Impulse and Momentum Sections Covered in the Text: Chapter 9, except 9.6 & 9.7

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1 Impulse and Momentum Sections Covered in the Text: Chapter 9, except 96 & 97 In this note we consider the physics of particles in collision Collisions form a class of problems in physics that involve two or more particles interacting in such a way that their states of motion are changed We shall simplify our description by defining another kinematic quantity: linear momentum 1 Isaac Newton called momentum an object s quantity of motion To appreciate the idea of momentum, suppose two objects, a car and a pingpong ball, are moving towards you both at a speed of 200 ms 1 You have the option of avoiding being hit by only one of the objects Which do you choose? Which object has the lesser quantity of motion? Even if you haven t studied physics, you would surely choose the ping-pong ball Being hit by a pingpong ball is likely to hurt less than being hit by a car moving at the same speed The car has the greater mass of the two and therefore the greater quantity of motion or momentum We begin by defining linear momentum We then develop the law of conservation of linear momentum for an isolated system We then describe a few cases of one-dimensional collisions We conclude with the concept of the center of mass Linear Momentum and its Conservation The linear momentum of a particle of mass m moving with velocity v is defined 2 p = mv [7-1] Linear momentum, being the product of a scalar and a vector, is a vector It has dimensions MLT 1 and units kgms 1 The direction of the momentum vector is the same as the velocity vector We can express Newton s second law in terms of linear momentum in this way: F = d p [7-2] 1 The use of collisions as a tool in physics research has been of immense value What is known about the structure of solids has been largely discovered with collision techniques The structure of elementary particles is studied by means of high-energy collisions 2 This is the non-relativistic definition, the definition that applies at speeds much less than the speed of light The relativistic definition that applies at any speed will be given in PHYA21H3S This statement allows for a mass m that can vary (which our earlier statement of the law in Note 04 did not) Eq[7-2] reduces to eq[4-1] in the special case where m is constant To see this we substitute eq[7-1] into eq[7-2] Simplification yields F = d p = d(mv) = m dv = ma The second line follows if m is constant 3 Thus the resultant force on an object (system) equals the time rate of change of linear momentum of the object (system) The definition of linear momentum enables us to put the second law into a more general powerful form If, in addition, the system is an isolated one then we can formulate a law of conservation of linear momentum as we shall see next Momentum and Isolated Systems The instantaneous state of an isolated system of two interacting particles may be depicted as in Figure 7-1 By isolated system we mean one in which the particles interact with one another but not with their environment This figure depicts two particles not in contact (for example interacting via the force of gravity or the electrostatic force), though we could easily modify it to describe two particles in contact Let the linear momentum of particles 1 and 2 be p 1 and p 2 respectively Applying the second law, eq[7-2], to each particle yields: F21 = d p 1 and F12 = d p 2 Here r F 12 is the force that particle 1 exerts on particle 2 The third law requires that the forces be equal and oppositely directed Their sum is therefore: F21 + F12 = 0, 3 The first line holds whether m is constant or not A system in which m might not be constant is a relativistic one This topic is discussed in PHYA21H3S 07-1

2 eq[7-4] are known, then the fourth can be calculated Eq[7-4] applies in a space of any dimension and can be extended to apply to any number of particles Let us consider an example Example Problem 7-1 Finding an Unknown Momentum Figure 7-1 Two particles making up an isolated system are shown interacting via non-contact forces or using eq[7-2], 07-2 d p 1 + d p 2 = d (p 1 + p 2 ) = 0 Momentum is a vector and the sum p 1 + p 2, the total momentum of the system, is also a vector 4 Since the time derivative of the total linear momentum is zero, then it follows that the total linear momentum is constant: r p tot = const Specifically, for a system of two particles, p 1i + p 2i = p 1 f + p 2 f, [7-3] [7-4] where i and f denote initial and final values of the linear momentum for the elapsed time over which the particles interact Eq[7-4] states in so many words that the total momentum of the final state of the system equals the total momentum of the initial state of the system Eq[7-3] is a mathematical statement of the law of conservation of linear momentum In words, in any interaction the total linear momentum of any isolated system is constant, or remains unchanged, or is conserved Obviously, if three of the four quantities in 4 We shall shorten the phrase linear momentum to just momentum in those cases when no confusion with angular momentum is likely to occur An isolated system consists of two particles that interact in some manner A sophisticated instrument enables us to measure the momentum of the particles before the interaction; they are found to be, respectively, p 1i =100 ) i ) j ) k, and p 2i = 200 ) i ) j ) k [7-5a] [7-5b] After the interaction, we measure the momentum of particle 1 and obtain p 1 f = 300 ) i ) j ) k [7-5c] Calculate the momentum of particle 2 after the interaction Solution: The system is an isolated one so the law of conservation of linear momentum applies From eq[7-4] we solve for p 2 f : p 2 f = p 1i + p 2i p 1 f Thus by substituting eqs[7-5] into the above expression we have what we seek: p 2 f = (000) ) i + (100) ) j + (200) ) k You can see that the x-component of the final momentum of particle 2 is zero This means that after the interaction particle 2 moves in the yz-plane The law of conservation of linear momentum is very powerful and enables us to solve certain collision problems more easily than would otherwise be the case Can you think of an easier way of solving this problem by applying Newton s laws or the kinematic equations developed in earlier notes? Clearly, we must add the law of conservation of linear momentum to our collection of physics tools

3 Let us consider a second, more real world example Example Problem 7-2 Conservation of Linear Momentum A 600 kg boy and a 400 kg girl, both wearing ice skates on an ice surface face each other initially at rest The girl pushes the boy, sending him eastward with a speed of 40 ms 1 Describe the subsequent motion of the girl (Ignore friction) Solution: We consider the system of boy and girl as an isolated one since the boy and girl interact only with each other and not with their environment Applying the law of conservation of momentum, the total momentum of the system before the push equals the total momentum of the system after the push Let us give the momentum vector pointing east a positive sign, the momentum vector pointing west a negative sign Thus we have Note 07 can be quite large Because the elapsed time is usually short, it is useful to define a quantity called impulse, which we do next Impulse and Momentum Impulse is another word used loosely in everyday speech In physics, the word impulse is defined as simply change in momentum In a collision between two particles (and especially a contact collision) the force of interaction might vary with time as is shown in Figure 7-2 The force is relatively short-lived, being zero before clock time t i and zero after clock time t f and having a relatively large value at maximum The elapsed time for the interaction is to a good approximation t = t f t i p before = 0 = p boy + p girl so p girl = p boy The girl s momentum vector points in a direction opposite the boy s, so as he moves eastward she moves westward Her speed is v = m v boy boy = (600kg)(40ms 1 ) = 60ms 1 m girl (400kg) A point to appreciate here is that the boy is heavier than the girl and both boy and girl are subject to the same magnitude of force (the push the third law) But because momentum is conserved here not speed, the girl moves away after the interaction at a higher speed than does the boy Recall that Figure 7-1 depicts two particles interacting without making apparent contact In the event that the particles do make contact then the elapsed time over which the forces act with appreciable magnitude might be quite short (A good example is the collision between two billiard balls) 5 During this time the force of interaction can vary widely, and at its maximum 5 A somewhat less ideal example is the collision of a glider with the elastic band at the end of an air track Figure 7-2 A force that varies over a relatively short elapsed time The area under the force curve is equal to the magnitude of the impulse Rearranging eq[7-2] we have for an infinitesmal change in momentum: d p = F(t) Notice that the sum of forces is a function of time To find the change in momentum we must integrate over the elapsed time for the interaction: t f Δ p = p f p i = F(t) t i 07-3

4 The change in momentum is defined as the impulse and given the symbol J r Thus r J = t f r F (t) = Δp r [7-5] t i Impulse has the same dimensions and units as momentum and is also a vector You should be able to spot from eq[7-5] and Figure 7-2a that the impulse has a magnitude equal to the area under the force curve between the two clock times, that is, over the elapsed time of the collision The direction of the impulse vector is the same as the direction of the change in momentum vector Collisions As implied above, two particles colliding may or may not make contact in the conventional sense Figure 7-3 shows how contact and non-contact collisions might be depicted when the force of interaction is repulsive contact force as shown in Figure 7-3a is, in essence, a collision involving a non-contact force such as the electromagnetic force 6 Collisions are generally classified as being inelastic, perfectly inelastic or elastic An inelastic collision is the most general and common type of collision, with elastic and perfectly inelastic collisions being special cases We consider the perfectly inelastic collision first in the next section because in many respects it is the simplest kind of collision to describe mathematically One-Dimensional Perfectly Inelastic Collision A perfectly inelastic collision is defined as a collision of two objects in which the objects stick together and move as one object after the collision Let the objects have mass and m 2 and initial velocity components v 1i and v 2i along a straight line (Figure 7-4) As before we assume that the system is isolated Figure 7-4 Velocity vectors for a perfectly inelastic head-on collision between two particles in an isolated system before the collision (a) and after the collision (b) Figure 7-3 A collision may or may not involve the conventional notion of contact In classical physics, collisions (a) and (b) are described by the same mathematics (allowing for the fact that (a) is a 1D collision and (b) is a 2D collision) if the force of interaction is repulsive Whether two particles actually make contact in a collision or not is irrelevant to the mathematical description of the problem An argument can be made that if we were to zero in on smaller and smaller distances (using some kind of super microscope) then even a collision involving what appears to be a 07-4 After the collision the objects stick together and move as one object with some common velocity component v f Because the total linear momentum of the twoobject isolated system before the collision equals the total linear momentum of the combined-object system after the collision we have v 1i v 2i = ( )v f, [7-6] 6 The visualization of this assertion requires a stretch of the imagination Its complete description is a part of most higher-level courses in classical mechanics

5 and therefore v f = v 1i v 2i [7-7] Thus if the masses and their initial velocities are known, then the final velocity of the combination can be calculated Let us consider a numerical example Example Problem 7-3 A Perfectly Inelastic Collision Two spherical lumps of plasticine of masses 050 kg and 100 kg approach each other with velocities 100 ms 1 and 050 ms 1, respectively, along the same line The first lump is moving to the right, the second to the left They collide in a perfectly inelastic collision Find the velocity of the combined lump after the collision Solution: Applying eq[7-7] and being careful to give the velocity of the second lump a negative sign, since it is moving to the left, we have v f = (050kg)(100ms 1 ) + (100kg)( 050ms 1 ) 050kg + 100kg = 0! After the collision the combined lump has a speed of zero and is therefore at rest Question: Because the velocity of the combination after the collision is zero, the total momentum after the collision is also zero But does this mean that the collision violates the law of conservation of linear momentum? One-Dimensional Elastic Collision An elastic collision is defined as a collision in which kinetic energy is conserved as well as linear momentum Consider the head-on elastic collision of two objects (Figure 7-5) Let the particles have mass and m 2, respectively, and initial velocity v1i and v2i, respectively, with initially moving to the right and m 2 initially moving to the left Let the particles collide head-on, and after the collision have velocity v1 f and v2 f respectively Applying the law of conservation of linear momentum we have v 1i v 2i = v 1f v 2 f [7-8] Figure 7-5 The velocity vectors for two objects in an isolated system undergoing a head-on elastic collision Using the fact that kinetic energy is also conserved we must also have 1 2 m v 2 1 1i m v 2 2 2i = 1 2 m v f m v f [7-9] Collecting over and m 2, eq[7-9] reduces to (v 2 1i v f ) = m 2 (v 2 f When factored, this equation becomes v 2 2i ) (v 1i v 1 f )(v 1i + v 1 f ) = m 2 (v 2 f v 2i )(v 2 f + v 2i ) [7-10] Now consider eq[7-8] Collecting over terms in and m 2 in eq[7-8] gives: (v 1i v 1 f ) = m 2 (v 2 f v 2i ) Dividing eq[7-10] by eq[7-11] yields which, when rearranged, is v 1i + v 1f = v 2 f + v 2i, v 1i v 2i = (v 1 f v 2 f ) [7-11] [7-12] Eq[7-12] states in so many words that the relative velocity of the two particles after the collision is equal to the negative of the relative velocity of the two particles before the collision or the relative velocity of the two particles undergoes a reversal in the collision 07-5

6 If the masses and the initial velocities of the two particles are known then the final velocities can be calculated from eqs[7-8] and [7-12] After some manipulation the results are: v 1 f = m m 1 2 2m 1i + 2 2i [7-13] 2m and v 2 f = 1 1i + m m 2 1 2i The Center of Mass In any object or collection of objects a special point exists called the center of mass The idea of the center of mass can be understood with the help of the object drawn in Figure 7-6, two balls connected by a rigid rod The center of mass (denoted CM) of the object is, as we shall see, a point located somewhere on the rod between the balls When using eqs[7-13], you must take care to assign the correct signs to the velocities For example, if particle 2 is initially moving to the left then v 2i must be given a negative sign Eqs[7-13] are general expressions in that they describe two simpler special cases: 1 If = m 2 then v if = v 2i and v 2f = v 1i In other words, if the particles have the same mass then they exchange speeds in the collision particle 1 moves off with the initial speed of particle 2 and vice versa 2 If m 2 is initially at rest then v 2i = 0 and eqs[7-13] reduce to v 1 f = m m 1 2 1i 2m and v 2 f = 1 1i [7-14] Figure 7-6 Illustration of the meaning of the center of mass If is very large in comparison to m 2 then eqs[7-14] show that v 1f v 1i and v 2f 2v 1i In other words, if a very large mass collides head-on with a very small mass then the large mass continues on afterwards with the same speed whereas the small mass takes on twice the initial speed of the large mass Thus far we have studied head-on collisions only, in other words, collisions in 1D space But a collision can be a glancing one, in 2D or 3D space, as was shown in Figure 7-3b Because of the complexity of the algebra and lack of time we shall leave 2D and 3D collisions to a higher-level course in classical mechanics We have established the physics, the rest is algebra To get the idea of the center of mass suppose we apply an external force horizontally from the left to a point in the object If the point is above the center of mass, then the object will tend to rotate in a clockwise direction If the point is below the center of mass, then the object will tend to rotate in a counter-clockwise direction If the point is at the center of mass, then the object will tend to move to the right with translational motion only It will not rotate at all Without resorting to a lot of mathematics we can state a number of attributes of the center of mass of certain objects One attribute, the most important one, is that if the object is a uniform rigid object then its center of mass is located at its geometric center Isaac Newton refused to publish his theory of gravitation in his Principia Mathematica until he had proved this attribute geometrically The task took him several years

7 To Be Mastered Note 07 Definitions: linear momentum, isolated system Definition: impulse Definition: center of mass Definitions: elastic collision, inelastic collision, perfectly inelastic collision Laws: Conservation of Linear Momentum Typical Quiz/Test/Exam Questions 1 Describe one example of each of the following: (a) an elastic collision, (b) a perfectly inelastic collision, (c) a collision in which linear momentum is not conserved 2 A 150 g bullet is fired horizontally into a 300 kg block of wood suspended by a long cord and initially at rest (see the figure) The bullet sticks in the block and the block swings 100 cm above its initial level Answer the following questions bullet 100 cm (a) What kind of collision is involved here? (b) What was the speed of the bullet in ms 1 at the time of impact? 07-7

8 3 A ball of mass 0500 kg is released from rest a height h = 300 cm above a hard smooth surface Assume the collision the ball makes with the surface is elastic Answer the following questions: m = 0500 kg h = 300 cm h' (a) Is h greater than h, less than h or equal to h? Explain your answer (b) How long does the ball take to reach the surface? (c) What is the speed of the ball when it collides with the surface? 4 An apparatus consists of two sections of straight metal track A and B as shown The track is grooved so that a hard rubber ball of mass 2000 g placed near one end of the track rolls down without falling off The ball rebounds from a bumper at C On one run, the ball is released from a height h = 300 cm in section A; it subsequently rolls through sections A and B and bounces back from C Kinetic friction is negligible Answer the following questions: h 45 C A B (a) What is the acceleration of the ball in section A? (b) What is the acceleration of the ball in section B? (c) What is the speed of the ball when it hits the bumper at C? (d) If the ball bounces back from C in an elastic collision, to what vertical height does the ball return to in section A? 07-8