Notes and Examples These notes contain subsections on model for friction Modelling with friction MEI Mechanics Model for Friction Section 1: Friction Look at the discussion point at the foot of page 1. s the motorcyclist was skidding, the wheels of the motorcycle were sliding along the road. Think about what forces were acting on the motorcycle, and what assumptions you need to make. model for friction It is important to realise that the size of the frictional force between two surfaces varies according to circumstances. For example, consider a block of mass m kg resting on a rough horizontal plane. R The only forces acting on the block are its weight and the reaction force. Even though the surface is rough, there is no frictional force. If a small horizontal force T is applied to the block, not enough to make it slide, a frictional force F opposes the motion. R mg F T In this case, as the block is not moving, F must be equal to T. mg If T is increased, F will also increase. However, eventually T will become large enough to move the block. In this case F has reached its maximum value of R. Once T exceeds the maximum value of F, there will be a net force in the direction of T, which causes the block to accelerate. MEI, 0/06/09 1/3
MEI M Friction Section 1 Notes and Examples Notice that when T is equal to R (the maximum value of F), there are two possibilities: the block may be stationary but on the point of moving, or it may be moving at constant speed (so that its acceleration is zero.) However, for the block to reach a constant speed it must first accelerate from an initial speed of zero to the required speed. To move the block at constant speed, therefore, T needs to be greater than R initially, so that the block accelerates, after which T can be reduced to a value equal to R. This shows, as you know from practical experience, that it is harder to start an object moving than it is to keep it moving once it has started! You can investigate a situation like this using the Flash resource Forces on a box. Ignore any cases in which the box tips instead of slides you will learn about this in chapter 3. The direction of the frictional force also varies according to circumstances. Consider a block of mass m kg at rest on a rough plane inclined at 30 to the horizontal. R F 30 mg Here the frictional force F acts to prevent the block from sliding down the plane. If a small force T is applied up the slope, then the frictional force becomes smaller, as the force T is helping to stop the block sliding down the plane. If T is increased, eventually no frictional force will be needed to stop the block from sliding down the plane. R F T 30 mg If T is increased further, the tendency of the block will no longer be to slide down the plane, but to slide up. This means that the frictional force F will now act in the opposite direction, to oppose the tendency to slide up the plane. R T F 30 mg MEI, 0/06/09 /3
MEI M Friction Section 1 Notes and Examples s T continues to increase, the value of F increases until it reaches its maximum value of R. ny further increase of T after this point will result in the block accelerating up the plane. Look at the discussion point at the top of page 4. In assumption 1, consider the effect on the calculation of both an upward slope and a downward slope. In assumption 3, consider the effect of a both a larger and a smaller value for the coefficient of friction. Modelling with friction The discussion point halfway down page 4 needs careful thought. The pedals cause the wheel to turn, but if there were no friction, which way would the wheel slide? Remember that friction always opposed the tendency to slide. Read the examples in the book carefully. These combine the use of Coulomb s model for friction with the use of Newton s Laws of motion. Many of the examples and the exercise questions deal with similar situations to problems you met in Mechanics 1. However, previously either surfaces were assumed to be smooth, or you were given a value for the frictional force. You are now in a position to improve your models for such situations by using the model for friction. Examples 1.1 and 1. address situations where all forces are either horizontal or vertical. Examples 1.3 and 1.4 introduce situations where some forces act at an angle, and it is therefore necessary to resolve forces into components. Make sure that you remember how to do this. Notice that when a slope is involved, as in example 1.4, it is easier to consider the forces parallel to the slope (i.e. in the direction of motion of the object) and perpendicular to the slope, rather than horizontally and vertically. You should always draw a clear force diagram when solving problems like these. MEI, 0/06/09 3/3
MEI Mechanics Moments of forces Section 1: The moment of a force Notes and Examples These notes contain subsections on Rigid bodies Moments Couples Equilibrium revisited Rigid bodies Read this section carefully. Notice that even quite a large body can be reasonably considered as a particle if all the forces on it act through the same point. This explains why you were able to use the particle model in previous work in Mechanics 1 and in chapter 1 of Mechanics, even for large objects such as a car! Moments In the discussion point at the foot of page 8, think about the forces you need to exert on each tool to undo the nut. Couples Couples are very important in real life, as we often need to turn objects without moving the object in any direction. The spider wrench in the discussion point on page 8 is one example where a couple is used. Notice that the greater the perpendicular distance between the two forces, the greater the turning force of the couple. Equilibrium revisited Read through Examples 3.1 and 3. carefully. Notice that where there is a hinge or fulcrum there is always some kind of reaction force at the hinge or fulcrum. This is why it often makes sense to take moments about the hinge or fulcrum, as the reaction force has no moment about that point. So in example 3.1, by considering the moments about the hinge, it is easy to find the force with which Peter pushes. MEI, 0/06/09 1/
MEI M Moments Section 1 Notes and Examples Example 3. shows how a problem can be solved by using a combination of resolving forces and taking moments. There are two unknown forces in this situation, R and S, so two equations are needed to solve the problem. s all the forces act vertically only one equation can be obtained by resolving forces. The second equation is obtained by taking moments about. In fact, you could solve this problem by taking moments about two different points, and not resolving forces at all. The example shows that S can be found directly by taking moments about. You should be able to see that you can find R directly by taking moments about. Either method is equally valid, and the choice in this case is really down to personal preference. Read the section on levers carefully. This is an important practical application of moments. Try the first discussion point on page 34. You can find the magnitude of the single force by considering the total vertical force in each case. To find the position of its line of action, you will need to consider the total moment of the forces P and Q in each case. Think carefully about the second discussion point on page 34. When you press hard, what is the effect on the frictional force which stops your fingers from slipping? MEI, 0/06/09 /
MEI Mechanics Moments of forces Section : Forces at an angle Notes and Examples These notes contain subsections on The moment of a force which acts at an angle Sliding and toppling The moment of a force which acts at an angle In this section of work you will solve problems by using a combination of resolving forces and taking moments. In previous work on equilibrium, you have solved two-dimensional problems involving two unknown forces. You do this by resolving in two perpendicular directions to obtain two equations describing the equilibrium. The directions involved are often horizontal and vertical, but you have also met situations, such as a particle on a slope, where it is easier to resolve in different directions, such as parallel to the slope and perpendicular to the slope. When solving a problem involving a rigid body, you must also consider possible turning motion, as well as possible movement in two dimensions. This means that you can find three equations to describe the equilibrium, and therefore you can solve problems involving three unknown quantities. This is the case in Examples 3.4 and 3.5. In both examples the unknown forces are found by resolving horizontally and vertically and taking moments about one point. You should be aware, however, that you can also solve these problems by resolving in one direction and taking moments about two points. Note, however, that even though you could write down four equations describing the equilibrium, by resolving in two directions and taking moments about two points, you cannot solve a problem involving four unknown forces. You would find that any one of your four equations could be obtained by combining the other three. When you are making a decision about how to solve a problem, remember that it is generally easier to take moments about a point which has the lines of action of several forces going through it. In the example on page 4, the two components of the reaction force, R V and R H, both act through O, so taking moments about O produces an equation involving only the weight and the tension T, allowing T to be calculated directly. Notice that you could also take moments about the other end of the rod to find R V directly, as an alternative to resolving vertically. MEI, 0/06/09 1/4
MEI M Moments Section Notes and Examples Example 3.6 shows a system which can be divided into two parts. Notice that you can resolve or take moments for just one part of the ladder, or for the system as a whole. It can be useful to treat the system as a whole, as this means the reaction force at the hinge and the tension in the string are not involved. In part (iv), by taking moments for the system as a whole the reaction force at the ground, R, can be found directly. Then the ladder has to be treated as two separate parts to find the reaction at the hinge and the tension in the string. Think about the discussion point at the top of page 46. Try to find the simplest possible method. Here are two further examples. Example 1 uniform plank of mass 5 kg is pivoted at and held at an angle of 30 to the vertical by a force applied at, perpendicular to. Find this force. Solution F R y 30 R x 5g Let the length of the plank be x. Taking moments about (as both R x and R y go through that point) 5gxcos60 F x 5g cos60 F F = 61.5 N Example uniform ladder of length 1.5 m and mass 48 kg rests with its top against a smooth wall and its foot on rough ground, 3.5 m from the base of the wall. Find the frictional and reaction forces at the base of the ladder. MEI, 0/06/09 /4
Solution MEI M Moments Section Notes and Examples R w 48g F R g 1 1.5 3.5 Resolving vertically: R g = 48g = 48 9.8 = 470.4 The reaction force at the ground is 470.4 N Resolving horizontally: R w = F Taking moments about the base of the ladder: 1.5 sin 6.5 48 g cos 0 R w 1 1.5F 6.5 48 9.8 1.5 F = 68.6 N The frictional force at the base of the ladder is 68.6 N 3.5 1.5 Sliding and toppling In previous work where you have used the particle model, motion is only possible if a resultant force acts on the particle. When you use the rigid body model, however, equilibrium can be broken either by acceleration caused by a resultant force, or by turning caused by a moment. Think about the discussion points on page 54 and the top of page 55. Consider the forces acting in each case, and decide whether there is likely to be a resultant force down the slope, causing sliding, or a resultant moment, causing toppling. In the discussion point at the foot of page 55, consider the moment of the forces about the point E. Read through the examples carefully, and look at the discussion point at the foot of page 57. Find a value for the coefficient of friction which would result in sliding occurring at the same angle as toppling occurs. Here is a further example. MEI, 0/06/09 3/4
MEI M Moments Section Notes and Examples Example 3 cubic block of mass 10 kg and side 1 cm, rests on a rough slope. The coefficient of friction between the block and the slope is 0.. The slope is gradually increased. Will the block topple or slide? Solution The block is on the point of toppling when its weight acts vertically through the its corner, as shown in the diagram. y looking at the geometry of the block it can be seen that this will occur when the angle reaches 45. 10g 45 The block is on the point of sliding when the frictional force up the slope is at its maximum possible value of R. R R 10g Resolving perpendicular to slope: Resolving parallel to slope: R = 10g cos R = 10g sin 0. 10g cos = 10g sin tan = 0. = 11.3 The block will slide when the slope is at an angle of 11.3 and topple at an angle of 45, so the block will slide. You can investigate sliding and toppling using the Flash resource Forces on a box. MEI, 0/06/09 4/4
Notes and Examples Mechanics Centre of Mass Section 1: Centre of Mass These notes contain subsections on Introduction to centre of mass Composite bodies Centre of mass for two- and three- dimensional bodies Introduction to centre of mass In your work on moments in chapter 3, you considered the whole weight of a rigid body such as a ladder to act at its centre. This is called the centre of mass. If the ladder were not uniform, the centre of mass would be at a different location. Look at the first discussion point on page 63. You need to consider whether the two short horizontal rods balance, as well as the longer one. You will need to measure the distances involved. In the second discussion point, you cannot determine the exact location of the centre of mass from the picture. s the gymnast is moving along the beam, you cannot assume that the centre of mass is directly above her hands. However, she is not (or should not be!) moving perpendicular to the beam. This gives some information on the position of the centre of mass. Read examples 4.1, 4. and 4.3 carefully. In example 4.1, the pivot is included to help you understand the concept that the centre of mass is the balance point of the object. This method is perfectly valid, but the method shown in Example 4. is the more usual way of solving a problem of this type. This uses the definition of centre of mass: moment of the whole mass at the centre of mass = sum of the moments of the individual masses ( m + m + m +...) x = mx + m x + m x +... 1 3 1 1 3 3 Here is a further example. Example 1 light rod lies along the x-axis and particles of mass 1 kg, kg and 5 kg lie at the points (, 0), (4, 0) and (7, 0) respectively. Find the coordinates of the centre of mass. MEI, 10/06/05 1/4
Mechanics Solution O 1 kg kg 5 kg (, 0) (4, 0) (7, 0) Relative to O: ( 1+ + 5) x = (1 ) + ( 4) + (5 7) 8 x = + 8 + 35 = 45 x = 5.65 The centre of mass lies at (5.65, 0). Composite bodies The positions of the centres of mass of the uniform objects given in the table at the bottom of page 67 are derived using calculus methods. This work is covered in Mechanics 3. Here is a further example involving a composite body. Example hammer has a uniform shaft 0.4 m long and a mass of 15 g. The shaft passes through the head of mass 1 kg whose centre of mass is 0.05 m from the end of the shaft. Find the position of the centre of mass of the whole hammer. Solution 0. m 0.15 kg 1 kg 0.05 m Relative to the end : ( 1+ 0.15) x = (1 0.05) + (0.15 0.) 1.15x = 0.05 + 0.05 x = 0.067 The centre of mass of the whole hammer is 0.067 m from the end of the head. Centre of mass for two- and three-dimensional bodies Working in two or three dimensions makes these problems feel more daunting than the one-dimensional work in the previous section. In three-dimensional work in particular it can be difficult to visualise the situation. It is important to draw a clear diagram in each case. MEI, 10/06/05 /4
Mechanics Read through all the examples in the book carefully. If you have not yet covered the work on vectors in Core 4, you may prefer to deal with each coordinate separately. However, it is still worth looking at the solutions which use vectors: in this context the vector is really just a convenient notation which allows you to write down the working for both, or all three, dimensions in one step. You do not need to know the mathematics of vectors covered in Core 4 in order to use this notation. Here are two further examples, one in two dimensions and one in three. In each case the solution is shown first by considering each coordinate separately, and secondly using vectors. Notice that the actual calculations are the same whichever method you use, but the vector method is more concise to write down. Example 3 Find the coordinates of the centre of mass of four particles of mass 5 kg, kg, kg and 3 kg situated at (3, 1), (4, 3), (5, ) and (-3, 1) respectively. Solution Method 1 For the x coordinate: ( 5 + + + 3) x = (5 3) + ( 4) + ( 5) + (3 3) 1 x = 4 x = For the y coordinate: ( 5 + + + 3) y = (5 1) + ( 3) + ( ) + (3 1) 1 y = 18 y = 1.5 The coordinates of the centre of mass are (, 1.5). Method (Vectors) x 3 4 5 3 1 = 5 + + + 3 y 1 3 1 x 4 1 = y 18 x = y 1.5 The coordinates of the centre of mass are (, 1.5). Example 4 Three pieces of uniform wood are stuck together to form two sides and the base of a box. O is the point where all 3 pieces meet, O is 1.5 m in the direction of the x-axis, O is 1 m in the direction of the y-axis and OC is 0.75 m in the direction of the z-axis. Find the coordinates of the centre of mass. MEI, 10/06/05 3/4
Solution z C 0.75 m O 1.5 m 1 m Mechanics rea 1.5 m² rea 1.15 m² rea 0.75 m² x y Since the wood has uniform density, the area is in proportion to the mass and therefore area may be considered as equivalent to mass. Shape O OC OC Total rea 1.5 1.15 0.75 3.375 Centre of mass (0.75, 0.5, 0) (0.75, 0, 0.375) (0, 0.5, 0.375) ( x, y, z) Method 1 Relative to OC 3.375x = 1.5 0.75 + 1.15 0. 75 x = 0.583 Relative to O 3.375y = 1.5 0.5 + 0.75 0. 5 y = 0.333 Relative to OC 3.375z = 1.15 0.375 + 0.75 0. 375 z = 0.08 The coordinates of the centre of mass are (0.583, 0.333, 0.08). Method (Vectors) x 0.75 0.75 0 3.375 y = 1.5 0.5 + 1.15 0 + 0.75 0.5 z 0 0.375 0.375 x 1.96875 3.375 y = 1.15 z 0.70315 x 0.583 y = 0.333 z 0.08 The coordinates of the centre of mass are (0.583, 0.333, 0.08). MEI, 10/06/05 4/4
MEI Mechanics Energy, Work and Power Section 1: Work and energy Notes and Examples These notes contain subsections on Work and energy Gravitational potential energy Work and kinetic energy of two-dimensional motion Work and energy The concept of work is introduced in this section. So far, in your study of Mechanics, you have met many situations in which bodies or particles are acted on by forces, but no reference to the source of such forces has been considered. When a force is applied to an object, causing it to move, then work has been done by the force. Read through the examples in the book carefully. Example 5.3 demonstrates the work-energy principle, which is the basis for the whole of this section of work. Exercise 5 focuses on applications of the work-energy principle. Notice that many of the problems can be solved using a combination of Newton s nd Law and the constant acceleration formulae. However, the workenergy principle is often quicker and more efficient. The example below shows both methods applied to the same problem. Example 1 car of mass 1000 kg travelling at 0 ms -1 applies its brakes and comes to rest over a distance of 10 m. Find the braking force, assumed to be constant. Solution 1 (using constant acceleration formula and Newton s nd Law) Constant acceleration formula v² = u² + as 0 = 0² + a 10 0a = -400 a = -0 Newton s nd Law F = ma F = 1000-0 = -0000 The braking force is 0000 N. Solution (using the work-energy principle) K.E. lost = ½ 1000 0² = 00000 J Work done by braking force = -00000 J F 10 = -00000 MEI, 08/0/10 1/5
MEI M Energy Section 1 Notes and Examples F = -0000 The braking force is 0000 N. (Notice that the braking force is negative, as it acts in the opposite direction to the motion.) Gravitational potential energy Gravitational potential energy is just one form of potential energy. You can think of potential energy as energy stored in an object, which gives it the potential to move when released. If a ball is dropped, the gravitational potential energy of the ball is converted into kinetic energy. s the height of the object decreases, the gravitational potential of the ball decreases, and its kinetic energy increases. nother form of potential energy, which you will meet if you study Mechanics 3, is elastic potential energy. This is the energy stored in a spring when it is stretched or compressed. n object attached to the spring has the potential to move when the spring is released. In the situations considered in this section, the work done by a force on an object will either change the speed or the vertical position of the object, or a combination of both. When solving problems involving a change in vertical position, it is often convenient to use the work-energy principle in a slightly different form. Remember: The total work done by the forces acting on a body is equal to the increase in the kinetic energy of the body. If a change in the vertical position is involved, then one of the forces acting on the body is its weight. The work done by this force is mgh, where h is the increase in height. (The minus sign is needed as the weight acts downwards.) The work-energy principle can therefore be rewritten as: Work done by external forces other than weight + work done by weight = increase in KE or: Work done by external forces other than weight mgh = increase in KE or: Work done by external forces other than weight = mgh + increase in KE = increase in GPE + increase in KE = increase in total mechanical energy MEI, 08/0/10 /5
MEI M Energy Section 1 Notes and Examples This version of the work-energy principle can sometimes be easier to use when solving problems. Notice that the principle of conservation of mechanical energy is a particular case of this principle: if there are no external forces other than weight, no work is done, so the increase in total mechanical energy is zero. Read the examples in this section carefully. Here are two additional examples. Example stone of mass 1 kg is dropped from a height of m onto a pond. The stone begins to fall through the water at ms -1. What is the energy lost in hitting the water? Solution In this example, no external force acts on the stone, so mechanical energy is conserved. Therefore loss in GPE = gain in KE Loss in GPE = mgh = 1 9.8 = 19.6 J Gain in KE = 19.6 J KE as the stone begins to fall through the water = ½ 1 ² = J Energy lost = 19.6 - = 17.6 J (Note that it is not necessary to work out the speed of the stone as it hits the water.) Example 3 body of mass 3 kg falls vertically against constant resistive forces of 15 N. The body passes through two points and with speeds of and 1 ms -1. Using energy considerations, find the distance. Solution Let h be the distance. KE gained = 0.5 3 (1² - ²) = 10 J GPE lost = 3 9.8h = 9.4h Total increase in mechanical energy = 10 9.4h Work done by resistive forces = -15h 10 9.4h = -15h 14.4h = 10 h = 14.6 The distance is 14.6 m. (The resistive forces act against the direction of the motion, so the work done is negative) MEI, 08/0/10 3/5
MEI M Energy Section 1 Notes and Examples Work and kinetic energy of two-dimensional motion The principles covered in the previous section can also be applied to twodimensional motion. The important thing to remember in a two-dimensional situation is that work done by a force is equal to the force distance moved in the direction of the force. So far, you have only needed to consider whether the work done is positive or negative, i.e. whether the force acts in the same or the opposite direction to the motion. Now, in two-dimensional work, you may need to resolve a force to find its component in the direction of the motion. The following example shows two slightly different approaches to the same problem. Example 4 particle of mass kg slides down a slope which makes an angle of 30 with the horizontal. frictional force of 4 N acts against the motion. Find the speed of the particle when it has travelled 5 m. R 4 5 m g 30 Solution 1 Total force in the direction of motion = g sin30 4 Work done by forces = 5(g sin30 4) = 9 J Gain in K.E. = ½ v² = 9 v 9 Solution Work done by frictional force = -4 5 = -0 J G.P.E. lost by particle g 5sin30 49 J K.E. gained by particle = ½ v² = v² Work done = total gain in mechanical energy 0 v 49 v 9 v 9 In Solution 1, all the external forces are considered in the work done. In Solution, the weight is not considered in the work done, but the change in G.P.E. is included in the calculation of the energy change. You can see that MEI, 08/0/10 4/5
MEI M Energy Section 1 Notes and Examples the two calculations are effectively the same. You can use whichever method you prefer. Notice that this example could also be solved using constant acceleration formulae and Newton s nd Law. MEI, 08/0/10 5/5
Notes and Examples Notice that the equation Power = force velocity MEI Mechanics Energy, Work and Power Section : Power refers to the velocity at a particular instant. The velocity of the vehicle does not need to be constant. If the vehicle is accelerating, and the engine is kept at the same power, the driving force must decrease as the velocity increases. ssuming that all other forces (such as resistance) remain constant, the net forward force also decreases, and hence the acceleration decreases. Eventually, the net forward force becomes zero, so that the acceleration becomes zero and the velocity is constant. The vehicle is then travelling at its maximum speed for that particular value of the power. When solving problems using power, remember that the force referred to in the equation P = Fv is the driving force (often of an engine, but sometimes the force produced by a person). s F is often used for frictional forces, and in Newton s nd Law F = ma, you may find it clearer to use another letter such as D to refer to the driving force, as shown in the second part of Example 5.13. Remember that in Newton s nd Law, F refers to the net force in the direction of the motion. This will usually be different from D, as you need to take into account resistance forces and sometimes a component of the weight. Here is a further example. Example 1 train of mass 100 tonnes, working at a maximum power of 300 kw, ascends a slope 1 which makes an angle of to the horizontal, where sin 50, at a constant speed of 15 ms -1. (i) Find the resistance to the motion of the train. (ii) The train s power is increased to 500 kw. ssuming that resistances remain the same, find the acceleration of the train at the instant that it is travelling at 0 ms -1. (iii) Find the maximum speed of the train on this slope at the new power of 500 kw. Solution R D F 100000g sin 1 50 MEI, 0/06/09 1/
MEI M Energy Section Notes and Examples (i) Power = Dv 300000 = 15D D = 0000 Newton s nd Law: D F 100000g sin = 0 1 0000 F 100000 9.8 50 F = 400 The resistance force is 400 N. (ii) Power = Dv 500000 = 0D D = 5000 Newton s nd Law: D F 100000g sin = 0 1 5000 400 100000 9.8 = 100000a 50 a = 0.05 The acceleration is 0.05 ms -. (iii) t maximum speed acceleration is zero. s all other forces are the same as in part (i), D must be 0000. Power = Dv 500000 = 0000v v = 5 The maximum speed is 5 ms -1. MEI, 0/06/09 /
MEI Mechanics Impulse and momentum Section 1: Introduction to impulse and momentum Notes and Examples These notes contain subsections on Impulse Conservation of momentum Impulse Look at the Discussion Point at the foot of page 118. To do this, you need to assume that a constant force is applied to the ball for the time where the racquet is in contact with the ball. You will need to calculate the acceleration of the ball during this time. Remember that the final velocity is in the opposite direction to the initial velocity! Notice that a greater force is required for a shorter impact. Think about the discussion point at the foot of page 119. Note in particular that the magnitude of the momentum of an object is thought of as its resistance to being stopped. For example, a large ship whose engines have been stopped takes a very long time to come to a halt, even if it is moving quite slowly, because of its large mass. small rowing boat, however, stops fairly quickly after rowing ceases, even if it has been moving quite fast. Read examples 6., 6.3, 6.4 and 6.5 carefully. In the examples in two dimension, notice that a greater impulse is required if the direction of motion changes. Conservation of momentum The law of conservation of momentum is fundamental to this chapter. Make sure that you understand the explanation on pages 18-19, which shows how Newton s 3 rd Law gives rise to the law of conservation of momentum. Read the examples carefully. It is very important in this work to draw a clear diagram showing the speeds and directions of motion of each body before and after the impact. Here are two further examples. The masses and initial speeds of the particles are the same in each of the two examples, but the outcome of the impact is different in each case. MEI, 3/11/10 1/3
MEI M Momentum Section 1 Notes and Examples Example 1 Two particles, and, of masses m and 4m respectively, are moving in the same direction with speeds of 5u and u respectively, so that is catching up with. fter they collide, the speed of is u. (i) Find the velocity of after the collision. (ii) Find the kinetic energy lost in the collision. Solution (i) efore impact fter impact 5u u v u m 4m m 4m Conservation of momentum: 5mu + 4mu = mv + 8mu mu = mv v = u The velocity of is u in the same direction as before. (ii) K.E. before impact = K.E. after impact = K.E. lost = 6mu. 1 1 m ( 5u) 4 mu = 1 1 mu 4m (u ) = 9 mu. 17 mu. Example Two particles, and, of masses m and 4m respectively, are moving in the same direction with speeds of 5u and u respectively, so that is catching up with. fter they collide, the speed of is.5u. (i) Find the velocity of after the collision. (ii) Find the kinetic energy lost in the collision. Solution (i) efore impact fter impact 5u u v.5u m 4m m 4m Conservation of momentum: 5mu + 4mu = mv + 10mu -mu = mv v = -u The speed of is u in the opposite direction to before. (ii) K.E. before impact = K.E. after impact = K.E. lost = 3 mu. 1 1 m ( 5u) 4 mu = 1 1 5 mu 4m ( u ) = 9 mu 13mu. MEI, 3/11/10 /3
MEI M Momentum Section 1 Notes and Examples These two examples show the same initial set of circumstances, with two different outcomes. You should be able to see that for the given masses and initial velocities, there are in theory an infinite number of possible sets of final velocities. The law of conservation of momentum only tells us that the final momentum of the system must equal 9mu as that is the initial momentum of the system. However, it is probably clear to you that not all theoretical solutions are possible in practice. For example, consider a final velocity for of -100u. This gives a final velocity for of 7.5u. These velocities satisfy the equation of conservation of momentum, but it seems obvious that they are not possible in practice. This can easily be justified by energy considerations: these final velocities would result in an increase in kinetic energy, which is not possible in the absence of any external forces. t present you can only solve examples like the ones above if you are given one of the final velocities, or if, as in Examples 6.7 and 6.8 in the textbook, the two bodies join together after impact. This seems unsatisfactory: in Mechanics we can usually predict what will happen in a given situation before it happens, if we have enough data. You may feel that we ought to be able to calculate the velocities of both particles after the impact. Clearly, to do this, we need to have further information. The outcome of the situation depends not only on the masses and speeds of the particles initially, but also on the type of materials of which they are made: how elastic ( bouncy ) they are. You can probably see that in the two examples given earlier, Example must involve particles made of a more elastic material than those in Example 1, as the first particle rebounds. This idea is backed up by the energy considerations: in Example less energy is lost during the impact than in Example 1. In the next section you will learn how the elasticity or bounciness of the objects involved in a collision affects the outcome of the collision. You will then, given the relevant information, be able to calculate the final velocities of both objects involved in a collision. MEI, 3/11/10 3/3
MEI Mechanics Impulse and momentum Section : Newton s Law of impact Notes and Examples These notes contain subsections on Newton s Law of impact Oblique impacts with a smooth plane Newton s Law of Impact The speed of approach and speed of separation are the relative velocities of the two objects before and after the impact. It is important to decide which direction is positive in order to apply the formula correctly. Read examples 6.10 and 6.11 carefully. Notice that in Example 6.11 the two objects are initially moving towards each other, so the initial speed of is taken to be negative. The speed of approach is therefore found by adding the magnitudes of the two speeds. In this case the solution shows that the objects both change direction after the collision, but this does not happen in every case. If the two objects coalesce and move off together after the impact, they both have the same speed and so the speed of separation is zero. The coefficient of restitution in such a case must therefore be zero. You can investigate collisions using varying initial speeds and coefficients of restitution using the Geogebra resource Collisions. The two further examples below are the same as those in the Notes and Examples for Section 1 of this chapter, with an additional part (iii) in each case, in which the coefficient of restitution between the particles is to be found. This illustrates how in two situations in which the masses and initial velocities of the particles are identical, a different coefficient of restitution results in a different outcome from the impact. Example 1 Two particles, and, of masses m and 4m respectively, are moving in the same direction with speeds of 5u and u respectively, so that is catching up with. fter they collide, the speed of is u. (i) Find the velocity of after the collision. (ii) Find the kinetic energy lost in the collision. (iii) Find the coefficient of restitution between the two particles. MEI, 0/11/09 1/4
MEI M Momentum Section Notes and Examples Solution (i) efore impact fter impact 5u u v u m 4m m 4m Conservation of momentum: 5mu + 4mu = mv + 8mu mu = mv v = u The velocity of is u in the same direction as before. (ii) K.E. before impact = K.E. after impact = 1 1 m ( 5u) 4 mu = 1 1 mu 4m (u ) = 9 17 mu 6mu. K.E. lost = 9 mu. 17 mu. (iii) Speed of approach = 5u u = 4u Speed of separation = u u = u u e = = 0.5. 4u Example Two particles, and, of masses m and 4m respectively, are moving in the same direction with speeds of 5u and u respectively, so that is catching up with. fter they collide, the speed of is.5u. (i) Find the velocity of after the collision. (ii) Find the kinetic energy lost in the collision. (iii) Find the coefficient of restitution between the two particles. Solution (i) efore impact fter impact Conservation of momentum: 5mu + 4mu = mv + 10mu -mu = mv v = -u The speed of is u in the opposite direction to before. (ii) K.E. before impact = K.E. after impact = K.E. lost = 5u u v.5u m 4m m 4m 3 mu. 1 1 m ( 5u) 4 mu = 1 1 5 mu 4m ( u ) = 9 mu 13mu. MEI, 0/11/09 /4
MEI M Momentum Section Notes and Examples (iii) Speed of approach = 5u u = 4u Speed of separation =.5u + u = 3.5u 3.5u e = = 0.875 4u In Example 1, the coefficient of restitution is lower, i.e. the particles are less elastic or bouncy. This results in a greater loss in kinetic energy. In Example, the coefficient of restitution is higher, so the particles are more elastic and less kinetic energy is lost. The combination of the law of conservation of momentum and Newton s Law of Impact makes it possible to predict the outcome of any collision involving two bodies with known masses and initial velocities and a known coefficient of restitution. In the general case, illustrated by the diagram below, efore impact fter impact u m m u v m m v it is possible to find expressions for v and v in terms of m, m, u, u and e. v m u m u m em m ( u u ) v m u m u m em m ( u u ) It is also possible to find an expression for the loss in kinetic energy caused by the collision. mm Loss in K.E. = ( u u )²(1 e²) ( m m ) You may like to derive these expressions for yourself, using the law of conservation of momentum and Newton s Law of Impact. Notice from these results that: If e = 0, the two velocities after the impact are the same. This shows that if e = 0 (a perfectly inelastic collision) the particles coalesce and move together. If e = 1, the loss in kinetic energy is zero. MEI, 0/11/09 3/4
MEI M Momentum Section Notes and Examples (Note that the expressions above are not formulae to be learnt and applied to problems: they are given to illustrate the points above.) Oblique impact with a smooth plane Read the text and example carefully. Notice that while the velocity parallel to the plane remains unchanged after a collision, the magnitude of the velocity perpendicular to the plane is always reduced by the collision (unless e = 1). This means that the angle between the plane and the direction of motion decreases after the collision. You can investigate impacts with a plane using the Geogebra resource Oblique impacts. The example below shows that the direction of motion after an impact is independent of the magnitude of the velocity before the impact. Example 3 n object is initially moving at 30 to a smooth plane. The coefficient of restitution between the ball and the plane is 0.8. Find the direction in which the object moves after colliding with the plane. Solution 30 u v Conservation of momentum parallel to plane: Newton s Law perpendicular to plane: u cos 30 vcos vsin 0.8u sin 30 Dividing: v sin 0.8sin 30 v cos cos 30 tan 0.8 tan 30 = 4.8 The object moves at an angle of 4.8 to the plane after the collision. MEI, 0/11/09 4/4
MEI Mechanics Frameworks Notes and Examples These notes contain subsections on Strategies for solving framework problems Worked example Strategies for solving framework problems The actual mathematics involved in solving a framework is quite straightforward: you only have to resolve forces and sometimes take moments. However, there are usually a large number of unknown forces involved, and a bewildering number of possible approaches. It can be tempting to write down a lot of equations by resolving horizontally and vertically at several different points, but if this is done without careful thought, the result may be a large number of equations involving several unknowns. However, if you think out a sensible strategy before you begin, the solution of the problem should be straightforward. Read the examples in the book carefully. Notice the general strategy used in these example (summarised in the Key Points on page 170): i) Consider external forces such as reaction forces and weights. Reactions are usually best represented as components. a) Resolve horizontally and vertically b) Take moments about some suitable point, if necessary. (Note: in some cases it may not be necessary to find the external forces.) ii) Put on the internal forces in each rod: draw a force from each end of the rod. (Note: in simple cases you may be able to deduce from the diagram whether each rod is in tension or thrust, in which case you can draw the arrows in the correct direction. If you cannot be certain of all the directions, you may find it easier to draw in the internal forces as if every rod were in tension: then any tensions which turn out to be negative are actually thrusts.) iii) Choose a point on the framework where there are only one or two unknown forces. Resolve horizontally and/or vertically to find these unknown forces. Go on to another point, and continue until you have found all the internal forces. Note: in most cases that you meet, the angles involved are likely to be simple ones such as 30, 45 and 60. When you calculate each force, you should write down the exact value using surds. If you need to use a result in another calculation, you can then use this exact value. nswers can then be given to a suitable degree of accuracy if required. MEI, 0/06/09 1/4
MEI M Frameworks Section 1 Notes and Examples This procedure is illustrated in the following example. Note in particular: That it is possible in this case to deduce whether each rod is in tension or compression before starting The reasoning behind the choice of points at which to resolve That answers are initially given in terms of surds, and these exact values are used in further calculations. Worked example Example 1 In the framework below,, C, CD and D are light, freely hinged rods of equal length. The framework is attached to a vertical wall at point and is resting on a smooth point, on the top of another wall. The rod is horizontal. (i) Find the reactions at and. (ii) Find the stresses in each rod, stating whether they are tensions or thrusts. D C 100 N Solution Y D R C 100 N Redraw the diagram showing the external forces. 60 X Let the length of each rod be a. Taking moments about : Ra 100(a + a cos 60 ) = 0 R = 150 The reaction at is 150 N. Resolve vertically: Y + R = 100 Y = 100 R = 100 150 Y = -50 Resolve horizontally: X = 0 The reaction at is 50 N vertically downwards. Notice that Y was drawn initially in the wrong direction MEI, 0/06/09 /4
MEI M Frameworks Section 1 Notes and Examples Redraw the diagram showing the internal forces and the external forces. T 50 D T 1 T 1 C T T 5 60 T 5 T 3 T 3 150 T 4 T 4 100 Notice that in this case it is possible to deduce the directions of all the internal forces. Look at C: T 4 must act upwards to counteract the 100 N force; and as this means that T 4 also acts to the right, T 1 must act to the left. Look at : you can deduce the directions of T and T 3 in a similar way. Look at D: you can now deduce the direction of T 5. You can now see that D and CD are in tension, and, C and D are in compression. Notice first that at both and C there are only two unknown forces, and furthermore at each of these two points one of these unknown forces can be found directly by resolving vertically. The other can then be found by resolving horizontally. Resolve vertically at : T sin 60 50 0 1 T 3 50 T 100 3 The stress in D is a tension of 57.7 N. Resolve horizontally at : T cos 60 T3 0 T 1 3 T T 50 3 3 The stress in is a thrust of 8.9 N. Resolve vertically at C: T sin 60 100 0 4 1 T 4 3 100 00 T 4 3 The stress in C is a thrust of 115.5 N. Resolve horizontally at C: T 4 cos 60 T1 0 T 1 1 T 4 T 1 100 3 MEI, 0/06/09 3/4
MEI M Frameworks Section 1 Notes and Examples The stress in CD is a tension of 57.7 N There are several possible ways to find the final force. Resolving vertically at D is the simplest as this involves only two forces. Resolve vertically at D: T 5 sin 60 T sin 60 0 T 5 T T 100 5 3 The stress in D is a thrust of 57.7 N. MEI, 0/06/09 4/4