February 06, Momentum Impulse.notebook. Momentum

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1 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 large momentum it takes a large and prolonged force to get the truck up to this speed, and it takes a large and prolonged force to bring it to a stop afterwards. If the truck were lighter, or moving slower, then it would have less momentum. Newton, first, expressed his Second Law in terms of momentum and not net force. He did not write the equation Fnet = ma, instead, he wrote that motion was proportional to the force impressed. What was meant by motion was the product of multiplying mass and velocity, and impressed force was an unbalanced net force. This quantity is known as linear momentum. Newton stated that a change in linear momentum was proportional to the impressed force. However, a small net force acting over a long period of time could produce the same change in linear momentum as a large force acting over a small time. This means that the relationship is time dependent. Which leads to the equation Fnet = p / t. 1

2 Momentum refers to the quantity of motion that an object has. If an object is in motion ("on the move") then it has momentum. Momentum, therefore, can be thought of as "mass in motion." All objects have mass; so if an object is moving, then it has momentum it has its mass in motion. The amount of momentum which an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object. Momentum = mass x velocity In physics, the symbol for the quantity momentum is the lower case "p"; thus, the above equation can be rewritten as: p = m v where m = mass and v = velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity. The units for momentum would be mass units times velocity 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 is where the p comes from. 2

3 The Definition of Momentum is: Momentum: (p) A property of a moving body that the body has by virtue of its mass and motion and that is equal to the product of the body's mass and velocity; broadly : a property of a moving body that determines the length of time required to bring it to rest when under the action of a constant force or moment. A vector quantity that is conserved in collisions between particles and in closed systems; in classical mechanics it is equal to the mass times the velocity of a body, or the vector sum of this product over all the components of a system. Momentum is a vector quantity. To fully describe the momentum of a 5 kg bowling ball moving westward (180 o ) at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. It is not enough to say that the ball has 10 kg m/s of momentum; the momentum of the ball is not fully described until information about its direction is given. The direction of the momentum vector is the same as the direction of the velocity of the ball. If the bowling ball is moving westward at 180 o, then its momentum can be fully described by saying that it is 10 kg m/s at 180 o. As a vector quantity, the momentum of an object is fully described by magnitude, direction, and unit. 3

4 From the definition of momentum, it becomes obvious that an object has a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum. Yet, if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest; for the momentum of any 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 eastward B) kgm/s northward C) 80 kgm/s southward 4

5 Momentum can only be changed by changing the object's velocity or its mass. Gravity has nothing to do with it Momentum can't be lost; it can only be transferred. If you catch a football, 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 the earth by a very small amount, but the change is canceled out by the opposite force from when the ball was thrown. This all leads up to conservation of momentum. Remember it has to go somewhere! This means that if you have several objects in a system, perhaps interacting with each other, but not being influenced by forces from outside of the system, then the total momentum of the system does not change over time. However, the separate momenta of each object within the system may change. One object might change momentum, say losing some momentum, as another object changes momentum in an opposite manner, picking up the momentum that was lost by the first. Law of conservation of linear momentum: The total momentum of the system prior to a collision is equal to the total momentum after the collision. The principle that a system under no external forces will maintain constant linear momentum. 5

6 Momentum is how Isaac Newton wrote his laws. Consider a change in momentum with time: dp dt since momentum is equal to mv dp dt = d(mv) dt = m dv dt dv since is equal to acceleration a. dt dp dt = ma = F net Thus, the derivative of momentum with respect to time is NET FORCE. 6

7 Impulse Impulse ( j ) is defined as the product of the average value of a force and the time during which it acts: the change in momentum produced by the force. A vector quantity given by the integral over time of the force acting on a body, usually in a collision in which the time interval is very brief; it is equal to the change in the momentum of the body. Impulse is calculated by multiplying the force and the time interval over which the force acts. Impulse ( j ) = F t = F (t) dt Impulse is a vector quantity whose direction is the direction of the force. Its SI unit is kg m/s or Ns. impulse equals the change in momentum. Impulse = change in momentum F t = j = p = m v The phrase "impulse equals change in momentum" is a handy phrase worth memorizing. 7

8 Suppose 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 with velocity v I, after some time t its velocity will be v F. This situation is diagrammed below. Start at t=0 V I at time t V F Fnet Fnet The acceleration of the object equals its change in velocity divided by the time it takes the velocity to change. In symbols: A = Δv/Δt = Δv/t Multiplying both sides of this equation by t gives: At = Δv = v F v I The right side of the equation above comes from the fact that the change in velocity equals the final velocity, v F, minus the starting velocity, v I. (Note: We could have just as well started with the kinematics equation v F = v I + at.) This is a valid kinematical statement about the motion. 8

9 At = Δv = v F v I To turn it into a dynamical statement about the motion, multiply both sides of the equation by the object's mass, m: mat = mδv = m(v F v I ) Since Newton's Second Law tells us that the net force on an object equals the product of the object's mass and acceleration, we can replace ma with Fnet in this equation. On the right side, the quantity mass times velocity is called momentum, p. Fnet t = mδv = mv F mv I Fnet t = p F p I = Δp The quantity on the left, Fnet t, is the impulse exerted on the object by the net force. The quantity on the right of the equation is the object's final momentum minus its starting momentum, which is its change in momentum. Thus: j = Δp This is the Impulse Momentum Equation 9

10 Collisions Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum change of the 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 are two ends of a spectrum (range) of collision types. Regardless of what type of collision occurs, the total momentum of the system will always be conserved if there are no external forces present. 10

11 Types of Collisions Collision Type Properties Example Perfectly Elastic KE is conserved No damage to either object Idealized collisions More Elastic Less damage and heating Less sound Super ball Golf ball Inelastic Less Elastic More damage and heating More sound Ball bearings Basket ball Tennis Ball Under inflated ball Perfectly Inelastic The colliding objects stick together and become one. Involves the greatest loss of Kinetic Energy. (keep in mind that if it loses KE then the energy will need to appear as thermal energy, and/or sound) The colliding objects usually suffer some kind of permanent damage Soft clay Automobile accidents Train coupling Remind yourself that for any of these collisions, the total momentum of the closed system remains the same and the total energy remains the same (it just may change form from KE to thermal energy, sound, etc.). Both the Momentum and the Energy are always conserved when dealing with a closed system. 11

12 Perfectly Inelastic Collision A perfectly inelastic collision occurs when the maximum amount of kinetic energy of a system is lost. In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy usually results in a maximum kinetic energy loss of the system. 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

13 Perfectly Inelastic Collision Next, we will derive an equation for a two body(body 1,Body 2) system in a one dimensional collision. In this example, momentum of the system is conserved because there is no friction between the sliding bodies and the surface. But the collision is perfectly inelastic because the two objects stick together. Before Collision After Collision v 2 = 0 V 1 V 1+2 m 1 M 2 frictionless surface frictionless surface m 1 M 2 Momentum Before Collision p 1 + p 2 m 1 V 1 + m 2 V 2 = = = Momentum after Collision p' 1+2 m 1+2 V' 1+2 Note the prime ' denotes post collision conditions Other notations for post collision velocity is to use the letter u 13

14 A 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 0 o after impact. Find the initial velocity of the bullet. at 0 o 14

15 A 10,000 Kg railroad car moving at 8.5 m/s at 0 o on a horizontal track, collides with a stationary railroad car. After the collision the two cars, which are attached to each other, move off with a velocity of 3.4 m/s at 0 o. What is the mass of the second car? M 1 = 10,000 Kg V 1 = 8.5 m/s at 0 o M 2 V 2 = 0 M 1 + M 2 V' 1+2 = 3.4 m/s at 0 o 15

16 Two blocks are moving towards each other on a frictionless surface. Block 1 has a mass of M and a velocity of 4 m/s right; and block 2 has a mass of 2M and a velocity of 1 m/s left. The 2 blocks collide in a perfectly inelastic collision, what is the velocity of the 2 stuck together blocks? Before Collision After Collision M 2M 3M V 1 = 4 m/s V 2 = 1 m/s V' 1+2 =? 16

17 Two Dimensional Perfectly Inelastic Collision Problems Two 1000 kg cars approach an intersection at a 90 o 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? Velocity of car 1 = 20 m/s east. The momentum of car 1 before the collision is 20,000 kg m/s East Velocity of car 2 = 10 m/s north.the momentum of car 2 before the collision is 10,000 kg m/s North In the collision between the two cars, total system momentum is conserved. Yet this might not be apparent without an understanding of the vector nature of momentum. 17

18 Momentum, like all vector quantities, has a magnitude (size), a direction, and unit. 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 these two vectors is not 30,000 kg m/s; this would only be the case if the two momentum vectors had the same direction. Instead, the sum of 20,000 kg m/s East and 10,000 kg m/s North is 22,360.7 kg m/s at an angle of 26.6 o North of East. Since the two momentum vectors are at right angles, their sum can be found using vector addition. Draw the 2 vectors so they have a common tail point. p 2 =10,000 kg m/s p 1 =20,000 kg m/s 18

19 We know if using the parallelogram method of addition of vectors the resultant would be the diagonal of the parallelogram formed be the 2 vectors as seen below. p 2 =10,000 kg m/s R p 1 =20,000 kg m/s Using numerical vector addition find the X and Y components for p 1 and p 2. p 2 =10,000 kg m/s X = 0 Y = 10,000 kg m/s p 1 =20,000 kg m/s X = 20,000 kg m/s Y = 0 19

20 p 2 =10,000 kg m/s X = 0 Y = 10,000 kg m/s p 1 =20,000 kg m/s X = 20,000 kg m/s Y = 0 Add all the X components to find Rx Rx = 20,000 kg m/s Add all the Y components to find Ry Ry = 10,000 kg m/s Find the magnitude of R by using the Pythagorean theorem. R = [(Rx) 2 + (Ry) 2 ] = 22,360.7 kg m/s Find θ using the arc tan function θ = / Tan 1 (Ry/Rx)/ = 26.6 o Find φ using the proper φ formula. φ = 26.6 o 20

21 The value 22,360.7 kg m/s at 26.6 o is the total momentum of the system before the collision; and since momentum is conserved, it is also the total momentum of the system after the collision. Since the cars have equal mass (and ONLY because they have equal mass), the total system momentum is shared equally by each individual car. In order to determine the momentum of either individual car, this total system momentum must be divided by two (approx. 11,200 kg m/s). Once the momentum of the individual cars are known, the after collision velocity is determined by simply dividing momentum by mass (v' = p/m) p' 1+2 = m 1+2 V' 1+2 V'1+2 = p' 1+2 / m 1+2 V'1+2 = (22,360.7 kg m/s at 26.6 o ) / 2000kg V' 1+2 = 11.2 m/s at 26.6 o 21

22 Two cars approach an intersection at a 90 o 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? M 1 = 1200 kg V 1 = 15 m/s east M 2 = 1400 kg V 2 = 25 m/s north 22

23 Two cars approach an intersection at a 90 o angle and collide inelastically, sticking together after the collision. What is the velocity of car 1, if the wreckage has a velocity of m/s at 29.1 o immediately after the collision? M 1 = 1200 kg V 1 =? east V' 1+2 = m/s at 29.1 o M 2 = 1000 kg V 2 = 20 m/s north 23

24 A 7,000 Kg truck, traveling with a velocity 5 m/s at 0 o, collides with a 1,200 Kg car moving 18 m/s at 220 o as in the figure below. After the collision, the two vehicles remain tangled together. With what velocity will the wreckage begin to move immediately after the crash? 220 o Truck Car 24

25 Momentum Conservation in Explosions As discussed in a previously, total system momentum is conserved for collisions between objects in an isolated system. For collisions occurring in isolated systems, there are no exceptions to this law. This same principle of momentum conservation can be applied to explosions. In an explosion, an internal impulse acts in order to propel the parts of a system (often a single object) into a variety of directions. After the explosion, the individual parts of the 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 could be added together to determine the total momentum after the explosion, then it should be the same as the total momentum before the explosion. Just like in collisions, total system momentum is conserved. 25

26 A 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 0 o. What is the speed of piece #2? 26

27 Two girls (masses m 1 and m 2 ) are on skates and at rest. The girls are close to each other, and are facing each other. Girl 1 pushes against girl 2 causing girl 2 to move backwards. A) Assuming the girls move freely on skates (no friction), find a formula to calculate the speed of girl 1. B) Calculate the speed of girl 1, If the mass of girl 1 is 54 kg, the mass of girl 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 the push was 0.8 s. 27

28 A 20.0 kg cannonball is fired from a kg cannon. If the cannon recoils with a velocity of 3.5 m/s backwards,what is the velocity of the cannonball? 28

29 A coal barge with a mass of kg drifts along a river. When it passes under a coal hopper, it is loaded with kg of coal. What is the speed of the unloaded barge if the barge after loading has a speed of 1.3 m/s? V V' = 1.3 m/s 29

30 A child jumps from a moving sled with a speed of 2.2 m/s and in the direction opposite the sled s motion. The sled continues to move in the forward direction, but with a new speed of 5.5 m/s. If the child has a mass of 38 kg and the sled has a mass 68 kg, what is the initial velocity of the sled? V c = 2.2 m/s at 180 o V s+c =? V' sled = 5.5 m/s at 0 o 30

31 An 80 kg stunt man jumps out of a window and falls 45 m to an airbag on the ground. A) How fast is he falling when he reaches the airbag? (Initial velocity is zero) B) He lands on a large air filled target, coming to rest in 1.5 s. What average force does he feel while coming to rest? C) What average force does he feel if he had fallen 45 m and landed on the ground (impact time = 10 ms =.001 s))? 45 m Air bag 31

32 An 85 kg stunt man jumps out of a window and falls 35 m to an airbag on the ground. A) How fast is he falling when he reaches the airbag? (Initial velocity is zero, ignore air resistance) B) He lands on a large air filled target, coming to rest in 2.5 s. What average force does he feel while coming to rest? C) What average force does he feel if he had fallen 35 m and landed on the ground (impact time = 10 ms = s))? 35 m (SHOW ALL WORK) Round ALL answers to the TENTHS place. Air bag 32

33 A railroad car of mass 2.50 x 10 4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.00 m/s. (a) What is the speed of the four cars after the collision? (b) How much energy is lost in the collision? 33

34 A railroad car of mass kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the oppsite direction with an initial speed of 2.00 m/s. (a) What is the speed of the four cars after the collision? (b) How much energy is lost in the collision? 34

35 Perfectly Elastic Collision A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic. Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy. Collisions in ideal gases approach perfectly elastic collisions, as do scattering interactions of sub atomic particles which are deflected by the electromagnetic force. Some largescale interactions like the slingshot type gravitational interactions between satellites and planets are perfectly elastic. 35

36 Collisions between hard spheres may be nearly elastic, so it is useful to calculate the limiting case of an elastic collision. The assumption of conservation of momentum as well as the conservation of kinetic energy makes possible the calculation of the final velocities in two body collisions. Momentum is Conserved p 1 + p 2 = p' 1 + p' 2 m1v 1 + m 2 V 2 = m 1 V' 1 + m 2 V' 2 36

37 Kinetic Energy is Conserved KE 1 + KE 2 = KE' 1 + KE' 2 1/2m 1 V /2m 2 V 2 2 = 1/2m 1 V' /2m 2 V' 2 2 Multiplying all terms by 2 m 1 V m 2 V 2 2 = m 1 V' m 2 V'

38 Consider two particles, m 1 and m 2, with velocities v 1 and v 2 respectively. They hit in a perfectly elastic collision at an angle, and both particles travel off at an angle to their original displacement, as shown below: Y Axis V' 1 V 1 at θ1 V 2 = 0 θ' 1 X Axis θ'2 M 1 M 2 V' 2 = 0 38

39 To solve this problem we again use our conservation laws to come up with 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: 1/2m 1 V /2m 2 V 2 2 = 1/2m 1 V' /2m 2 V' 2 2 m 1 V m 2 V 2 2 = m 1 V' m 2 V' 2 2 Whereas in the one dimensional problem we could only generate one equation for the conservation of linear momentum, in two dimensional problems we can generate two equations: one for the x component and one for the y component. 39

40 X Direction Let's start with the X component. Our initial momentum in the X direction is given by: m 1 V 1x + m 2 V 2x. (Note V 1x is the X component of V 1, and V 2x is the X component of V 2 ). After the collision, each particle maintains a component of their velocity in the X direction, which can be calculated using trigonometry. Our final momentum in the X direction is given by: m 1 V' 1x + m 2 V' 2x (Note V' 1x is the X component of V' 1, and V' 2x is the X component of V' 2 ). Thus our equation for the conservation of linear momentum in the X direction is: m 1 V 1x + m 2 V 2x = m 1 V' 1x + m 2 V' 2x m 1 V 1 cosθ 1 + m 2 V 2 cosθ 2 = m 1 V' 1 cosθ' 1 + m 2 V' 2 cosθ' 2 40

41 Y Direction Let's now look at the Y component. Our initial momentum in the Y direction is given by: m 1 V 1y + m 2 V 2y. (Note V 1y is the Y component of V 1, and V 2y is the Y component of V 2 ). After the collision, each particle maintains a component of their velocity in the Y direction, which can be calculated using trigonometry. Our final momentum in the Y direction is given by: m 1 V' 1y + m 2 V' 2y (Note V' 1y is the Y component of V' 1, and V' 2y is the Y component of V' 2 ). Thus our equation for the conservation of linear momentum in the Y direction is: m 1 V 1y + m 2 V 2y = m 1 V' 1y + m 2 V' 2y m 1 V 1 sinθ 1 + m 2 V 2 sinθ 2 = m 1 V' 1 sinθ' 1 + m 2 V' 2 sinθ' 2 41

42 We now have three equations: conservation of kinetic energy, and conservation of momentum in both the X and Y directions. With this information, is this problem solvable? Recall that if we are given only the initial masses 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 an additional variable. Perhaps we are trying to make a pool shot, and can tell the angle of the ball being hit by where the hole is, but would like to know where the cue ball will end up. This equation would be solvable, since with the angle the ball will take to hit the pocket we have specified another variable. m 1 V m 2 V 2 2 = m 1 V' m 2 V' 2 2 m 1 V 1x + m 2 V 2x = m 1 V' 1x + m 2 V' 2x m 1 V 1y + m 2 V 2y = m 1 V' 1y + m 2 V' 2y V1x = V 1 cos θ 1 V 1y = V 1 sin θ 1 V' 1x = V' 1 cos θ' 1 V' 1y = V' 1 sin θ' 1 V 2x = V 2 cos θ 2 V 2y = V 2 sin θ 2 V' 2x = V' 2 cos θ' 2 V' 2y = V' 2 sin θ' 2 42

43 Consider two billiard balls, M 1 and M 2, with velocities V 1 and V 2 respectively. They hit in a perfectly elastic off center collision, and both balls travel off at an angle to their original displacement. Determine V'1, V'2, and θ' 1. Y Axis V' 1 V 1 = 2 m/s at 0 o M 1 = 1.5 kg V 2 = 0 θ' 1 M 2 =.5 kg θ' 2 = 40 o X Axis V' 2 43

44 The solution to this problem is extremely difficult involving trigonometry and extensive algebra. The resulting equations for its solution are below courtesy of wikipedia. In a center of momentum frame at any time the velocities of the two bodies are in opposite directions, with magnitudes inversely proportional to the masses. In an elastic collision these magnitudes do not change. The directions may change depending on the shapes of the bodies and the point of impact. For example, in the case of spheres the angle depends on the distance between the (parallel) paths of the centers of the two bodies. Any non zero change of direction is possible: if this distance is zero the velocities are reversed in the collision; if it is close to the sum of the radii of the spheres the two bodies are only slightly deflected. Assuming that the second particle is at rest before the collision, the angles of deflection of the two particles, θ 1 and θ 2, are related to the angle of deflection θ in the system of the center of mass by: The velocities of the particles after the collision are: 44

45 Consider two billiard balls, denoted by subscripts 1 and 2. Let m 1 and m 2 be the masses, V1 and V 2 the velocities before collision, and V' 1 and V' 2 the velocities after collision. If the collision is perfectly elastic, determine the final velocities of the two balls m 1 = 2 kg m 2 = 4 kg m 1 = 2 kg m 2 = 4 kg V 1 = 2 m/s V 2 = 0 V' 1 =? V' 2 =? 45

46 Consider two billiard balls, M 1 and M 2, with velocities V 1 and V 2 respectively. They hit in an elastic off center collision, and both balls travel off at an angle to their original displacement. Determine V'1, V'2, and θ' 1. Y Axis V' 1 V 1 = 2 m/s at 0 o M 1 = 1.5 kg V 2 = 0 M 2 =.5 kg θ' 1 = 15.6 o θ' 2 = 40 o X Axis V' 2 = 1.96 m/s 46

47 Elastic Collision Two billiard balls collide as shown below: What is the velocity of ball #2 after the collision? m 1 = 0.17 kg V' 1 =? m 1 = 0.17 kg V 1 = 2.5 m/s at 0 o 30 o m 3 = 0.16 kg V 3 = 1.0 m/s at 180 o m 3 = 0.16 kg V' 3 = 0.62 m/s at 210 o 47

48 Ballistic Pendulum In a ballistic pendulum an object of mass m is fired with an initial speed v 0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement θ as shown. The first part of the problem deals with the conservation of momentum. The second part of the problem deals with the conservation of mechanical energy. 48

49 A 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 inside V1 = 150 m/s m 1 = 0.05 kg V 2 = 0 M 2 = 3 kg V' 1+2 =? m 1 + M 2 = 3.05 kg Start the problem by finding V' 1+2 using the conservation of momentum. p 1 + p 2 = p' 1+2 Since the pendulum is initially at rest p 2 equals zero. Also, V 1+2 acts in the positive X direction so the problem reduces to a one dimensional perfectly inelastic collision problem. 49

50 Solving for V' 1+2 p 1 + p 2 = p' p 1 = p' 1+2 m 1 V 1 = m 1+2 V' 1+2 (0.05 kg)(150 m/s) = (3.05 kg)v' kgm/s = (3.05 kg)v' m/s = V' 1+2 Next we apply our knowledge of energy. The pendulum is frictionless, therefore, this is a conservative system. 50

51 B C V C = 0 h C =? h B = 0 At B all the mechanical energy of the pendulum is in KE since h B = 0. The velocity at b is the velocity we calculated earlier V 1+2 = V B. ME B = KE B = 1/2 M 1+2 V B 2 At C all the mechanical energy of the pedulum is in GPE since V C = 0. ME C = GPE C = M 1+2 gh C ME B = ME C 1/2 M 1+2 V B 2 = M 1+2 gh C 1/2 V B 2 = gh C V B 2 2g = h C V B = (2gh C ) 51

52 Solving for h C : V B 2 2g (2.46 m/s) 2 2(10 m/s 2 ) = h C = h C m = h C 52

53 Solve the following: A 0.06 kg bullet with velocity 250 m/s is shot into a 3.5 kg ballistic pendulum. Find how high the pendulum rises after the bullet gets stuck inside V1 = 250 m/s m 1 = 0.06 kg V 2 = 0 M 2 = 3.5 kg V' 1+2 =? m 1 + M 2 =? 53

54 A pendulum consisting of a 0.5 kg ball is released from the position as shown below and strikes a block of mass 5 kg. The block slides a distance S before stopping due to a frictional force (The coefficient of friction is 0.2). Find the distance S if the ball rebounds to an angle of 15 degrees. L = 1 m 37 o S 54

55 Solution Strategies 1) Break down the problem to smaller parts 1. Pendulum 2. Collision 3. Sliding 2) Ask what the key concept of each part might be 1. Pendulum Oscillations, Conservation of Mechanical Energy, Simple Harmonic Motion. 2. Collision Momentum and Impulse 3. Sliding Kinematics (the Big 3), Dynamics, Non conservative Energy, Work. 3) Isolate each part and determine the given values. Organize these values with a picture. Figure out what the unknowns of interest are. 4) Determine a conceptual path to the solution, and devise a solution method. 5) Solve. 55

56 Pendulum The first part of the problem deals with the swinging of a pendulum. It strikes an object and rebounds to a different height. The collision between the pendulum bob and the block is obviously a momentum problem so we must know the velocity of the bob just before and just after the collision. We can do this by using the concept of conservation of mechanical energy. Before Collision V A = 0 h A =? A L = 1 m 37 o B H h A V B =? h B = 0 ME A = ME B 0 0 ME A = KE A + GPE A + EPE A ME A = mgh A ME B = KE B + GPE B + EPE B 0 ME B = 1/2 mv B 2 0 mgh A = 1/2 mv B 2 V B = 2gh A We are missing h A, but this can be found using trigonometry. 56

57 Finding h A L = 1 m 37 o H L L θ H H = L cosθ A h A B h A = L H h A = L L cosθ h A = L(1 cosθ) h A = L(1 cosθ) h A = 1m(1 cos37 o ) h A = m V B = 2gh A V B = 2(10m/s 2 )(0.201 m) V B = 2.0 m/s 57

58 After Collision V D = 0 h D =? D L = 1 m 15 o C H h D V C =? h C = 0 ME D = ME C 0 0 ME D = KE D + GPE D + EPE D ME D = mgh D 0 0 ME C = KE C + GPE C + EPE C ME C = 1/2 mv 2 C mgh D = 1/2 mv C 2 V C = 2gh D We are missing h D, but this can be found using trigonometry. h D = L(1 cosθ) h D = 1m(1 cos15 o ) h D = m V D = 2gh A V B = 2(10m/s 2 )(0.034 m) V B = m/s V B is negative because of the direction 58

59 Before The Collision After A B A B m A = 0.5 kg V A = 2.0 m/s m B = 5.0 kg V B = 0 m A = 0.5 kg V' A = m/s m B = 5.0 kg V' B =? 0 p A + p B = p' A + p' B m A V A = m A V' A + m B V' B (0.5 kg)(2.0 m/s) = (0.5 kg)( m/s) + (5.0 kg) V' B 1 kgm/s = kgm/s + (5.0 kg) V' B kgm/s = (5.0 kg) V' B V' B = m/s 59

60 The Slide m = 5.0 kg V A = m/s S m = 5.0 kg V B = 0 Find the frictional force. F N f F g = mg Rx = F Net Ry = 0 F N = mg F Net = f f = μ K F N = μ K mg F Net = 10 N f = 0.2 (5.0 kg)(10 m/s 2 ) f = 10 N 60

61 The Slide F Net = 10 N at 180 o m = 5.0 kg V A = m/s S m = 5.0 kg V B = 0 W Net = ΔKE F Net S cos θ = 1/2 m (V 2 F V I 2) 10 N S cos 180 o = 1/2 (5 kg) (0 2 ( m/s) 2 ) ( 10 N) S = J S = m S = 20 cm 61

62 Two discs of masses m 1 = 2 kg and m 2 = 8 kg are placed on a horizontal frictionless surface. Disc m 1 moves at a constant speed of 8 m/s in +x direction and disc m 2 is initially at rest. The collision of two discs is elastic and the directions of two velocities are presented by the diagram. Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 v 1 m 1 v 1 ' A) What is the x component of the initial momentum of disc m 1? B) What is the y component of the initial momentum of disc m 1? C) What is the x component of the initial momentum of disc m 2? D) What is the y component of the initial momentum of disc m 2? E) What is the initial momentum of the system? F) What is the x component of the final momentum of disc m 1? G) What is the x component of the final momentum of disc m 2? H) What is the y component of the final momentum of disc m 2? I) What is the y component of the final momentum of disc m 1? J) What is the final vector velocity of disc m 1? K) What is the final vector velocity of m 2? L) Determine if the collision was perfectly elastic. 62

63 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 V 1y ' = V 1y ' sin 270 A) What is the x component of the initial momentum (p 1x ) of disc m 1? p 1 = mv 1 p 1x = mv 1x p 1x = (2 kg)(8 m/s) p 1x = 16 kg m/s positive because its in the postive X direction B) What is the y component of the initial momentum (p 1y ) of disc m 1? p 1 = mv 1 p 1y = mv 1y p 1y = (8 kg)(0 m/s) p 1y = 0 kg m/s 63

64 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 V 1y ' = V 1y ' sin 270 C) What is the x component of the initial momentum (p 2x ) of disc m 2? p 2 = mv 2 p 2x = mv 2x p 2x = (2 kg)(0 m/s) p 2x = 0 D) What is the y component of the initial momentum (p 2y ) of disc m 2? p 2 = mv 2 p 2y = mv 2y p 2y = (8 kg)(0 m/s) p 2y = 0 kg m/s 64

65 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 V 1y ' = V 1y ' sin 270 E) What is the initial momentum (p initial ) of the system? p totalx = Σp ix = p 1x + p 2x p totalx = 16 kg m/s + 0 = 16 kg m/s p totaly = Σp iy = p 1y + p 2y p totaly = = 0 p initial = p totalx + p totaly Using VECTOR ADDITION p initial = 16 kg m/s at 0 o 65

66 Since momentum is conserved total momentum before the collision = total momentum after the collision p = p' 16 kg m/s at 0 o = p' The components of momentum before and after the collision must also be equal p totalx = p' totalx 16 kg m/s = p' totalx positive because its in the postive X direction p totaly = p' totaly 0 = p' totaly 66

67 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 = 0 V 1y ' = V 1y ' sin 270 F) What is the x component of the final momentum (p 1x ')of disc m 1? Because the direction of m 1 after the collision is 270 o which causes V 1x ' to equal 0, (since p = mv) then momentum also MUST equal zero! Thus p 1x ' = 0 67

68 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 = 0 V 1y ' = V 1y ' sin 270 G) What is the x component of the final momentum (p' 2x ) of disc m 2? p' totalx = 16 kg m/s p' totalx = p' 1x + p' 2x Since p 1x ' = 0 Then p' totalx = 0 + p' 2x p' totalx = p' 2x 16 kg m/s = p' 2x positive because its in the postive X direction 68

69 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 = 0 V 1y ' = V 1y ' sin 270 H) What is the y component of the final momentum (p 2y ') of disc m 2? p 2 ' 30 o p 2x ' = 16 kg m/s p 2y ' Solve using trigonometry. tan 30 o = p 2y '/p 2x ' p 2y ' = p 2x ' tan 30 o p 2y ' = 16 kg m/s tan 30 o p 2y ' = 9.2 kg m/s 69

70 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 = 0 V 1y ' = V 1y ' sin 270 I) What is the y component of the final momentum (p' 1y ) of disc m 1? p' totaly = 0 p' totaly = p' 1y + p' 2y Then 0 = p' 1y + p' 2y p' 1y = p' 2y p' 1y = 9.2 kg m/s 70

71 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 = 0 V 1y ' = V 1y ' sin 270 J) What is the final vector velocity (V 1 ') of disc m 1? Find p 1 ' p 1 ' = p 1x ' + p 1y ' p 1 ' = 0 + ( 9.2 kg m/s) Using VECTOR ADDITION p 1 ' = 9.2 kg m/s at 270 o 71

72 p 1 ' = m 1 V 1 ' V 1 ' = p 1' m 1 V 1 ' = 9.2 kg m/s at 270o 2 kg V 1 ' = 4.6 m/s at 270 o 72

73 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 = 0 V 1y ' = V 1y ' sin 270 J) What is the final vector velocity (V 2 ') of disc m 1? Find p 2 ' p 2 ' = p 2x ' + p 2y ' p 2 ' = 16 kg m/s kg m/s Using VECTOR ADDITION p 2 ' = 18.5 kg m/s at 30 o 73

74 p 2 ' = m 2 V 2 ' V 2 ' = p 2' m 2 V 2 ' = 18.5 kg m/s at 30o 8 kg V 2 ' = 2.3 m/s at 270 o 74

75 Before Collision m 2 After Collision v 2 ' m 2 30 o m 1 V 1x = 8 m/s V 1y = 0 v 1 m 1 V 2x ' = V 2x ' cos 30 V 2y ' = V 2y ' sin 30 V 2x = 0 v 1 ' V 2y = 0 V 1x ' = V 1x ' cos 270 = 0 V 1y ' = V 1y ' sin 270 L) Determine if the collision was perfectly elastic. If a collision is perfectly elastic then kinetic energy is conserved. KE 1 + KE 2 = KE 1 ' + KE 2 ' ½m 1 V ½m 2 V 2 2 = ½m 1 V 1 2' + ½m 2 V 2 2' ½(2 kg)(8 m/s) 2 + ½(8 kg)(0) 2 = ½(2 kg)(4.6 m/s) 2 + ½(8 kg)(2.3 m/s) 2 64 J + 0 = J J 64 J J Kinetic Energy is not conserved. 75

76 Ft 2 Ft kg 76

77 Draw an FBD Ft 2 Ft 1 X = F T2 cos 70 Ft Y = F T2 sin Ft 1 X = F T1 Y = 0 Fg = 150N Fg = 150N X = 75.0 N Y = N Rx = 0 F T1 F T2 cos N = 0 F T1 = F T2 cos N F T1 = N Ry = 0 F T2 sin N = 0 F T2 sin 70 = N F T2 = N sin 70 F T2 = N 77

78 Ft 2 Ft kg 78

79 An 8,000 Kg truck, traveling with a velocity 12 m/s at 0 o, collides with a 1,400 Kg car moving 22 m/s at 220 o as in the figure below. After the collision, the two vehicles remain tangled together. With what velocity will the wreckage begin to move immediately after the crash? 220 o Truck Car 79

80 Two girls (masses m 1 and m 2 ) are on skates and at rest. The girls are close to each other, and are facing each other. Girl 1 pushes against girl 2 causing girl 2 to move backwards. A) Assuming the girls move freely on skates (no friction), find a formula to calculate the speed of girl 1. B) Calculate the speed of girl 1, If the mass of girl 1 is 48 kg, the mass of girl 2 is 62 kg, and girl 2 s speed is 1.8 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 the push was 0.7 s. 80

81 An 65 kg stunt man jumps out of a window and falls 65 m to an airbag on the ground. A) How fast is he falling when he reaches the airbag? (Initial velocity is zero) B) He lands on a large air filled target, coming to rest in 2.1 s. What average force does he feel while coming to rest? C) What average force does he feel if he had fallen 65 m and landed on the ground (impact time = 10 ms =.001 s))? 65 m Air bag 81

82 A railroad car of mass 30,000 kg is moving with a speed of 9.00 m/s to the right. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the opposite direction with an initial speed of 3.00 m/s. (a) What is the speed of the four cars after the collision? (b) How much energy is lost in the collision? 9 m/s 3 m/s 82