Momentum - HHS Physics
Momentum Impulse.notebookFebruary 06, 2014MomentumIn classical mechanics, linear momentum or translational momentum (pluralmomenta; SI unit kg m/s, or, equivalently, N s) is the product of the massand velocity of an object. For example, a heavy truck moving fast has alarge momentum—it takes a large and prolonged force to get the truck up tothis speed, and it takes a large and prolonged force to bring it to a stopafterwards. If the truck were lighter, or moving slower, then it would haveless momentum.Newton, first, expressed his Second Law in terms of momentum and not netforce. He did not write the equation Fnet mA, instead, he wrote thatmotion was proportional to the force impressed. What was meant bymotion was the product of multiplying mass and velocity, and impressedforce was an unbalanced net force. This quantity is known as linearmomentum. Newton stated that a change in linear momentum wasproportional to the impressed force. However, a small net force acting overa long period of time could produce the same change in linear momentumas a large force acting over a small time. This means that the relationshipis time dependent. Which leads to the equation Fnet p / t.1
Momentum Impulse.notebookFebruary 06, 2014Momentum refers to the quantity of motion that an object has. If an object isin motion ("on the move") then it has momentum. Momentum, therefore, canbe thought of as "mass in motion." All objects have mass; so if an object ismoving, then it has momentum it has its mass in motion. The amount ofmomentum which an object has is dependent upon two variables: how muchstuff is moving and how fast the stuff is moving. Momentum depends uponthe variables mass and velocity. In terms of an equation, the momentum ofan object is equal to the mass of the object times the velocity of the object.Momentum mass x velocityIn physics, the symbol for the quantity momentum is the lower case "p";thus, the above equation can be rewritten as:p mvwhere m mass and v velocity. The equation illustrates that momentum isdirectly proportional to an object's mass and directly proportional to theobject's velocity. The units for momentum would be mass units timesvelocity units. The standard metric unit of momentum is the kg m/s.Why is p used to represent momentum?Before the word "momentum" was decided upon, they used "impetus".Impetus comes from the latin, "petere," to go towards or rush upon. Petere iswhere the p comes from.2
Momentum Impulse.notebookFebruary 06, 2014The Definition of Momentum is:Momentum: (p) A property of a moving body that the body has by virtue ofits mass and motion and that is equal to the product of the body's mass andvelocity; broadly : a property of a moving body that determines the length oftime required to bring it to rest when under the action of a constant forceor moment. A vector quantity that is conserved in collisions between particlesand in closed systems; in classical mechanics it is equal to the mass times thevelocity of a body, or the vector sum of this product over all the components of asystem.Momentum is a vector quantity. To fully describe the momentum of a 5 kgbowling ball moving westward (180o) at 2 m/s, you must include informationabout both the magnitude and the direction of the bowling ball. It is not enoughto say that the ball has 10 kg m/s of momentum; the momentum of the ball is notfully described until information about its direction is given. The direction of themomentum vector is the same as the direction of the velocity of the ball. If thebowling ball is moving westward at 180o, then its momentum can be fullydescribed by saying that it is 10 kg m/s at 180o. As a vector quantity, themomentum of an object is fully described by magnitude, direction, and unit.3
Momentum Impulse.notebookFebruary 06, 2014From the definition of momentum, it becomes obvious that an object has alarge momentum if either its mass or its velocity is large. Both variables areof equal importance in determining the momentum of an object. Consider aMack truck and a roller skate moving down the street at the same speed.The considerably greater mass of the Mack truck gives it a considerablygreater momentum. Yet, if the Mack truck were at rest, then the momentumof the least massive roller skate would be the greatest; for the momentum ofany object which is at rest is 0. Objects at rest do not have momentum they do not have any "mass in motion." Both variables mass and velocity are important in comparing the momentum of two objects.1. Determine the momentum of the following:A) 60 kg halfback moving eastward at 9 m/s.B) 1000 kg car moving northward at 20 m/s.C) 40 kg student moving southward at 2 m/s.A) 540 kgm/s eastwardB) 20000 kgm/s northwardC) 80 kgm/s southward4
Momentum Impulse.notebookFebruary 06, 2014Momentum can only be changed by changing the object's velocity or itsmass. Gravity has nothing to do with itMomentum can't be lost; it can only be transferred. If you catch afootball, then the football's momentum goes through you and into the earth(or else you fall down). This will actually change the rotational speed of theearth by a very small amount, but the change is canceled out by the oppositeforce from when the ball was thrown. This all leads up to conservation ofmomentum. Remember it has to go somewhere! This means that if youhave several objects in a system, perhaps interacting with each other, but notbeing influenced by forces from outside of the system, then the totalmomentum of the system does not change over time. However, the separatemomenta of each object within the system may change. One object mightchange momentum, say losing some momentum, as another object changesmomentum in an opposite manner, picking up the momentum that was lostby the first.Law of conservation of linear momentum: The total momentum of thesystem prior to a collision is equal to the total momentum after thecollision. The principle that a system under no external forces willmaintain constant linear momentum.5
Momentum Impulse.notebookFebruary 06, 2014Momentum is how Isaac Newton wrote his laws. Consider a change inmomentum with time:dpdtdpdtsincedpdtsince momentum is equal to mV d(mV)dtdVdt mdVdtis equal to acceleration a. ma FnetThus, the derivative of momentum with respect to time isNET FORCE.6
Momentum Impulse.notebookFebruary 06, 2014ImpulseImpulse ( j ) is defined as the product of the average value of a force andthe time during which it acts: the change in momentum produced by theforce. A vector quantity given by the integral over time of the force acting ona body, usually in a collision in which the time interval is very brief; it is equalto the change in the momentum of the body. Impulse is calculated bymultiplying the force and the time interval over which the force acts.Impulse ( j ) F t F(t)dtImpulse is a vector quantity whose direction is the direction of the force. ItsSI unit is kg m/s or Ns.impulse equals the change in momentum.Impulse change in momentumF t j p m vThe phrase "impulse equals change in momentum" is a handy phrase worthmemorizing.7
Momentum Impulse.notebookFebruary 06, 2014Suppose you apply a constant net force, Fnet, to an object of mass m.Newton's Second Law tells you that the object will accelerate, so if it starts withvelocity vI, after some time t its velocity will be vF. This situation is diagrammedbelow.Start at t 0at time tVIVFFnetFnetThe acceleration of the object equals its change in velocity divided by the timeit takes the velocity to change. In symbols:A Δv/Δt Δv/tMultiplying both sides of this equation by t gives:At Δv vF vIThe right side of the equation above comes from the fact that the change invelocity equals the final velocity, vF, minus the starting velocity, vI. (Note: Wecould have just as well started with the kinematics equation vF vI at.) Thisis a valid kinematical statement about the motion.8
Momentum Impulse.notebookFebruary 06, 2014At Δv vF vITo turn it into a dynamical statement about the motion, multiply both sides ofthe equation by the object's mass, m:mAt mΔv m(vF vI)Since Newton's Second Law tells us that the net force on an object equalsthe product of the object's mass and acceleration, we can replace ma withFnet in this equation. On the right side, the quantity mass times velocity iscalled momentum, p.Fnet t mΔv mvF mvIFnet t pF pI ΔpThe quantity on the left, Fnet t, is the impulse exerted on the object by the netforce. The quantity on the right of the equation is the object's final momentumminus its starting momentum, which is its change in momentum. Thus:j ΔpThis is the Impulse Momentum Equation9
Momentum Impulse.notebookFebruary 06, 2014CollisionsCollisions between objects are governed by laws of momentum andenergy. When a collision occurs in an isolated system, the total momentumof the system of objects is conserved. Provided that there are no net externalforces acting upon the objects, the momentum of all objects before thecollision equals the momentum of all objects after the collision. If thereare only two objects involved in the collision, then the momentum change ofthe individual objects are equal in magnitude and opposite in direction.Most text books describe three different types of collisions (perfectly elastic,inelastic, and perfectly inelastic). In reality, it is more true to say there aretwo ends of a spectrum (range) of collision types. Regardless of what type ofcollision occurs, the total momentum of the system will always beconserved if there are no external forces present.10
Momentum Impulse.notebookFebruary 06, 2014Types of PropertiesKE is conservedNo damage to either objectLess damage and heatingLess soundExample Idealized collisions Super ball Golf ballInelastic Ball bearings Basket ball Tennis BallLessElasticPerfectlyInelasticMore damage and heatingMore sound The colliding objects sticktogether and become one. Involves the greatest loss ofKinetic Energy. (keep in mindthat if it loses KE then theenergy will need to appear asthermal energy, and/or sound) The colliding objects usuallysuffer some kind ofpermanent damage Under inflated ball Soft clay Automobileaccidents Train couplingRemind yourself that for any of these collisions, the total momentum of the closedsystem remains the same and the total energy remains the same (it just maychange form from KE to thermal energy, sound, etc.). Both the Momentum andthe Energy are always conserved when dealing with a closed system.11
Momentum Impulse.notebookFebruary 06, 2014Perfectly Inelastic CollisionA perfectly inelastic collision occurs when the maximum amount of kineticenergy of a system is lost. In a perfectly inelastic collision, i.e., a zerocoefficient of restitution, the colliding particles stick together. In such acollision, kinetic energy is lost by bonding the two bodies together. Thisbonding energy usually results in a maximum kinetic energy loss of thesystem.In a perfectly inelastic collision there are certain points to be kept in mind:1. The coefficient of restitution is zero.2. The colliding objects stick together after the collision.3. The momentum is conserved.4. Kinetic energy is not conserved.12
Momentum Impulse.notebookFebruary 06, 2014Perfectly Inelastic CollisionNext, we will derive an equation for a two body(Body 1,Body 2) system in aone dimensional collision. In this example, momentum of the system isconserved because there is no friction between the sliding bodies and thesurface. But the collision is perfectly inelastic because the two objects sticktogether.Before CollisionAfter Collisionv2 0V1 2V1M2m1m1frictionless surfaceMomentum Before Collisionp1 p2m1V1 m2V2M2frictionless surface Momentum after Collisionp'1 2m1 2V'1 2Note the prime ' denotes post collision conditionsOther notations for post collision velocity is to use the letter u13
Momentum Impulse.notebookFebruary 06, 2014A 15 g bullet is fired into an 10 kg block of wood and lodges in it. The block,which is free to move on a frictionless surface, has a velocity of 0.65 m/s at 0oafter impact. Find the initial velocity of the bullet.at 0o14
Momentum Impulse.notebookFebruary 06, 2014A 10,000 Kg railroad car moving at 8.5 m/s at 0o on a horizontal track, collideswith a stationary railroad car. After the collision the two cars, which areattached to each other, move off with a velocity of 3.4 m/s at 0o. What is themass of the second car?M1 10,000 KgV1 8.5 m/s at 0oM2V2 0M1 M2V'1 2 3.4 m/s at 0o15
Momentum Impulse.notebookFebruary 06, 2014Two blocks are moving towards each other on a frictionless surface. Block 1has a mass of M and a velocity of 4 m/s right; and block 2 has a mass of 2Mand a velocity of 1 m/s left. The 2 blocks collide in a perfectly inelasticcollision, what is the velocity of the 2 stuck together blocks?Before CollisionM1V1 4 m/sAfter Collision3M2M2V2 1 m/s12V'1 2 ?16
Momentum Impulse.notebookFebruary 06, 2014Two Dimensional Perfectly Inelastic Collision ProblemsTwo 1000 kg cars approach an intersection at a 90o angle and collideinelastically, sticking together after the collision. What is the velocity (speedand direction) of the two car clump of twisted metal immediately after thecollision?Velocity of car 1 20m/s east. Themomentum of car 1before the collision is20,000 kg m/s EastVelocity of car 2 10m/s north.Themomentum of car 2before the collision is10,000 kg m/s NorthIn the collision between the two cars, total system momentum is conserved.Yet this might not be apparent without an understanding of the vector natureof momentum.17
Momentum Impulse.notebookFebruary 06, 2014Momentum, like all vector quantities, has a magnitude (size), a direction, andunit. When considering the total momentum of the system before the collision,the individual momentum of the two cars must be added as vectors. That is,20,000 kg m/s East must be added to 10,000 kg m/s North. The sum of thesetwo vectors is not 30,000 kg m/s; this would only be the case if the twomomentum vectors had the same direction. Instead, the sum of 20,000 kg m/sEast and 10,000 kg m/s North is 22,360.7 kg m/s at an angle of 26.6o North ofEast.Since the two momentum vectors are at right angles, their sum can be foundusing vector addition. Draw the 2 vectors so they have a common tail point.p2 10,000 kg m/sp1 20,000 kg m/s18
Momentum Impulse.notebookFebruary 06, 2014We know if using the parallelogram method of addition of vectors theresultant would be the diagonal of the parallelogram formed be the 2vectors as seen below.p2 10,000 kg m/sRp1 20,000 kg m/sUsing numerical vector addition find the X and Y components for p1 and p2.p2 10,000 kg m/sX 0Y 10,000 kg m/sp1 20,000 kg m/sX 20,000 kg m/sY 019
Momentum Impulse.notebookFebruary 06, 2014p2 10,000 kg m/sX 0Y 10,000 kg m/sp1 20,000 kg m/sX 20,000 kg m/sY 0Add all the X components to find RxRx 20,000 kg m/sAdd all the Y components to find RyRy 10,000 kg m/sFind the magnitude of R by using the Pythagorean theorem.R [(Rx)2 (Ry)2] 22,360.7 kg m/sFind θ using the arc tan functionθ / Tan 1(Ry/Rx)/ 26.6oFind φ using the proper φ formula. φ 26.6o20
Momentum Impulse.notebookFebruary 06, 2014The value 22,360.7 kg m/s at 26.6o is the total momentum of the systembefore the collision; and since momentum is conserved, it is also the totalmomentum of the system after the collision. Since the cars have equal mass(and ONLY because they have equal mass), the total system momentum isshared equally by each individual car. In order to determine the momentum ofeither individual car, this total system momentum must be divided by two(approx. 11,200 kg m/s). Once the momentum of the individual cars areknown, the after collision velocity is determined by simply dividing momentumby mass (v' p/m)p'1 2 m1 2 V'1 2V'1 2 p'1 2 / m1 2V'1 2 (22,360.7 kg m/s at 26.6o) / 2000kgV'1 2 11.2 m/s at 26.6o21
Momentum Impulse.notebookFebruary 06, 2014Two cars approach an intersection at a 90o angle and collide inelastically,sticking together after the collision. What is the velocity (speed and direction)of the two car clump of twisted metal immediately after the collision?M1 1200 kgV1 15 m/s eastM2 1400 kgV2 25 m/s north22
Momentum Impulse.notebookFebruary 06, 2014Two cars approach an intersection at a 90o angle and collide inelastically,sticking together after the collision. What is the velocity of car 1, if thewreckage has a velocity of 18.72 m/s at 29.1o immediately after the collision?V'1 2 18.72 m/s at 29.1oM1 1200 kgV1 ? eastM2 1000 kgV2 20 m/s north23
Momentum Impulse.notebookFebruary 06, 2014A 7,000 Kg truck, traveling with a velocity 5 m/s at 0o, collides with a 1,200 Kgcar moving 18 m/s at 220o as in the figure below. After the collision, the twovehicles remain tangled together. With what velocity will the wreckage beginto move immediately after the crash?220oTruckCar24
Momentum Impulse.notebookFebruary 06, 2014Momentum Conservation in ExplosionsAs discussed in a previously, total system momentum is conserved forcollisions between objects in an isolated system. For collisions occurring inisolated systems, there are no exceptions to this law. This same principle ofmomentum conservation can be applied to explosions. In an explosion, aninternal impulse acts in order to propel the parts of a system (often a singleobject) into a variety of directions. After the explosion, the individual parts ofthe system (that is often a collection of fragments from the original object)have momentum. If the vector sum of all individual parts of the system couldbe added together to determine the total momentum after the explosion, thenit should be the same as the total momentum before the explosion. Just like incollisions, total system momentum is conserved.25
Momentum Impulse.notebookFebruary 06, 2014A 32 kg bomb which is initially at rest explodes into 2 pieces as shown below.Piece #2 has a mass of 12 kg and piece #1 has a speed of 57.8 m/s at 0o.What is the speed of piece #2?26
Momentum Impulse.notebookFebruary 06, 2014Two girls (masses m1 and m2) are on skates and at rest. The girls are close toeach other, and are facing each other. Girl 1 pushes against girl 2 causing girl2 to move backwards.A) Assuming the girls move freely on skates (no friction), find a formula tocalculate the speed of girl 1.B) Calculate the speed of girl 1, If the mass of girl 1 is 54 kg, the mass ofgirl 2 is 42 kg, and girl 2’s speed is 2.4 m/s.C) Calculate the impulse felt by girl 1 and girl 2. (Use the values in B)D) Calculate the force felt by both girl 1 and girl 2, if the duration of thepush was 0.8 s.27
Momentum Impulse.notebookFebruary 06, 2014A 20.0 kg cannonball is fired from a 2.40 103 kg cannon. If the cannonrecoils with a velocity of 3.5 m/s backwards,what is the velocity of thecannonball?28
Momentum Impulse.notebookFebruary 06, 2014A coal barge with a mass of 1.36 104 kg drifts along a river. When itpasses under a coal hopper, it is loaded with 8.4 103 kg of coal. What isthe speed of the unloaded barge if the barge after loading has a speed of1.3 m/s?V' 1.3 m/sV29
Momentum Impulse.notebookFebruary 06, 2014A child jumps from a moving sled with a speed of 2.2 m/s and in the directionopposite the sled’s motion. The sled continues to move in the forwarddirection, but with a new speed of 5.5 m/s. If the child has a mass of 38 kg andthe sled has a mass 68 kg, what is the initial velocity of the sled?Vc 2.2 m/s at 180oVs c ?V'sled 5.5 m/s at 0o30
Momentum Impulse.notebookFebruary 06, 2014An 80 kg stunt man jumps out of a window and falls 45 m to an airbag on theground.A) How fast is he falling when he reaches the airbag? (Initial velocity iszero)B) He lands on a large air filled target, coming to rest in 1.5 s. Whataverage force does he feel while coming to rest?C) What average force does he feel if he had fallen 45 m and landed onthe ground (impact time 10 ms .001 s))?45 mAir bag31
Momentum Impulse.notebookFebruary 06, 2014An 85 kg stunt man jumps out of a window and falls 35 m to an airbag on theground.A) How fast is he falling when he reaches the airbag? (Initial velocity iszero, ignore air resistance)B) He lands on a large air filled target, coming to rest in 2.5 s. Whataverage force does he feel while coming to rest?C) What average force does he feel if he had fallen 35 m and landed onthe ground (impact time 10 ms 0.001 s))?(SHOW ALL WORK)Round ALL answers to the TENTHSplace.35 mAir bag32
Momentum Impulse.notebookFebruary 06, 2014A railroad car of mass 2.50 x 104 kg is moving with a speed of 4.00 m/s. Itcollides and couples with three other coupled railroad cars, each of the samemass as the single car and moving in the same direction with an initial speed of2.00 m/s.(a) What is the speed of the four cars after the collision?(b) How much energy is lost in the collision?33
Momentum Impulse.notebookFebruary 06, 2014A railroad car of mass 30000 kg is moving with a speed of 4.00 m/s. It collidesand couples with three other coupled railroad cars, each of the same mass asthe single car and moving in the oppsite direction with an initial speed of 2.00m/s.(a) What is the speed of the four cars after the collision?(b) How much energy is lost in the collision?34
Momentum Impulse.notebookFebruary 06, 2014Perfectly Elastic CollisionA perfectly elastic collision is defined as one in which there is no loss ofkinetic energy in the collision. An inelastic collision is one in which part ofthe kinetic energy is changed to some other form of energy in the collision.Any macroscopic collision between objects will convert some of the kineticenergy into internal energy and other forms of energy, so no large scaleimpacts are perfectly elastic. Momentum is conserved in inelastic collisions,but one cannot track the kinetic energy through the collision since some of itis converted to other forms of energy. Collisions in ideal gases approachperfectly elastic collisions, as do scattering interactions of sub atomicparticles which are deflected by the electromagnetic force. Some large scale interactions like the slingshot type gravitational interactions betweensatellites and planets are perfectly elastic.35
Momentum Impulse.notebookFebruary 06, 2014Collisions between hard spheres may be nearly elastic, so it is useful tocalculate the limiting case of an elastic collision. The assumption ofconservation of momentum as well as the conservation of kinetic energymakes possible the calculation of the final velocities in two body collisions.Momentum is Conservedm1p1 p2 p'1 p'2V1 m2V2 m1V'1 m2V'236
Momentum Impulse.notebookFebruary 06, 2014Kinetic Energy is ConservedKE1 KE2 KE'1 KE'21/2m1V12 1/2m2V22 1/2m1V'12 1/2m2V'22Multiplying all terms by 2m1V12 m2V22 m1V'12 m2V'2237
Momentum Impulse.notebookFebruary 06, 2014Consider two particles, m1 and m2, with velocities v1 and v2 respectively.They hit in a perfectly elastic collision at an angle, and both particlestravel off at an angle to their original displacement, as shown below:Y AxisV'1V1 at θ1M1V2 0θ'1θ'2X AxisM2V'2 038
Momentum Impulse.notebookFebruary 06, 2014To solve this problem we again use our conservation laws to come upwith equations that we hope to be able to solve. In terms of kinetic energy,since energy is a scalar quantity, we need not take direction into account,and may simply state:1/2m1V12 1/2m2V22 1/2m1V'12 1/2m2V'22m1V12 m2V22 m1V'12 m2V'22Whereas in the one dimensional problem we could only generate oneequation for the conservation of linear momentum, in two dimensionalproblems we can generate two equations: one for the x component andone for the y component.39
Momentum Impulse.notebookFebruary 06, 2014X DirectionLet's start with the X component. Our initial momentum in the X direction isgiven by: m1 V1x m2 V2x . (Note V1x is the X component of V1, and V2x isthe X component of V2). After the collision, each particle maintains acomponent of their velocity in the X direction, which can be calculated usingtrigonometry. Our final momentum in the X direction is given by: m1 V'1x m2 V'2x (Note V'1x is the X component of V'1, and V'2x is the Xcomponent of V'2). Thus our equation for the conservation of linearmomentum in the X direction is:m1V1x m2V2x m1V'1x m2V'2xm1V1cosθ1 m2V2cosθ2 m1V'1cosθ'1 m2V'2cosθ'240
Momentum Impulse.notebookFebruary 06, 2014Y DirectionLet's now look at the Y component. Our initial momentum in the Y directionis given by: m1 V1y m2 V2y . (Note V1y is the Y component of V1, and V2yis the Y component of V2). After the collision, each particle maintains acomponent of their velocity in the Y direction, which can be calculated usingtrigonometry. Our final momentum in the Y direction is given by: m1 V'1y m2 V'2y (Note V'1y is the Y component of V'1, and V'2y is the Y componentof V'2). Thus our equation for the conservation of linear momentum in the Ydirection is:m1V1y m2V2y m1V'1y m2V'2ym1V1sinθ1 m2V2sinθ2 m1V'1sinθ'1 m2V'2sinθ'241
Momentum Impulse.notebookFebruary 06, 2014We now have three equations: conservation of kinetic energy, andconservation of momentum in both the X and Y directions. With thisinformation, is this problem solvable? Recall that if we are given only the initialmasses and velocities we are working with four unknowns: V'1, V'2, θ'1 and θ'2.We cannot solve for four unknowns with three equations, and must specify anadditional variable. Perhaps we are trying to make a pool shot, and can tell theangle of the ball being hit by where the hole is, but would like to know wherethe cue ball will end up. This equation would be solvable, since with the anglethe ball will take to hit the pocket we have specified another variable.m1V12 m2V22 m1V'12 m2V'22m1V1x m2V2x m1V'1x m2V'2xm1V1y m2V2y m1V'1y m2V'2y V1 cos θ1V1y V1 sin θ1V'1x V'1 cos θ'1V'1y V'1 sin θ'1V2x V2 cos θ2V2y V2 sin θ2V'2x V'2 cos θ'2V'2y V'2 sin θ'2V1x42
Momentum Impulse.notebookFebruary 06, 2014Consider two billiard balls, M1 and M2, with velocities V1 and V2respectively. They hit in a perfectly elastic off center collision, and bothballs travel off at an angle to their original displacement. Determine V'1,V'2, and θ'1.Y AxisV'1V1 2 m/s at 0oV2 0θ'1X Axisθ'2 40oM1 1.5 kgM2 .5 kgV'243
Momentum Impulse.notebookFebruary 06, 2014The solution to this problem is extremely difficult involving trigonometry andextensive algebra. The resulting equations for its solution are below courtesyof wikipedia.In a center of momentum frame at any time the velocities of the two bodiesare in opposite directions, with magnitudes inversely proportional to themasses. In an elastic collision these magnitudes do not change. Thedirections may change depending on the shapes of the bodies and the pointof impact. For example, in the case of spheres the angle depends on thedistance between the (parallel) paths of the centers of the two bodies. Anynon zero change of direction is possible: if this distance is zero the velocitiesare reversed in the collision; if it is close to the sum of the radii of the spheresthe two bodies are only slightly deflected.Assuming that the second particle is at rest before the collision, the angles ofdeflection of the two particles, θ1 and θ2, are related to the angle of deflectionθ in the system of the center of mass by:The velocities of the particles after the collision are:44
Momentum Impulse.notebookFebruary 06, 2014Consider two billiard balls, denoted by subscripts 1 and 2. Let m1 and m2 bethe masses, V1 and V2 the velocities before collision, and V'1 and V'2 thevelocities after collision. If the collision is perfectly elastic, determine the finalvelocities of the two balls. 1212m1 2 kgm2 4 kgm1 2 kgm2 4 kgV1 2 m/sV2 0V'1 ?V'2 ?45
Momentum Impulse.notebookFebruary 06, 2014Consider two billiard balls, M1 and M2, with velocities V1 and V2respectively. They hit in an elastic off center collision, and both ballstravel off at an angle to their original displacement. Determine V'1, V'2,and θ'1.Y AxisV'1V1 2 m/s at 0oV2 0θ'1 15.6oθ'2 40X AxisoM1 1.5 kgM2 .5 kgV'2 1.96 m/s46
Momentum Impulse.notebookFebruary 06, 2014Elastic CollisionTwo billiard balls collide as shown below: What is the velocity of ball #2 afterthe collision?m1 0.17 kgV'1 ?m1 0.17 kgV1 2.5 m/s at 0o30om3 0.16 kgV3 1.0 m/s at 180om3 0.16 kgV'3 0.62 m/s at 210o47
Momentum Impulse.notebookFebruary 06, 2014Ballistic PendulumIn a ballistic pendulum an object of mass m is fired with an initial speed v0 at apendulum bob. The bob has a mass M, which is suspended by a rod of lengthL and negligible mass. After the collision, the pendulum and object sticktogether and swing to a maximum angular displacement θ as shown.The first part of the problemdeals with the conservation ofmomentum.The second part of the problemdeals with the conservation ofmechanical energy.48
Momentum Impulse.notebookFebruary 06, 2014A 0.05 kg bullet with velocity 150 m/s is shot into a 3 kg ballistic pendulum.Find how high the pendulum rises after the bullet gets stuck insideV1 150 m/sm1 0.05 kgV2 0M2 3 kgV'1 2 ?m1 M2 3.05 kgStart the problem by finding V'1 2 using the conservation of momentum.p1 p2 p'1 2Since the pendulum is initially at rest p2 equals zero. Also, V1 2 acts in thepositive X direction so the problem reduces to a one dimensional perfectlyinelastic collision problem.49
Momentum Impulse.notebookSolving for V'1 2February 06, 20140p1 p2 p'1 2p1 p'1 2m1V1 m1 2V'1 2(0.05 kg)(150 m/s) (3.05 kg)V'1 27.5 kgm/s (3.05 kg)V'1 22.46 m/s V'1 2Next we apply our knowledge of energy. The pendulum is friction
Momentum Impulse.notebook 1 February 06, 2014 Momentum In classical mechanics, linear momentum or translational momentum (plural momenta; SI unit kg m/s, or, equivalently, N s) is the product of the mass and velocity of an object. For example, a heavy truck moving fast has a
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: 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.
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The Health and Human Services (HHS) Enterprise Architecture (EA) Governance Plan describes the major activities of the HHS EA Prog ram and, especially, the interaction of HHS EA Program activities with related functions, processes, and initiatives within and outside HHS. The Governance Plan defines . 3.2.5 HHS Data Architecture Work Group .
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 (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
