- S217_1Collisions and conservation laws
**About this free course**This free course is an adapted extract from the Open University course S217 *Physics: from classical to quantum* www.open.ac.uk/courses/modules/s217.This version of the content may include video, images and interactive content that may not be optimised for your device.You can experience this free course as it was originally designed on OpenLearn, the home of free learning from The Open University – www.open.edu/openlearn/science-maths-technology/collisions-and-conservation-laws/content-section-0 There you’ll also be able to track your progress via your activity record, which you can use to demonstrate your learning.Copyright © 2017 The Open University**Intellectual property**Unless otherwise stated, this resource is released under the terms of the Creative Commons Licence v4.0 http://creativecommons.org/licenses/by-nc-sa/4.0/deed.en_GB . Within that The Open University interprets this licence in the following way: www.open.edu/openlearn/about-openlearn/frequently-asked-questions-on-openlearn . Copyright and rights falling outside the terms of the Creative Commons Licence are retained or controlled by The Open University. Please read the full text before using any of the content.We believe the primary barrier to accessing high-quality educational experiences is cost, which is why we aim to publish as much free content as possible under an open licence. If it proves difficult to release content under our preferred Creative Commons licence (e.g. because we can’t afford or gain the clearances or find suitable alternatives), we will still release the materials for free under a personal end-user licence.This is because the learning experience will always be the same high quality offering and that should always be seen as positive – even if at times the licensing is different to Creative Commons.When using the content you must attribute us (The Open University) (the OU) and any identified author in accordance with the terms of the Creative Commons Licence.The Acknowledgements section is used to list, amongst other things, third party (Proprietary), licensed content which is not subject to Creative Commons licensing. Proprietary content must be used (retained) intact and in context to the content at all times.The Acknowledgements section is also used to bring to your attention any other Special Restrictions which may apply to the content. For example there may be times when the Creative Commons Non-Commercial Sharealike licence does not apply to any of the content even if owned by us (The Open University). In these instances, unless stated otherwise, the content may be used for personal and non-commercial use.We have also identified as Proprietary other material included in the content which is not subject to Creative Commons Licence. These are OU logos, trading names and may extend to certain photographic and video images and sound recordings and any other material as may be brought to your attention.Unauthorised use of any of the content may constitute a breach of the terms and conditions and/or intellectual property laws.We reserve the right to alter, amend or bring to an end any terms and conditions provided here without notice.All rights falling outside the terms of the Creative Commons licence are retained or controlled by The Open University.Head of Intellectual Property, The Open University978 1 4730 2530 1 (.kdl)

978 1 4730 2531 8 (.epub)IntroductionThis free course, *Collisions and conservation laws*, is about collisions and how they may be understood using concepts referred to as the conservation of linear momentum and the conservation of kinetic energy. We’ll begin by defining some important quantities that will be used in what follows. A collision is a brief, but often powerful, interaction between two bodies in close proximity; we often idealise the situation in physics problems to consider collisions of pointlike objects travelling along a line or in a plane. Linear momentum is a physical property of a body in motion which is equivalent to its mass multiplied by its velocity. It is a vector quantity so possesses both a magnitude and a direction, which is the same as the direction of the body’s velocity. The kinetic energy of a body is a measure of the energy it possesses by virtue of its motion. It is a scalar quantity, possessing a magnitude only, which is equivalent to half the body’s mass multiplied by the square of its speed.This OpenLearn course is an adapted extract from the Open University course S217 *Physics: from classical to quantum*.After studying this course, you should be able to: understand the meaning of all the newly defined (emboldened) terms introduced describe the essential features of elastic and inelastic collisions, and give examples of each use the law of conservation of momentum, and (when appropriate) the law of conservation of kinetic energy, to solve a variety of simple collision problems. 1 Analysing collisionsThe analysis of collisions is of fundamental importance in physics, particularly in nuclear and particle physics, and the techniques used to analyse collisions are well established and widely used. They are also very firmly rooted in the basic conservation principles (or ‘conservation laws’ as they are sometimes known), particularly those of momentum and energy.2 Elastic and inelastic collisionsWhen starting to investigate collision problems, we usually consider situations that either start or end with a single body. The reason for this self-imposed limitation is that such problems can be solved by applying momentum conservation alone, namely the result that the total linear momentum of an isolated system is constant. The analysis of more general collisions requires the use of other principles in addition to momentum conservation. To illustrate this, we now consider a one-dimensional problem in which two colliding bodies with known masses m_1 m sub 1 and m_2 m sub 2 , and with known initial velocities u_{1x} u subscript 1 x end and u_{2x} u subscript 2 x end collide and then separate with final velocities v_{1x} v subscript 1 x end and v_{2x} v subscript 2 x end . The problem is that of finding the two unknowns v_{1x} v subscript 1 x end and v_{2x} v subscript 2 x end . Conservation of momentum in the x x -direction provides only one equation linking these two unknowns:m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x}m sub 1 u subscript 1 x end + m sub 2 u subscript 2 x end = m sub 1 v subscript 1 x end + m sub 2 v subscript 2 x end which is insufficient to determine both unknowns.In the absence of any detailed knowledge about the forces involved in the collision, the usual source of an additional relationship between v_{1x} v subscript 1 x end and v_{2x} v subscript 2 x end comes from some consideration of the translational kinetic energy involved. The precise form of this additional relationship depends on the nature of the collision.Collisions may be classified by comparing the total (translational) kinetic energy of the colliding bodies before and after the collision. If there is no change in the total kinetic energy, then the collision is an **elastic collision** . If the kinetic energy after the collision is less than that before the collision then the collision is an **inelastic collision** . In some situations (e.g. where internal potential energy is released) the total kinetic energy may even increase in the collision; in which case the collision is said to be a **superelastic collision** .In the simplest case, when the collision is elastic, the consequent conservation of kinetic energy means that \tfrac12 m_1 u_{1x}^2 + \tfrac12 m_2 u_{2x}^2 = \tfrac12 m_1 v_{1x}^2 + \tfrac12 m_2 v_{2x}^2 fraction 1 over 2 end m sub 1 u squared subscript 1 x end + fraction 1 over 2 end m sub 2 u squared subscript 2 x end = fraction 1 over 2 end m sub 1 v squared subscript 1 x end + fraction 1 over 2 end m sub 2 v squared subscript 2 x end. This equation, together with Equation 1 will allow v_{1x} v subscript 1 x end and v_{2x} v subscript 2 x end to be determined provided the masses and initial velocities have been specified. We consider this situation in more detail in the next section.Real collisions between macroscopic objects are usually inelastic but some collisions, such as those between steel ball bearings or between billiard balls, are very nearly elastic. Collisions between subatomic particles, such as electrons and/or protons, commonly are elastic. The kinetic energy which is lost in an inelastic collision appears as energy of a different form (e.g. thermal energy, sound energy, light energy, etc.), so that the total energy is conserved. Collisions in which the bodies stick together on collision and move off together afterwards, are examples of **completely inelastic collisions** . In these cases the maximum amount of kinetic energy, consistent with momentum conservation, is lost. (Momentum conservation usually implies that the final body or bodies must be moving and this inevitably implies that there must be some final kinetic energy; it is the remainder of the initial kinetic energy, after this final kinetic energy has been subtracted, that is lost in a completely inelastic collision.)3 Elastic collisions in one dimensionIn this section you will examine the outcomes of various elastic collisions in one dimension. These are essentially particular cases of the general elastic collision described by Equation 1 and Equation 2. Cataloguing one-dimensional elastic collisions may sound like a rather esoteric pastime, but as you will see, you have probably witnessed many collisions of this type, and may even have paid handsomely for the privilege.3.1 Elastic collisions with a stationary targetWe begin with an example of a one-dimensional elastic collision between two particles of identical mass, one of which is initially stationary. Our aim now is to find the final velocity of each particle after the collision.Activity 1A particle of mass m m moves along the x x -axis with velocity u_{1x} u subscript 1 x end and collides *elastically* with an identical particle at rest. What are the velocities of the two particles after the collision?Let the final velocities be v_{1x} v subscript 1 x end and v_{2x} v subscript 2 x end . Conservation of momentum along the x x -axis givesm u_{1x} = m v_{1x} + m v_{2x}m u subscript 1 x end = m v subscript 1 x end + m v subscript 2 x end and conservation of kinetic energy for this elastic collision gives \frac{1}{2}m u_{1x}^2 = \frac{1}{2}m v_{1x}^2 + \frac{1}{2}m v_{2x}^2. fraction 1 over 2 end m u squared subscript 1 x end = fraction 1 over 2 end m v squared subscript 1 x end + fraction 1 over 2 end m v squared subscript 2 x end. By eliminating common factors, Equation 3 can be simplified to giveu_{1x} = v_{1x} + v_{2x}u subscript 1 x end = v subscript 1 x end + v subscript 2 x end \text{ i.e. }\quad v_{2x} = u_{1x} - v_{1x}i.e. v subscript 2 x end = u subscript 1 x end minus v subscript 1 x end and Equation 4 can be treated similarly to giveu_{1x}^2 = v_{1x}^2 + v_{2x}^2.u squared subscript 1 x end = v squared subscript 1 x end + v squared subscript 2 x end. Rearranging Equation 6 givesv_{2x}^2 = u_{1x}^2 - v_{1x}^2,v squared subscript 2 x end = u squared subscript 1 x end minus v squared subscript 1 x end comma the right-hand side of which may be rewritten using the general identity a^2 - b^2 = (a - b)(a + b) a squared minus b squared = open bracket a minus b close bracket open bracket a+ b close bracket , thusv_{2x}^2 = (u_{1x} - v_{1x})(u_{1x} + v_{1x}).v squared subscript 2 x end = open bracket u subscript 1 x end minus v subscript 1 x end close bracket open bracket u subscript 1 x end + v subscript 1 x end close bracket. Dividing both sides of this last equation by v_{2x} v subscript 2 x end , v_{2x} = \frac{1}{v_{2x}}(u_{1x} - v_{1x})(u_{1x} + v_{1x}) v subscript 2 x end = fraction 1 over v subscript 2 x end end open bracket u subscript 1 x end minus v subscript 1 x end close bracket open bracket u subscript 1 x end + v subscript 1 x end close bracket and using Equation 5 to simplify the resulting right-hand side givesv_{2x} = u_{1x} + v_{1x}.v subscript 2 x end = u subscript 1 x end + v subscript 1 x end. Comparing this expression for v_{2x} v subscript 2 x end with that in Equation 5 shows that v_{1x} = 0\quad \text{ and thus} \quad v_{2x} = u_{1x}. v subscript 1 x end = 0 and thus v subscript 2 x end = u subscript 1 x end. The result of Example 1 will be familiar to anyone who has seen the head-on collision of two bowls on a bowling green. The moving one stops, and the one that was initially stationary moves off with the original velocity of the first. In effect, the bowls exchange velocities. Activity 2 Predict qualitatively (i.e. without calculation) what would happen when a body of mass m_1 m sub 1 collides with another body of mass m_2 m sub 2 that is initially at rest if:(a) m_1 \gg m_2 m sub 1 gg m sub 2 (The symbol \gg gg should be read as ‘is very much greater than’.)Experience should tell you that a high-mass projectile fired at a low-mass target would be essentially unaffected by the collision.(b) m_2 \gg m_1 m sub 2 gg m sub 1 .A low-mass projectile fired at a massive target would bounce back with unchanged speed.3.2 Elastic collisions in generalWe now consider the general one-dimensional elastic collision between particles of mass m_1 m sub 1 and m_2 m sub 2 which move with initial velocities u_{1x} u subscript 1 x end and u_{2x} u subscript 2 x end before the collision and final velocities v_{1x} v subscript 1 x end and v_{2x} v subscript 2 x end after the collision. As stated earlier, the outcome of collisions of this kind is determined by Equation 1 and Equation 2. We shall not write down the details (though you might like to work them out for yourself) but by arguments similar to those used in Example 1, the following result can be obtained.If the initial velocity of particle 1 1 *relative* to particle 2 2 is taken to beu_{12x} = u_{1x} - u_{2x}u subscript 12 x end = u subscript 1 x end minus u subscript 2 x end and if the final velocity of particle 1 1 *relative* to particle 2 2 is taken to bev_{12x} = v_{1x} - v_{2x},v subscript 12 x end = v subscript 1 x end minus v subscript 2 x end comma then, as a result of an elastic collisionv_{12x} = - u_{12x}.v subscript 12 x end = minus u subscript 12 x end. In other words:In a one-dimensional *elastic* collision between two particles the relative velocity of approach is the negative of the relative velocity of separationv_{1x} - v_{2x} = - (u_{1x} - u_{2x}).v subscript 1 x end minus v subscript 2 x end = minus open bracket u subscript 1 x end minus u subscript 2 x end close bracket. Combining this result (which incorporates the conservation of kinetic energy) with Equation 1 (which expresses conservation of momentum), leads to the following expressions for the final velocities: v_{1x} = \frac{u_{1x} (m_1 - m_2) + 2 m_2 u_{2x}}{m_1 + m_2}\quad (\text{elastic}) v subscript 1 x end = fraction u subscript 1 x end open bracket m sub 1 minus m sub 2 close bracket + 2 m sub 2 u subscript 2 x end over m sub 1 + m sub 2 end open bracket elastic close bracket v_{2x} = \frac{u_{2x} (m_2 - m_1) + 2 m_1 u_{1x}}{m_1 + m_2}\quad (\text{elastic}). v subscript 2 x end = fraction u subscript 2 x end open bracket m sub 2 minus m sub 1 close bracket + 2 m sub 1 u subscript 1 x end over m sub 1 + m sub 2 end open bracket elastic close bracket. In the next two exercises you can use Equations 8 and 9. However, these equations are complicated so *you are not expected to memorise them*. You should be able to solve this type of question starting from the equations of conservation of momentum (Equation 1) and kinetic energy (Equation 2). Activity 3 A neutron of mass mm rebounds elastically in a head-on collision with a gold nucleus of mass 197m 197 m that is initially at rest. What fraction of the neutron’s initial kinetic energy is transferred to the recoiling gold nucleus? Repeat this calculation when the target is a carbon nucleus at rest and of mass 12m 12 m .The initial kinetic energy isE_{\text{trans}} = \frac{1}{2}m u_{1x}^2uppercase E sub trans = fraction 1 over 2 end m u subscript 1 x end squared and the final kinetic energy isE_{\text{trans}} = \frac{1}{2}m v_{1x}^2,uppercase E sub trans = fraction 1 over 2 end m v subscript 1 x end squared comma therefore the loss in energy is \Delta E_{\text{trans}} = \frac{1}{2}m u_{1x}^2 - \frac{1}{2}m v_{1x}^2 = \frac{1}{2}m(u_{1x}^2 - v_{1x}^2) Delta uppercase E sub trans = fraction 1 over 2 end m u subscript 1 x end squared minus fraction 1 over 2 end m v subscript 1 x end squared = fraction 1 over 2 end m open bracket u subscript 1 x end squared minus v subscript 1 x end squared close bracket and the fractional loss is \frac{\Delta E_{\text{trans}}}{E_{\text{trans}}} = \frac{u_{1x}^2 - v_{1x}^2}{u_{1x}^2} = 1 - \left(\frac{v_{1x}^2}{u_{1x}^2}\right). fraction Delta uppercase E sub trans over uppercase E sub trans end = fraction u subscript 1 x end squared minus v subscript 1 x end squared over u subscript 1 x end squared end = 1 minus open bracket fraction v subscript 1 x end squared over u subscript 1 x end squared end close bracket. With u_{2x} = 0 u subscript 2 x end = 0 in Equation 8,v_{1x} = u_{1x} \frac{m_1 - m_2}{m_1 + m_2},v subscript 1 x end = u subscript 1 x end fraction m sub 1 minus m sub 2 over m sub 1 + m sub 2 end comma we can write the fractional loss as \frac{\Delta E_{\text{trans}}}{E_{\text{trans}}} = 1 - \left(\frac{v_{1x}^2}{u_{1x}^2}\right) = 1 - \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2. fraction Delta uppercase E sub trans over uppercase E sub trans end = 1 minus open bracket fraction v subscript 1 x end squared over u subscript 1 x end squared end close bracket = 1 minus open bracket fraction m sub 1 minus m sub 2 over m sub 1 + m sub 2 end close bracket squared. For gold \frac{\Delta E_{\text{trans}}}{E_{\text{trans}}} = 1 - \left(\frac{196}{198}\right)^2 = 0.02,\quad \text{ i.e. } 2.0%. fraction Delta uppercase E sub trans over uppercase E sub trans end = 1 minus open bracket fraction 196 over 198 end close bracket squared = 0.02 comma mathrm. For carbon \frac{\Delta E_{\text{trans}}}{E_{\text{trans}}} = 1 - \left(\frac{11}{13}\right)^2 = 0.28,\quad \text{ i.e. } 28%. fraction Delta uppercase E sub trans over uppercase E sub trans end = 1 minus open bracket fraction 11 over 13 end close bracket squared = 0.28 comma mathrm. So, a low-mass nucleus is much more effective than a more massive nucleus when it comes to slowing down fast neutrons by elastic collisions. It is because of this fact that carbon is used in a nuclear reactor for just this purpose. Activity 4 A tennis player returns a service in the direction of the server. The ball of mass 50 \, \mathrm{g} 50 mathrm arrives at the racket of mass 350\, \mathrm{g} 350 mathrm with a speed of 45\, \mathrm{m}\, \mathrm{s}^{-1} 45 mathrm mathrm superscript minus 1 end and the racket is travelling at 10\, \mathrm{m}\, \mathrm{s}^{-1} 10 mathrm mathrm superscript minus 1 end at impact. Calculate the velocity of the returning ball, assuming elastic conditions.We designate the ball as particle 1 1 and the racket as particle 2 2 , with the ball initially travelling along the positive x x -direction. From Equation 8 v_{1x} = \frac{u_{1x} (m_1 - m_2) + 2 m_2 u_{2x}}{(m_1 + m_2)} v subscript 1 x end = fraction u subscript 1 x end open bracket m sub 1 minus m sub 2 close bracket + 2 m sub 2 u subscript 2 x end over open bracket m sub 1 + m sub 2 close bracket end so v_{1x} = \frac{[45\, \mathrm{m}\, \mathrm{s}^{-1} \times (- 0.30\, \mathrm{kg}\,) + 2 \times 0.350\, \mathrm{kg}\, \times (- 10\, \mathrm{m}\, \mathrm{s}^{-1})]}{0.400\, \mathrm{kg}\, } v subscript 1 x end = fraction [45 mathrm mathrm superscript minus 1 end times open bracket minus 0.30 mathrm close bracket + 2 times 0.350 mathrm times open bracket minus 10 mathrm mathrm superscript minus 1 end close bracket ] over 0.400 mathrm end i.e. v_{1x} = \frac{(- 13.5\, \mathrm{kg}\, \mathrm{m}\, \mathrm{s}^{-1} - 7.0\, \mathrm{kg}\, \mathrm{m}\, \mathrm{s}^{-1})}{0.4\, \mathrm{kg}\, } = - 51\, \mathrm{m}\, \mathrm{s}^{-1}. v subscript 1 x end = fraction open bracket minus 13.5 mathrm mathrm mathrm superscript minus 1 end minus 7.0 mathrm mathrm mathrm superscript minus 1 end close bracket over 0.4 mathrm end = minus 51 mathrm mathrm superscript minus 1 end. 3.3 Four special cases of general elastic collisions It is interesting to examine these results for v_{1x} v subscript 1 x end and v_{2x} v subscript 2 x end in a few special cases, including some that have been mentioned earlier. The cases are illustrated in Animation 1, and have many familiar sporting applications.Animation 1 Animation showing body 1 1 with mass m_1 m sub 1 and initial speed u_1 u sub 1 moving in one dimension and colliding with body 2 2 with mass m_2 m sub 2 and speed u_2 u sub 2 initially moving in the opposite direction. The controls in the animation allow you to change the values of m_1 m sub 1 , m_2 m sub 2 , u_1 u sub 1 and u_2 u sub 2 . Press the play button to see what happens during the elastic collision. Initially body 1 1 is moving along a straight line from left to right towards body 2 2 moving in the opposite direction. Once the bodies collide, several things can happen depending on the values of m_1 m sub 1 , m_2 m sub 2 and u_1 u sub 1 : both bodies may move towards the right, body 1 1 may start to move backwards, with body 2 2 either stationary or moving towards the right, or both bodies may stand still. Activity 5 Using Animation 1, explore what happens in the cases listed below and find, using Equation 8 and Equation 9, expressions for v_{1x} v subscript 1 x end and v_{2x} v subscript 2 x end :(a) m_1 = m_2 m sub 1 = m sub 2 With the masses equal, we can see that if particle 1 1 is moving slowly and approaching particle 2, 2 comma which is moving quickly, then after the collision, the particles rebound but now particle 1 1 is moving quickly and particle 2 2 is moving slowly. In fact, Equation 8 and Equation 9 give v_{1x} = u_{2x} v subscript 1 x end = u subscript 2 x end and v_{2x} = u_{1x} v subscript 2 x end = u subscript 1 x end so the velocities of the particles are simply exchanged.(b) m_1 \gg m_2 \text{ and } u_{2x} = 0 m sub 1 gg m sub 2 and u subscript 2 x end = 0 The motion of the high-mass particle is virtually unchanged by the collision. If m_2 m sub 2 is very small compared with m_1 m sub 1 , then m_1 + m_2 \approx m_1 m sub 1 + m sub 2 approx m sub 1 and m_1 - m_2 \approx m_1 m sub 1 minus m sub 2 approx m sub 1 . With u_{2x} = 0 u subscript 2 x end = 0 , Equation 8 and Equation 9 then give v_{1x} \approx u_{1x} v subscript 1 x end approx u subscript 1 x end and v_{2x} \approx 2 u_{1x} v subscript 2 x end approx 2 u subscript 1 x end . The low-mass particle moves off with a velocity of *twice* that of the high-mass particle.(c) m_2 \gg m_1 \text{ and } u_{2x} = 0 m sub 2 gg m sub 1 and u subscript 2 x end = 0 The low-mass particle rebounds with almost unchanged speed while the high-mass particle remains essentially at rest. If m_1 m sub 1 is very small compared with m_2 m sub 2 (which is stationary), then Equation 8 and Equation 9 lead to v_{1x} \approx - u_{1x} v subscript 1 x end approx minus u subscript 1 x end and v_{2x} \approx 0 v subscript 2 x end approx 0 .(d) m_2 \gg m_1 \text{ and } u_{2x} \approx - u_{1x} m sub 2 gg m sub 1 and u subscript 2 x end approx minus u subscript 1 x end The low-mass particle bounces back with higher speed, while the high-mass particle continues essentially unaffected by the collision. If m_1 m sub 1 is negligible compared with m_2 m sub 2 , and the two bodies approach head-on with equal speeds then Equation 8 and Equation 9 lead to v_{1x} \approx - 3 u_{1x} v subscript 1 x end approx minus 3 u subscript 1 x end and v_{2x} \approx u_{2x} v subscript 2 x end approx u subscript 2 x end .The results for the four special cases in the exercises accord with common experience. Let’s summarise them: m_1 = m_2 m sub 1 = m sub 2 : the particles simply exchange velocities. We saw this result earlier in this section; it is a familiar occurrence in bowls and snooker. m_1 \gg m_2 m sub 1 gg m sub 2 ; u_{2x} = 0 u subscript 2 x end = 0 : the low-mass particle moves off with a velocity of *twice* that of the high-mass particle. Tennis players will be familiar with this case from serving. m_2 \gg m_1 m sub 2 gg m sub 1 ; u_{2x} = 0 u subscript 2 x end = 0 : the low-mass particle rebounds with almost unchanged speed while the high-mass particle remains essentially at rest. Golfers whose ball hits a tree will recognise this situation. m_2 \gg m_1 m sub 2 gg m sub 1 ; u_{2x} \approx - u_{1x} u subscript 2 x end approx minus u subscript 1 x end : the low-mass particle bounces back with *three* times the initial speed, while the high-mass particle continues essentially unaffected by the collision. This case will be recognised by a batsman playing cricket or by a tennis player returning a serve.The results quoted above under points 2, 3 and 4 give an upper limit to the speed that can be imparted to a ball hit by a club, bat or racket. Activity 6 The ‘Newton’s cradle’ executive toy shown below performs repeated collisions between one ball and a row of four identical balls. You may assume the collisions are perfectly elastic.(a) Explain how this toy fits into the framework of four special cases enumerated above.While at first glance, we have the situation m_2 = 4 m_1 m sub 2 = 4 m sub 1 ; u_{2x} = 0 u subscript 2 x end = 0 , most closely corresponding to point 3 above, the behaviour of the toy is not as predicted by point 3. In fact, the toy executes a rapid series of four repeated collisions corresponding to point 1 above: m_1 = m_2 m sub 1 = m sub 2 with the case u_{2x} = 0 u subscript 2 x end = 0 . The second ball moves instantaneously with velocity v_{2x} = u_{1x} v subscript 2 x end = u subscript 1 x end , colliding immediately with the third ball, which moves instantaneously with velocity v_{3x} = u_{1x} v subscript 3 x end = u subscript 1 x end , and so on until the fifth ball carries the momentum away with velocity v_{5x} = u_{1x} v subscript 5 x end = u subscript 1 x end .(b) Gravitational potential energy is energy possessed by an object by virtue of its height – objects further from the surface of the Earth will have a greater gravitational potential energy. As an object falls, its gravitational potential energy will be converted into kinetic energy, and as an object rises, its kinetic energy will be converted into gravitational potential energy. What role does the exchange of gravitational potential energy and kinetic energy play in the dynamics of the first and fifth balls?After the four collisions, the fifth ball moves in a pendulum-like trajectory until its kinetic energy is converted to gravitational potential energy. It comes instantaneously to rest, then moves back to collide with the fourth ball with velocity v_{5x} = - u_{1x} v subscript 5 x end = minus u subscript 1 x end (neglecting air resistance encountered during the motion). The same thing happens when the first ball is set in motion by the second ball, it begins to move as a pendulum with velocity v_{1x} = - u_{1x} v subscript 1 x end = minus u subscript 1 x end . This kinetic energy is again converted to gravitational energy until the first ball comes instantaneously to rest, then moves back to collide with the second ball with velocity u_{1x} u subscript 1 x end , repeating the cycle.4 Elastic collisions in two or three dimensionsThe laws of conservation of momentum and energy that we used to analyse elastic collisions in one dimension are also used to analyse elastic collisions in two or three dimensions. We simply treat the motions in each dimension as independent, and apply conservation of momentum separately along each Cartesian coordinate axis. Kinetic energy conservation continues to provide one additional equation relating the squares of the particle speeds. Since we have been careful to use vector notation throughout, this extension to two or three dimensions is easily made.Consider the elastic collision of two identical bodies of mass m m , one at rest and the other approaching with velocity \boldsymbol{u}_1 bold u sub 1 . The particles are no longer confined to move in one dimension, so our x x -component equation (Equation 1), embodying conservation of momentum, becomes a full vector equation: m \boldsymbol{u}_1 = m \boldsymbol{v}_1 + m \boldsymbol{v}_2. m bold u sub 1 = m bold v sub 1 + m bold v sub 2.The law of conservation of energy (Equation 2) does not change, so \tfrac12 m u_1^2 = \tfrac12 m v_1^2 + \tfrac12 m v_2^2 fraction 1 over 2 end m u sub 1 squared = fraction 1 over 2 end m v sub 1 squared + fraction 1 over 2 end m v sub 2 squared. These can be simplified to: \boldsymbol{u}_1 = \boldsymbol{v}_1 + \boldsymbol{v}_2 bold u sub 1 = bold v sub 1 + bold v sub 2andu_1^2 = v_1^2 + v_2^2.u sub 1 squared = v sub 1 squared + v sub 2 squared. These equations are most easily interpreted by a diagram. Figure 2 shows how the three vectors \boldsymbol{u}_1 bold u sub 1 , \boldsymbol{v}_1 bold v sub 1 and \boldsymbol{v}_2 bold v sub 2 are related to one another. Equation 10 tells us that * all three velocity vectors must lie in a single plane *, and that they must form a closed triangle. Equation 11 tells us that the triangle must be a *right-angled triangle*, since its sides obey Pythagoras’ theorem. The implication of this is striking, it means that the angle between \boldsymbol{v}_1 bold v sub 1 and \boldsymbol{v}_2 bold v sub 2 must be 90^{\circ} 90 super degrees .Following the elastic collision of two identical particles, one of which is initially at rest, the final velocities of the two particles will be at right-angles.This is a simplifying feature of equal-mass collisions in two or three dimensions, analogous to the simple result of the exchange of velocities, which we found in one dimension.You may have noticed that this result does not tell us exactly where the bodies go after the collision. *Any* pair of final velocities which can be represented by Figure 2 will be equally satisfactory, and there are an infinite number of these. The reason for this is that we have said nothing about the shape or size of the bodies, or just how they collide. We usually need to have additional information of this kind if we are to determine unique final velocities in such cases. Figure 3 shows the outcome of a particular collision in which spherical bodies make contact at a specific point. The location of this point is the sort of additional information required to determine unique values for v_1 v sub 1 and v_2 v sub 2 . Activity 7 For the case illustrated in Figure 2 (two bodies of equal mass, one of which is initially at rest), if the moving body has an initial speed of 10\, \mathrm{m}\, \mathrm{s}^{-1} 10 mathrm mathrm superscript minus 1 end , and is deflected through 20\degree 20 degrees in the collision, find the magnitudes and directions of the velocities \boldsymbol{v}_1 bold v sub 1 and \boldsymbol{v}_2 bold v sub 2 .We draw a vector triangle like the one shown in Figure 2bWe can now see that v_1 = 10\cos 20\degree\, \mathrm{m}\, \mathrm{s}^{-1} = 9.4\, \mathrm{m} \, \mathrm{s}^{-1} v sub 1 =10 cosine 20 degrees mathrm mathrm superscript minus 1 end = 9.4 mathrm mathrm superscript minus 1 end and is at the given angle of 20\degree 20 degrees to the x x -axis; \boldsymbol{v}_2 bold v sub 2 has a magnitude v_2 = 10\sin 20\degree \, \mathrm{m}\, \mathrm{s}^{-1} = 3.4\, \mathrm{m}\, \mathrm{s}^{-1} v sub 2 = 10 sine 20 degrees mathrm mathrm superscript minus 1 end = 3.4 mathrm mathrm superscript minus 1 end and must be at 70\degree 70 degrees to the x x -axis so that the two angles add up to 90\degree 90 degrees . Activity 8 In the same situation (Figure 2), if, instead of the outcome specified in Activity 7, the speed of the moving body is reduced from 10.0\, \mathrm{m}\, \mathrm{s}^{-1} 10.0 mathrm mathrm superscript minus 1 end to 6.0\, \mathrm{m}\, \mathrm{s}^{-1} 6.0 mathrm mathrm superscript minus 1 end by the collision, find the final velocities.Using the triangle in Figure 2 Pythagoras’ theorem tells us that v_1^2 + v_2^2 = u_1^2 v sub 1 squared + v sub 2 squared = u sub 1 squared , so v_2^2 = u_1^2 - v_1^2 = ({10.0}^2 - {6.0}^2)\, \mathrm{m}^2 \, \mathrm{s}^{-2} = 64\, \mathrm{m}^2 \, \mathrm{s}^{-2} v sub 2 squared = u sub 1 squared minus v sub 1 squared = open bracket 10.0 squared minus 6.0 squared close bracket mathrm squared mathrm superscript minus 2 end = 64 mathrm squared mathrm superscript minus 2 end so that v_2 = 8.0\, \mathrm{m}\, \mathrm{s}^{-1} v sub 2 = 8.0 mathrm mathrm superscript minus 1 end .Now \boldsymbol{v}_1 bold v sub 1 is at an angle \arccos(6.0/10.0) = 53\degree arccos open bracket 6.0/10.0 close bracket = 53 super degrees and \boldsymbol{v}_2 bold v sub 2 at an angle \arccos(8.0 / 10.0) = 37\degree arccos open bracket 8.0/10.0 close bracket = 37 degrees to the x x -axis. You will observe that the two angles add up to 90\degree 90 degrees , as they should.When the masses of the two colliding particles are unequal the algebraic manipulations required to solve elastic collision problems become rather complicated, but no new physics is involved in the solution so we will not pursue such problems here.5 Inelastic collisionsWe now extend our discussion to include inelastic cases, where the total kinetic energy changes during the one-dimensional collision.First we consider the case where the two particles stick together on impact; this is an example of a completely inelastic collision, which occurs with the maximum loss of kinetic energy consistent with conservation of momentum. As a simple example, suppose we have two bodies of equal mass, with one initially at rest. If the initial velocity of the other is u_x u sub x and the initial momentum is mu_x m u sub x , the final momentum must be the same so, since the mass has been doubled, the final velocity is u_x/2 u sub x /2 and the final kinetic energy is therefore \frac{1}{2} \times 2m {(\frac{u_x}{2})}^2 = \frac{m u_x^2}{4}. fraction 1 over 2 end times 2 m open bracket fraction u sub x over 2 end close bracket squared = fraction m u sub x squared over 4 end. Since the initial kinetic energy was twice as large as this, it follows that half the original kinetic energy has been lost (mainly as thermal energy), during the collision.For the more general case where the colliding masses are unequal, but they stick together at collision, we still have v_{1x} = v_{2x} = v_x v subscript 1 x end = v subscript 2 x end = v sub x and so momentum conservation implies thatm_1 u_{1x} + m_2 u_{2x} = (m_1 + m_2) v_xm sub 1 u subscript 1 x end + m sub 2 u subscript 2 x end = open bracket m sub 1 + m sub 2 close bracket v sub x and provides a full solution of the problem (a value for v_x v sub x ), without recourse to energy.To complete the picture, let us mention the general case where two particles collide but where the transfer of kinetic energy into other forms is less than that for the completely inelastic case. This problem has no general solution without more information, such as the fraction of kinetic energy converted. Such problems have solutions which lie between those for the two extremes of elastic and completely inelastic collisions but they must be tackled on an individual basis, using the general principles of conservation of momentum and energy.You will see that in all these calculations we have not needed to invoke the rather complicated forces involved in the interaction of the two particles, but rather have been able to solve the problems using only the principles of conservation of momentum and energy. This is a great simplification and illustrates the power of using conservation principles whenever possible.6 Collisions all around usCollisions are occurring around us all the time and on all size scales, from the very largest to the very smallest. In the following sections, you will listen to and watch various audios and videos which describe how these collisions may be understood.6.1 Collisions in spaceCollisions in space are a frequent occurrence. In the following audio, we describe how such collisions may be understood in terms of the concepts introduced in this course.Audio 1 Collisions in spaceOn an astronomical scale, entire galaxies can collide and merge. Our own galaxy, the Milky Way, in which the Sun is just one of about 10^{11} 10 super 11 stars, has absorbed a number of small companion galaxies. One such ‘victim’, a dwarf galaxy in Sagittarius, was discovered in the late 1990s.A large body of scientific evidence now exists that supports the idea that a major asteroid or comet impact occurred in the Caribbean region at the boundary of the Cretaceous and Tertiary periods in Earth’s geological history (about 65 million years ago). Such an impact is suspected of being responsible for the mass extinction of many species of plants and animals, including the large dinosaurs. The correctness of this hypothesis is still not certain, but the possibility of such a collision is very real; it is becoming increasingly clear that the Earth orbits the Sun in a sort of cosmic shooting gallery. The collision of the fragments of comet Shoemaker–Levy 9 with the planet Jupiter in 1994 was just one of the many side-shows in this gallery. The most dangerous asteroids and comets, those capable of causing major regional or global disasters, are extremely rare, impacting on the Earth perhaps once every 250 000 years or so. Nevertheless, a great deal of media and scientific attention has been focused on strategies of defence. These include the use of nuclear devices and high-speed collisions to deflect or fragment such an object heading for Earth. Calculations using the principles we have discussed in this course have shown that high-speed *inelastic* collisions, at 20000km\, h^{-1} 20000 mathrm mathrm superscript minus 1 end or so, of 15m 15 mathrm wide projectiles with a kilometre-wide asteroid would only change the speed of the main body of the asteroid by about 1km\, h^{-1} 1 mathrm mathrm superscript minus 1 end . Such collisions would have to take place years before impact in order for enough deflection to take place to avert catastrophe.6.2 Collisions on the roadsThe growing number of cars on today’s roads makes it increasingly likely that each driver will be involved in at least one collision during their lifetime, making this a matter of personal interest for all of us (Video 1).Video 1 The physics of colliding cars.ROBERT LLEWELLYN: All collisions obey the rules of nature. If police detectives know these laws, they can piece together what happened in a crash from the bits that are left. It’s the same whether it’s cars or atoms colliding. While the subatomic world may be strange and complex, scientists can always rely on some fundamental laws of collisions. Laws that apply from the mysterious atom to the world of the all too familiar.ADRIAN HOBBS: More people are killed in road accidents than have been killed in all the wars and other types of accidents put together. We’re having in excess of one road accident per second worldwide. In this country, we have something like three and a half thousand people killed every year. It’s absolutely imperative that, in trying to understand what’s going on, that we do understand the physics. Just saying, well, I have a car, and it crushes the front, and not understanding why doesn’t tell me how I change things. So it is fundamental to all our work that we have to understand the laws of physics.JON NEADES: Accident investigation is looking at the physical evidence that’s been left at the scene of a collision and establishing what has actually happened in the collision itself.Okay, let’s have a look at the marks that we’ve got.I’m Jon Neades, and I’m an ex-police officer. And now I teach accident investigation to police officers. You can see the black marks that have been produced as the tyre has run over the surface of the road. What that black mark is, certainly over this portion, a mark...What is actually happening in the collision, what’s happened to the various objects, why we have marks on the road surface, why a vehicle behaves in a particular way. And it’s all based on the laws of physics.ROBERT LLEWELLYN: In a skid, you may be out of control. But the laws of physics aren’t. Behind the chaos and confusion of a crash is the order and certainty of nature. If you understand these rules, you can work backwards and discover just how the metal got mangled.It’s over a hundred years since the first person was killed in a road accident in Britain. This car may look lovely but it’s also deadly. Both car and driver would be written off in a crash.ADRIAN HOBBS: I think that if we look at what’s happened in the whole development of the cars, we can say that by understanding how energy is absorbed, this has enabled us to move forward so that, in the unhappy event that you have an accident, then there’s a much greater chance that you’ll be protected in that accident. And without our understanding of physics, we would never have got to that point.If we’re looking at energy, we have energy. We have kinetic energy if anything’s moving. Of course, when two cars crash into each other, they have a lot of kinetic energy. And we have to absorb that kinetic energy in the front structure of the car. If we don’t absorb it in the front structure of the car, it will have to be absorbed somewhere. And it’ll be absorbed by collapse of the passenger compartment. So we want to have the softest front structure that you can have but would absorb sufficient impact energy.If you could design a car that had its front end, that would be like a spring, so that when cars collided, they crushed and then recovered. The problem there is the cars would then bounce back. And so the change in velocity on the cars would be much greater. There’s no point in stopping somebody and then saying, I’m now going to accelerate you backwards and increase your injuries. Now, you can’t do that perfectly. Cars will recover. But the ideal is cars which collapse and stay collapsed.ROBERT LLEWELLYN: So car designers have a stark choice. In a crash, the energy either deforms the car or the people. Energy always has to go somewhere. It can’t just disappear. That’s what crumple zones are about. They allow the energy to go into bending metal. The front of the car crumples so the people don’t.The total amount of energy in the Universe always remains the same. It’s conserved. Nature’s law on the conservation of energy is never broken. Scientists rely on it absolutely. But energy isn’t the only thing that obeys the law of conservation.Motion is the key to understanding crashes. If my toy car were travelling at the same velocity as a real car, the real car would do a lot more damage. That’s because it’s got more momentum. Now momentum is the velocity of a vehicle multiplied by its mass. If two vehicles are travelling at the same velocity, the one with more mass has more momentum.ADRIAN HOBBS: When we consider two cars hitting each other, then the first thing we have to consider is the momentum. So if you have one car that is twice the mass of the other car, then the momentum that that car has is its mass times its velocity. Both the cars are going at the same velocity. Then one car has twice as much momentum as the other car.If you were to imagine something like a Mini hitting a truck - a very extreme example - you can see the very simple situation is if a truck is coming along at 30 miles an hour into a Mini at 30 miles an hour, the Mini will basically be going back at nearly 30 miles an hour. So it’s had a change in velocity of something like 60 miles an hour. And the truck will be going along still at almost 30 miles an hour. So it’s had a velocity change of virtually nothing.JON NEADES: Well, momentum is one of those fundamental features that’s always conserved in every collision. The momentum is going to remain constant in a closed system. And we use that to establish, with two vehicles colliding, whatever momentum they had before the collision, exactly the same amount of momentum will come out of the collision at the other end. In fact, we use it in reverse. We know what happened after the accident, and we use that to predict what has happened and what the speeds of the vehicles are before the accident.ROBERT LLEWELLYN: Just like energy, momentum is conserved. It stays the same. In a collision, momentum has to go somewhere. You can depend on that.The police certainly do, as they attempt to work out the cause of the accident, and who’s to blame from the debris that’s left. Whether it’s on tarmac or green baize, behind every collision are the laws of conservation of momentum and energy. It’s these rules that let us predict how things behave during an impact.6.3 Collisions on a small-scaleIn case you think it’s a long time since you personally were involved in a collision, you should be aware that even the air that you breathe has its properties regulated by the innumerable collisions that occur every second between the molecules in the air. The air pressure that helps to keep your lungs inflated and enables you to breathe is a result of the rate at which momentum is transferred between the molecules in the air and lung tissue.Collisions continue to be of importance in nuclear physics, but they are even more significant in subnuclear physics. Collision experiments, usually at very high energies, are almost synonymous with the experimental investigation of elementary particles such as protons, and their supposedly fundamental constituents, the quarks and gluons. These investigations are carried out with the aid of purpose-built particle accelerators, such as the ones at The European Centre for Particle Physics (CERN) (Video 2) or Brookhaven National Laboratory in the USA. Sophisticated detectors allow the energies and momenta of the emerging particles to be measured, aiding the identification of the particles and the analysis of their behaviour. The results give an indication of the underlying structure of the colliding particles, and have revealed the existence of forms of matter that would still be unknown and possibly even unsuspected were it not for collision experiments. The most recent of these results is the observation of a particle consistent with the Higgs boson announced by CERN in July 2012.Video 2 A look inside CERN.BRIAN COX (VOICEOVER): Every civilisation has its own creation story. The ancient Chinese, Indian mystics and Christian theologians all place a divine creator at the heart of their creation stories. Science, too, has an elaborate story that describes the Universe’s genesis. It tells us how the fundamental constituents of the cosmos took on their form.The difference with this story is that we can test it. We can find out if it’s true by tearing matter apart and looking at the pieces. All you need is a machine powerful enough to restage the first moments after creation.In the beginning, there was nothing. No space, no time, just endless nothing. Then, 13.7 billion years ago, from nothing... came everything. The Universe exploded into existence. From that fireball of energy emerged the simplest building blocks of matter. Finding experimental evidence of these fundamental entities has become the holy grail of physics.PROFESSOR ALVARO DE RUJULA: Well, the Universe is an object that is not stable. It is expanding and cooling. It’s doing things. And it was therefore different in the past and it will be different in the future. It has a history. It has a life. It has an evolution.BRIAN COX (VOICEOVER): As the early Universe grew, its mysterious primeval constituents transformed themselves into atoms, then molecules and, eventually, stars and planets. Now, billions of years on from the big bang, the Universe is so complex that all traces of the enigmatic building blocks are lost.PROFESSOR ALVARO DE RUJULA: Understanding the evolution of the Universe requires understanding what it is made of. As it turns out, most of that of which the Universe is made are things that we do not understand at all.BRIAN COX (VOICEOVER): But we hope that the LHC is about to bridge this profound gap in our knowledge, by peering further back in time than ever before. The LHC is truly colossal. Its accelerator ring is 27 kilometres long - big enough to encircle a small city. And around it, we’ve built four enormous experiments that will investigate the big bang in exquisite new detail.BRIAN COX: This is my experiment, the experiment that I work on - ATLAS. And what you can see is just the surface buildings. The experiment is actually a hundred metres below the ground, which is where the LHC is. And, basically, this is just a building that covers cranes, where we winch everything down.And this is pretty much the last time that not only TV crews but me and the people that built it will be able to go down. Because, once it starts operating, the whole area becomes a radiation area. It becomes mildly radioactive.You’ve always got to be worried when you see those things. One of the most expensive bits, if not the most expensive, bit of ATLAS actually, was digging the cavern. We even have iris scanners. So, a little bit of science fiction.BRIAN COX (VOICEOVER): It’s down here, in caverns brimming with the latest technology, that the big bangs will be made.BRIAN COX: We just take little bits of matter, little bits of this stuff, and accelerate them to as close to the speed of light as we can get, and then smash them together, right in the middle of that detector, to re-create the conditions that were present back at the beginning of time.BRIAN COX (VOICEOVER): The bits of matter we’re going to fire around the LHC are called protons. They come from a family of particles that give the collider its name - the hadrons.BRIAN COX: Protons are going to fly around here, so close to the speed of light that they go round this 27-kilometre tunnel 11 000 times a second.BRIAN COX (VOICEOVER): The ring has two barrels that will shoot beams of protons around in opposite directions. When they collide, they’ll have the energy equivalent to an aircraft carrier steaming at 30 knots. All this energy will be focused into a space just a fraction of the width of a human hair. The resulting explosion will be so intense that no one’s quite sure what will happen.BRIAN COX: This machine really is a leap into the unknown. I mean, it’s often said with scientific experiments, but I think, in this case, it’s absolutely right. We’re a step - something like a factor of ten in energy. So it’s a huge jump up in energy. It’s a huge jump up in the number of times we can smash particles together per second. It collides protons together so often that your chances of seeing something incredibly interesting and profound are increased way beyond anything that we’ve found before. And I can think of no better place to be, actually, at the moment. This is exciting.BRIAN COX (VOICEOVER): The dream of understanding the building blocks from which the Universe is constructed has inspired the greatest minds for over two millennia.7 Relativistic collisionsThe high-energy collision experiments carried out at CERN, Brookhaven National Laboratory and other such facilities, involve particles that travel at speeds close to that of light. Under such circumstances the definitions of momentum and translational kinetic energy, that play such an important role in Newtonian mechanics, reveal certain shortcomings. It is still the case that translational kinetic energy is conserved in an elastic collision and that the momentum of an isolated system is always conserved, but the Newtonian expressions \boldsymbol{p} = m\boldsymbol{v}\quad and\quad E_{\text{trans}} = \frac{1}{2}m v^2 bold p = m bold v and uppercase E sub trans = fraction 1 over 2 end m v squared are now recognised as approximations, valid only at low speeds (i.e. at speeds much less than the speed of light), to more complicated expressions that work at any speed, up to the speed of light. The breakthrough that led to this realisation was the development of Einstein’s special theory of relativity in 1905. Here we shall quote a few of its well-established results concerning momentum and energy.7.1 Relativistic momentumAccording to Einstein’s theory the **relativistic momentum** of a particle with mass m m and velocity \boldsymbol{v} bold v is given byp = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}}bold p = fraction m bold v over square root 1 minus fraction v squared over c squared end end end where v v is the speed of the particle and c c is the speed of light in a vacuum. The speed of light in a vacuum, c = 3.0 \times {10}^8\, \mathrm{m}\, \mathrm{s}^{-1} c = 3.0 times 10 super 8 mathrm mathrm superscript minus 1 end , plays an important role throughout special relativity. Among other things it represents an upper limit to the speed of any particle.Equation 12 implies that the momentum of a particle increases more rapidly with increasing speed than the Newtonian relation ( \boldsymbol{p} = m\boldsymbol{v} bold p = m bold v ) predicts. This is shown in Figure 6, where the behaviour of the Newtonian and relativistic definitions of momentum magnitude are compared. You can see the good agreement at low speed, but you can also see the increasing discrepancy as the speed increases. Note that the relativistic definition does not extend beyond v = c v = c . This reflects the fact that in special relativity it is impossible to accelerate a particle with mass to the speed of light, as you will soon see.7.2 Relativistic kinetic energyOne of the most celebrated aspects of special relativity is Einstein’s discovery of **mass energy** , the energy that a particle has by virtue of its mass. The mass energy of a particle of mass m m (sometimes called the *rest mass* in this context) is given byE_{\mathrm{mass}} = m c^2.uppercase E sub mass = m c squared.The mass energy is also known as *rest energy*.The reason for mentioning this relation here is that it plays a part in determining the kinetic energy of a particle. How is this? Well, according to special relativity the total energy (including the mass energy) of a particle of mass m m travelling with speed v v is E_{\mathrm{tot}} = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}}. uppercase E sub tot = fraction m c squared over square root 1 minus fraction v squared over c squared end end end. Since this quantity is the sum of the translational kinetic energy and the mass energy of the particle it follows that, according to the theory of relativity, the translational kinetic energy of a particle of mass m m and speed v v is E_{\text{trans}} = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m c^2. uppercase E sub trans = fraction m c squared over square root 1 minus fraction v squared over c squared end end end minus m c squared. Unlikely as it may seem, this expression actually agrees very closely with the Newtonian expression for translational kinetic energy (\frac{1}{2}m v^2) open bracket fraction 1 over 2 end m v squared close bracket when v v is small compared with c c . The relativistic and Newtonian definitions of translational kinetic energy are compared in Figure 7. The figure also indicates one reason why it is impossible to accelerate a particle to the speed of light; doing so would require the transfer of an unlimited amount of energy to the particle.In analysing high-speed **relativistic collisions** , it is the relativistic expressions for momentum and energy that must be used rather than their Newtonian counterparts. In an elastic collision, all of the quantities we have just defined will be conserved:momentummass energykinetic energytotal energy.However, many high-energy collisions are actually inelastic, and in a high-energy inelastic collision the only quantities that are certain to be conserved are the momentum and total energy. In a general inelastic collision, neither kinetic energy nor mass energy is necessarily conserved. This means that in an inelastic collision it is quite possible for particles to be created or destroyed, thereby increasing or decreasing the mass energy. However, the conservation of total energy means that any change in mass energy must be accompanied by a compensating change in the kinetic energy. Thus particles may be created, but only at the expense of kinetic energy.The need for kinetic energy in order to create particles explains why advances in particle physics often require the construction of powerful new particle accelerators. Increasing the kinetic energy of the colliding particles increases the mass of the particles that may be created in the collision and thus opens up the possibility of creating previously undiscovered forms of matter. Figure 8 shows the tracks of particles created in one such ‘ultra-relativistic’ collision.This section has made much use of the phrase ‘high-speed collision’. The meaning of the term ‘high speed’ obviously depends on context. However, if we simply take it to refer to speeds that are sufficiently high that there is a clear discrepancy between the Newtonian and relativistic values of kinetic energy and momentum, then we can say that high speed means greater than about 0.1c = 3 \times {10}^7\, \mathrm{m}\, \mathrm{s}^{-1} 0.1 c = 3 times 10 super 7 mathrm mathrm superscript minus 1 end . This may not be obvious from the curves in Figure 6 and Figure 7 because of the scale that has been used to draw them, but 0.1 c 0.1 c is the threshold used by physicists.ConclusionHaving completed this free course, *Collisions and conservation laws*, you should now be able to state the law of conservation of momentum and describe the essential features of elastic and inelastic collisions. You should also be able to use the law of conservation of momentum and (when appropriate) the law of conservation of kinetic energy to solve a variety of simple collision problems. In addition, you should recognise the expressions for momentum and energy that arise in special relativity and explain their implications for the creation of new particles in high-speed inelastic collisions at CERN.This OpenLearn course is an adapted extract from the Open University course S217 *Physics: from classical to quantum* .Keep on learningStudy another free courseThere are more than **800 courses on OpenLearn** for you to choose from on a range of subjects.Find out more about all our free courses.Take your studies furtherFind out more about studying with The Open University by visiting our online prospectus.If you are new to university study, you may be interested in our Access Courses or Certificates.What’s new from OpenLearn?Sign up to our newsletter or view a sample.For reference, full URLs to pages listed above:OpenLearn – www.open.edu/openlearn/free-coursesVisiting our online prospectus – www.open.ac.uk/coursesAccess Courses – www.open.ac.uk/courses/do-it/accessCertificates – www.open.ac.uk/courses/certificates-heNewsletter – www.open.edu/openlearn/about-openlearn/subscribe-the-openlearn-newsletter **collision** A brief interaction between two or more particles or bodies in close proximity. **completely inelastic collision** A *collision* in which the colliding bodies stick together, resulting in the maximum loss of *kinetic energy* consistent with *conservation of momentum*. **conservation of kinetic energy** The principle that the total kinetic energy of any isolated system is constant. **conservation of linear momentum** The principle that the total *linear momentum* of any isolated system is constant. **elastic collision** A collision in which *kinetic energy* is conserved. **inelastic collision** A collision in which *kinetic energy* is not conserved. **internal force** In the context of a given system, an internal force is a *force* that acts within the system and which has a *reaction* that also acts within the system. **isolated system** A *system* which cannot exchange *matter* or energy with its *environment*. In the context of *mechanics*, an isolated system is one that is subject only to *internal forces*. **kinetic energy** The energy that a body possesses by virtue of its motion. **law of conservation of linear momentum** See *conservation of linear momentum*. **linear momentum** The *momentum* associated with the *translational motion* of a body. For a *particle* of *mass* m m travelling with *velocity* \boldsymbol{v} bold v , the linear momentum is \boldsymbol{p} = m \boldsymbol{v} bold p = m bold v . **mass energy** The *energy* that a body possesses by virtue of its *mass*, as given by E_{\mathrm{mass}} = {mc}^2 uppercase E sub mass = m c squared , where c c is the speed of light in a vacuum. The existence of mass energy is one of the many implications of the *special theory of relativity*. The mass energy of a free particle is the difference between its (total) *relativistic energy* and its *relativistic translational kinetic energy*. Mass energy is also known as *rest energy*. **momentum** A *vector* quantity, useful in various situations as a measure of a body's tendency to continue in its existing state of *rotational* or *translational motion*. **principle of conservation of linear momentum** See *conservation of linear momentum*. **relativistic collision** A collision involving sufficiently high speeds that its analysis requires the use of the relativistic relations for momentum and energy rather than their Newtonian counterparts. Relativistic collisions are often *inelastic* and are characterised by the creation of new particles and an associated increase in *mass energy* (at the expense of *kinetic energy*). **relativistic energy** According to the theory of special relativity the total energy (including the *mass energy*) of a particle of mass m m travelling with speed v v is E_{tot} = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}}. uppercase E sub mathrm = fraction m c squared over square root 1 minus fraction v squared over c squared end end end. **relativistic kinetic energy** According to the theory of special relativity, the translational kinetic energy of a particle of mass m m and speed v v is equal to its total relativistic energy minus its mass energy E_{\text{trans}} = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m c^2. uppercase E sub mathrm = fraction m c squared over square root 1 minus fraction v squared over c squared end end end minus m c squared. **relativistic momentum** The *momentum* of a body according to the *special theory of relativity*. For a particle of (rest) mass m m , travelling with velocity \boldsymbol{v} bold v , the relativistic momentum is p = \frac{mv}{\sqrt{1 - (v^2 ∕ c^2)}}. bold p = fraction m bold v over square root 1 minus open bracket v squared / c squared close bracket end end. At speeds which are small compared with the speed of light, c c , this reduces to the Newtonian expression \boldsymbol{p} = m\boldsymbol{v} bold p = m bold v . **superelastic collision** A collision in which the *kinetic energy* increases, typically as a result of the release of *potential energy*. **system** That part of the Universe which is the subject of an investigation. This free course was written by Professor Andrew Norton.Except for third party materials and otherwise stated (see terms and conditions), this content is made available under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 Licence .The material acknowledged below is Proprietary and used under licence (not subject to Creative Commons Licence). Grateful acknowledgement is made to the following sources for permission to reproduce material in this free course:ImagesCourse image: D. M. Eigler, IBM Research Division.Figure 1: DemonDeLuxe (Dominique Toussaint). This file is licensed under the Creative Commons Attribution-Share Alike LicenceFigure 5: gl0ck33 / 123RFFigure 8: courtesy © CERNAudio-visualAnimation 1: © The Open UniversityAudio 1: © The Open UniversityVideo 1: (6.2): from: Mother of All Collisions (2000) by The BBC for The Open University © The Open University and its licensorsVideo 2: (6.3): from: Big Bang Night, The Big Bang Machine, BBC4 3 September 2000 © The BBCEvery effort has been made to contact copyright owners. 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