PHYS 2.5: Momentum and collisions	PPLATO @
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1 Opening items

1.1 Module introduction

When two billiard balls collide they remain in contact for only a very short time, during which the forces between them vary rapidly. To predict the outcome of such a collision by calculating the acceleration of each ball at each instant of time we would need to know exactly how the forces vary with time, and this is not an easy matter. However, we may predict the outcome in a simpler way, using the concept of linear momentum – the subject of this module.

The total linear momentum involved in a collision is important because, under certain conditions, it has the same value both before and after the collision. In other words, it is a conserved quantity. Interestingly, when appropriately interpreted, the principle of conservation of linear momentum extends beyond the confines of classical Newtonian mechanics, into the realms of relativistic mechanics and even quantum mechanics. It is, therefore, along with the conservation of energy, one of the most wide ranging and far reaching principles in the whole of physics.

We introduce linear momentum in Section 2 and show how Newton’s second law of motion may be expressed in terms of linear momentum. The principle of conservation of linear momentum is then derived from Newton’s second and third laws of motion and it is illustrated with some simple one–dimensional applications.

In Section 3 we extend this to study collisions, and introduce the idea of energy, translational kinetic energy and the principle of conservation of energy. Collisions are classified as elastic or inelastic, according to whether kinetic energy is conserved. Elastic and inelastic collisions are then analysed in terms of the conservation principles, using examples that involve one– and two–dimensional motions. Finally, brief mention is made of a revised definition of momentum, necessary for the extension of momentum conservation into relativistic mechanics.

Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to the Subsection 1.3Ready to study? Subsection.

1.2 Fast track questions

Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 5.1Module summary and the Subsection 5.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 5.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.

1.3 Ready to study?

Study comment In order to study this module you will need to be familiar with the following terms: acceleration, components_of_a_vectorcomponent (of a vector), force, mass, Newton’s laws of motion, scalar, speed, vector, vector addition and velocity. Also, you will need to be familiar with the use of Cartesian coordinate systems, Pythagoras’s theorem, trigonometric functions, the sine rule, and the cosine rule. Additionally, this module makes some use of the derivativederivative notation of basic differentiation to represent rates of change and assumes you are familiar with symbols such as dυ/dt. If you are uncertain of any of these terms you can review them now by referring to the Glossary which will indicate where in FLAP they are developed. The following questions will allow you to establish whether you need to review some of the topics before embarking on this module.

2 Linear momentum and Newton’s second law

2.1 Definition of linear momentum

It is easy to imagine that a lorry travelling at 40 m s⁻¹ is more difficult to stop than the same lorry travelling at 20 m s⁻¹. Similarly, a lorry travelling at 40 m s⁻¹ is more difficult to stop than a car travelling at 40 m s⁻¹. From this it is apparent that the difficulty in stopping a body depends on both the mass and the velocity of the body. The scientific quantity identifying this property of a moving body, which we have loosely described in terms of the ‘difficulty in stopping’ is called linear momentum (more often simply momentum), p. In Newtonian mechanics, a body of mass m moving with velocity υ, has a momentum p defined by:

As velocity is a vector quantity, linear momentum is also a vector quantity, with magnitude mυ and a direction

which is the same as that of the velocity vector. The SI units of linear momentum are kg m s⁻¹.

It is often convenient to express the momentum i vector p in terms of its components. If the components of the velocity vector υ are υ_x, υ_y and υ_z, then we write: υ = (υ_x, υ_y, υ_z) and p = (p_x, p_y, p_z), where the components of the momentum vector are given by:

p_x = mυ_x    p_y = mυ_y    p_z = mυ_z(1a)

In dealing with momentum we frequently encounter problems in which motion is not restricted to a single line. Bodies moving in different directions may collide and may change directions. Since momentum is a vector quantity, a change in direction represents a change in momentum, even if the magnitude of the momentum is unchanged.

2.2 Relationship between force and rate of change of momentum

Momentum is closely associated with Newton’s second law of motion, F = ma. Indeed, if we assume that the mass m is constant, then we can express Newton’s second law in terms of momentum. To see this first note that:

$\boldsymbol{F} = m\boldsymbol{a} = m\dfrac{d\boldsymbol{\upsilon}}{dt}$

Then note that if m is constant

$m\dfrac{d\boldsymbol{\upsilon}}{dt} = \dfrac{d(m\boldsymbol{\upsilon})}{dt} = \dfrac{d\boldsymbol{p}}{dt}$ i

i.e. the resultant force acting on a body is equal to the rate of change of its momentum.

Equation 2 is a vector equation and so represents the three equations, one for each of the three components of force:

$F_x = \dfrac{dp_x}{dt} \quad F_y = \dfrac{dp_y}{dt} \quad F_z = \dfrac{dp_z}{dt}$(2a)

We now have two forms of Newton’s second law when m is constant:

$\boldsymbol{F} = m\boldsymbol{a} \quad\text{and}\quad F = \dfrac{d\boldsymbol{p}}{dt}$

Faced with two ways of expressing Newton’s second law it is legitimate to ask if one is more general or more fundamental than the other. In fact, the general view is that Equation 2 is the more fundamental. It can be applied directly to systems of particles as well as to single bodies, and, in some cases, it can even be applied to bodies with changing mass, such as a rocket that accelerates by expelling the fuel that it carries.

Given experimental fact that F = dp/dt is more generally applicable throughout classical mechanics you might wonder how F = ma fails in various contexts, such as that of varying mass. In that particular case the inadequacy of F = ma is easy to demonstrate provided you are familiar with the product_rule_of_differentiationproduct rule of differentiation which enables us to write the derivative of a product as a sum of products.

If we start from Equation 2,

$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2)

but do not assume constant mass, then:

$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt} = \dfrac{d(m\boldsymbol{\upsilon})}{dt} = m\dfrac{d\boldsymbol{\upsilon}}{dt} + \boldsymbol{\upsilon}\dfrac{dm}{dt}$(3)

From this we see that if m is not constant then we have to add a further term υdm/dt to ma to obtain the correct expression for F. There are several surprising aspects of this. First, a resultant force acting on a body does not always cause acceleration; it could be that the mass of the body is increasing with time at just the correct rate to allow constant velocity. Second, even when there is no resultant force acting on the body it may nevertheless accelerate, if its mass is changing.

The identification of ‘force’ with ‘rate of change of momentum’ sanctioned by Equation 2,

$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2)

also has another important consequence; if F = 0 i then dp/dt = 0 thus:

If no resultant force acts on a body then its momentum does not change.

As you will see in Subsection 2.3 a generalization of this result leads to the important principle of momentum conservation.

Impulse

There are many situations where the force on a body acts only for a very short time (e.g. bats on balls) and we are concerned only with the consequent changes in the motion of the body. If the force F is constant and acts for a short time ∆t, it follows from Equation 2 that the momentum of the body will change by an amount

∆p = F∆t(4) i

The quantity F∆t is called the impulse of the force. i

In many situations where varying forces act for short times it is only their effects (i.e. the momentum changes they cause) that are of real interest. In such cases Equation 4 may be used to find the constant force that would produce the same overall effect as the varying force that actually acted. This procedure can give surprising insights into the rapidly varying forces that arise in various impulsive interactions. For example, in American baseball the bat and ball are typically in contact for one or two thousandths of a second, during which the effective average force has a magnitude of about 6000 N. Such large forces are not unusual.

2.3 Conservation of momentum

The concept of momentum is particularly useful when dealing with situations where there are two or more interacting bodies which interact with each other but not with their surroundings. We call such a system an isolated system. Any forces which act on parts of an isolated system arise from the mutual interactions between bodies within the system. Each force is then one member of a Newton’s third law pair of action–reaction forces. Such forces are called internal forces, to distinguish them from the external forces which act on the system or parts of the system from outside. We will begin by considering an isolated system in which there are only two bodies and they only interact with one another.

Let p₁ = m₁υ₁ and p₂ = m₂υ₂ be the momenta of the two bodies in the system. These two momenta will change with time, due to the mutual interaction. The total momentum of the system, p, is given by the vector sum of the momenta of the two bodies:

p = p₁ + p₂

The rate of change of p with time is therefore determined by the rates of change of p₁ and p₂:

$\dfrac{d\boldsymbol{p}}{dt} = \dfrac{d\boldsymbol{p}_1}{dt} + \dfrac{d\boldsymbol{p}_2}{dt}$ i

At a given instant let F₁₂ be the force on the first body due to the second and F₂₁ be the force on the second body due to the first. Equation 2,

$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2)

gives:$\boldsymbol{F}_{12} = \dfrac{d\boldsymbol{p}_1}{dt}\quad\text{and}\quad\boldsymbol{F}_{21} = \dfrac{d\boldsymbol{p}_2}{dt}$

Newton’s third law of motion tells us that:

F₁₂ = −F₂₁  or  F₁₂ + F₂₁ = 0

Thus p = p₁ + p₂ is constant with time, although both p₁ and p₂ may be varying. At every instant of time the total momentum is the same (since p does not depend on t). We can summarize the situation by saying that the total momentum of the isolated system is a conserved quantity.

Note that in deriving Equation 5 we have assumed that the only forces acting on the bodies are F₁₂ and F₂₁. If an external force F had been acting on the bodies then the system would not be isolated, Equation 5 would not be valid and the total momentum of the system would not be conserved; rather the total momentum would vary according to Equation 2,

$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2)

Although we have considered only two bodies in deriving Equation 5, the conservation of momentum result can be extended to any number of bodies in an isolated system. The same reasoning as before may be used: the rate of change of total momentum of the isolated system is zero because the vector sum of the impulses from each action–reaction pair of forces is zero. This result constitutes one of the most important principles of physics:

2.4 Symmetry and conservation

If an object or a system of objects can be altered in some way without changing its overall behaviour or appearance, that object or system is said to exhibit symmetry under the alteration. For instance, a uniform sphere may be rotated through an arbitrary angle about its centre point without changing its properties or appearance in any way; such a sphere is consequently said to exhibit rotational symmetry and to be invariant (i.e. unchanging) under arbitrary rotations about its centre.

Symmetries and their related invariances are common features of physical systems and of the laws that describe them. Some of these symmetries are so readily taken for granted that it is easy to overlook them. For example, it is generally assumed amongst physicists that the outcome of a fundamental experiment does not depend on where or when it is performed. There is, of course, a good deal of evidence in support of this belief but it is impossible to test every case. Consequently, our expectation that the fundamental behaviour of physical systems will not vary from country to country or from day to day is largely based on our belief in the symmetry of nature under translations (i.e. ‘shifts’ or ‘transfers’) through space and time. In fact, much of modern science is based on the following assumptions:

• The laws of physics must be invariant under arbitrary translation through space.
• The laws of physics must be invariant under arbitrary translation through time.

If either of these fundamental invariances was violated, if momentum was only conserved on Tuesdays say, or only in Western Europe, then physics would be a much less interesting subject and even harder to master.

The reason for embarking on this discussion of symmetry and invariance is that it has been known for a long time that there is a deep link between the symmetries of nature and the existence of conserved physical quantities. It can be argued, for instance, that the invariance of physical laws under translations through time implies the existence of a conserved scalar quantity which may be identified as energy (about which more is said later in the module). Similarly, the invariance of physical laws under translation through space implies the existence of a conserved vector quantity which may be identified as momentum. Thus the principle of momentum conservation may be regarded as one of the fundamental principles of physics and you should not be surprised to learn that its validity extends beyond the confines of classical Newtonian mechanics.

The mathematical techniques needed to establish the link between symmetry and conservation are beyond the scope of FLAP but the importance of the result would be hard to overestimate. The search for fundamental symmetries, of which we have mentioned only two, has been one of the central pillars of modern physics and has fuelled much of the progress made in the last 40 years.

2.5 Worked examples involving conservation of momentum

In a scientific context, the term collision means a brief but powerful interaction between two particles or bodies in close proximity. To illustrate the value of the principle of conservation of momentum in solving problems, let us begin with the simple case of two bodies moving in one dimension which collide and stick together on impact. Later, we will extend our discussion to two– and three–dimensional motion and to collisions where the bodies do not stick together.

Collisions in one dimension, where the two bodies stick together

Collisions in two dimensions, where the two bodies stick together

In problems in which motion is in more than one dimension we have to take care over the directions of motion of the bodies involved. This means that in applying conservation of momentum we need to remember the vector nature of momentum. In our discussions so far we have prepared the ground for this extension by stressing the vector approach throughout and by insisting on the use of component notation, even with one–dimensional problems where there was the temptation to drop the subscripts! For two– or three–dimensional problems there are two possible approaches which may be used: vector summation or use of components. We will consider these alternative approaches in the following worked examples.

Study comment Whilst you should practice using both the vector summation method and the component method in solving problems on momentum involving more than one dimension, the component approach is easier to use with more complicated examples. A good general strategy is to set up a clearly labelled diagram of the collision, with appropriately chosen Cartesian coordinate axes x, y, z shown. These axes are chosen so as to simplify the solution – perhaps by arranging them in such a way that one of the incoming particles is moving only along the x–axis. Then momentum conservation is applied for components along each of the axes in turn, with each direction treated entirely independently. Finally, the components are recombined to give the magnitude and direction of the required vector.

3 Momentum and kinetic energy in collisional problems

In the collision problems tackled so far we have always either finished up with a single body, or started with a single body (e.g. in α–decay). The reason for this limitation is that only such problems have a unique solution, obtainable by applying momentum conservation alone. For all other collisions we need to involve other principles in addition to momentum conservation. To illustrate this we can consider a one–dimensional problem in which two colliding bodies with known masses m₁ and m₂, and with known initial velocities u_1x and u_2x collide and then separate with final velocities υ_1x and υ_2x. In general, when the initial conditions are given, the problem involves finding the two unknowns υ_1x and υ_2x. However, consideration of momentum conservation along the x–direction:

m₁u_1x + m₂u_2x = m₁υ_1x + m₂υ_2x

gives only one equation connecting these two unknowns and so the problem is insoluble without more information. i

The additional information is provided by introducing the idea of energy. i Energy can exist in several forms, but the energy that a body has due to its instantaneous linear motion is called its translational kinetic energy. In classical Newtonian mechanics the translational kinetic energy of a body of mass m and speed υ is defined as:

Energy is a scalar quantity. In the SI system, the unit of energy is the joule (J) where 1 J = 1 kg m² s⁻².

Translational kinetic energy i is only one of several forms in which energy can appear. A body may also have potential energy due to its position or internal state, as in the case of a mass suspended above the surface of the Earth, or a compressed spring. Quite apart from these various forms of mechanical energy we can also associate energy with other physical phenomena, so we may speak of electrical energy, light energy, sound energy and thermal energy. Energy may be transformed between these different forms by physical and chemical processes but it is generally accepted that, for an isolated system, the sum of all these different forms of energy remains constant. This fundamental principle is known as the principle of conservation of energy.

Collisions may be categorized by comparing the total 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 increase in the collision; this type of collision is a superelastic collision.

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, protons, commonly are elastic. The kinetic energy which is lost in an inelastic collision appears as energy in a different form, e.g. thermal energy, sound energy, light energy, so that the total energy is conserved. The collisions which we have dealt with so far, 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 converted into other forms of energy.

3.1 Elastic collisions in one dimension

We begin this discussion with an example of a simple one–dimensional elastic collision between two identical masses, where one incoming projectile strikes another, initially stationary target. We wish to know the final velocities of each mass after the collision.

This result is familiar to anyone who has seen the head–on collision of two bowls on a bowling green. The moving one stops, and the stationary one moves off with the original velocity of the first.

Now we consider the mathematical solution of the general one–dimensional case of an elastic collision between particles of mass m₁ and m₂ which move with speeds u_1x and u_2x before the collision and speeds υ_1x and υ_2x after the collision. Since the collision is elastic the kinetic energy is conserved and so we can write:

½ m₁u_1x² + ½ m₂u_2x² = ½ m₁υ_1x² + ½ m₂υ_2x²(11)

This equation, together with the momentum conservation equation,

m₁u_1x + m₂u_2x = m₁υ_1x + m₂υ_2x(12)

allows υ_1x and υ_2x to be found, if we know the values of m₁, m₂ and the initial velocities u_1x and u_2x. The algebra is rather lengthy but it is worth the trouble because on the way we derive another useful expression relating the velocities. From Equation 11 we have:

m₁(u_1x − υ_1x)(u_1x + υ_1x) = m₂(υ_2x − u_2x)(υ_2x + u_2x)(13)

and from Equation 12:

m₁(u_1x − υ_1x) = m₂(υ_2x − u_2x)(14)

If we divide each side of Equation 13 by the corresponding side of Equation 14 we obtain:

u_1x + υ_1x = υ_2x + u_2x

i.e.u_1x − u_2x = −(υ_1x − υ_2x) = υ_2x − υ_1x(15)

Equation 15 shows that the following general result holds true:

In an elastic collision between two masses the relative velocity of approach is the negative of the relative velocity of separation.

Equation 15 can be combined with Equation 14 to solve simultaneously for υ_1x and υ_2x. We multiply Equation 15 by m₂ to obtain:

m₂u_1x − m₂u_2x = m₂υ_2x − m₂υ_1x

If we subtract this equation from Equation 14 we have:

u_1x(m₁ − m₂) + 2m₂u_2x = υ_1x(m₁ + m₂)

Similarly we can derive the following expression for υ_2x:

It is interesting to examine these results for υ_1x and υ_2x in a few special cases, including some that have been mentioned earlier.

1

m₁ = m₂

If two objects of equal mass collide, Equations 16 and 17 give υ_1x = u_2x and υ_2x = u_1x. This means that the particles simply exchange velocities on collision. For example, if particle 2 is at rest initially (u_2x = 0), then finally particle 1 is at rest (υ_1x = 0) and particle 2 moves with the initial velocity of particle 1 (υ_2x = u_1x). We saw this result earlier in this subsection; it is a familiar tactic in bowls.

2

m₁ ≫ m₂; u_2x = 0

If m₂ is very small compared with m₁, and u_2x = 0, Equations 16 and 17 give υ_1x ≈ u_1x and υ_2x ≈ 2u_1x. Thus the motion of the high mass particle is virtually unchanged by the collision but the low mass particle moves off with a velocity of twice that of the high mass particle. Tennis players serving will be familiar with this case.

3

m₂ ≫ m₁; u_2x = 0

If m₁ is very small compared with m₂ (which is stationary), then Equations 16 and 17 lead to υ_1x ≈ −u_1x and υ_2x ≈ 0. i Therefore, 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 recognize this situation.

4

m₂ ≫ m₁; u_2x ≈ −u_1x

If m₁ is negligible compared with m₂, and the two bodies approach head–on with equal speeds then Equations 16 and 17 lead to υ_1x ≈ −3u_1x and υ_2x ≈ u_2x. This shows that the low mass particle bounces back with three times its initial speed, while the high mass particle continues essentially unaffected by the collision. This case will be recognized by a batsman playing cricket or by a tennis player returning a serve; a less familiar example is covered in the Exit test.

The results for these four special cases accord with common experience. The results quoted above in cases 2, 3 and 4 give an upper limit to the speed that can be imparted to a ball hit by a club, bat or racquet.

3.2 Inelastic collisions in one dimension

We now extend our discussion to include cases where the total kinetic energy changes during the one–dimensional collision.

Completely inelastic collisions

First we return to our original 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 momentum conservation. 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, the initial momentum is mu_x; the final momentum must be the same so, since the mass has been doubled, the final velocity is u_x/2 and the final kinetic energy is 2m(u_x/2)²/2 = mu_x2/4. Half the original kinetic energy has been transformed into other forms, mainly heat, during the collision.

For the more general case where the colliding masses are unequal, but they stick together at collision, we still have υ_1x = υ_2x = υ_x and so Equation 12 simplifies to our earlier Equation 6,

m₁u_1x + m₂u_2x = (m₁ + m₂)υ_x(Eqn 6)

and provides a full solution of the problem, without recourse to energy.

The general case of an inelastic collision in one dimension

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.

3.3 Collisions in two or three dimensions

The same principles apply when we look at collisions in two or three dimensions. We have already considered this in Subsection 2.5, for the case of completely inelastic collisions. The extension to include elastic collisions is straightforward. We simply treat the motions in each dimension as independent motions, as we did in Subsection 2.5, and apply momentum conservation separately along each Cartesian coordinate axis, with the scalar quantity of kinetic energy and its conservation providing one additional equation relating the squares of the particle speeds. Once again, since we have been careful to use vector notation throughout, this extension to two or three dimensions is easily made.

Elastic collisions between two equal masses, with one mass at rest

Consider the elastic collision between two identical bodies of mass m, one at rest and the other approaching with velocity u₁. The particles are no longer confined to move in one dimension, so our x–component equations (Equations 9 and 10), embodying momentum and kinetic energy conservation,

mu_1x = mυ_1x + mυ_2x(Eqn 9)

½ mu_1x² = ½ mυ_1x² + ½ mυ_2x²(Eqn 10)

become full vector equations:

mu₁ = mυ₁ + mυ₂(18)

and½ mu₁² = ½ mυ₁² + ½ mυ₂²(Eqn 19)

These can be simplified to:

u₁ = υ₁ + υ₂(20)

andu₁² = υ₁² + υ₂²(21)

Equation 20 tells us that all three velocity vectors must lie in a single plane, in order for their vector sum to be zero. These equations are most easily interpreted by a diagram. Figure 2 shows how the three vectors u₁, υ₁ and υ₂ are related to one another. They form a triangle which must be a right–angled triangle, since the sides obey Pythagoras’s theorem. The implication of this is striking; when two equal masses collide elastically they move away at 90° to each other. This is the simplifying feature of equal–mass collisions in two or three dimensions, corresponding to the simple result of the exchange of velocities, which we found in one dimension. i

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, and just how they collide. If we have any information about how the bodies collide, as for example in the situation shown in Figure 3, then we will usually be able to find the final velocities.

When the two masses are unequal the algebraic manipulations required to solve collision problems become rather complex, but no new physics is involved in the solution and we will not pursue such problems here.

3.4 Relativistic momentum

It is now quite easy to study collisions in which the particles are travelling at a significant fraction of c, the speed of light in vacuum. These experiments fall mainly into the realm of nuclear and particle physics and an example might be the elastic collision between a very fast electron from a particle accelerator and an electron in the material of a stationary target. From what we have said in Subsection 3.3 we would predict that the angle between the outgoing tracks of the scattered electron and the target electron ought to be 90°. The prediction proves correct at low impact speeds but as this speed is increased the angle becomes less than 90° and the reduction in the angle increases with the impact speed. At the highest attainable speeds the incident electron is barely deflected at all by its collision with the target electron while the target electron also moves off in about this same direction. The collision essentially reverts to a one–dimensional bat–ball type of collision. From our discussions in Subsection 3.1Subsections 3.1 and Subsection 3.33.3 we recognize that this is exactly what would happen if the incoming electron had a mass which increases with speed, so that its mass eventually becomes much greater than that of the target electron and it is essentially unaffected by the collision. This suggestion, that the mass of a particle is not a constant but rather increases with speed is one of the possible interpretations of this experiment. However, this experiment does not measure mass but rather momentum and so its interpretation is best made in terms of the momentum, rather than in terms of either mass or speed separately. All we can be sure of is that if momentum conservation is valid then the magnitude of the momentum at high speeds cannot simply be given by the product of mass and speed.

The magnitude of the momentum of a particle with mass m and speed υ must always be greater than mυ. Although mυ is a good approximation at sufficiently low speeds, it underestimates the momentum progressively as the speed increases.

In 1905 Albert Einstein (1879–1955) postulated his special theory of relativity and amongst its consequences was a new definition for the momentum of a moving particle. The new definition of the relativistic momentum of a mass m when moving with velocity υ is:

$\boldsymbol{p} = \dfrac{m\boldsymbol{\upsilon}}{\sqrt{1-\dfrac{\upsilon^2}{c^2}}}$(22) i

This definition incorporates the features required to explain our experiment.

The result for part (c) (i.e. where υ/c = 1) illustrates why it is not possible to accelerate a body to c, the speed of light in vacuum; to do so would require either an infinite force or an infinite time. Note, however, that it is the speed of light in a vacuum which is the limiting speed, not the speed of light in any other material, which is always less than c. So, it is perfectly possible to have a body travelling faster than the speed of light in glass or in water or even in air! i This discussion is only the briefest mention of relativity, about which more is said elsewhere is FLAP.

4 Closing items

4.1 Module summary

1

The momentum of a body, a vector quantity, is defined as the product of its mass and its velocity in classical Newtonian mechanics

p = mυ(Eqn 1)

At high speeds this definition must be modified into that for relativistic momentum

$\boldsymbol{p} = \dfrac{m\boldsymbol{\upsilon}}{\sqrt{1-\dfrac{\upsilon^2}{c^2}}}$(Eqn 22)

2

The force acting on a body is equal to the rate of change of momentum

$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2) i

This statement is a more fundamental statement of Newton’s second law of motion than that in which force is equated to the product of mass and acceleration, since the latter form is applicable only for situations in which mass is constant.

3

The principle of conservation of momentum for an isolated system can be derived from Newton’s second and third laws of motion.

4

The principle of conservation of momentum plays an essential part in the solution of collision problems.

5

Translational kinetic energy is the energy that a body has as a result of its motion. In classical Newtonian mechanics

E_tran = ½ mυ²(Eqn 8)

6

One-dimensional problems involving elastic collisions, in which kinetic energy is conserved, may be solved completely.

7

Problems involving motion in more than one dimension can only be solved fully if some details of the collision are known. One example is the elastic collision between two equal masses, where one mass initially is at rest. Here, the departing trajectories always make an angle of 90°.

4.2 Achievements

Having completed this module, you should be able to:

A1

Define the terms that are emboldened and flagged in the margins of the module.

A2

Determine the momentum of a body or system of bodies, given the mass and velocity of each body.

A3

State and apply Newton’s second law in terms of momentum.

A4

Show that the force acting on a body can be expressed as the sum of two terms, one involving rate of change of velocity and the other involving the rate of change of mass.

A5

Derive the principle of conservation of momentum from Newton’s second and third laws of motion.

A6

Describe the essential features of elastic and inelastic collisions.

A7

Use the principle of conservation of momentum to solve problems involving completely inelastic collisions in one– or two–dimensional motion or ones involving the disruption of a single mass by explosion or decay.

A8

Use the principles of conservation of momentum and kinetic energy to solve elastic collision problems.

Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.

4.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions, each of which tests one or more of the Achievements.

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