Momentum. Conservation of Linear Momentum. Slide 1 / 140 Slide 2 / 140. Slide 3 / 140. Slide 4 / 140. Slide 6 / 140. Slide 5 / 140.

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1 Slide 1 / 140 Slide 2 / 140 Moentu Slide 3 / 140 Slide 4 / 140 Table of Contents Click on the topic to go to that section Conservation of Linear Moentu Ipulse - Moentu Equation Collisions in One Diension Collisions in Two Diensions Center of Mass It is Rocket Science Ballistic Pendulu Conservation of Linear Moentu Return to Table of Contents Slide 5 / 140 Conservation of Moentu The ost powerful concepts in science are called "conservation principles". Without worrying about the details of a process, conservation principles can be used to solve probles. If we were to take a snapshot of the initial and final syste, coparing the two would provide a lot of inforation. The last unit presented the Conservation of Energy, which proved helpful in solving probles where the situation was too coplex for Newton's Laws to be effectively used. Conservation of Energy is not enough. Slide 6 / 140 Conservation of Moentu If two ice skaters are holding hands, but then push away fro each other, Conservation of Energy will not be able to deterine each skater's velocity after the push. Assue a closed syste with no external forces. We need soething new. Let's start with Newton's Third Law. Assue the ice is frictionless. Then the force exerted by one skater on the second is equal and opposite to the force exerted by the second skater on the first and there are no external forces. Stating Newton's Third Law and doing a little bit of substitution

2 Slide 7 / 140 Conservation of Moentu Slide 8 / 140 Conservation of Moentu Make the assuption that we're not dealing with relativistic conditions (the skaters are oving way slower than the speed of light), and the skaters don't lose any ass in the process. Then, the constant ass ters can be put within the derivative. The initial linear oentu is then equal to the final linear oentu of the syste. Thus, linear oentu is conserved in a closed syste with no external forces. This is good, as it looks like we can now say soething about the skaters's velocities. The tie rate of change of the su of each skater's ass ties velocity is zero - it doesn't change. The quantity is defined as linear oentu and will be represented as - it is a vector. For two particles,, and since the initial and final oentu reain constant, we can write: Slide 9 / 140 Conservation of Moentu The Energy chapter of this course discussed how potential energy could be calculated for a syste that had only conservative forces. The Conservation of Moentu does not put this restriction on forces - as long as the forces are internal, oentu is conserved. We will leave the Conservation of Moentu for now, and apply it to probles involving collisions between objects in an upcoing chapter. Many of the great discoveries in nuclear and particle physics involve sashing atos and nuclei into each other and seeing what coes out. The Conservation of Moentu is key in these experients. Slide 10 / 140 Newton's Second Law restated Let's take a brief detour and exaine Newton's Second Law as it is taught today, and use the definition of oentu. The su of forces on a particle is equal to a. Since we assue the ass is constant, we can bring it inside the derivative. This is how Newton presented his Second Law in Principia - the su of forces on an object changes its oentu over tie. The ore general stateent of the Second Law has the benefit that it can be extended to cases where the ass of the objects change - not just the velocity. More on this later. Slide 11 / 140 Moentu is a Vector Quantity A key difference between oentu and energy is that energy is a scalar, while oentu is a vector. When there is ore than one object in a syste, the total oentu of the syste is found by the vector addition of the each object's oentu. Another key difference is that oentu coes in only one flavor (there is no kinetic, potential, elastic, etc.). The unit for oentu is kg-/s. There is no special unit for oentu, which presents an excellent opportunity to honor another physicist. Slide 12 / What is the oentu of a 20 kg object with a velocity of +5.0 /s? A -100 kg-/s B -50 kg-/s C 0 kg-/s D 50 kg-/s E 100 kg-/s

3 Slide 12 () / What is the oentu of a 20 kg object with a velocity of +5.0 /s? Slide 13 / What is the oentu of a 20 kg object with a velocity of -5.0 /s? A -100 kg-/s B -50 kg-/s A -100 kg-/s B -50 kg-/s C 0 kg-/s D 50 kg-/s E C 0 kg-/s D 50 kg-/s E 100 kg-/s E 100 kg-/s Slide 13 () / What is the oentu of a 20 kg object with a velocity of -5.0 /s? Slide 14 / 140 Moentu of a Syste of Objects A -100 kg-/s If a syste contains ore than one object, its total oentu is the vector su of each object's individual oentu: B -50 kg-/s C 0 kg-/s A D 50 kg-/s E 100 kg-/s Slide 15 / 140 Moentu of a Syste of Objects Slide 16 / 140 Moentu of a Syste of Objects In order to deterine the total oentu of a syste : Select a direction to be positive for each diension that's being considered. Assign positive values to each oentu in that direction. Assign negative values to each oentu in the opposite direction. Add the oenta together independently for each diension. Vectorially add each diension's oentu together to get the total oentu - the Pythagoras theore is used to find the agnitude, and trigonoetry is used to find its direction. Let's work an exaple in one diension. Deterine the oentu of a syste of two objects: 1, has a ass of 15 kg and a velocity of 16 /s towards the east and 2,has a ass of 42 kg and a velocity of 6.0 /s towards the west. or Choose East as positive. to the west

4 Slide 17 / Deterine the agnitude of the oentu of a syste of two objects: 1, which has a ass of 6.0 kg and a velocity of 13 /s north and 2, which has a ass of 14 kg and a velocity of 7.0 /s south. Assue north is positive. Slide 17 () / Deterine the agnitude of the oentu of a syste of two objects: 1, which has a ass of 6.0 kg and a velocity of 13 /s north and 2, which has a ass of 14 kg and a velocity of 7.0 /s south. Assue north is positive. A 10 kg /s A 10 kg /s B 15 kg /s B 15 kg /s E C 20 kg /s C 20 kg /s D -15 kg /s D -15 kg /s E -20 kg /s E -20 kg /s Slide 18 / Deterine the oentu of a syste of 3 objects: 1, which has a ass of 7.0 kg and a velocity of 23 /s north, 2, which has a ass of 9.0 kg and a velocity of 7.0 /s north and 3, which has a ass of 5.0 kg and a velocity of 42 /s south. Assue north is positive. Slide 18 () / Deterine the oentu of a syste of 3 objects: 1, which has a ass of 7.0 kg and a velocity of 23 /s north, 2, which has a ass of 9.0 kg and a velocity of 7.0 /s north and 3, which has a ass of 5.0 kg and a velocity of 42 /s south. Assue north is positive. A -12 kg /s A -12 kg /s B 12 kg /s C -14 kg /s B 12 kg /s C -14 kg /s D D 14 kg /s D 14 kg /s E 15 /s E 15 /s Slide 19 / 140 Slide 20 / 140 Ipulse-Moentu Theore Ipulse - Moentu Equation The Conservation of Linear Moentu applies to an isolated syste of particles. The overall oentu is conserved, but what about the oentu of each particle? Start with Newton's Second Law, as expressed in Principia, where we look at all the forces on one of the particles. Assue the force acts over a tie interval t 0 to t f, and integrate this expression. Return to Table of Contents The particle's oentu will change.

5 Slide 21 / 140 Ipulse-Moentu Theore We will exaine the specific case where a single, constant, very large force acts on the particle for a very short tie, so all other forces need not be considered. The equation siplifies to: Slide 22 / 140 Ipulse-Moentu Theore The force is not always constant - for exaple when a tennis racquet strikes a tennis ball, the force starts out sall, and increases as the ball increases its contact tie with the racquet, then decreases as it leaves. The large force at the peak results in a deforation of the ball. Define Ipulse as: F(N) The Ipulse-Moentu equation is then: Ipulse is a vector, and it is in the sae direction as the change of oentu or velocity of the particle acted on by the force. t (s) Slide 23 / 140 Ipulse-Moentu Theore Slide 24 / 140 Ipulse-Moentu Theore F(N) F(N) The shaded areas are equal in agnitude. F(N) F(N) Favg Favg t (s) The force - tie graph can be used to find the Ipulse delivered by the racquet in two ways: Find the area underneath the curve, either by integration if the force is specified as a function of tie, or by nuerical ethods. Find the average force delivered by the racquet and ultiply if by the tie interval. t (s) t (s) Force known as a function of tie: Average force known: t (s) But what if the force is not easily expressed in ters of tie or you don't have a nuerical integration capability? Slide 25 / 140 Ipulse-Moentu Theore Slide 26 / 140 Iplications of Ipulse F(N) F(N) Ipulse tells us that we can get the sae change in oentu with a large average force acting for a short tie, or a sall average force acting for a longer tie. Favg t (s) The equation also works in reverse (of course). If you have the change in oentu of the object, and the tie over which it occurs, the average force can be found. t (s) For a given change of oentu (when a person stops oving because of an ipact, like a car accident or falling), the force to the person can be iniized by extending the duration of the velocity reducing event. This is why one should bend their knees during a parachute landing, why airbags are used, and why landing on a pillow hurts less than landing on concrete.

6 Slide 27 / An external force of 25 N acts on a syste for 10 s. What is the agnitude of the Ipulse delivered to the syste? Slide 27 () / An external force of 25 N acts on a syste for 10 s. What is the agnitude of the Ipulse delivered to the syste? A 25 N-s B 100 N-s A 25 N-s B 100 N-s C 150 N-s D 200 N-s C 150 N-s D 200 N-s E E 250 N-s E 250 N-s Slide 28 / An external force of 25 N acts on a syste for 10 s. What is the change in oentu of the syste? Slide 28 () / An external force of 25 N acts on a syste for 10 s. What is the change in oentu of the syste? A 25 N-s B 100 N-s A 25 N-s B 100 N-s C 150 N-s D 200 N-s C 150 N-s D 200 N-s E E 250 N-s E 250 N-s Slide 29 / An average force of 5,000 N acts for 0.03 s on a 2.5 kg object that is initially at rest. What is its velocity after the application of the force? Slide 29 () / An average force of 5,000 N acts for 0.03 s on a 2.5 kg object that is initially at rest. What is its velocity after the application of the force? A 80 /s A 80 /s B 70 /s C 60 /s B 70 /s C 60 /s C D 50 /s D 50 /s E 40 /s E 40 /s

7 Slide 30 / An object at rest experiences a net horizontal force in the +x direction and begins oving. Use the force - tie graph below to find the net ipulse delivered by the force after 6 s. Slide 30 () / An object at rest experiences a net horizontal force in the +x direction and begins oving. Use the force - tie graph below to find the net ipulse delivered by the force after 6 s. A 12 N-s B 14 N-s C 16 N-s F (N) A 12 N-s B 14 N-s C 16 N-s F (N) A D 18 N-s D 18 N-s E 20 N-s t (s) E 20 N-s t (s) Slide 31 / A 2 kg object at rest experiences a net horizontal force in the +x direction and begins oving. Use the force - tie graph below to find the net ipulse delivered by the force after 6 s. Slide 31 () / A 2 kg object at rest experiences a net horizontal force in the +x direction and begins oving. Use the force - tie graph below to find the net ipulse delivered by the force after 6 s. A 4 N-s B 6 N-s C 8 N-s F (N) A 4 N-s B 6 N-s C 8 N-s F (N) C D 10 N-s D 10 N-s E 12 N-s t (s) E 12 N-s t (s) Slide 32 / A 2 kg object at rest experiences a net horizontal force in the +x direction and begins oving. Use the force - tie graph below to find the object's velocity after 6 s. Slide 32 () / A 2 kg object at rest experiences a net horizontal force in the +x direction and begins oving. Use the force - tie graph below to find the object's velocity after 6 s. A 10 /s B 8 /s C 6 /s F (N) A 10 /s B 8 /s C 6 /s F (N) E D 5 /s D 5 /s E 4 /s t (s) E 4 /s t (s)

8 Slide 33 / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball. Assue the bat loses contact with the ball when the force decreases to zero. Over what tie interval does this force act? Slide 33 () / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball. Assue the bat loses contact with the ball when the force decreases to zero. Over what tie interval does this force act? The force acts fro t = 0 to t = 1.01s Slide 34 / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball. The force acts over a tie interval of 1.01s. What is the agnitude of the axiu force delivered to the ball? Slide 34 () / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball. The force acts over a tie interval of 1.01s. What is the agnitude of the axiu force delivered to the ball? Slide 35 / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball. The force acts over a tie interval of 1.01s. What is the agnitude of the ipulse delivered to the ball? Slide 35 () / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball. The force acts over a tie interval of 1.01s. What is the agnitude of the ipulse delivered to the ball?

9 Slide 36 / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball. The force acts over a tie interval of 1.01s. What is the agnitude of the average force delivered to the ball? Slide 36 () / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball. The force acts over a tie interval of 1.01s. What is the agnitude of the average force delivered to the ball? Slide 37 / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball and delivers an ipulse of 32 N/s. What is the velocity of the ball at t = 1.01 s, assuing it started fro rest? Slide 37 () / A force described by F(t) = 190t - 189t 2 is applied by a bat to a kg ball and delivers an ipulse of 32 N/s. What is the velocity of the ball at t = 1.01 s, assuing it started fro rest? Slide 38 / 140 Slide 39 / 140 Types of Collisions Collisions in One Diension Objects in an isolated syste can interact with each other in two basic ways: They can collide. If they are stuck together, they can explode (push apart). In an isolated syste both oentu and total energy are conserved. But the energy can change fro one for to another. Return to Table of Contents Conservation of oentu and change in kinetic energy can help predict what will happen in these events.

10 Slide 40 / 140 Types of Collisions Slide 41 / 140 Inelastic Collisions We differentiate collisions and explosions by the way the energy changes or does not change for. inelastic collisions: two objects collide, converting soe kinetic energy into other fors of energy such as potential energy, heat or sound. elastic collisions: two objects collide and bounce off each other while conserving kinetic energy - energy is not transfored into any other type. explosions: an object or objects breaks apart because potential energy stored in one or ore of the objects is transfored into kinetic energy. There are two types of Inelastic Collisions. perfect inelastic collisions: two objects collide, stick together and ove as one ass after the collision, transferring kinetic energy into other fors of energy. general inelastic collisions: two objects collide and bounce off each other, transferring kinetic energy into other fors of energy. Slide 42 / 140 Elastic Collisions There is really no such thing as a perfect elastic collision. During all collisions, soe kinetic energy is always transfored into other fors of energy. But soe collisions transfor so little energy away fro kinetic energy that they can be dealt with as perfect elastic collisions. In cheistry, the collisions between olecules and atos are odeled as perfect elastic collisions to derive the Ideal Gas Law. Other exaples include a steel ball bearing dropping on a steel plate, a rubber "superball" bouncing on the ground, and billiard balls bouncing off each other. Slide 43 / 140 Explosions A firecracker is an exaple of an explosion. The cheical potential energy inside the firecracker is transfored into kinetic energy, light and sound. A cart with a copressed spring is a good exaple. When the spring is against a wall, and it is released, the cart starts oving - converting elastic potential energy into kinetic energy and sound. Think for a oent - can you see a reseblance between this phenoenon and either an elastic or inelastic collision? Slide 44 / 140 Explosions In both an inelastic collision and an explosion, kinetic energy is transfored into other fors of energy - such as potential energy. But they are tie reversed! An inelastic collision transfors kinetic energy into other fors of energy, such as potential energy. An explosion changes potential energy into kinetic energy. Thus, the equations to predict their otion will be inverted. The next slide suarizes the four types of collisions and explosions. Slide 45 / 140 Collisions and Explosions Event Description Moentu Conserved? General Inelastic Collision Perfect Inelastic Collision Elastic Collision Explosion Objects bounce off each other Objects stick together Objects bounce off each other Object or objects break apart Yes Yes Yes Yes Kinetic Energy Conserved? No. Kinetic energy is converted to other fors of energy No. Kinetic energy is converted to other fors of energy Yes No. Release of potential energy increases kinetic energy

11 Slide 46 / Moentu is conserved in which of the following types of collis Slide 46 () / Moentu is conserved in which of the following types of collis A Perfect Inelastic A Perfect Inelastic B Inelastic C Elastic B Inelastic C Elastic E D Explosions D Explosions E All of the Above E All of the Above Slide 47 / Kinetic energy is conserved in which of the following types of collisions? Slide 47 () / Kinetic energy is conserved in which of the following types of collisions? A Perfect Inelastic A Perfect Inelastic B Inelastic C Elastic B Inelastic C Elastic C D Explosions D Explosions E All of the Above E All of the Above Slide 48 / 140 Slide 49 / 140 Conservation of Moentu During a collision or an explosion, easureents show that the total oentu of a closed syste does not change. The diagra below shows the objects approaching, colliding and then separating. A Av A Bv B B +x the prie eans "after" A B Av A' A B Bv B' If the easureents don't show that the oentu is conserved, then this would not be a valid law. Fortunately they do, and it is!

12 Slide 50 / A 13,500 kg railroad freight car travels on a level track at a speed of 4.5 /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 syste. What is the speed of the two cars after colliding? Slide 50 () / A 13,500 kg railroad freight car travels on a level track at a speed of 4.5 /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 syste. What is the speed of the two cars after colliding?,500+25,000)kg st car's initial velocity ull tab] Slide 51 / 140 Slide 51 () / A cannon ball with a ass of kg flies in horizontal direction with a speed of 250 /s and strikes a ship initially at rest. The ass of the ship is 15,000 kg. Find the speed of the ship after the ball becoes ebedded in it. 19 A cannon ball with a ass of kg flies in horizontal direction with a speed of 250 /s and strikes a ship initially at rest. The ass of the ship is 15,000 kg. Find the speed of the ship after the ball becoes ebedded in it. +15,000)kg on ball's initial velocity ull tab] Slide 52 / A 40 kg girl skates at 5.5 /s on ice toward her 70 kg friend who is standing still, with open ars. As they collide and hold each other, what is their speed after the collision? Slide 52 () / A 40 kg girl skates at 5.5 /s on ice toward her 70 kg friend who is standing still, with open ars. As they collide and hold each other, what is their speed after the collision? )kg girls's initial b]

13 Slide 53 / 140 Slide 54 / 140 Explosions In an explosion, one object breaks apart into two or ore pieces (or coupled objects break apart), oving afterwards as separate objects. To ake the probles solvable at this ath level, we will assue: the object (or a coupled pair of objects) breaks into two pieces. the explosion is along the sae line as the initial velocity. Slide 55 / 140 Slide 55 () / A 5 kg cannon ball is loaded into a 300 kg cannon. When the cannon is fired, it recoils at 5 /s. What is the cannon ball's velocity after the explosion? 21 A 5 kg cannon ball is loaded into a 300 kg cannon. When the cannon is fired, it recoils at 5 /s. What is the cannon ball's velocity after the explosion? 1 = 5kg 2 = 300kg v1 = v2 = v = 0 v2' = 5/s (1+2) v = 1v1'+2v2' 0 = 1v1'+2v2' 1v1' = -2v2' v1' = -2v2'/1 = -(300kg)(5/s) / (5kg) = - 300/s Slide 56 / Two railcars, one with a ass of 4000 kg and the other with a ass of 6000 kg, are at rest and stuck together. To separate the a sall explosive is set off between the. The 4000 kg car is easured travelling at 6 /s. How fast is the 6000 kg car going? Slide 56 () / Two railcars, one with a ass of 4000 kg and the other with a ass of 6000 kg, are at rest and stuck together. To separate the a sall explosive is set off between the. The 4000 kg car is easured travelling at 6 /s. How fast is the 6000 kg car going? v = 1v1'+2v2' 1'+2v2' -2v2' 2v2'/1 4000kg)(6/s) / (6000kg) /s object is a pull tab]

14 Slide 57 / 140 Slide 58 / 140 Elastic Collisions In an elastic collision, two objects collide and bounce off each other, as shown below, and both oentu and kinetic energy are conserved. This will give us two siultaneous equations to solve to predict their otion after the collision. Before (oving towards) After (oving apart) p A= Av A p B= Bv B p A'= Av A' p B'= Bv B' A B A B Slide 59 / 140 Slide 60 / 140 Elastic Collision Siultaneous Equations Conservation of Moentu 1v 1 + 2v 2 = 1v 1' + 2v 2' 1v 1-1v 1' = 2v 2' - 2v 2 1(v 1 - v 1') = 2(v 2' - v 2) Conservation of Kinetic Energy ½ 1v ½ 2v 2 2 = ½ 1v 1' 2 +½ 2v 2' 2 1v v 2 2 = 1v 1' 2 + 2v 2' 2 1v 2 1-1v 1' 2 = 2v 2' 2-2v 2 2 1(v v 1' 2 ) = 2(v 2' 2 - v 22 ) 1(v 1 + v 1')(v 1 - v 1') = 2(v 2' + v 2)(v 2' - v 2) 1(v 1 + v 1')(v 1 - v 1') = 2(v 2' + v 2)(v 2' - v 2) 1(v 1 - v 1') = 2(v 2' - v 2) v 1 + v 1' = v 2' + v 2 v 1 - v 2 = -(v 1' - v' 2) Slide 61 / 140 Properties of Elastic Collisions By solving the conservation of oentu and constant kinetic energy equations siultaneously, the following result appeared: v 1 - v 2 = -(v 1' - v' 2) Slide 62 / Two objects have an elastic collision. Before they collide, they are approaching with a velocity of 4 /s relative to each other. With what velocity do they ove apart fro one another after the collision? Do you recognize the ters on the left and right of the above equation? And, what does it ean? The ters are the relative velocities of the two objects before and after the collision. It eans that for all elastic collisions - regardless of ass - the relative velocity of the objects is the sae before and after the collision.

15 Slide 62 () / Two objects have an elastic collision. Before they collide, they are approaching with a velocity of 4 /s relative to each other. With what velocity do they ove apart fro one another after the collision? Slide 63 / Two objects have an elastic collision. Object 1, has an initial velocity of +4.0 /s and 2 has a velocity of -3.0 /s. After the collision, 1 has a velocity of 1.0 /s. What is the velocity of 2? 4 /s away fro each other v1-v2 = -(v1'-v2') The difference in the initial velocities is the sae as the negative difference of the final velocities. Or, the relative velocity between the objects before the collision is equal to the negative of the relative velocity between the objects after the collision. Slide 63 () / Two objects have an elastic collision. Object 1, has an initial velocity of +4.0 /s and 2 has a velocity of -3.0 /s. After the collision, 1 has a velocity of 1.0 /s. What is the velocity of 2? Slide 64 / Two objects have an elastic collision. Object 1, has an initial velocity of +6.0 /s and 2 has a velocity of 2.0 /s. After the collision, 1 has a velocity of 1.0 /s. What is the velocity of 2? v1 = 4 /s v2 = -3 /s v1' = 1 /s v2' =? v1-v2 = -(v1'-v2') v2' = v1+v1'-v2 v2' = 4/s + 1/s - (-3/s) v2' = 8 /s Slide 64 () / 140 Slide 65 / Two objects have an elastic collision. Object 1, has an initial velocity of +6.0 /s and 2 has a velocity of 2.0 /s. After the collision, 1 has a velocity of 1.0 /s. What is the velocity of 2? v1 = 6 /s v2 = 2 /s v1' = 1 /s v2' =? v1-v2 = -(v1'-v2') v2' = v1+v1'-v2 v2' = 6/s + 1/s - (2/s) v2' = 5 /s

16 Slide 66 / 140 Slide 67 / 140 Slide 68 / 140 Slide 69 / 140 Slide 70 / 140 Slide 71 / 140

17 Slide 72 / 140 Slide 73 / 140 Slide 74 / 140 Slide 75 / A 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 75 () / 140 Slide 76 / A 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? 27 A 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. How fast is the baseball going after it's hit? v1 = +v v2 = 0 v1' = +v v2' =? v1-v2 = -(v1'-v2') v2' = v1+v1'-v2 v2' = v + v - 0 v2' = 2 v (ping pong ball's speed is twice that of the bowling ball)

18 Slide 76 () / A 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. How fast is the baseball going after it's hit? Slide 77 / Two objects with identical asses have an elastic collision: the initial velocity of 1 is +6.0 /s and 2 is -3.0 /s. What is the velocity of 1 after the collision? v1 = +v v2 = -2v v1' = +v v2' =? v1-v2 = -(v1'-v2') v2' = v1+v1'-v2 v2' = v + v - (-2v) v2' = 4 v Slide 77 () / Two objects with identical asses have an elastic collision: the initial velocity of 1 is +6.0 /s and 2 is -3.0 /s. What is the velocity of 1 after the collision? Slide 78 / Two objects with identical asses have an elastic collision: the initial velocity of 1 is +6.0 /s and 2 is -3.0 /s. What is the velocity of 2 after the collision? v1 = +6/s v2 = -3/s v1' =? v2' =? When identical ass objects experience an elastic collision, they swap their initial velocities: v1' = v2 = -3.0 /s v2' = v1 = 6.0 /s So the velocity of 1 is -3.0 /s. Slide 78 () / Two objects with identical asses have an elastic collision: the initial velocity of 1 is +6.0 /s and 2 is -3.0 /s. What is the velocity of 2 after the collision? Slide 79 / A golf ball is hit against a solid ceent wall, and experiences an elastic collsion. The golf ball strikes the wall with a velocity of +35 /s. What velocity does it rebound with? v1 = +6/s v2 = -3/s v1' =? v2' =? When identical ass objects experience an elastic collision, they swap their initial velocities: v1' = v2 = -3.0 /s v2' = v1 = 6.0 /s So the velocity of 1 is 6.0 /s.

19 Slide 79 () / 140 Slide 80 / A golf ball is hit against a solid ceent wall, and experiences an elastic collsion. The golf ball strikes the wall with a velocity of +35 /s. What velocity does it rebound with? When a light object strikes a very assive object, the light object rebounds with the opposite velocity - in this case, the golf ball will leave the wall with a velocity of -35 /s. Collisions in Two Diensions Return to Table of Contents Slide 81 / 140 Slide 82 / 140 Conservation of Moentu in Two Diensions Moentu vectors (like all vectors) can be expressed in ters of coponent vectors relative to a reference frae This eans that the oentu conservation equation p = p' can be solved independently for each coponent: This, of course also applies to three diensions, but we'll stick with two for this chapter! Slide 83 / 140 Exaple: Collision with a Wall Consider the case of a golf ball colliding elastically with a hard wall, rebounding with the sae velocity, where its angle of incidence equals its angle of reflection. Is oentu conserved in this proble? Slide 84 / 140 p y' p y p x' p x p' The solid lines represent the oentu of the ball (blue - prior to collision, red - after the collision). The dashed lines are the x and y coponents of the oentu vectors. p θ θ Exaple: Collision with a Wall Moentu is not conserved! An external force fro the wall is being applied to the ball in order to reverse its direction in the x axis. However, since we have an elastic collision, the ball bounces off the wall with the sae speed that it struck the wall. Hence, the agnitude of the initial oentu and the final oentu is equal: p' p y' p x' p x θ θ p y p Now it's tie to resolve oentu into coponents along the x and y axis.

20 Slide 85 / 140 Slide 86 / A tennis ball of ass strikes a wall at an angle θ relative to noral then bounces off with the sae speed as it had initially. What was the change in oentu of the ball? A -v B -2v C -v cosθ D -2v cosθ Slide 87 / A tennis ball of ass strikes a wall an an angle θ relative to noral and then bounces off with the sae speed as it had initially. What is the change in oentu of the ball? Slide 87 () / A tennis ball of ass strikes a wall an an angle θ relative to noral and then bounces off with the sae speed as it had initially. What is the change in oentu of the ball? A 0 A 0 B -v C -2v D -v cosθ E -2v cosθ p' p y' p x' θ p y' p x' θ p B -v C -2v D -v cosθ E -2v cosθ p' p y' p x' θ p y' p x' θ p E Slide 88 / A tennis ball of ass strikes a wall an an angle θ relative to noral and then bounces off with the sae speed as it had initially. What is the change in oentu of the ball in the y direction? Slide 88 () / A tennis ball of ass strikes a wall an an angle θ relative to noral and then bounces off with the sae speed as it had initially. What is the change in oentu of the ball in the y direction? A 0 A 0 B -v C -2v D v E 2v p' p y' p x' θ p y' p x' θ p B -v C -2v D v E 2v p' p y' p x' θ A p y' p x' θ p

21 Slide 89 / 140 General Two Diensional Collisions Slide 90 / 140 General Two Diensional Collisions 1 2 Before 1 2 p 2 = 0 After 1 2 p 2 =? We'll now consider the ore general case of two objects oving in rando directions in the x-y plane and colliding. Since there is no absolute reference frae, we'll line up the x-axis with the velocity of one of the objects. This is not a head on collision - note how 1 heads off with a y coponent of velocity after it strikes 2. Also, did you see how we rotated the coordinate syste so the x axis is horizontal? To siplify the proble, we will specify that ass 2 is at rest. The proble now is to find the oentu of 2 after the collision. Slide 91 / 140 General Two Diensional Collisions Before 1 2 p 2 = 0 After 1 2 p 2 =? This will be done by looking at the vectors first - oentu ust be conserved in both the x and the y directions. Since the oentu in the y direction is zero before the collision, it ust be zero after the collision. And, the value that 1 has for oentu in the x direction ust be shared between both objects after the collision - and not equally - it will depend on the asses and the separation angle. Slide 93 / After the collision shown below, which of the following is the ost likely oentu vector for the blue ball? Slide 92 / 140 General Two Diensional Collisions Here is the oentu vector breakdown of ass 1 after the collision: 1 2? 2 needs to have a coponent in the y direction to su to zero with 1's final y oentu. And it needs a coponent in the x direction to add to 1's final x oentu to equal the initial x oentu of 1: 2 Slide 93 () / 140 and this is the final oentu for ass 2 by vectorially adding the final p x and p y. 34 After the collision shown below, which of the following is the ost likely oentu vector for the blue ball? 1 before 2 after 2? 1 before 2 after 2? A 1 A 1 B C B C D D D E E

22 Slide 94 / 140 General Two Diensional Collisions Slide 95 / 140 General Two Diensional Collisions Now that we've seen the vector analysis, let's run through the algebra to find the exact velocity (agnitude and direction) that 2 leaves with after the collision kg-/s before after 20.0 kg-/s There is a bowling ball with oentu 20.0 kg-/s that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above. What is the final velocity of the pin? θ 12.0 kg-/s before after 20.0 kg-/s Given: Find: θ 60.0 Slide 96 / 140 Slide 97 / 140 General Two Diensional Collisions General Two Diensional Collisions 12.0 kg-/s before after 20.0 kg-/s kg-/s before after 20.0 kg-/s θ 2 2 θ Use Conservation of Moentu in the x and y directions. x direction y-direction Now that the x and y coponents of the oentu of ass 2 have been found, the total final oentu is calculated. Slide 98 / A 5.0 kg bowling ball strikes a stationary bowling pin. After the collision, the ball and the pin ove in directions as shown and the agnitude of the pin's oentu is 18.0 kg-/s. What was the velocity of the ball before the collision? Slide 98 () / A 5.0 kg bowling ball strikes a stationary bowling pin. After the collision, the ball and the pin ove in directions as shown and the agnitude of the pin's oentu is 18.0 kg-/s. What was the velocity of the ball before the collision? before? ball 53.1 pin kg-/s after before before p1x =? p2x = 0 p1y = 0 p2y = 0 1 = 5 kg y-direction p1y + p2y = p'1y + p'2y = p'1y +18sin(-30 ) 0 = p'1y - 9 p'1y = 9 kg-/s after p'2x = 18cos(30 )kg-/s p'1x =? p'2y = 18sin(30 )kg-/s p'1y =?? ball 53.1 x-direction p1x + p2x = p'1x + p'2x tan(53.1 ) = p1y' / p1x' pin kg-/s after p1x = p'1y/tan(53.1 ) + p'2x p1x = 9/tan(53.1 ) + 18cos(30 ) p1x = 9/ = 22.4 kg-/s v1x = p1x /1 = 22.4/5 = 4.48 /s

23 Slide 99 / 140 Perfect Inelastic Collisions in Two Diensions Slide 100 / 140 Perfect Inelastic Collisions in Two Diensions One coon kind of inelastic collision is where two cars collide and stick at an intersection. In this situation the two objects are traveling along paths that are perpendicular just prior to the collision. Before 1 p 1 2 p After θ p' Before 2 p 2 1 p 1 p-conservation in x: in y: final oentu: final velocity: final direction: 1 2 After p' x θ p' p' y Slide 101 / Object A with ass 20.0 kg travels to the east at 10.0 /s and object B with ass 5.00 kg travels south at 20.0 /s. They collide and stick together. What is the velocity (agnitude and direction) of the objects after the collision? Slide 101 () / Object A with ass 20.0 kg travels to the east at 10.0 /s and object B with ass 5.00 kg travels south at 20.0 /s. They collide and stick together. What is the velocity (agnitude and direction) of the objects after the collision? south of east Slide 102 / 140 Slide 103 / 140 Center of Mass Center of Mass Return to Table of Contents All the probles we've solved have assued that we're dealing with point particles - this is shown ost clearly when free body diagras were used to find the acceleration of objects due to ultiple forces. The object was actually represented by a point! Other than aking the proble sipler (which is good), why did the coputed answers actually atch what happened to the larger, ore extended object? Siple solutions are no good unless they represent reality. To answer this question, the center of ass will be defined, and its relationship to the conservation of oentu will be shown.

24 Slide 104 / 140 Center of Mass There is a collection of particles, with different asses, in space, each with an x and y coponent (x, y) specifying their location. Define the center of ass as having coordinates (x c, y c), where: Slide 105 / 140 Center of Mass A particle's position is described by a position vector, r: Therefore r c is expressed as: This is a "weighted average" of each particle - ore assive particles contribute ore to the coordinates of the center of ass What's next? Think Kineatics. Slide 106 / 140 Slide 107 / 140 Center of Mass Let's derive the velocity and the acceleration of the center of ass by taking the tie derivatives position. First the velocity. Slide 108 / 140 Center of Mass It shows that the total oentu of a syste of objects is equal to the total ass of the syste ties the velocity of its center of ass. The syste acts the sae as if all of its ass was concentrated at the center of ass. Slide 109 / 140 Center of Mass One ore step to finalize the description of the otion of the center of ass. Take the derivative of the oentu vector of the center of ass with respect to tie. Rewriting: Newton's Second Law If the oentu of the syste is constant (no net external forces acting), then no atter what the individual objects are doing, the velocity of their center of ass reains constant. This shows that conservation of oentu applies equally to a solid object as well as a syste of particles.

25 Slide 110 / 140 Center of Mass In a syste of objects, both internal and external forces are present. Internal, fro the interaction of the objects, and external coing fro outside the syste, so we write: Newton's Third Law - the internal forces are action reaction forces and their net effect on the syste is zero, so ΣF int = 0. This enables us to work with solid, acroscopic bodies - the body accelerates as if all of the external forces are acting on the center of ass of all its coponent parts. Slide 112 / A wire is bent into the below shape. What are the coordinates for the center of ass of the wire? Slide 111 / 140 Separating Masses The below photograph shows 3 different fireworks rockets that were sent up into the air and then exploded. The center of ass of each rocket kept oving up, and the particles syetrically spread out fro each one. Slide 112 () / 140 For each rocket, the velocity of the center of ass is equal to the weighted average of the velocities of each particle, resulting in the starburst pattern. The acceleration of each particle (after the explosion) and the center of ass all equal -g. 37 A wire is bent into the below shape. What are the coordinates for the center of ass of the wire? Slide 113 / A issile is launched with velocity v 0, and explodes id flight into over 1000 fragents. What is the velocity of the center of ass of the syste after the explosion? Slide 113 () / A issile is launched with velocity v 0, and explodes id flight into over 1000 fragents. What is the velocity of the center of ass of the syste after the explosion? A 0 A 0 B v 0 C 2v 0 B v 0 C 2v 0 B D -v 0 D -v 0 E -2v 0 E -2v 0

26 Slide 114 / 140 Explosions in Two Diensions Slide 115 / 140 Explosions in Two Diensions The Black object explodes into 3 pieces (blue, red and green). We want to deterine the oentu of the third piece. p' 3 p' 2 p' 1 During an explosion, the total oentu is unchanged, since no EXTERNAL force acts on the syste. By Newton's Third Law, the forces that occur between the particles within the object will add up to zero, so they don't affect the oentu. If the initial oentu is zero, the final oentu is zero. The third piece ust have equal and opposite oentu to the su of the other two. Move the dashed box to see the third piece's oentu. before: p x = p y = 0 p' 2 p' 1 p' 3 # after: p' 1x + p' 2x + p' 3x = 0 p' 1y + p' 2y + p' 3y = 0 The Black object explodes into 3 pieces (blue, red and green). We want to deterine the oentu of the third piece. In this case the blue and red pieces are oving perpendicularly to each other, so: Slide 116 / A stationary cannon ball explodes into three pieces. The oenta of two of the pieces is shown below. What is the direction of the oentu of the third piece? Slide 116 () / A stationary cannon ball explodes into three pieces. The oenta of two of the pieces is shown below. What is the direction of the oentu of the third piece? A A B B C D C D C E E Slide 117 / A stationary 10.0 kg bob explodes into three pieces. A 2.00 kg piece oves west at /s. Another piece with a ass of 3.00 kg oves north with a velocity of /s. What is the velocity (speed and direction) of the third piece? Slide 117 () / A stationary 10.0 kg bob explodes into three pieces. A 2.00 kg piece oves west at /s. Another piece with a ass of 3.00 kg oves north with a velocity of /s. What is the velocity (speed and direction) of the third piece?

27 Slide 118 / 140 Slide 119 / 140 It is Rocket Science It is Rocket Science Return to Table of Contents On July 20, 1969, the Aerican Apollo 11 spacecraft landed on the oon, and six hours later, Neil Arstrong becae the first huan to walk on the oon. While the spacecraft was enroute to the oon, the New York Ties published the following article: JULY 17, 1969: On Jan. 13, 1920, Topics of The Ties, an editorial-page feature of The New Yo Ties, disissed the notion that a rocket could function in a vacuu and coented on the idea of Robert H. Goddard, the rocket pioneer, as follows: ''That Professor Goddard, with his 'chair' in Clark College and the countenancing of the Sithsonian Institution, does not know the relation o action to reaction, and of the need to have soething better than a vacuu against which to reac to say that would be absurd. Of course he only sees to lack the knowledge ladled out daily in h schools.'' Further investigation and experientation have confired the findings of Isaac Newton in the 17 century and it is now definitely established that a rocket can function in a vacuu as well as in an atosphere. The Ties regrets the error. Slide 120 / 140 It is Rocket Science Slide 121 / 140 It is Rocket Science The 1920 article referenced Newton's Third Law, but how did they isinterpret it? And how is Newton's Third Law related to the Conservation of Moentu? A rocket operates by expelling burned fuel through its tail. The fuel has both ass and velocity, so it has oentu. Before launch, the rocket was at rest, so its oentu was zero. Define the syste as the rocket and its fuel (the onboard and the expelled fuel), and since there are no external forces acting on it, oentu ust be conserved. The rocket is launched by burning the fuel and expelling it. It then has a oentu opposite the expelled fuel and oves forward to keep the total oentu of the syste equal to zero. Slide 122 / 140 It is Rocket Science The action - reaction forces described by Newton's Third Law and the free body diagras fro his Second Law are: The rocket exerts a force on the burned fuel as it expels it out the tail. Slide 123 / 140 It is Rocket Science This is a slightly ore coplex proble than the skaters pushing off of each other or an archer shooting an arrow. In those cases, the asses of each object reains the sae (the arrow's ass is negligible copared to the archer). The burned fuel exerts a force on the rocket ship as it leaves. Burned fuel Rocket And that's how Newton's Third Law relates to the Conservation of Moentu!

28 Slide 124 / 140 Slide 125 / 140 It is Rocket Science For the ore general case, assue the rocket is already oving at a constant veloicty and has a certain oentu. It then lights its engines and starts burning fuel. There is no external force so the oentu right before the burn ust equal the oentu at any tie. The rocket and the burned fuel both increase oentu but the oentu at any tie during the fuel burn is constant. Soe algebraic anipulation: Slide 126 / 140 It is Rocket Science Slide 127 / 140 It is Rocket Science Integrate fro v 0 to v f and fro M r0 (initial ass of rocket with fuel) to M rf (final ass of rocket with reaining, unburned fuel). v ef is the velocity of the expelled fuel which is kept constant by the pilot: Take the liit of Δv and Δ ef as Δt approaches zero: As the ass of the expelled fuel increases, the ass of the rocket plus unburned fuel decreases by the sae aount. Rocket equation Slide 128 / A issile is launched fro rest. The initial ass of the issile and its fuel is 125 kg. The fuel's exhaust velocity is 2500 /s. How uch fuel is used to accelerate the rocket to a speed of 926 /s? Slide 128 () / A issile is launched fro rest. The initial ass of the issile and its fuel is 125 kg. The fuel's exhaust velocity is 2500 /s. How uch fuel is used to accelerate the rocket to a speed of 926 /s? A 7.8 kg A 7.8 kg B 20 kg B 20 kg C 39 kg D 50 kg C 39 kg D 50 kg C E 86 kg E 86 kg

29 Slide 129 / 140 Slide 130 / 140 Ballistic Pendulu Ballistic Pendulu Before they were rendered obsolete by odern sensors, Ballistic Pendulus were used to find the velocities of projectiles (bullets). But, they still are used in Physics Labs, where a cobination of Conservation of Moentu, and Conservation of Total Mechanical Energy can be used to find the approxiate velocity of a projectile. l l Return to Table of Contents v M vm = 0 v' M+ A ore exact solution requires the use of rotational dynaics to account for the oent of inertia of the block. h Slide 131 / 140 Slide 132 / 140 Ballistic Pendulu A bullet of ass is fired into a block of ass M and reains ebedded in the block. The block is attached to a stand with a fixed length, l, of string/wire and oves as shown below. What type of collision is this? What is conserved? Ballistic Pendulu This is a perfectly inelastic collision, so linear oentu is conserved, but kinetic energy is not conserved - soe of the bullet's energy goes into non conservative forces, such as friction and aterial deforation of the block and bullet. After the collision, assue that there are zero non conservative external forces acting on the syste. What conservation law can now be used? l l v M vm = 0 M+ v' h v l M vm = 0 l M+ v' h Slide 133 / 140 Ballistic Pendulu Conservation of Total Mechanical Energy. After the collision, the bullet-block syste rises a height h, above its initial position, and the string akes an angle θ with the vertical. That is shown in the picture on the right where one of the strings has been reoved to see the angle θ ore clearly. Slide 134 / 140 Ballistic Pendulu A little trigonoetry to find the change in gravitational potential energy due to the block rising after the bullet's ipact. l l θ l lcosθ θ v M vm = 0 M+ v' h l - lcosθ M+ v' h

30 Slide 135 / 140 Slide 136 / 140 Ballistic Pendulu Substitute the first into the second equation: l v M vm = 0 l lcosθ θ l - lcosθ M+ v' h Slide 137 / 140 Slide 138 / A 13 g bullet is fired at 460 /s into a stationary 1.0 kg block and ebeds within the block (copletely inelastic collision). Find the velocity of the bullet - block syste. A 3.2 /s B 4.7 /s C 5.0 /s v D 5.9 /s E 6.3 /s Slide 138 () / A 13 g bullet is fired at 460 /s into a stationary 1.0 kg block and ebeds within the block (copletely inelastic collision). Find the velocity of the bullet - block syste. Slide 139 / A 13 g bullet is fired at 460 /s into a stationary 1.0 kg block attached to a 2.0 long string and ebeds within the block (copletely inelastic collision). Find the axiu height that the bullet - block syste oves to. A 3.2 /s B 4.7 /s C 5.0 /s D 5.9 /s E 6.3 /s v D A 1.0 B 1.5 C 1.8 D 2.2 E 2.3 Before v After 2 h

31 Slide 139 () / A 13 g bullet is fired at 460 /s into a stationary 1.0 kg block attached to a 2.0 long string and ebeds within the block (copletely inelastic collision). Find the axiu height that the bullet - block syste oves to. A 1.0 B 1.5 C 1.8 D 2.2 E 2.3 Before v After 2 h C Slide 140 / A 13 g bullet is fired at 460 /s into a stationary 1.0 kg block attached to a 2.0 long string and ebeds within the block (copletely inelastic collision). Find the angle θ that is subtended by the string at the axiu height of the bullet - block syste. A 57 0 B 62 0 C 65 0 D 77 0 E h Slide 140 () / A 13 g bullet is fired at 460 /s into a stationary 1.0 kg block attached to a 2.0 long string and ebeds within the block (copletely inelastic collision). Find the angle θ that is subtended by the string at the axiu height of the bullet - block syste. A 57 0 B 62 0 C E D 77 0 h E 84 0