Physics
Collisions and Momentum Conservation
In physics, collisions involve the interaction between two or more objects, resulting in a change in their velocities. Momentum conservation states that in the absence of external forces, the total momentum of a system remains constant before and after a collision. This principle is a fundamental concept in understanding the behavior of objects during collisions.
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11 Key excerpts on "Collisions and Momentum Conservation"
eBook - PDF- Paul Peter Urone, Roger Hinrichs(Authors)
- 2012(Publication Date)
- Openstax(Publisher)
• The conservation of momentum principle is valid when considering systems of particles. 8.4 Elastic Collisions in One Dimension • An elastic collision is one that conserves internal kinetic energy. • Conservation of kinetic energy and momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one dimensional two-body collisions. 8.5 Inelastic Collisions in One Dimension • An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). • A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision. • Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations. 8.6 Collisions of Point Masses in Two Dimensions • The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the x -axis parallel to the velocity of the incoming particle. • Two-dimensional collisions of point masses where mass 2 is initially at rest conserve momentum along the initial direction of mass 1 (the x -axis), stated by m 1 v 1 = m 1 v′ 1 cos θ 1 + m 2 v′ 2 cos θ 2 and along the direction perpendicular to the initial direction (the y -axis) stated by 0 = m 1 v′ 1y +m 2 v′ 2y . • The internal kinetic before and after the collision of two objects that have equal masses is Chapter 8 | Linear Momentum and Collisions 307 1 2 mv 1 2 = 1 2 mv′ 1 2 + 1 2 mv′ 2 2 + mv′ 1 v′ 2 cos ⎛ ⎝ θ 1 − θ 2 ⎞ ⎠ . • Point masses are structureless particles that cannot spin. 8.7 Introduction to Rocket Propulsion • Newton’s third law of motion states that to every action, there is an equal and opposite reaction. • Acceleration of a rocket is a = v e m Δm Δt − g . • A rocket’s acceleration depends on three main factors.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
7.3 | Collisions in One Dimension As discussed in the previous section, the total linear momentum is conserved when two objects collide, provided they constitute an isolated system. When the objects are atoms or subatomic particles, the total kinetic energy of the system is often conserved also. In other words, the total kinetic energy of the particles before the collision equals the total kinetic energy of the particles after the collision, so that kinetic energy gained by one particle is lost by another. In contrast, when two macroscopic objects collide, such as two cars, the total kinetic energy after the collision is generally less than that before the collision. Kinetic energy is lost mainly in two ways: (1) It can be converted into heat because of friction, and (2) it is spent in creating permanent distortion or damage, as in an automobile collision. With very hard objects, such as a solid steel ball and a marble floor, the permanent distortion suffered upon collision is much smaller than with softer objects and, consequently, less kinetic energy is lost. Collisions are often classified according to whether the total kinetic energy changes during the collision: 1. Elastic collision: One in which the total kinetic energy of the system after the col- lision is equal to the total kinetic energy before the collision. 2. Inelastic collision: One in which the total kinetic energy of the system is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic. Problem-Solving Insight As long as the net external force is zero, the conservation of linear momentum applies to any type of collision. This is true whether the collision is elastic or inelastic. 7.3 | Collisions in One Dimension 163 The boxcars coupling together in Figure 7.8 provide an example of a completely inelastic collision. When a collision is completely inelastic, the greatest amount of kinetic energy is lost.
eBook - PDF- Michael Tammaro(Author)
- 2019(Publication Date)
- Wiley(Publisher)
Ine- lastic collisions, on the other hand, are those in which the kinetic energy of the system changes. In a completely inelastic collision, the objects do not “rebound” from one another—they stick together and essentially become a single physical object. If the 202 | Chapter 7 colliding objects constitute an isolated system, then linear momentum is conserved. For a one-dimensional collision between two objects, 1 and 2, conservation of momentum reads as + = + m v m v m v m v 1 1i 2 2i 1 1f 2 2f (7.4.2) where i and f denote initial and final, respectively. In a one-dimensional elastic collision, where object 2 is initially at rest, the final velocities are given by v m m m m v v m m m v and 2 1f 1 2 1 2 1i 2f 1 1 2 1i = − + = + (7.4.3) Equations 7.4.3 only apply if the collision is elastic. 7.5 Two-Dimensional Collisions In a two-dimensional collision, the motion of the colliding objects is constrained to be in the same plane both before and after the collision. If we choose an x–y coordinate system to be in the plane of the velocities, then the x and y components of the total initial momentum must be equal to the x and y components of the total final momentum, which is expressed by Equations 7.5.2: m v m v m v m v m v m v m v m v and x x x x 1 1i 2 2i 1 1f 2 2f 1 1iy 2 2iy 1 1fy 2 2fy + = + + = + (7.5.2) 7.6 Center of Mass The center of mass of an object or system of objects is the point that represents the aver- age position of the mass in the system. An object will be balanced if it is supported at its center of mass. Consider two small objects with masses m 1 and m 2 . The coordinates x 1 and x 2 define the position of the centers of the objects.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler, Heath Jones, Matthew Collins, John Daicopoulos, Boris Blankleider(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
180 Physics REASONING STRATEGY Applying the principle of conservation of linear momentum 1. Decide which objects are included in the system. 2. Relative to the system that you have chosen, identify the internal forces and the external forces. 3. Verify that the system is isolated. In other words, verify that the sum of the external forces applied to the system is zero. Only if this sum is zero can the conservation principle be applied. If the sum of the average exter- nal forces is not zero, consider a different system for analysis. 4. Set the total final momentum of the isolated system equal to the total initial momentum. Remember that linear momentum is a vector. If necessary, apply the conservation principle separately to the various vector components. 7.3 Collisions in one dimension LEARNING OBJECTIVE 7.3 Analyse one-dimensional collisions. As discussed in the previous section, the total linear momentum is conserved when two objects collide, provided they constitute an isolated system. When the objects are atoms or subatomic particles, the total kinetic energy of the system is often conserved also. In other words, the total kinetic energy of the particles before the collision equals the total kinetic energy of the particles after the collision, so that kinetic energy gained by one particle is lost by another. In contrast, when two macroscopic objects collide, such as two cars, the total kinetic energy after the collision is generally less than that before the collision. Kinetic energy is lost mainly in two ways: (1) It can be converted into heat because of friction, and (2) it is spent in creating permanent distortion or damage, as in an automobile collision. With very hard objects, such as a solid steel ball and a marble floor, the permanent distortion suffered upon collision is much smaller than with softer objects and, consequently, less kinetic energy is lost.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
When the objects are atoms or subatomic particles, the total kinetic energy of the system is often conserved also. In other words, the total kinetic energy of the particles before the collision equals the total kinetic energy of the particles after the collision, so that kinetic energy gained by one particle is lost by another. In contrast, when two macroscopic objects collide, such as two cars, the total kinetic energy after the collision is generally less than that before the collision. Kinetic energy is lost mainly in two ways: (1) It can be converted into heat because of friction, and (2) it is spent in creating permanent distortion or damage, as in an automobile collision. With very hard objects, such as a solid steel ball and a marble floor, the permanent distortion suffered upon collision is much smaller than with softer objects and, consequently, less kinetic energy is lost. Collisions are often classified according to whether the total kinetic energy changes during the collision: 1. Elastic collision: One in which the total kinetic energy of the system after the col- lision is equal to the total kinetic energy before the collision. 2. Inelastic collision: One in which the total kinetic energy of the system is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic. Problem-Solving Insight As long as the net external force is zero, the conservation of linear momentum applies to any type of collision. This is true whether the collision is elastic or inelastic. 7.3 | Collisions in One Dimension 183 The boxcars coupling together in Figure 7.8 provide an example of a completely inelastic collision. When a collision is completely inelastic, the greatest amount of kinetic energy is lost. Example 7 shows how one particular elastic collision is described using the conserva- tion of linear momentum and the fact that no kinetic energy is lost.
- David Halliday, Jearl Walker, Patrick Keleher, Paul Lasky, John Long, Judith Dawes, Julius Orwa, Ajay Mahato, Peter Huf, Warren Stannard, Amanda Edgar, Liam Lyons, Dipesh Bhattarai(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
If the system is closed and isolated, the total linear momentum of the system must be conserved, which we can write in vector form as p 1i + p 2i = p 1f + p 2f , where subscripts i and f refer to values just before and just after the collision, respectively. Pdf_Folio:155 CHAPTER 9 Centre of mass and linear momentum 155 • If the motion of the bodies is along a single axis, the collision is one dimensional and we can write the equation in terms of velocity components along that axis: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f . • If the bodies stick together, the collision is a completely inelastic collision and the bodies have the same fnal velocity V (because they are stuck together). • The centre of mass of a closed, isolated system of two colliding bodies is not affected by a collision. In particular, the velocity v com of the centre of mass cannot be changed by the collision. In module 9.4, we considered the collision of two particle‐like bodies but focused on only one of the bodies at a time. For the next several modules we switch our focus to the system itself, with the assumption that the system is closed and isolated. In module 9.5, we discussed a rule about such a system: the total linear momentum of the system cannot change because there is no net external force to change it. This is a very powerful rule because it can allow us to determine the results of a collision without knowing the details of the collision (such as how much damage is done). We will also be interested in the total kinetic energy of a system of two colliding bodies. If that total happens to be unchanged by the collision, then the kinetic energy of the system is conserved (it is the same before and after the collision). Such a collision is called an elastic collision. In everyday collisions of common bodies, such as two cars or a ball and a bat, some energy is always transferred from kinetic energy to other forms of energy, such as thermal energy or energy of sound.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
9.6.2 Identify a one-dimensional collision as one where the objects move along a single axis, both before and after the collision. 9.6.3 Apply the conservation of momentum for an isolated one-dimensional collision to relate the initial momenta of the objects to their momenta after the collision. 9.6.4 Identify that in an isolated system, the momentum and velocity of the center of mass are not changed even if the objects collide. Key Ideas ● In an inelastic collision of two bodies, the kinetic energy of the two-body system is not conserved. If the system is closed and isolated, the total linear momen- tum of the system must be conserved, which we can write in vector form as p → 1i + p → 2i = p → 1f + p → 2f , where subscripts i and f refer to values just before and just after the collision, respectively. ● If the motion of the bodies is along a single axis, the collision is one-dimensional and we can write the equation in terms of velocity components along that axis: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f . ● If the bodies stick together, the collision is a completely inelastic collision and the bodies have the same final velocity V (because they are stuck together). ● The center of mass of a closed, isolated system of two colliding bodies is not affected by a collision. In particular, the velocity v → com of the center of mass cannot be changed by the collision. Momentum and Kinetic Energy in Collisions In Module 9.4, we considered the collision of two particle-like bodies but focused on only one of the bodies at a time. For the next several modules we switch our focus to the system itself, with the assumption that the system is closed and isolated. In Module 9.5, we discussed a rule about such a system: The total linear momentum P → of the system cannot change because there is no net external force to change it.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
In an elastic collision, is the kinetic energy of each object the same before and after the collision? 206 CHAPTER 7 Impulse and Momentum 17. Concept Simulation 7.2 at www.wiley.com/college/cutnell illustrates the concepts that are involved in this question. Also review Multiple-Concept Example 8. Suppose two objects collide head on, as in Example 8, where initially object 1 (mass = m 1 ) is moving and object 2 (mass = m 2 ) is stationary. Now assume that they have the same mass, so m 1 = m 2 . Which one of the following statements is true? (a) Both objects have the same velocity (magnitude and direction) after the collision. (b) Object 1 rebounds with one-half its initial speed, while object 2 moves to the right, as in Figure 7.11, with one-half the speed that object 1 had before the collision. (c) Object 1 stops completely, while object 2 acquires the same velocity (magnitude and direction) that object 1 had before the collision. 7.4 Collisions in Two Dimensions The collisions discussed so far have been head-on, or one-dimensional, because the velocities of the objects all point along a single line before and after contact. Collisions often occur, however, in two or three dimensions. Figure 7.14 shows a two-dimensional case in which two balls collide on a horizontal frictionless table. For the system consisting of the two balls, the external forces include the weights of the balls and the corresponding normal forces produced by the table. Since each weight is balanced by a normal force, the sum of the external forces is zero, and the total momentum of the system is conserved, as Equation 7.7b indicates. Momentum is a vector quantity, however, and in two dimensions the x and y components of the total momentum are conserved separately.
eBook - PDF- David Agmon, Paul Gluck;;;(Authors)
- 2009(Publication Date)
- WSPC(Publisher)
282 Classical and Relativistic Mechanics 9.3 Momentum conservation in many-body systems So far we have dealt with momentum conservation for a single particle. We now extend our treatment to a system of many particles: We start with two and then generalize to N particles. Consider an isolated system consisting of two particles on which no 12 * w ~~~~* F n external forces act. The particles may interact n}l among themselves (with gravitational, electromagnetic or other forces). Denote by F2 the force on particle 2 exerted by particle 1, and by F 2 the force on particle 1 exerted by particle 2. By Newton's third law F 1 2 +F 2 1 =0 (9.6) During a short time interval At particle 1 exerts an impulse J X2 = F i2 At on particle 2, causing a momentum change Ap 2 = F l2 At. Similarly, particle 2 exerts an impulse J 2 = F 2X At on particle 1, causing a momentum change Api = F 2i At. The momentum of each particle changes, but the total vector momentum of the two particles remains unchanged, Ap tot =Ap l +Ap 2 =(F l2 +F 2l )At = 0 Equivalently, (Pi+P 2 )f=(Pi+P 2 )i Generalization to a closed system of N particles is straightforward: (9.7) (9.8) (9.9) Ptot=Y,Pk= const k= Note that a system defined as closed with respect to an inertial frame will certainly not be closed relative to an accelerating frame. An observer in the latter will discern an inertial (or d'Alembert) force whose direction is opposite to that of the acceleration. For him momentum will not be conserved in this direction. An impressive example of momentum conservation is a shell at rest in space suddenly exploding into a myriad fragments. The initial momentum being zero, the vector sum of the momenta of all the fragments must vanish at all times subsequent to the explosion. We conclude with a comment about Newton's third law. It is not valid for the electric interaction of charged particles accelerated in an electromagnetic field, F 2 ± -F 2X .
eBook - PDF- William Moebs, Samuel J. Ling, Jeff Sanny(Authors)
- 2016(Publication Date)
- Openstax(Publisher)
Identify a closed system. 2. Write down the equation that represents conservation of momentum in the x-direction, and solve it for the desired quantity. If you are calculating a vector quantity (velocity, usually), this will give you the x-component of the vector. 3. Write down the equation that represents conservation of momentum in the y-direction, and solve. This will give you the y-component of your vector quantity. 4. Assuming you are calculating a vector quantity, use the Pythagorean theorem to calculate its magnitude, using the results of steps 3 and 4. Example 9.14 Traffic Collision A small car of mass 1200 kg traveling east at 60 km/hr collides at an intersection with a truck of mass 3000 kg that is traveling due north at 40 km/hr (Figure 9.23). The two vehicles are locked together. What is the velocity of the combined wreckage? Figure 9.23 A large truck moving north is about to collide with a small car moving east. The final momentum vector has both x- and y-components. Strategy First off, we need a closed system. The natural system to choose is the (car + truck), but this system is not closed; friction from the road acts on both vehicles. We avoid this problem by restricting the question to finding the velocity at the instant just after the collision, so that friction has not yet had any effect on the system. With that 436 Chapter 9 | Linear Momentum and Collisions This OpenStax book is available for free at http://cnx.org/content/col12031/1.5 restriction, momentum is conserved for this system. Since there are two directions involved, we do conservation of momentum twice: once in the x-direction and once in the y-direction. Solution Before the collision the total momentum is p → = m c v → c + m T v → T . After the collision, the wreckage has momentum p → = (m c + m T ) v → w . Since the system is closed, momentum must be conserved, so we have m c v → c + m T v → T = (m c + m T ) v → w . We have to be careful; the two initial momenta are not parallel.
eBook - PDF- Robert Resnick, David Halliday, Kenneth S. Krane(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
According to this observer, the initial momentum of the first spaceship (before the separation) is zero, because the relative velocity of the first ship is zero. After the separation, the total final momentum of the two craft is zero: as you can show. According to this observer, the initial and final momenta are both zero, and thus mo- mentum is conserved. If momentum is conserved in one inertial reference frame, then it is conserved in every inertial reference frame. Often it is convenient to solve a problem in one reference frame and then transform the results to another. In the remainder of this section we discuss how the second reference frame used in this problem, in which the total momentum is zero, can often yield insight into the analysis of collisions. P f x m 1 v 1f x m 2 v 2f x 0, (m)(8.45 km /s) (m/4)(11.63 km /s) 3m/4 7.39 km /s. v 2f x (m 1 m 2 )v i x m 1 v 1f x m 2 (m 1 m 2 )v i x m 1 v 1f x m 2 v 2f x . 6-5 Two-Body Collisions 127 v ix x 3m/4 m/4 m 2 m 1 Figure 6-14. Sample Problem 6-7. Sample Problem 6-8. A puck is sliding without friction on the ice at a speed of 2.48 m/s. It collides with a second puck of mass 1.5 times that of the first and moving initially with a velocity of 1.86 m/s in a direction 40° away from the direction of the first puck (Fig. 6-15). After the collision, the first puck moves at a ve- locity of 1.59 m/s in a direction at an angle of 50° from its initial direction (as shown in Fig. 6-15). Find the speed and direction of the second puck after the collision. Solution In this problem we must use the law of conservation of momentum in its two-dimensional vector form. We define the x axis as the direction of the initial motion of the first puck. Let the second puck move with velocity at an angle with the x axis.
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