Chapter 11 Collision Theory
|
|
- Marcia Hodge
- 5 years ago
- Views:
Transcription
1 Chapter Collision Theory Introduction. Center o Mass Reerence Frame Consider two particles o masses m and m interacting ia some orce. Figure. Center o Mass o a system o two interacting particles Choose a coordinate system (Figure.) in which the position ector o body is gien by r and the position ector o body is gien by r. The relatie position o body with respect to body is gien by r, = r r. Figure. Center o mass coordinate system. During the course o the interaction, body is displaced by dr and body is displaced by dr, so the relatie displacement o the two bodies during the interaction is gien by dr = dr dr. The relatie elocity between the particles is, dr, dr dr, = = =. (..) dt dt dt
2 We shall now show that the relatie elocity between the two particles is independent o the choice o reerence rame. Let R be the ector rom the origin o rame S to the origin o reerence rame S. Denote the position ector o particle i with respect to origin o reerence rame S by r i and similarly, denote the position ector o particle i with respect to origin o reerence rame S by r i (Figure.4). Figure.4 Position ector o th i particle in two reerence rames. The position ectors are related by r = r + R i i. (..) The relatie elocity (call this the boost elocity) between the two reerence rames is gien by V = d R. (..3) dt Assume the boost elocity between the two reerence rames is constant. Then, the relatie acceleration between the two reerence rames is zero, A= dv = 0. (..4) dt th Suppose the i particle in Figure.4 is moing; then obserers in the dierent th reerence rames will measure dierent elocities. Denote the elocity o i particle in rame S by i = dr i dt, and the elocity o the same particle in rame S by i = dr i dt. Since the deriatie o the position is elocity, the elocities o the particles in two dierent reerence rames are related according to = + V i i. (..5) In rame S, the relatie elocity is
3 , = (..6) The relatie elocity in reerence rame S can be determined rom using Equation (..5) to express Equation (..6) in terms o the primed quantities, = = + V + V = = ( ) ( ),, (..7) and is equal to the relatie elocity in rame S. For a two-particle interaction, the relatie elocity between the two ectors is independent o the choice o reerence rame. In Appendix 8.B, we showed that when two particles o masses m and m interact, the change o kinetic energy between the inal state B and the initial state A due to the interaction orce only is equal to Δ K = μ ( B A ) (..8) where μ = mm / m + m ( ) is the reduced mass o the two-particle system. (I Equation (..4) did not hold, Equation (..8) would not be alid in all rames.) In Equation (..8), the square o the inal relatie elocity ( ) ( ) B B is gien by ( ) = (( ) ( ) ), (( ) ( ) ) B B B B B (..9) A and the square o the initial relatie elocity ( ) ( ) A is gien by ( ) = (( ) ( ) ), (( ) ( ) ) A A A A A. (..0) By expressing the change o kinetic energy in terms o the relatie elocity, a quantity that is independent o the reerence rame, the change in kinetic energy is thereore independent o the choice o reerence rame. Characterizing Collisions In a collision, the ratio o the magnitudes o the initial and inal relatie elocities is called the coeicient o restitution and denoted by the symbol e, e B =. (..) A 3
4 I the magnitude o the relatie elocity does not change during a collision, e =, then the change in kinetic energy is zero, (Equation (..8)). Collisions in which there is no change in kinetic energy are called elastic collisions, ΔK = 0, elastic collision. (..) I the magnitude o the inal relatie elocity is less than the magnitude o the initial relatie elocity, e <, then the change in kinetic energy is negatie. Collisions in which the kinetic energy decreases are called inelastic collisions, ΔK < 0, inelastic collision. (..3) I the two objects stick together ater the collision, then the relatie inal elocity is zero, e = 0. Such collisions are called totally inelastic. The change in kinetic energy can be ound rom Equation (..8), Δ = μ = mm K A A, totally inelastic collision m+ m. (..4) I the magnitude o the inal relatie elocity is greater than the magnitude o the initial relatie elocity, e >, then the change in kinetic energy is positie. Collisions in which the kinetic energy increases are called superelastic collisions,. Worked Examples.. Example: Elastic One-Dimensional Collision ΔK > 0, superelastic collision. (..5) Consider the elastic collision o two carts along a track; the incident cart has mass m and moes with initial speed,0. The target cart has mass m = m and is initially at rest,,0 = 0. Immediately ater the collision, the incident cart has inal speed and the target cart has inal speed o the initial speed. Solution,0,,. Calculate the inal elocities o the carts as a unction Draw a momentum low diagram or the objects beore (initial state) and ater (inal state) the collision (Figure.5, with a greatly simpliied rendering o a cart ). 4
5 Figure.5 Momentum low diagram or elastic one-dimensional collision There are no external orces acting on the system, so the component o the momentum along the direction o the collision is the same beore and ater the collision, m,0 = m, + m,. (..) Note that in the aboe igure and Equation (..), the incident cart is taken to be moing backwards i, > 0. Equation (..) simpliies to, 0 =, +,. (..) The collision is elastic; the kinetic energy is the same beore and ater the collision, m =, 0 m, + m,, (..3) which simpliies to, 0 =, +,. (..4) There are many ways to manipulate Equations (..) and (..4) to sole or the inal elocities in terms o. One straightorward way is to sole Equation (..) or the inal speed o cart,,0, =,, 0 (..5) and substitute Equation (..5) into Equation (..4) yielding ( ) Equation (..6) simpliies to = + = (..6),0,,0,,,0,,0, 5
6 0 = 6, 4,, 0. (..7) We can now sole or solution is,. There are two solutions to this quadratic equation. One, = 3, 0. (..8) The inal elocity o the incident cart is then, =,, 0 = 4 3, 0,0 = 3,0. (..9) The other solution is, = 0, in which the inal state o the target is the same as the initial state; in addition, Equation (..5) would gie, =,0, corresponding to the incident cart moing at its initial elocity. That is, the carts did not collide. This situation is not physical or carts on a track o inite length, but could certainly happen in other circumstances, as beginning pool players know well. Or, i in the top igure in Figure.5 the initially moing cart were placed to the right o the stationary cart, Equations (..) and (..3) would still be alid or no collision. A technique that physicists know well is to rewrite Equations (..) and (..4) as,0 +, =,,0, =,. (..0) I,0 +, 0 ( 0 + = 0,, is the second case o a miss, discussed aboe), the second equation in (..0) can be diided by the irst to yield,0, =,. (..) Adding Equation (..) to the irst expression in (..0) yields Equation (..8). The more general case o a one-dimensional collision is discussed in Appendix.A... Example: Elastic Two-Dimensional Collision Object with mass is initially moing with a speed,0 = 3.0m s and collides m elastically with object that has the same mass, m = m, and is initially at rest. Ater the collision, object moes with an unknown speed, at an angle θ, = 30 o with respect 6
7 to its initial direction o motion and object moes with an unknown speed unknown angle θ, objects and the angle., at an (as shown in the Figure.6). Find the inal speeds o each o the θ,, Solution: Figure.6 Momentum low diagram or two-dimensional elastic collision Choose a set o positie unit ectors or the initial and inal states as shown in Figure.7. We designate the respectie speeds o each o the particles on the momentum low diagrams. Figure.7 Choice o unit ectors or momentum low diagram Initial State: The components o the total momentum state are gien by total p x,0 = m, 0 total p y,0 = 0. p = m + m,0 total 0,0 in the initial (..) Final State: The components o the momentum p total = m, + m, are gien by in the inal state 7
8 p = m cosθ + m cosθ total x,,,,, p = m sinθ m sin θ. total y,,,,, (..3) There are no any external orces acting on the system, so each component o the total momentum remains constant during the collision, total total p x,0 = p x, (..4) p total total y,0 = p y,. (..5) These two equations become m = m cosθ + m cosθ,0,,,, 0= m sinθ m sin θ.,,,, (..6) The collision is elastic; the kinetic energy is the same beore and ater the collision, or K total total 0 = K, (..7) m =, 0 m, + m,. (..8) We hae three equations, two momentum equations and one energy equation, with three unknown quantities,,,, and θ, since we are already gien that, 0 = 3.0 m s and θ, = 30 o. We irst rewrite the expressions in Equation (..6), canceling the actors o, as m cosθ =,,,0,,,,,, cosθ sinθ = sin θ. (..9) Add the squares o the expressions in Equation (..9), yielding ( ), cos θ,, sin θ,,0, cosθ,, sin, + = + θ. (..0) We can use the identities cos θ, + sin θ, = and cos θ, + sin θ, = to simpliy Equation (..0), yielding 8
9 = cosθ +. (..),,0,0,,, Substituting Equation (..) into Equation (..8) yields m,0= m, + m(,0,0, cosθ, +, ). (..) Equation (..) simpliies to 0= cosθ, (..3),,0,, which may be soled or the inal speed o object,,,0, ( ) = cosθ = 3.0 m s cos30 =.6 m s. (..4) Diide the expressions in Equation (..9), yielding sinθ = sinθ,,,,, cosθ,,0, cosθ,. (..5) Equation (..5) simpliies to tanθ, = sinθ,, cosθ,0,,. (..6) Thus object moes at an angle θ θ, tan sinθ,, = = tan cosθ,0,, (.6 m s ) sin30 ( ) 3.0 m s.6 m s cos30, = 60. (..7) The aboe results or, and θ, may be substituted into either o the expressions in Equation (..9), or Equation (..8), to ind., =.5m s Beore going on, it must be noted that the act that θ, + θ, = 90, that is, the objects moe away rom the collision point at right angles, is not a coincidence. A ector 9
10 deriation is presented below. We can see this result algebraically rom the aboe result. Using the result o Equation (..4),, =,0 cosθ,, in Equation (..6) yields tanθ cosθ sinθ = = cotθ,,,, cosθ, ; (..8) the angles θ, and θ, are complements. It should be noted that Equation (..3) also has the solution, = 0, which would correspond to the incident particle missing the target completely. We can proe that the particles emerge rom the collision at right angles by making explicit use o the act that momentum is a ector quantity. Since there are no external orces acting on the two objects during the collision (the collision orces are all internal), momentum is constant. Thereore p = p total total 0 (..9) which becomes m = m + m,0,, (..30) Equation (..30) simpliies to = +, 0,,. (..3) Recall the ector identity that the square o the speed is gien by the dot product =. (..3) With this identity in mind, we take the dot product o each side o Equation (..3) with itsel, This becomes, 0, 0 = (, +, ) (, +, ) (..33) = + +.,,,,,, = + +, 0,,,,. (..34) 0
11 Recall that kinetic energy is the same beore and ater an elastic collision, and the masses o the two objects are equal, so Equation (..8) simpliies to = +. (..35), 0,, Comparing Equation (..34) with Equation (..35), we see that,, = 0. (..36) The dot product o two nonzero ectors is zero when the two ectors are at right angles to each other. Since = 30 o, the inal angle o object must be θ, θ =, 60. (..37) A geometric demonstration or obtaining the result o Equation (..37) is gien in Appendix.B to this chapter. Note that Equation (..36) allows the algebraic solutions, = 0 and, = 0. Had we not been gien a nonzero alue or, the, = 0 solution would represent a collision, in which all o the momentum and energy o object is transerred to object. In this case, the gien angle o θ = is not well deined, but is immaterial or a stationary, 30 particle. Had we not been gien a nonzero alue or θ, the, = 0 solution would indicate a complete miss...3 Example: Bouncing Superballs Two superballs are dropped rom a height aboe the ground, one on top o the other. The ball on top has a mass m, and the ball on the bottom has a mass m. Assume that the when the lower ball collides with the ground there is no loss o kinetic energy. Then, as the lower ball starts to moe upward, it collides with the upper ball that is still moing downwards. Assume again that the total energy o the two balls remains the same ater the collision. How high will the upper ball rebound in the air? Assume m >> m. Hint: Consider this set o collisions rom an inertial reerence rame that moes upward with the same speed as the lower ball has ater it collides with ground. What speed does the upper ball hae in this reerence rame ater it collides with the lower ball?,
12 Figure.8 Fie stages o motion (not necessarily to scale) as seen by obserer at rest on ground. Solution: The system consists o the two superballs. There are ie special states or this motion (Figure.8). Initial State (time t 0 ): the superballs are released rom rest at a height h 0 aboe the ground. State (time ): the superballs just reach the ground with the same speed =. t,0,0 State (time t ): immediately beore the collision o the large and small superballs but ater the larger superball has collided with the ground and reersed direction with the same speed,,0 =. The smaller superball is still moing down with speed,0., 0 State 3 (time t 3 ): immediately ater the collision o the superballs. The smaller superball moes upward with speed,. The larger superball moes upward with speed., Final State (time ): the smaller superball reaches maximum height aboe the t ground. Choice o Reerence Frame: This collision is best analyzed rom the reerence rame o the obserer moing upward with speed,0 =,0, the elocity o the larger superball just ater it rebounded with the ground. In this rame immediately beore the collision, the smaller superball is moing h
13 downward with a speed, 0 that is twice the speed as seen by an obserer at rest on the ground (lab reerence rame),,0 =, 0. (..38) The momentum low diagrams between States and 3 or the two dierent reerence rames are shown in Figure.9. Figure.9 Momentum low diagrams or States and 3 as determined by an obserer at rest on ground (let igure) and an obserer moing upwards with speed o large superball,0 (right igure). Model: The mass o the larger superball is much larger than the mass o the smaller superball, m >> m. This enables us to consider the collision (between States and 3) to be equialent to the small superball bouncing o a hard wall, while the larger ball experiences irtually no recoil. Hence the large superball remains at rest in the reerence rame moing upwards with speed with respect to obserers at rest on ground.,0 Beore the collision, the smaller superball has speed, 0 =, 0. Since there is no loss o kinetic energy during the collision, the result o the collision is that the smaller superball changes direction but maintains the same speed, =, 0 =, 0 (..39) Howeer, according to an obserer at rest on the ground, ater the collision the smaller superball is moing upwards with speed. (..40), =,0 +, 0 = 3, 0 Ater the collision, the mechanical energy o the smaller superball is constant and hence between State 3 and the Final State, 3
14 ΔK + ΔU = 0. (..4) The change in kinetic energy is Δ K = m( 3,0). (..4) The change in potential energy is ΔU = m gh. (..43) Equation (..4), the condition that mechanical energy is constant, becomes m( 3,0) + mgh = 0 mgh = 9 m(,0). (..44) Recall that we can also use the act that the mechanical energy doesn t change between the Initial State and State, yielding an equation similar to Equation, (..44), m m ( ),0 0 ( ) m gh = 0 = m gh.,0 0 (..45) Comparing the second expressions in (..44) and (..45), the smaller superball is seen to reach a maximum height h = 9h. (..46) 0..4 Example Pendulums and Collisions A simple pendulum consists o a bob o mass m that is suspended rom a piot by a string o negligible mass. The bob is pulled out and released rom a height h 0 as measured rom the bob s lowest point directly under the piot point and then swings downward in a circular orbit (Figure.0a). At the bottom o the swing, the bob collides with a block o mass m that is initially at rest on a rictionless table. Assume that there is no riction at the piot point. a) What is the speed o the bob immediately beore the collision at the bottom o the swing? 4
15 b) Assume that the kinetic energy o the bob beore the collision is equal to the kinetic energy o the bob and the block ater the collision (the collision is elastic). Also assume that the bob and the block moe in opposite directions but with the same speed ater the collision (Figure.0b). What is the mass o the block? m c) Suppose the bob and block stick together ater the collision due to some putty that is placed on the block. What is the speed o the combined system immediately ater the collision? (Assume now that is the combined mass o the block and putty.) m d) What is the change in kinetic energy o the block and bob due to the collision in part c)? What is the ratio o the change in kinetic energy to the kinetic energy beore the collision? e) Ater the collision in part d), the bob and block moe together in circular motion. What is the height h aboe the low point o the bob s swing when they both irst come to rest ater the collision (Figure.0c)? Ignore any air resistance. (Hint: The relatie sizes o the heights h and h in Figure.0 (a), (c) are not correct.) 0 (a) (b) (c) Figure.0 (a) Simple pendulum released on a collision course with a stationary block (b) Collision at the bottom o the swing (c) Bob and block rise to the top o the swing Solution: a) The mechanical energy o the bob is constant between when it is released and the bottom o the swing. We can use m = mgh (..47),0 0 to calculate the speed o the bob at the low point o the swing just beore the collision,, 0 = gh 0. (..48) 5
16 b) Consider the bob and the block as the system. Although tension in the string and the graitation orce are now acting as external orces, both are particular to the motion o the bob and block during the collision. I we additionally assume that the collision is nearly instantaneous, then the momentum is constant in the direction o the bob s motion, m,0 = m, m,, (..49) where, is the speed o the block immediately ater the collision. Since the bob and block are gien to hae the same speeds ater the collision, deine, =, and rewrite Equation (..49) as ( ) m = m m. (..50),0 Sole Equation (..50) or the speed o the bob and block ater the collision, =,0 m m m (..5) (at this point we see explicitly what we might hae guessed, that m > m ). The collision is gien to be elastic, m,0 = ( m+ m). (..5) Substituting Equation (..5) into Equation (..5) yields m m = ( m+ m),0,0 m m. (..53) Canceling the common actor o m, 0 rom both sides o Equation (..53) and rearranging gies ( ) ( ) m m = m + m m. (..54) Expanding the square and canceling m yields m (m 3 m ) = 0, (..55) 6
17 and so the block has mass m = 3m (..56) and the inal speed is gh,0 0 = =. (..57) c) The bob and block stick together and moe with a speed ater the collision. The external orces are still perpendicular to the motion, and i we assume that the collision time is negligible, then the momentum in the direction o the motion is constant, ( ) m = m+ m (..58),0 The speed immediately ater the collision is (recalling that m ) = 3m m,0 = = m+ m,0. (..59) 4 Using Equation (..48) in Equation (..59) yields gh gh =,0 = 0 =. (..60) d) The change in kinetic energy o the bob and block due to the collision in part c) is gien by Δ K = K K = ( m + m ) m ater beore,0 Using Equation (..59), (..48) and (..59) in Equation (..6),. (..6) gh0 Δ K = ( 4m) mg 8 3 = mgh 0. 4 h 0 (..6) The kinetic energy beore the collision was mgh 0, and so the ratio o the change in kinetic energy to the kinetic energy beore the collision is 7
18 ΔK K beore = 3/4. (..63) See Appendix.C or a more general result, ΔK m = K m + m beore (..64) or completely inelastic collisions when the target object (the block o mass example) is initially stationary. m in this e) Ater the collision, the tension is acting on the bob-block system but the tension orce is perpendicular to the motion so does no work on the bob-block system and the mechanical energy ater the collision is the same as when the bob-block combination reaches its highest point, K = ( m + m ) gh mgh 4 ater 0 = 4mgh h 6 (..65) 0 h =. (..66) 8
Chapter 15 Collision Theory
Chapter 15 Collision Theory Chapter 15 Collision Theory 151 Introduction 15 Reference Frames Relative and Velocities 151 Center of Mass Reference Frame 3 15 Relative Velocities 4 153 Characterizing Collisions
More informationModule 18: Collision Theory
Module 8: Collision Theory 8 Introduction In the previous module we considered examples in which two objects collide and stick together, and either there were no external forces acting in some direction
More informationCJ57.P.003 REASONING AND SOLUTION According to the impulse-momentum theorem (see Equation 7.4), F t = mv
Solution to HW#7 CJ57.CQ.003. RASONNG AND SOLUTON a. Yes. Momentum is a ector, and the two objects hae the same momentum. This means that the direction o each object s momentum is the same. Momentum is
More informationChapter 15 Collision Theory
Chapter 5 Collision Theory 5 Introduction 5 Reference Frames and Relative Velocities 5 Relative Velocities 3 5 Center-of-mass Reference Frame 4 53 Kinetic Energy in the Center-of-Mass Reference Frame 5
More informationFOCUS ON CONCEPTS Section 7.1 The Impulse Momentum Theorem
WEEK-6 Recitation PHYS 3 FOCUS ON CONCEPTS Section 7. The Impulse Momentum Theorem Mar, 08. Two identical cars are traeling at the same speed. One is heading due east and the other due north, as the drawing
More informationCollision Theory Challenge Problems Solutions
Collision Theory Challenge Problems Solutions Problem 1 Estimate the energy loss in a completely inelastic collision between two identical cars that collide head-on traveling at highway speeds! Solution:
More informationLinear Momentum and Collisions Conservation of linear momentum
Unit 4 Linear omentum and Collisions 4.. Conseration of linear momentum 4. Collisions 4.3 Impulse 4.4 Coefficient of restitution (e) 4.. Conseration of linear momentum m m u u m = u = u m Before Collision
More informationCollision Theory Challenge Problems
Collision Theory Challenge Problems Problem 1 Estimate the energy loss in a completely inelastic collision between two identical cars that collide head-on traveling at highway speeds. Problem 2 You just
More informationOne-Dimensional Motion Review IMPORTANT QUANTITIES Name Symbol Units Basic Equation Name Symbol Units Basic Equation Time t Seconds Velocity v m/s
One-Dimensional Motion Review IMPORTANT QUANTITIES Name Symbol Units Basic Equation Name Symbol Units Basic Equation Time t Seconds Velocity v m/s v x t Position x Meters Speed v m/s v t Length l Meters
More informationUNDERSTAND MOTION IN ONE AND TWO DIMENSIONS
SUBAREA I. COMPETENCY 1.0 UNDERSTAND MOTION IN ONE AND TWO DIMENSIONS MECHANICS Skill 1.1 Calculating displacement, aerage elocity, instantaneous elocity, and acceleration in a gien frame of reference
More information2. REASONING According to the impulse-momentum theorem, the rocket s final momentum mv f
CHAPTER 7 IMPULSE AND MOMENTUM PROLEMS. REASONING According to the ipulse-oentu theore, the rocket s inal oentu diers ro its initial oentu by an aount equal to the ipulse ( ΣF ) o the net orce eerted on
More informationWork Up an Incline. Work = Force x Distance. Push up: 1500J. What is the PE at the top? mg = 500N. An incline is a simple machine!
Quick Question Work Up an Incline The block o ice weighs 500 Newtons. How much work does it take to push it up the incline compared to liting it straight up? Ignore riction. Work Up an Incline Work = Force
More informationLesson 6: Apparent weight, Radial acceleration (sections 4:9-5.2)
Beore we start the new material we will do another Newton s second law problem. A bloc is being pulled by a rope as shown in the picture. The coeicient o static riction is 0.7 and the coeicient o inetic
More informationModule 27: Rigid Body Dynamics: Rotation and Translation about a Fixed Axis
Module 27: Rigid Body Dynamics: Rotation and Translation about a Fixed Axis 27.1 Introduction We shall analyze the motion o systems o particles and rigid bodies that are undergoing translational and rotational
More informationYour Thoughts. What is the difference between elastic collision and inelastic collision?
Your Thoughts This seemed pretty easy...before we got the checkpoint questions What is the difference between elastic collision and inelastic collision? The most confusing part of the pre lecture was the
More informationSo now that we ve mentioned these terms : kinetic, potential, work we should try to explain them more. Let s develop a model:
Lecture 12 Energy e are now at the point where we can talk about one of the most powerful tools in physics, energy. Energy is really an abstract concept. e hae indicators of energy (temperature, elocity
More informationPhysics Test VI Chapter 7 Impulse and Momentum
Physics Test VI Chapter 7 Impulse and Momentum Name: Date: Period: Honor Pledge On my honor as a student I have neither given nor received aid on this test Sign Below HW Grade: Test Grade / Mr. Stark Loudoun
More informationPhysics 101 Lecture 12 Equilibrium and Angular Momentum
Physics 101 Lecture 1 Equilibrium and Angular Momentum Ali ÖVGÜN EMU Physics Department www.aovgun.com Static Equilibrium q Equilibrium and static equilibrium q Static equilibrium conditions n Net external
More informationPhysics 4A Solutions to Chapter 4 Homework
Physics 4A Solutions to Chapter 4 Homework Chapter 4 Questions: 4, 1, 1 Exercises & Problems: 5, 11, 3, 7, 8, 58, 67, 77, 87, 11 Answers to Questions: Q 4-4 (a) all tie (b) 1 and tie (the rocket is shot
More informationCHAPTER 7 IMPULSE AND MOMENTUM
CHAPTER 7 IMPULSE AND MOMENTUM PROBLEMS 1. SSM REASONING The ipulse that the olleyball player applies to the ball can be ound ro the ipulse-oentu theore, Equation 7.4. Two orces act on the olleyball while
More information(a) Taking the derivative of the position vector with respect to time, we have, in SI units (m/s),
Chapter 4 Student Solutions Manual. We apply Eq. 4- and Eq. 4-6. (a) Taking the deriatie of the position ector with respect to time, we hae, in SI units (m/s), d ˆ = (i + 4t ˆj + tk) ˆ = 8tˆj + k ˆ. dt
More informationPage 1. t F t m v. N s kg s. J F t SPH4U. From Newton Two New Concepts Impulse & Momentum. Agenda
SPH4U Agenda Fro Newton Two New Concepts Ipulse & oentu Ipulse Collisions: you gotta consere oentu! elastic or inelastic (energy consering or not) Inelastic collisions in one diension and in two diensions
More informationUnit 11: Vectors in the Plane
135 Unit 11: Vectors in the Plane Vectors in the Plane The term ector is used to indicate a quantity (such as force or elocity) that has both length and direction. For instance, suppose a particle moes
More informationPractice Problems for Exam 2 Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall Term 008 Practice Problems for Exam Solutions Part I Concept Questions: Circle your answer. 1) A spring-loaded toy dart gun
More informationPhysics Momentum: Collisions
F A C U L T Y O F E D U C A T I O N Departent o Curriculu and Pedagogy Physics Moentu: Collisions Science and Matheatics Education Research Group Supported by UBC Teaching and Learning Enhanceent Fund
More informationDepartment of Physics PHY 111 GENERAL PHYSICS I
EDO UNIVERSITY IYAMHO Department o Physics PHY 111 GENERAL PHYSICS I Instructors: 1. Olayinka, S. A. (Ph.D.) Email: akinola.olayinka@edouniersity.edu.ng Phone: (+234) 8062447411 2. Adekoya, M. A Email:
More informationPhysics 40 Chapter 8 Homework Q: 12, 13 P: 3, 4, 7, 15, 19, 24, 32, 34, 39, 54, 55, 58, 59, 62, 64
Physics 40 Chapter 8 Homework Q:, 3 P: 3, 4, 7, 5, 9, 4, 3, 34, 39, 54, 55, 58, 59, 6, 64 Conceptual Questions *Q8. We hae (/)m = μ k mgd so d = /μ k g. The quantity /μ k controls the skidding distance.
More informationa by a factor of = 294 requires 1/T, so to increase 1.4 h 294 = h
IDENTIFY: If the centripetal acceleration matches g, no contact force is required to support an object on the spinning earth s surface. Calculate the centripetal (radial) acceleration /R using = πr/t to
More informationFs (30.0 N)(50.0 m) The magnitude of the force that the shopper exerts is f 48.0 N cos 29.0 cos 29.0 b. The work done by the pushing force F is
Chapter 6: Problems 5, 6, 8, 38, 43, 49 & 53 5. ssm Suppose in Figure 6. that +1.1 1 3 J o work is done by the orce F (magnitude 3. N) in moving the suitcase a distance o 5. m. At what angle θ is the orce
More information, remembering that! v i 2
Section 53: Collisions Mini Inestigation: Newton s Cradle, page 34 Answers may ary Sample answers: A In Step, releasing one end ball caused the far ball on the other end to swing out at the same speed
More informationFirst Year Physics: Prelims CP1 Classical Mechanics: DR. Ghassan Yassin
First Year Physics: Prelims CP1 Classical Mechanics: DR. Ghassan Yassin MT 2007 Problems I The problems are divided into two sections: (A) Standard and (B) Harder. The topics are covered in lectures 1
More informationCollisions. Of classical objects like collisions of motor vehicles. Of subatomic particles collisions allow study force law.
Collision Theory Collisions Any interaction between (usually two) objects which occurs for short time intervals Δt when forces of interaction dominate over external forces. Of classical objects like collisions
More informationA. unchanged increased B. unchanged unchanged C. increased increased D. increased unchanged
IB PHYSICS Name: DEVIL PHYSICS Period: Date: BADDEST CLASS ON CAMPUS CHAPTER B TEST REVIEW. A rocket is fired ertically. At its highest point, it explodes. Which one of the following describes what happens
More informationWORK ENERGY AND POWER
WORK ENERGY AND POWER WORK PHYSICAL DEINITION When the point of application of force moves in the direction of the applied force under its effect then work is said to be done. MATHEMATICAL DEINITION O
More informationChapter (3) Motion. in One. Dimension
Chapter (3) Motion in One Dimension Pro. Mohammad Abu Abdeen Dr. Galal Ramzy Chapter (3) Motion in one Dimension We begin our study o mechanics by studying the motion o an object (which is assumed to be
More informationApplications of Forces
Chapter 10 Applications of orces Practice Problem Solutions Student Textbook page 459 1. (a) rame the Problem - Make a sketch of the ector. - The angle is between 0 and 90 so it is in the first quadrant.
More information( ) Momentum and impulse Mixed exercise 1. 1 a. Using conservation of momentum: ( )
Momentum and impulse Mixed exercise 1 1 a Using conseration of momentum: ( ) 6mu 4mu= 4m 1 u= After the collision the direction of Q is reersed and its speed is 1 u b Impulse = change in momentum I = (3m
More informationWork and Energy Problems
09//00 Multiple hoice orce o strength 0N acts on an object o ass 3kg as it oes a distance o 4. I is perpendicular to the 4 displaceent, the work done is equal to: Work and Energy Probles a) 0J b) 60J c)
More informationFrames of Reference, Energy and Momentum, with
Frames of Reference, Energy and Momentum, with Interactie Physics Purpose: In this lab we will use the Interactie Physics program to simulate elastic collisions in one and two dimensions, and with a ballistic
More informationElastic collisions in two dimensions 5B
Elastic collisions in two dimensions 5B a First collision: e=0.5 cos α = cos30 () sin α = 0.5 sin30 () Squaring and adding equations () and () gies: cos α+ sin α = 4cos 30 + sin 30 (cos α+ sin α)= 4 3
More informationPhysics 1: Mechanics
Physics 1: Mechanics Đào Ngọc Hạnh Tâm Office: A1.53, Email: dnhtam@hcmiu.edu.n HCMIU, Vietnam National Uniersity Acknowledgment: Most of these slides are supported by Prof. Phan Bao Ngoc credits (3 teaching
More informationDynamics ( 동역학 ) Ch.2 Motion of Translating Bodies (2.1 & 2.2)
Dynamics ( 동역학 ) Ch. Motion of Translating Bodies (. &.) Motion of Translating Bodies This chapter is usually referred to as Kinematics of Particles. Particles: In dynamics, a particle is a body without
More informationPhysics 11 HW #6 Solutions
Physics HW #6 Solutions Chapter 6: Focus On Concepts:,,, Probles: 8, 4, 4, 43, 5, 54, 66, 8, 85 Focus On Concepts 6- (b) Work is positive when the orce has a coponent in the direction o the displaceent.
More informationN10/4/PHYSI/SPM/ENG/TZ0/XX PHYSICS STANDARD LEVEL PAPER 1. Monday 8 November 2010 (afternoon) 45 minutes INSTRUCTIONS TO CANDIDATES
N1/4/PHYSI/SPM/ENG/TZ/XX 881654 PHYSICS STANDARD LEVEL PAPER 1 Monday 8 Noember 21 (afternoon) 45 minutes INSTRUCTIONS TO CANDIDATES Do not open this examination paper until instructed to do so. Answer
More informationdifferent formulas, depending on whether or not the vector is in two dimensions or three dimensions.
ectors The word ector comes from the Latin word ectus which means carried. It is best to think of a ector as the displacement from an initial point P to a terminal point Q. Such a ector is expressed as
More informationA possible mechanism to explain wave-particle duality L D HOWE No current affiliation PACS Numbers: r, w, k
A possible mechanism to explain wae-particle duality L D HOWE No current affiliation PACS Numbers: 0.50.-r, 03.65.-w, 05.60.-k Abstract The relationship between light speed energy and the kinetic energy
More informationPhysics 111. Lecture 18 (Walker: 8.3-4) Energy Conservation I March 11, Conservation of Mechanical Energy
Physics 111 Lecture 18 (Walker: 8.3-4) Energy Conservation I March 11, 2009 Lecture 18 1/24 Conservation o Mechanical Energy Deinition o mechanical energy: (8-6) I the only work done in going rom the initial
More informationMotion in Two and Three Dimensions
PH 1-1D Spring 013 Motion in Two and Three Dimensions Lectures 5,6,7 Chapter 4 (Halliday/Resnick/Walker, Fundamentals of Physics 9 th edition) 1 Chapter 4 Motion in Two and Three Dimensions In this chapter
More informationConcept Question: Normal Force
Concept Question: Normal Force Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is 1. larger than 2. identical
More informationNEWTONS LAWS OF MOTION AND FRICTIONS STRAIGHT LINES
EWTOS LAWS O OTIO AD RICTIOS STRAIGHT LIES ITRODUCTIO In this chapter, we shall study the motion o bodies along with the causes o their motion assuming that mass is constant. In addition, we are going
More informationConservation of Mechanical Energy 8.01
Conservation o Mechanical Energy 8.01 Non-Conservative Forces Work done on the object by the orce depends on the path taken by the object Example: riction on an object moving on a level surace F riction
More informationPhysics 231 Lecture 9
Physics 31 Lecture 9 Mi Main points o today s lecture: Potential energy: ΔPE = PE PE = mg ( y ) 0 y 0 Conservation o energy E = KE + PE = KE 0 + PE 0 Reading Quiz 3. I you raise an object to a greater
More informationGeneral Lorentz Boost Transformations, Acting on Some Important Physical Quantities
General Lorentz Boost Transformations, Acting on Some Important Physical Quantities We are interested in transforming measurements made in a reference frame O into measurements of the same quantities as
More informationCHAPTER 7 IMPULSE AND MOMENTUM
CHAPTER 7 IMPULSE AND MOMENTUM CONCEPTUAL QUESTIONS 1. REASONING AND SOLUTION The linear oentu p o an object is the product o its ass and its elocity. Since the autoobiles are identical, they hae the sae
More informationStatus: Unit 2, Chapter 3
1 Status: Unit, Chapter 3 Vectors and Scalars Addition of Vectors Graphical Methods Subtraction of Vectors, and Multiplication by a Scalar Adding Vectors by Components Unit Vectors Vector Kinematics Projectile
More informationYour Name: PHYSICS 101 MIDTERM. Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon
1 Your Name: PHYSICS 101 MIDTERM October 26, 2006 2 hours Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon Problem Score 1 /13 2 /20 3 /20 4
More informationRELEASED. Go to next page. 2. The graph shows the acceleration of a car over time.
1. n object is launched across a room. How can a student determine the average horizontal velocity of the object using a meter stick and a calculator? The student can calculate the object s initial potential
More informationPhysics 4A Solutions to Chapter 9 Homework
Physics 4A Solutions to Chapter 9 Homework Chapter 9 Questions:, 10, 1 Exercises & Problems: 3, 19, 33, 46, 51, 59, 86, 90, 100, 104 Answers to Questions: Q 9- (a) ac, cd, bc (b) bc (c) bd, ad Q 9-10 a,
More informationVISUAL PHYSICS ONLINE RECTLINEAR MOTION: UNIFORM ACCELERATION
VISUAL PHYSICS ONLINE RECTLINEAR MOTION: UNIFORM ACCELERATION Predict Obsere Explain Exercise 1 Take an A4 sheet of paper and a heay object (cricket ball, basketball, brick, book, etc). Predict what will
More informationPHY131 Summer 2011 Class 9 Notes 6/14/11
PHY131H1F Summer Class 9 Today: Hooke s Law Elastic Potential Energy Energy in Collisions Work Calories Conservation of Energy Power Dissipative Forces and Thermal Energy Ch.10 Reading Quiz 1 of 3: Two
More informationWelcome back to Physics 211
Welcome back to Physics 211 Today s agenda: Circular motion Impulse and momentum 08-2 1 Current assignments Reading: Chapter 9 in textbook Prelecture due next Thursday HW#8 due NEXT Friday (extension!)
More informationLecture 12! Center of mass! Uniform circular motion!
Lecture 1 Center of mass Uniform circular motion Today s Topics: Center of mass Uniform circular motion Centripetal acceleration and force Banked cures Define the center of mass The center of mass is a
More informationAP Physics C. Momentum. Free Response Problems
AP Physics C Momentum Free Response Problems 1. A bullet of mass m moves at a velocity v 0 and collides with a stationary block of mass M and length L. The bullet emerges from the block with a velocity
More informationPre-AP Physics Chapter 1 Notes Yockers JHS 2008
Pre-AP Physics Chapter 1 Notes Yockers JHS 2008 Standards o Length, Mass, and Time ( - length quantities) - mass - time Derived Quantities: Examples Dimensional Analysis useul to check equations and to
More informationFeb 6, 2013 PHYSICS I Lecture 5
95.141 Feb 6, 213 PHYSICS I Lecture 5 Course website: faculty.uml.edu/pchowdhury/95.141/ www.masteringphysics.com Course: UML95141SPRING213 Lecture Capture h"p://echo36.uml.edu/chowdhury213/physics1spring.html
More informationSimple Harmonic Motion Practice Problems PSI AP Physics B
Simple Harmonic Motion Practice Problems PSI AP Physics B Name Multiple Choice 1. A block with a mass M is attached to a spring with a spring constant k. The block undergoes SHM. Where is the block located
More informationChapter 1: Kinematics of Particles
Chapter 1: Kinematics of Particles 1.1 INTRODUCTION Mechanics the state of rest of motion of bodies subjected to the action of forces Static equilibrium of a body that is either at rest or moes with constant
More informationChapter 9 Centre of Mass and Linear Momentum
Chater 9 Centre o Mass and Linear Moentu Centre o ass o a syste o articles / objects Linear oentu Linear oentu o a syste o articles Newton s nd law or a syste o articles Conseration o oentu Elastic and
More informationDoppler shifts in astronomy
7.4 Doppler shift 253 Diide the transformation (3.4) by as follows: = g 1 bck. (Lorentz transformation) (7.43) Eliminate in the right-hand term with (41) and then inoke (42) to yield = g (1 b cos u). (7.44)
More informationLast Name First Name Date
Last Name irst Name Date 16.1 The Nature of Waes 16.2 Periodic Waes 16.3 The Speed of a Wae in a String Conceptual Questions 1,2,3,7, 8, 11 page 503 Problems 2, 4, 6, 12, 15, 16 page 501-502 Types of Waes
More informationChapter 2 Motion Along a Straight Line
Chapter Motion Along a Straight Line In this chapter we will study how objects moe along a straight line The following parameters will be defined: (1) Displacement () Aerage elocity (3) Aerage speed (4)
More informationReview. acceleration is the rate of change of velocity (how quickly the velocity is changing) For motion in a line. v t
Accelerated Motion Reiew acceleration is the rate o change o elocity (how quickly the elocity is changing) For motion in a line a i t t When an object is moing in a straight line, a positie acceleration
More informationDemonstrating the Quantization of Electrical Charge Millikan s Experiment
Demonstrating the Quantization o Electrical Charge Millikan s Experiment Objecties o the experiment To demonstrate that electrical charge is quantized, and to determine the elementary electron charge by
More informationMOTION OF FALLING OBJECTS WITH RESISTANCE
DOING PHYSICS WIH MALAB MECHANICS MOION OF FALLING OBJECS WIH RESISANCE Ian Cooper School of Physics, Uniersity of Sydney ian.cooper@sydney.edu.au DOWNLOAD DIRECORY FOR MALAB SCRIPS mec_fr_mg_b.m Computation
More informationImpulse and Momentum continued
Chapter 7 Impulse and Momentum continued 7.2 The Principle of Conservation of Linear Momentum External forces Forces exerted on the objects by agents external to the system. Net force changes the velocity
More information11. (7 points: Choose up to 3 answers) What is the tension,!, in the string? a.! = 0.10 N b.! = 0.21 N c.! = 0.29 N d.! = N e.! = 0.
A harmonic wave propagates horizontally along a taut string of length! = 8.0 m and mass! = 0.23 kg. The vertical displacement of the string along its length is given by!!,! = 0.1!m cos 1.5!!! +!0.8!!,
More informationChapter 2: 1D Kinematics Tuesday January 13th
Chapter : D Kinematics Tuesday January 3th Motion in a straight line (D Kinematics) Aerage elocity and aerage speed Instantaneous elocity and speed Acceleration Short summary Constant acceleration a special
More informationPhysics 107 TUTORIAL ASSIGNMENT #7
Physics 07 TUTORIL SSIGNMENT #7 Cutnell & Johnson, 7 th edition Chapter 6: Problems 5, 65, 79, 93 Chapter 7: Problems 7,, 9, 37, 48 Chapter 6 5 Suppose that sound is emitted uniormly in all directions
More informationThere are two types of forces: conservative (gravity, spring force) nonconservative (friction)
Chapter 8: Conservation o Energy There are two types o orces: conservative (gravity, spring orce) nonconservative (riction) Conservative Forces Conservative Force the work done by the orce on an object
More informationPhysics 2A Chapter 3 - Motion in Two Dimensions Fall 2017
These notes are seen pages. A quick summary: Projectile motion is simply horizontal motion at constant elocity with ertical motion at constant acceleration. An object moing in a circular path experiences
More information21.60 Worksheet 8 - preparation problems - question 1:
Dynamics 190 1.60 Worksheet 8 - preparation problems - question 1: A particle of mass m moes under the influence of a conseratie central force F (r) =g(r)r where r = xˆx + yŷ + zẑ and r = x + y + z. A.
More information= v 0 x. / t = 1.75m / s 2.25s = 0.778m / s 2 nd law taking left as positive. net. F x ! F
Multiple choice Problem 1 A 5.-N bos sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the bos sliding to the right at 1.75 m/s
More information(a) On the dots below that represent the students, draw and label free-body diagrams showing the forces on Student A and on Student B.
2003 B1. (15 points) A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on
More informationChapter 8 Conservation of Energy and Potential Energy
Chapter 8 Conservation o Energy and Potential Energy So ar we have analyzed the motion o point-like bodies under the action o orces using Newton s Laws o Motion. We shall now use the Principle o Conservation
More informationWork and Kinetic Energy
Work Work an Kinetic Energy Work (W) the prouct of the force eerte on an object an the istance the object moes in the irection of the force (constant force only). W = " = cos" (N " m = J)! is the angle
More informationChapter 14 Thermal Physics: A Microscopic View
Chapter 14 Thermal Physics: A Microscopic View The main focus of this chapter is the application of some of the basic principles we learned earlier to thermal physics. This will gie us some important insights
More informationThe Magnetic Force. x x x x x x. x x x x x x. x x x x x x q. q F = 0. q F. Phys 122 Lecture 17. Comment: What just happened...?
The Magnetic orce Comment: I LOVE MAGNETISM q = qe + q q Comment: What just happened...? q = 0 Phys 122 Lecture 17 x x x x x x q G. Rybka Magnetic Phenomenon ar magnet... two poles: N and S Like poles
More informationDO PHYSICS ONLINE. WEB activity: Use the web to find out more about: Aristotle, Copernicus, Kepler, Galileo and Newton.
DO PHYSICS ONLINE DISPLACEMENT VELOCITY ACCELERATION The objects that make up space are in motion, we moe, soccer balls moe, the Earth moes, electrons moe, - - -. Motion implies change. The study of the
More informationMultiple choice questions [60 points]
Multiple choice questions [60 points] Answer all of the following questions. Read each question carefully. Fill the correct bubble on your scantron sheet. Each correct answer is worth 4 points. Each question
More informationCHAPTER 4 NEWTON S LAWS OF MOTION
62 CHAPTER 4 NEWTON S LAWS O MOTION CHAPTER 4 NEWTON S LAWS O MOTION 63 Up to now we have described the motion of particles using quantities like displacement, velocity and acceleration. These quantities
More informationPHYSICS FORMULAS. A. B = A x B x + A y B y + A z B z = A B cos (A,B)
PHYSICS FORMULAS A = A x i + A y j Φ = tan 1 A y A x A + B = (A x +B x )i + (A y +B y )j A. B = A x B x + A y B y + A z B z = A B cos (A,B) linear motion v = v 0 + at x - x 0 = v 0 t + ½ at 2 2a(x - x
More informationF = q v B. F = q E + q v B. = q v B F B. F = q vbsinφ. Right Hand Rule. Lorentz. The Magnetic Force. More on Magnetic Force DEMO: 6B-02.
Lorentz = q E + q Right Hand Rule Direction of is perpendicular to plane containing &. If q is positie, has the same sign as x. If q is negatie, has the opposite sign of x. = q = q sinφ is neer parallel
More informationModule 17: Systems, Conservation of Momentum and Center of Mass
Module 17: Systems, Conservation of Momentum and Center of Mass 17.1 External and Internal Forces and the Change in Momentum of a System So far we have restricted ourselves to considering how the momentum
More informationYou are given two carts, A and B. They look identical, and you are told they are made of the same material. You put A at rest on a low-friction
You are given two carts, A and B. They look identical, and you are told they are made of the same material. You put A at rest on a low-friction track, then send B towards it to the right. After the collision,
More information(A) 10 m (B) 20 m (C) 25 m (D) 30 m (E) 40 m
PSI AP Physics C Work and Energy (Algebra Based) Multiple Choice Questions (use g = 10 m/s 2 ) 1. A student throws a ball upwards from the ground level where gravitational potential energy is zero. At
More informationMomentum and Energy. Relativity and Astrophysics Lecture 24 Terry Herter. Energy and Momentum Conservation of energy and momentum
Momentum and Energy Relatiity and Astrohysics Lecture 4 Terry Herter Outline Newtonian Physics Energy and Momentum Conseration of energy and momentum Reading Sacetime Physics: Chater 7 Homework: (due Wed.
More informationPHYS 131 MIDTERM October 31 st, 2008
PHYS 131 MIDTERM October 31 st, 2008 The exam comprises two parts: 8 short-answer questions, and 4 problems. Calculators are allowed, as well as a formula sheet (one-side of an 8½ x 11 sheet) of your own
More informationCenter of Mass & Linear Momentum
PHYS 101 Previous Exam Problems CHAPTER 9 Center of Mass & Linear Momentum Center of mass Momentum of a particle Momentum of a system Impulse Conservation of momentum Elastic collisions Inelastic collisions
More informationMotion in Two and Three Dimensions
PH 1-A Fall 014 Motion in Two and Three Dimensions Lectures 4,5 Chapter 4 (Halliday/Resnick/Walker, Fundamentals of Physics 9 th edition) 1 Chapter 4 Motion in Two and Three Dimensions In this chapter
More informationPSI AP Physics I Work and Energy
PSI AP Physics I Work and Energy Multiple-Choice questions 1. A driver in a 2000 kg Porsche wishes to pass a slow moving school bus on a 4 lane road. What is the average power in watts required to accelerate
More information