Momentum - NJCTL
Slide 1 / 140Slide 2 / 140Momentumwww.njctl.orgTable of ContentsClick on the topic to go to that section· Conservation of Linear Momentum· Impulse - Momentum Equation· Collisions in One Dimension· Collisions in Two Dimensions· Center of Mass· It is Rocket Science· Ballistic PendulumSlide 3 / 140
Slide 4 / 140Conservation of LinearMomentumReturn to Tableof ContentsConservation of MomentumSlide 5 / 140The most powerful concepts in science are called "conservationprinciples". 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.The last unit presented the Conservation of Energy, whichproved helpful in solving problems where the situation was toocomplex for Newton's Laws to be effectively used.Conservation of Energy is not enough.Conservation of MomentumIf two ice skaters are holding hands, but then push away fromeach other, Conservation of Energy will not be able to determineeach skater's velocity after the push. Assume a closed systemwith no external forces.We need something new. Let's start with Newton's Third Law.Assume the ice is frictionless. Then the force exerted by oneskater on the second is equal and opposite to the force exerted bythe second skater on the first and there are no external forces.Stating Newton's Third Law anddoing a little bit of substitutionSlide 6 / 140
Conservation of MomentumSlide 7 / 140Make the assumption that we're not dealing with relativisticconditions (the skaters are moving way slower than the speed oflight), and the skaters don't lose any mass in the process.Then, the constant mass terms can be put within the derivative.This is good, as it looks like wecan now say something about theskaters's velocities. The time rateof change of the sum of eachskater's mass times velocity iszero - it doesn't change. Thequantityis defined as linearmomentum and will berepresented as - it is a vector.Conservation of MomentumSlide 8 / 140The initial linear momentum is then equal to the final linearmomentum of the system. Thus, linear momentum is conservedin a closed system with no external forces.For two particles,, and since the initial and finalmomentum remain constant, we can write:Conservation of MomentumThe Energy chapter of this course discussed how potential energycould be calculated for a system that had only conservative forces.The Conservation of Momentum does not put this restriction onforces - as long as the forces are internal, momentum is conserved.We will leave the Conservation of Momentum for now, and apply itto problems involving collisions between objects in an upcomingchapter.Many of the great discoveries in nuclear and particle physicsinvolve smashing atoms and nuclei into each other and seeingwhat comes out. The Conservation of Momentum is key in theseexperiments.Slide 9 / 140
Newton's Second Law restatedSlide 10 / 140Let's take a brief detour and examine Newton's Second Law as it istaught today, and use the definition of momentum.The sum of forces on a particle is equalto ma. Since we assume the mass isconstant, we can bring it inside thederivative.This is how Newton presented his Second Law in Principia - thesum of forces on an object changes its momentum over time.The more general statement of the Second Law has the benefit thatit can be extended to cases where the mass of the objects change not just the velocity. More on this later.Slide 11 / 140Momentum is a Vector QuantityA key difference between momentum and energy is that energyis a scalar, while momentum is a vector.When there is more than one object in a system, the totalmomentum of the system is found by the vector addition of theeach object's momentum.Another key difference is that momentum comes in only oneflavor (there is no kinetic, potential, elastic, etc.).The unit for momentum is kg-m/s. There is no special unit formomentum, which presents an excellent opportunity to honoranother physicist.1 What is the momentum of a 20 kg object with a velocity of 5.0 m/s?A -100 kg-m/sB -50 kg-m/sC 0 kg-m/sD 50 kg-m/sE 100 kg-m/sSlide 12 / 140
1 What is the momentum of a 20 kg object with a velocity of 5.0 m/s?Slide 12 (Answer) / 140A -100 kg-m/sC 0 kg-m/sD 50 kg-m/sAnswerB -50 kg-m/sEE 100 kg-m/s[This object is a pull tab]2 What is the momentum of a 20 kg object with a velocity of-5.0 m/s?Slide 13 / 140A -100 kg-m/sB -50 kg-m/sC 0 kg-m/sD 50 kg-m/sE 100 kg-m/s2 What is the momentum of a 20 kg object with a velocity of-5.0 m/s?B -50 kg-m/sC 0 kg-m/sAnswerA -100 kg-m/sAD 50 kg-m/sE 100 kg-m/s[This object is a pull tab]Slide 13 (Answer) / 140
Momentum of a System of ObjectsSlide 14 / 140If a system contains more than one object, its total momentumis the vector sum of each object's individual momentum:Momentum of a System of ObjectsSlide 15 / 140In order to determine the total momentum of a system :· Select a direction to be positive for each dimension that'sbeing considered.· Assign positive values to each momentum in that direction.· Assign negative values to each momentum in the oppositedirection.· Add the momenta together independently for eachdimension.· Vectorially add each dimension's momentum together to getthe total momentum - the Pythagoras theorem is used tofind the magnitude, and trigonometry is used to find itsdirection.Momentum of a System of ObjectsLet's work an example in one dimension. Determine the momentumof a system of two objects: m1, has a mass of 15 kg and a velocity of16 m/s towards the east and m2,has a mass of 42 kg and a velocityof 6.0 m/s towards the west.Choose East as positive.orto the westSlide 16 / 140
3 Determine the magnitude of the momentum of a system oftwo objects: m1, which has a mass of 6.0 kg and a velocity of13 m/s north and m2, which has a mass of 14 kg and avelocity of 7.0 m/s south. Assume north is positive.Slide 17 / 140A 10 kg m/sB 15 kg m/sC 20 kg m/sD -15 kg m/sE -20 kg m/sA 10 kg m/sB 15 kg m/sAnswer3 Determine the magnitude of the momentum of a system oftwo objects: m1, which has a mass of 6.0 kg and a velocity of13 m/s north and m2, which has a mass of 14 kg and avelocity of 7.0 m/s south. Assume north is positive.Slide 17 (Answer) / 140EC 20 kg m/sD -15 kg m/s[This object is a pull tab]E -20 kg m/s4 Determine the momentum of a system of 3 objects: m1, whichhas a mass of 7.0 kg and a velocity of 23 m/s north, m2, whichhas a mass of 9.0 kg and a velocity of 7.0 m/s north and m3,which has a mass of 5.0 kg and a velocity of 42 m/s south.Assume north is positive.A -12 kg m/sB 12 kg m/sC -14 kg m/sD 14 kg m/sE 15 m/sSlide 18 / 140
4 Determine the momentum of a system of 3 objects: m1, whichhas a mass of 7.0 kg and a velocity of 23 m/s north, m2, whichhas a mass of 9.0 kg and a velocity of 7.0 m/s north and m3,which has a mass of 5.0 kg and a velocity of 42 m/s south.Assume north is positive.Slide 18 (Answer) / 140B 12 kg m/sAnswerA -12 kg m/sDC -14 kg m/sD 14 kg m/s[This object is a pull tab]E 15 m/sSlide 19 / 140Impulse - MomentumEquationReturn to Tableof ContentsImpulse-Momentum TheoremThe Conservation of Linear Momentum applies to an isolatedsystem of particles. The overall momentum is conserved, but whatabout the momentum of each particle?Start with Newton's Second Law, as expressed in Principia, wherewe look at all the forces on one of the particles.Assume the force acts over atime interval t0 to tf, and integratethis expression.The particle's momentum will change.Slide 20 / 140
Impulse-Momentum TheoremSlide 21 / 140We will examine the specific case where a single, constant, verylarge force acts on the particle for a very short time, so all otherforces need not be considered. The equation simplifies to:Define Impulse as:The Impulse-Momentum equation is then:Impulse is a vector, and it is in the same direction as the change ofmomentum or velocity of the particle acted on by the force.Impulse-Momentum TheoremSlide 22 / 140The force is not always constant - for example when a tennisracquet strikes a tennis ball, the force starts out small, andincreases as the ball increases its contact time with the racquet,then decreases as it leaves. The large force at the peak results in adeformation of the ball.F(N)t (s)Impulse-Momentum TheoremF(N)F(N)The shaded areas areequal in magnitude.Favgt (s)t (s)The force - time graph can be used to find the Impulse delivered bythe racquet in two ways:· Find the area underneath the curve, either by integration if theforce is specified as a function of time, or by numerical methods.· Find the average force delivered by the racquet and multiply if bythe time interval.Slide 23 / 140
Impulse-Momentum TheoremF(N)Slide 24 / 140F(N)Favgt (s)t (s)· Force known as a function of time:· Average force known:But what if the force is not easily expressedin terms of time or you don't have anumerical integration capability?Impulse-Momentum TheoremF(N)Slide 25 / 140F(N)Favgt (s)t (s)The equation also works in reverse (of course). If you have thechange in momentum of the object, and the time over which itoccurs, the average force can be found.Implications of ImpulseImpulse tells us that we can get the same change in momentumwith a large average force acting for a short time, or a smallaverage force acting for a longer time.For a given change of momentum (when a person stops movingbecause of an impact, like a car accident or falling), the force tothe person can be minimized by extending the duration of thevelocity reducing event.This is why one should bend their knees during a parachutelanding, why airbags are used, and why landing on a pillow hurtsless than landing on concrete.Slide 26 / 140
5 An external force of 25 N acts on a system for 10 s. What isthe magnitude of the Impulse delivered to the system?Slide 27 / 140A 25 N-sB 100 N-sC 150 N-sD 200 N-sE 250 N-s5 An external force of 25 N acts on a system for 10 s. What isthe magnitude of the Impulse delivered to the system?Slide 27 (Answer) / 140A 25 N-sC 150 N-sD 200 N-sAnswerB 100 N-sEE 250 N-s[This object is a pull tab]6 An external force of 25 N acts on a system for 10 s. What isthe change in momentum of the system?A 25 N-sB 100 N-sC 150 N-sD 200 N-sE 250 N-sSlide 28 / 140
6 An external force of 25 N acts on a system for 10 s. What isthe change in momentum of the system?Slide 28 (Answer) / 140A 25 N-sC 150 N-sD 200 N-sAnswerB 100 N-sEE 250 N-s[This object is a pull tab]7 An average force of 5,000 N acts for 0.03 s on a 2.5 kg objectthat is initially at rest. What is its velocity after the applicationof the force?Slide 29 / 140A 80 m/sB 70 m/sC 60 m/sD 50 m/sE 40 m/s7 An average force of 5,000 N acts for 0.03 s on a 2.5 kg objectthat is initially at rest. What is its velocity after the applicationof the force?B 70 m/sC 60 m/sAnswerA 80 m/sCD 50 m/sE 40 m/s[This object is a pull tab]Slide 29 (Answer) / 140
8 An object at rest experiences a net horizontal force in the xdirection and begins moving. Use the force - time graphbelow to find the net impulse delivered by the force after 6 s.A 12 N-sB 14 N-sF(N)Slide 30 / 14064C 16 N-s2D 18 N-s0E 20 N-s246t (s)8 An object at rest experiences a net horizontal force in the xdirection and begins moving. Use the force - time graphbelow to find the net impulse delivered by the force after 6 s.B 14 N-sF(N)64AnswerA 12 N-sSlide 30 (Answer) / 140C 16 N-sA2D 18 N-s0E 20 N-s246t (s)[This object is a pull tab]9 A 2 kg object at rest experiences a net horizontal force in the x direction and begins moving. Use the force - time graphbelow to find the net impulse delivered by the force after 6 s.A 4 N-sB 6 N-sC 8 N-sF(N)642D 10 N-sE 12 N-s0246t (s)Slide 31 / 140
9 A 2 kg object at rest experiences a net horizontal force in the x direction and begins moving. Use the force - time graphbelow to find the net impulse delivered by the force after 6 s.B 6 N-sF(N)6AnswerA 4 N-sSlide 31 (Answer) / 1404C 8 N-sC2D 10 N-s0E 12 N-s246t (s)[This object is a pull tab]10 A 2 kg object at rest experiences a net horizontal force in the x direction and begins moving. Use the force - time graphbelow to find the object's velocity after 6 s.A 10 m/sB 8 m/sF(N)Slide 32 / 14064C 6 m/s2D 5 m/s0E 4 m/s246t (s)10 A 2 kg object at rest experiences a net horizontal force in the x direction and begins moving. Use the force - time graphbelow to find the object's velocity after 6 s.B 8 m/sC 6 m/sF(N)AnswerA 10 m/s64E2D 5 m/sE 4 m/s0246t (s)[This object is a pull tab]Slide 32 (Answer) / 140
11 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball. Assume the bat loses contact with the ballwhen the force decreases to zero. Over what time intervaldoes this force act?Slide 33 (Answer) / 140Answer11 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball. Assume the bat loses contact with the ballwhen the force decreases to zero. Over what time intervaldoes this force act?Slide 33 / 140The force acts from t 0 to t 1.01s[This object is a pull tab]12 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball. The force acts over a time interval of 1.01s.What is the magnitude of the maximum force delivered to theball?Slide 34 / 140
Slide 34 (Answer) / 140Answer12 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball. The force acts over a time interval of 1.01s.What is the magnitude of the maximum force delivered to theball?[This object is a pull tab]13 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball. The force acts over a time interval of 1.01s.What is the magnitude of the impulse delivered to the ball?Answer13 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball. The force acts over a time interval of 1.01s.What is the magnitude of the impulse delivered to the ball?[This object is a pull tab]Slide 35 / 140Slide 35 (Answer) / 140
14 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball. The force acts over a time interval of 1.01s.What is the magnitude of the average force delivered to theball?Slide 36 (Answer) / 140Answer14 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball. The force acts over a time interval of 1.01s.What is the magnitude of the average force delivered to theball?Slide 36 / 140[This object is a pull tab]15 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball and delivers an impulse of 32 N/s. What is thevelocity of the ball at t 1.01 s, assuming it started from rest?Slide 37 / 140
Slide 37 (Answer) / 140Answer15 A force described by F(t) 190t - 189t2 is applied by a bat to a0.145 kg ball and delivers an impulse of 32 N/s. What is thevelocity of the ball at t 1.01 s, assuming it started from rest?[This object is a pull tab]Slide 38 / 140Collisions in OneDimensionReturn to Tableof ContentsSlide 39 / 140Types of CollisionsObjects in an isolated system can interact with each other in twobasic ways:· They can collide.· If they are stuck together, they can explode (push apart).In an isolated system both momentum and total energy areconserved. But the energy can change from one form to another.Conservation of momentum and change in kinetic energy canhelp predict what will happen in these events.
Slide 40 / 140Types of CollisionsWe differentiate collisions and explosions by the way the energychanges or does not change form.· inelastic collisions: two objects collide, converting some kineticenergy into other forms of energy such as potential energy, heat orsound.· elastic collisions: two objects collide and bounce off each otherwhile conserving kinetic energy - energy is not transformed intoany other type.· explosions: an object or objects breaks apart because potentialenergy stored in one or more of the objects is transformed intokinetic energy.Slide 41 / 140Inelastic CollisionsThere are two types of Inelastic Collisions.· perfect inelastic collisions: two objects collide, stick togetherand move as one mass after the collision, transferring kinetic energyinto other forms of energy.· general inelastic collisions: two objects collide and bounce offeach other, transferring kinetic energy into other forms of energy.Slide 42 / 140Elastic CollisionsThere is really no such thing as a perfect elastic collision. Duringall collisions, some kinetic energy is always transformed intoother forms of energy.But some collisions transform so little energy away from kineticenergy that they can be dealt with as perfect elastic collisions.In chemistry, the collisions between molecules and atoms aremodeled as perfect elastic collisions to derive the Ideal Gas Law.Other examples include a steel ball bearing dropping on a steelplate, a rubber "superball" bouncing on the ground, and billiardballs bouncing off each other.
Slide 43 / 140ExplosionsA firecracker is an example of an explosion. The chemicalpotential energy inside the firecracker is transformed into kineticenergy, light and sound.A cart with a compressed spring is a good example. When thespring is against a wall, and it is released, the cart starts moving- converting elastic potential energy into kinetic energy andsound.Think for a moment - can you see a resemblance between thisphenomenon and either an elastic or inelastic collision?Slide 44 / 140ExplosionsIn both an inelastic collision and an explosion, kinetic energy istransformed into other forms of energy - such as potentialenergy. But they are time reversed!An inelastic collision transforms kinetic energy into other forms ofenergy, such as potential energy.An explosion changes potential energy into kinetic energy.Thus, the equations to predict their motion will be inverted.The next slide summarizes the four types of collisions andexplosions.Slide 45 / 140Collisions and c unceoff each otherYesNo. Kinetic energy isconverted to otherforms of energyPerfectInelasticCollisionObjects sticktogetherYesNo. Kinetic energy isconverted to otherforms of energyElasticCollisionObjectsbounceoff each otherYesYesExplosionObject orobjects breakapartYesNo. Release ofpotential energyincreases kineticenergy
16 Momentum is conserved in which of the following types of collisions?Slide 46 / 140A Perfect InelasticB InelasticC ElasticD ExplosionsE All of the Above16 Momentum is conserved in which of the following types of collisions?Slide 46 (Answer) / 140B InelasticC ElasticAnswerA Perfect InelasticED ExplosionsE All of the Above[This object is a pull tab]17 Kinetic energy is conserved in which of the following typesof collisions?A Perfect InelasticB InelasticC ElasticD ExplosionsE All of the AboveSlide 47 / 140
17 Kinetic energy is conserved in which of the following typesof collisions?Slide 47 (Answer) / 140A Perfect InelasticAnswerB InelasticC ElasticCD ExplosionsE All of the Above[This object is a pull tab]Slide 48 / 140Conservation of MomentumDuring a collision or an explosion, measurements show that thetotal momentum of a closed system does not change. The diagrambelow shows the objects approaching, colliding and thenseparating.AmAvAm Bv BB xthe prime means "after"A Bm Bv B 'm Av A'ABIf the measurements don'tshow that the momentum isconserved, then this would notbe a valid law. Fortunately theydo, and it is!Slide 49 / 140
18A 13,500 kg railroad freight car travels on a level track at a speed of4.5 m/s. It collides and couples with a 25,000 kg second car, initiallyat rest and with brakes released. No external force acts on thesystem. What is the speed of the two cars after colliding?18A 13,500 kg railroad freight car travels on a level track at a speed of4.5 m/s. It collides and couples with a 25,000 kg second car, initiallyat rest and with brakes released. No external force acts on thesystem. What is the speed of the two cars after colliding?Slide 50 / 140Slide 50 (Answer) / 140m1v1' m2v2' m2) v' m2)0kg)(4.5m/s) / (13,500 25,000)kgirection as the first car's initial velocity[This object is a pull tab]Slide 51 / 14019A cannon ball with a mass of 100.0 kg flies in horizontal directionwith a speed of 250 m/s and strikes a ship initially at rest. The massof the ship is 15,000 kg. Find the speed of the ship after the ballbecomes embedded in it.
Slide 51 (Answer) / 14019A cannon ball with a mass of 100.0 kg flies in horizontal directionwith a speed of 250 m/s and strikes a ship initially at rest. The massof the ship is 15,000 kg. Find the speed of the ship after the ballbecomes embedded in it.m1v1' m2v2' m2) v' m2))(250m/s) / (100 15,000)kgdirection as cannon ball's initial velocity[This object is a pull tab]20A 40 kg girl skates at 5.5 m/s on ice toward her 70 kg friend who isstanding still, with open arms. As they collide and hold each other,what is their speed after the collision?20A 40 kg girl skates at 5.5 m/s on ice toward her 70 kg friend who isstanding still, with open arms. As they collide and hold each other,what is their speed after the collision?m1v1' m2v2' m2) v' m2)5.5m/s) / (40 70)kgtion as the 40kg girls's initialThis object is a pull tab]Slide 52 / 140Slide 52 (Answer) / 140
Slide 53 / 140ExplosionsIn an explosion, one object breaks apart into two or more pieces(or coupled objects break apart), moving afterwards as separateobjects.To make the problems solvable at this math level, we will assume:· the object (or a coupled pair of objects) breaks into two pieces.· the explosion is along the same line as the initial velocity.Slide 54 / 140Slide 55 / 14021A 5 kg cannon ball is loaded into a 300 kg cannon. When thecannon is fired, it recoils at 5 m/s. What is the cannon ball'svelocity after the explosion?
Slide 55 (Answer) / 14021A 5 kg cannon ball is loaded into a 300 kg cannon. When thecannon is fired, it recoils at 5 m/s. What is the cannon ball'svelocity after the explosion?m1 5kgAnswerm2 300kgv1 v2 v 0v2' 5m/s(m1 m2) v m1v1' m2v2'0 m1v1' m2v2'm1v1' -m2v2'v1' -m2v2'/m1 -(300kg)(5m/s) / (5kg) - 300m/s[This object is a pull tab]22Two railcars, one with a mass of 4000 kg and the other with amass of 6000 kg, are at rest and stuck together. To separatethem a small explosive is set off between them. The 4000 kg caris measured travelling at 6 m/s. How fast is the 6000 kg cargoing?22Two railcars, one with a mass of 4000 kg and the other with amass of 6000 kg, are at rest and stuck together. To separatethem a small explosive is set off between them. The 4000 kg caris measured travelling at 6 m/s. How fast is the 6000 kg cargoing?(m1 m2) v m1v1' m2v2'0 m1v1' m2v2'm1v1' -m2v2'v1' -m2v2'/m1 -(4000kg)(6m/s) / (6000kg) -4m/s[This object is a pull tab]Slide 56 / 140Slide 56 (Answer) / 140
Slide 57 / 140Elastic CollisionsIn an elastic collision, two objects collide and bounce off each other,as shown below, and both momentum and kinetic energy areconserved.This will give us two simultaneous equations to solve to predict theirmotion after the collision.Before (moving towards)pA mAvAApB mBvBBAfter (moving apart)pA' mAvA'ApB' mBvB'BSlide 58 / 140Slide 59 / 140
Slide 60 / 140Elastic Collision Simultaneous EquationsConservation of MomentumConservation of Kinetic Energym 1 v 1 m 2 v 2 m 1 v 1 ' m 2 v 2 '½m 1 v 1 2 ½m 2 v 2 2 ½m 1 v 1 '2 ½m 2 v 2 '2m 1v 1 - m 1v 1' m 2v 2' - m 2v 2m 1 v 1 2 m 2 v 2 2 m 1 v 1 '2 m 2 v 2 '2m 1 (v 1 - v 1 ') m 2 (v 2 ' - v 2 )m 1 v 1 2 - m 1 v 1 '2 m 2 v 2 '2 - m 2 v 2 2m 1 (v 1 2 - 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 ')(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 2v 1 - v 2 -(v 1 ' - v'2 )Slide 61 / 140Properties of Elastic CollisionsBy solving the conservation of momentum and constant kineticenergy equations simultaneously, the following result appeared:v 1 - v 2 -(v 1 ' - v'2 )Do you recognize the terms on the left and right of the aboveequation? And, what does it mean?The terms are the relative velocities of the two objectsbefore and after the collision. It means that for allelastic collisions - regardless of mass - the relativevelocity of the objects is the same before and after thecollision.23Slide 62 / 140Two objects have an elastic collision. Before they collide, they areapproaching with a velocity of 4 m/s relative to each other. Withwhat velocity do they move apart from one another after thecollision?
23Slide 62 (Answer) / 140Two objects have an elastic collision. Before they collide, they areapproaching with a velocity of 4 m/s relative to each other. Withwhat velocity do they move apart from one another after thecollision?Answer4 m/s away from each otherv1 -v2 -(v1 '-v2 ')The difference in the initial velocities is thesame as the negative difference of the finalvelocities. Or, the relative velocity between theobjects before the collision is equal to thenegative of the relative velocity between theobjects after the collision.[This object is a pull tab]Two objects have an elastic collision. Object m1, has an initialvelocity of 4.0 m/s and m2 has a velocity of -3.0 m/s. After thecollision, m1 has a velocity of 1.0 m/s. What is the velocity ofm2?24Two objects have an elastic collision. Object m1, has an initialvelocity of 4.0 m/s and m2 has a velocity of -3.0 m/s. After thecollision, m1 has a velocity of 1.0 m/s. What is the velocity ofm2?Answer24v1 4 m/sv2 -3 m/sv1' 1 m/sv2' ?v1-v2 -(v1'-v2')v2' v1 v1'-v2v2' 4m/s 1m/s - (-3m/s)v2' 8 m/s[This object is a pull tab]Slide 63 / 140Slide 63 (Answer) / 140
25Two objects have an elastic collision. Object m1, has an initialvelocity of 6.0 m/s and m2 has a velocity of 2.0 m/s. After thecollision, m1 has a velocity of 1.0 m/s. What is the velocity ofm2?25Two objects have an elastic collision. Object m1, has an initialvelocity of 6.0 m/s and m2 has a velocity of 2.0 m/s. After thecollision, m1 has a velocity of 1.0 m/s. What is the velocity ofm2?Slide 64 / 140Slide 64 (Answer) / 140Answerv1 6 m/sv2 2 m/sv1 ' 1 m/sv2 ' ?v1-v2 -(v1'-v2')v2 ' v1 v1 '-v2v2 ' 6m/s 1m/s - (2m/s)v2 ' 5 m/s[This object is a pull tab]Slide 65 / 140
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Slide 75 / 14026A bowling ball has a velocity of v when it collides with a pingpong ball that is at rest. The velocity of the bowling ball isvirtually unaffected by the collision. What will be the speed ofthe ping pong ball?26A bowling ball has a velocity of v when it collides with a pingpong ball that is at rest. The velocity of the bowling ball isvirtually unaffected by the collision. What will be the speed ofthe ping pong ball?AnswerSlide 75 (Answer) / 140v1 vv2 0v1' vv2' ?v1-v2 -(v1'-v2')v2' v1 v1'-v2v2 ' v v - 0v2 ' 2 v(ping pong ball's speed is twice that of thebowling ball)[This object is a pull tab]27Slide 76 / 140A baseball bat has a velocity of v when it collides with a baseballthat has a velocity of -2v. The bat barely changes velocity duringthe collision. How fast is the baseball going after it's hit?
Slide 76 (Answer) / 140A baseball bat has a velocity of v when it collides with a baseballthat has a velocity of -2v. The bat barely changes velocity duringthe collision. How fast is the baseball going after it's hit?Answer27v1 vv2 -2vv1' vv2' ?v1-v2 -(v1'-v2')v2 ' v1 v1 '-v2v2 ' v v - (-2v)v2 ' 4 v[This object is a pull tab]28Two objects with identical masses have an elastic collision: theinitial velocity of m1 is 6.0 m/s and m2 is -3.0 m/s. What is thevelocity of m1 after the collision?28Two objects with identical masses have an elastic collision: theinitial velocity of m1 is 6.0 m/s and m2 is -3.0 m/s. What is thevelocity of m1 after the collision?Answerv1 6m/sv2 -3m/sv1 ' ?v2 ' ?When identical mass objects experience anelastic collision, they swap their initial velocities:v1' v2 -3.0 m/sv2' v1 6.0 m/sSo the velocity of m1 is -3.0 m/s.[This object is a pull tab]Slide 77 / 140Slide 77 (Answer) / 140
29Two objects with identical masses have an elastic collision: theinitial velocity of m1 is 6.0 m/s and m2 is -3.0 m/s. What is thevelocity of m2 after the collision?29Two objects with identical masses have an elastic collision: theinitial velocity of m1 is 6.0 m/s and m2 is -3.0 m/s. What is thevelocity of m2 after the collision?Slide 78 / 140Slide 78 (Answer) / 140Answerv1 6m/sv2 -3m/sv1 ' ?v2 ' ?When identical mass objects experience anelastic collision, they swap their initial velocities:v1' v2 -3.0 m/sv2' v1 6.0 m/sSo the velocity of m1 is 6.0 m/s.[This object is a pull tab]30 A golf ball is hit against a solid cement wall, and experiences anelastic collsion. The golf ball strikes the wall with a velocity of 35m/s. What velocity does it rebound with?Slide 79 / 140
Answer30 A golf ball is hit against a solid cement wall, and experiences anelastic collsion. The golf ball strikes the wall with a velocity of 35m/s. What velocity does it rebound with?Slide 79 (Answer) / 140When a light object strikes a verymassive object, the light objectrebounds with the opposite velocity in this case, the golf ball will leavethe wall with a velocity of -35 m/s.[This object is a pull tab]Slide 80 / 140Collisions in TwoDimensionsReturn to Tableof ContentsConservation of Momentum in TwoDimensionsMomentum vectors (like all vectors) can be expressed in terms ofcomponent vectors relative to a reference frameThis, of course also applies tothree dimensions, but we'll stickwith two for this chapter!This means that the momentumconservation equation p p' can be solvedindependently for each component:Slide 81 / 140
Slide 82 / 140Example: Collision with a WallmConsider
Define Impulse as: The Impulse-Momentum equation is then: Impulse is a vector, and it is in the same direction as the change of momentum or velocity of the particle acted on by the force. Slide 21 / 140 Impulse-Momentum Theorem The force is not always constant - for example when a tennis racquet strikes a tennis ball, the force starts out small .
CHAPTER 3 MOMENTUM AND IMPULSE prepared by Yew Sze Ling@Fiona, KML 2 3.1 Momentum and Impulse Momentum The linear momentum (or "momentum" for short) of an object is defined as the product of its mass and its velocity. p mv & & SI unit of momentum: kgms-1 or Ns Momentum is vector quantity that has the same direction as the velocity.
Momentum ANSWER KEY AP Review 1/29/2018 Momentum-1 Bertrand Momentum How hard it is to stop a moving object. Related to both mass and velocity. For one particle p mv For a system of multiple particles P p i m i v Units: N s or kg m/s Momentum is a vector! Problem: Momentum (1998) 43. The magnitude of the momentum of the
momentum is kg·m/s. Linear Momentum Linear momentum is defined as the product of a system's mass multiplied by its velocity: p mv. (8.2) Example 8.1Calculating Momentum: A Football Player and a Football (a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player's momentum with the
1. Impulse and Momentum: You should understand impulse and linear momentum so you can: a. Relate mass, velocity, and linear momentum for a moving body, and calculate the total linear momentum of a system of bodies. Just use the good old momentum equation. b. Relate impulse to the change in linear momentum and the average force acting on a body.
Momentum: Unit 1 Notes Level 1: Introduction to Momentum The Definition Momentum is a word we sometime use in everyday language. When we say someone has a lot of momentum, it means they are on a roll, difficult to stop, really moving forward. In physics, momentum means "mass in motion". The more mass an object has, the more momentum it has.
Momentum (p mv)is a vector, so it always depends on direction. Sometimes momentum is if velocity is in the direction and sometimes momentum is if the velocity is in the direction. Two balls with the same mass and speed have the same kinetic energy but opposite momentum. Momentum vs. Kinetic Energy A B Kinetic Energy Momentum
momentum), it is optimal for investors to hold less or may short the momentum asset and hence sufier less or even beneflt from momentum crashes. Key words: Portfolio selection, momentum crashes, dynamic optimal momentum strategy. JEL Classiflcation: C32, G11 Date: March 11, 2016. 1
Section Review 9.1 Impulse and Momentum pages 229–235 page 235 6. Momentum Is the momentum of a car traveling south different from that of the same car when it travels north at the same speed? Draw the momentum vectors to sup-port your answer. Yes, momentum is a vector quantity, and the momenta of the two cars are in opposite directions. 7.8 .File Size: 806KBPage Count: 31
