net force on a system of particles is not zero, them momentum is not conserved and the velocity of the center of mass changes. Further deriving the equation can be used to describe acceleration. The sum of all forces on all the particles is then Since Newton's third law dictates the internal forces cancel in pairs, and. Only the external force are left. When a body or a collection of particles is acted on by external forces, the center of mass moves just as though all the mass were concentrated at the point and it were acted on by a net force equal to the sum of the external forces on a system. Slide 68 / 69 Separating Masses Suppose an artillery shell was flying in a parabolic path and then explodes in the middle of air and splits into two parts. The two fragments will now follow different, parabolic paths. However the center of mass continues on the original trajectory. This property of the center of mass is important when we analyze the motion of rigid bodies. y using of particles with another method. we can describe motion of a system y further substituting and deriving, we can find another equation to describe motion. Slide 69 / 69