Exam Question 5: Work, Energy, Impacts and Collisions. June 18, Applied Mathematics: Lecture 5. Brendan Williamson.

Transcription

1 Exam Question 5: Work, Energy, Impacts and June 18, 016

2 In this section we will continue our foray into forces acting on objects and objects acting on each other. We will first discuss the notion of energy, specifically the principle of conservation of energy, before moving on to how impacts between objects affect their respective velocities. Some equations learned here should be familiar to those studying the Leaving Cert Physics course, e.g. those related to work, power etc., and although not all of them will be used extensively in tackling Q5 of the paper, they are useful in other chapters, and I have mentioned them here, all at once, for the sake of both clarity and brevity. As a result this lecture may be referred back to when I discuss other topics.

3 Work Work and power and not particularly difficult concepts, nor are they tools that you will learn how to use, unlike the equations of motion, or the principle of conservation of momentum, which we will learn later. Simply put, they are measurements that you will see appear in numerous topics. The work done by a force is given by the distance it moves a given object multiplied by the component of the force in that direction, i.e. W = Fs. Work done is a scalar and is measured in joules. It might be counter-intuitive to think of work done as a scalar as the equation appears to be a vector (F ) multiplied by a scalar (s). However F in this case is the component of the force acting in a given direction, and therefore appears as a scalar in this equation.

4 Work Example: If a ship travels 100m North with the engine providing a force of 10N, then the work done is W = Fs = 100(10) = 1000 joules. But what if there was wind coming from the East so that the ship needed to aim North-East so as to travel directly North? The force coming from the engine would now be 5 i + 5 j, and so the work done in travelling 100m North would be W = Fs = 5 (100) = 500 joules, which is less than before. It may seem counter-intuitive to define work like this, after all the same conclusion was achieved. However work done was originally defined as weight lifted through a height, and so battling other forces just to keep steady is not seen as work. Also if you are lifting something at a constant velocity, the force applied is proportional to the gravitational pull of the planet that you are on, and so that is why work is proportional not only to distance travelled, but also to the force applied. As we will see later, this definition of work also ties in nicely with our notions of kinetic and potential energy.

5 Power Power is simply defined as the rate at which work is done, given by the equation P = W /t. So when a system moves objects from place to place, it would be the speed at which the objects are moved which determines the power in the system. We can see this more clearly by replacing W with Fs in the equation for P, which gives us P = Fs/t = Fv, where v is speed. Power is measured in watts. For example, if a machine was pushing an object over a rough surface, with just enough force so as to counteract friction but not to cause any acceleration, then the speed at which the machine pushes the object is proportional to the power. Example: A car weighing 1,000kg travelling up a hill of incline A with sin A = 7/5 has an engine whose maximum power output is 100,000 watts. The coefficient of friction between the car and the road is µ = 0.1. Assume there is no air resistance or any other forces other than gravity. What is the maximum speed of the car?

6 Power The following are the forces acting on the car, F is the force coming from the engine and M = As sin A = 7/5, cos A = 4/5, and so we can show that R = 960g, so that µr = 96g. As there is no acceleration when the car is travelling at its maximum speed, F = 96g + 80g = 376g at this time. We also know that P = 100, 000, and as P = Fv we get that v 7.14m/s.

7 Energy Energy is closely related to work. The energy of a body is its capacity for doing work. Some bodies have energy because they are high above the ground. This is called potential energy. Other objects have energy because of their speed. This is called kinetic energy. There are of course other types of energy, but they won t be discussed here. The potential energy of an object is measured by the amount of work it can do by moving from its position to some fixed height (usually the ground). If the distance between the object and the ground is h, then since the force that would move the object to the ground is F = mg where m is the mass of the object, the work done, and hence the potential energy of the object is W = Fs = mgh. The kinetic energy of an object is the work it does before it comes to rest. This seems ambigious, but it is not. If deceleration is constant (say it s equal to a where a > 0) then by our fourth equation of motion 0 = u as as = u /. But a = F /m where F is the force causing the deceleration, so that Fs = mu /. So the work done is the same regardless of the deceleration involved.

8 of Energy A specific case of a more general principle in physics is that when no other forces are involved the sum of the potential and kinetic energy of an object is constant. This is called the of Energy. We can prove this using our equations of motion. As no other forces are involved, the motion of the object is directly downward with acceleration g. See that if an object moves from height h 1 to h and in doing so changes velocity from u to v, then v = u + g(h 1 h ) P.E. + K.E. = mgh + mv = m (gh 1 + u = P.E. 1 + K.E. 1. = m (gh + u ) = mgh 1 + mu ) + g(h 1 h ) This means that we can find the work done by an object in freefall either by finding the decrease in potential energy or the increase in kinetic energy, depending on whether or not we are given the loss in height or the increase in speed.

9 Impulses Remember that the momentum of an object is the product of its mass and velocity, and is measured in N s (Newton seconds, we will see why later). When two objects collide, the velocities of both will change. We define the impulse imparted onto a body by an impact as the change in its momentum. Mathematically, we have that I = mv mu. Note that the impulse I is a vector quantity, and hence it is important to use the right signs for u and v if the direction of motion changes. For example, if a tennis ball of mass 50g starts at rest, is hit by a racquet and leaves the racquet with velocity 30 m/s, then the impulse imparted to the ball is I = mv mu = 0.05(30) 0.05(0) = 1.5 N s. Now assume the ball is coming at the tennis player with a velocity of 5 im/s and the impulse remains the same. Then we must apply a direction, note that the impulse occurs in the negative i direction. Then we have 1.5 = 0.05v 0.05(5) v = 5 im/s.

10 Direct Say two objects collide. Both objects change velocity, and hence they have an acceleration during impact. The time for this impact is very small, but as the velocity changes so fast, the acceleration is very large. See that for one of the objects, I 1 = m 1 (v 1 u 1 ) = m 1 a 1 t = F 1 t where F 1 is the force causing the acceleration. For the other object, by Newton s Third Law of Motion F = F 1, so that the impulses are equal in magnitude but in the opposite direction. Therefore I 1 = I and m 1 (v 1 u 1 ) = m (v u ) m 1 u 1 + m u = m 1 v 1 + m v. In other words, the total momentum of the objects before impact is equal to the total momentum after impacts. This is referred to as the principle of conservation of momentum.

11 Direct Example: A snooker ball is at rest, and is hit by a second snooker ball travelling at 5 m/s. After impact the second snooker ball is travelling in the same direction at 1 m/s, what is the velocity of the first snooker ball if both balls have mass 0.kg? What if the first snooker ball has mass 0.4 kg? In the first case m 1 = m = 0., u 1 = 0, u = 5, v = 1 and we are looking for v 1. We therefore have that 0.(0) + 0.(5) = 0.v (1) v 1 = 4m/s. Note that the mass of the balls was unimportant, as long as the masses were equal we would get the same answer. If m 1 = 0.4, then 0.4(0) + 0.(5) = 0.4v (1) v 1 = m/s.

12 Direct Example: A bullet of mass 0g hits a block of wood of mass 980g, which is at rest on a smooth flat table, at a speed of 490m/s. After impact the bullet becomes embedded in the block of wood. At what speed does the joint mass travel after impact? What is the total loss in kinetic energy? Because the bullet has become embedded in the block of wood, we know that v 1 = v. Also, m 1 =.0, m =.98, u 1 = 490 and u = 0 so that.0(490) +.98(0) = ( )v v = 9.8m/s. To find the loss in kinetic energy, we simply look at the sum of the kinetic energy of the bullet and block before and after impact. K.E. Before = (0.0)490 K.E. After = (1)9.8 = 401 joules. = 48.0 joules. So the loss in kinetic energy is joules.

13 Direct and of Energy We can also combine the principle of conservation of momentum and the principle of conservation of energy to answer what appear to be intractable questions. Such questions have fallen out of favour in recent years but are still worth thinking about. Example: Say the block in the previous question was hanging from a light 10m string when it was hit by the bullet. How high would the joint mass travel? What angle did the block swing through before coming to rest? As before the velocity of the joint mass immediately after impact is 9.8 m/s. After impact, the only vertical force applied to the joint mass is gravity, and so by the principle of conservation of energy the kinetic energy lost equals the potential energy gained. The kinetic energy immediately after impact equals 48.0 joules, just like before. The potential energy when the block comes to rest is mgh = 9.8h, where h is the height gained by the block. Setting these equal we get h = 4.9m. A quick sketch and some trigonometry yields that the angle the block swings through is cos 1 ( ) degrees.

14 The way in which the velocities of objects change during impact can be described using a quantity called the coefficient of restitution, which measures the elasticity between two objects. Let s first consider one object hitting a stationary surface. Say a snooker ball hits the cushion of the table perpendicularly at a speed of 5 m/s, and after impact moves in the opposite direction at a speed of 3 m/s. The ratio between these speeds is 3/5, which we refer to as the coefficient of restitution, e. More formally, if we take the cushions of the table as the i and j axes, and the snooker ball travels at u i before impact and v i afterwards, then e = v/u. Note that if u > 0, then v < 0 so the negative sign serves to make e positive. It holds that e is always between 0 and 1.

15 Now say that the snooker ball travels obliquely, with velocity 4 i + j, and it hits the cushion parallel to the j axis. What is its velocity after impact if the coefficient of restitution between the ball and the cushion e = 3/4? The following is a diagram of the motion of the ball. As there are no forces acting on the ball in the j direction, the j velocity remains unchanged. The relationship between the new and old velocity in the i direction is dictated by the coefficient of restitution, so if v i + j is the final velocity, then e = v/u 3/4 = v/4, so that v = 3. So the new velocity is 3 i + j.

16 and Direct We can also apply this principle to two moving objects. Say two snooker balls are travelling in a straight line, one at a velocity im/s and the other at a velocity of 5 im/s, when they collide. After collision their velocities are 4 im/s and 3 im/s respectively. What is the coefficient of restitution? It is best to think about this in terms of relative velocity. From the perspective of the nd ball, the first ball is travelling towards it with a speed of 5 = 3 m/s. More formally, we have that u ab = u b u a = 3 im/s. So the impact experienced is like that of an object travelling at 3 m/s hitting a stationary object. After impact, from the perspective of the second ball, it appears that the first ball is moving away with a velocity of 1 m/s. More formally we have that v ab = v b v a = im/s. These are the new and old velocities we use to find e. In this case, e = ( 1)/3 = 1/3.

17 and Direct In general, if the initial velocities of the balls were u 1 i and u i and the final velocities were v 1 i and v i, our relative velocities would be u u 1 and v v 1 and so we would have that e = v v 1 u u 1. The fact that the coefficient of restitution is always the same for two objects regardless of their velocities is referred to as Netwon s Experimental Law of Restitution. If one of the objects is stationary, such as a wall, a snooker cushion or the ground, then by setting v 1 = u 1 = 0 we get the equation for one object that we discussed earlier. Before when we dealt with the collision of two moving objects we needed the masses and initial velocities of both objects and the final velocity of one of them so that we could find the final velocity of the fourth. Now if we are given the coefficient of restitution between both objects we only need the mass and initial velocities of both objects.

18 and Direct Example: Two smooth spheres have velocities 7 i and 4 i respectively before they collide. Their masses are 1kg and kg respectively, and their coefficient of restitution is e = /3. Find their velocities after collision. From the principle of conservation of momentum we know that 1(7) + (4) = v 1 + v 15 = v 1 + v. From the equation for e we get that 3 = v v = v v 1 3 = v v 1. We therefore have simultaneous equations in v 1 and v, and we can easily solve them to get v 1 = 11/3 m/s, v = 17/3 m/s. As expected, the velocity of the faster object in the i direction decreased and that of the slower object increased.

19 and Direct What is the percentage loss in kinetic energy due to the collision? K.E. Before = m 1u 1 K.E. After = m 1v 1 + m u + m v = 40.5 joules. = joules. So the total loss in kinetic energy is 1.67 joules, and therefore the percentage loss is %.

20 So far we have only considered objects that travel along the same one-dimensional path. This way we can say that they both move only in the i direction and also we know that they collide directly. What if they collide obliquely? For example, what happens when two smooth spheres moving purely in the i and j direction respectively collide is a significantly more complicated question than what we have studied thus far. Among other things, we need to know at what angle they collide, as this surely affects the trajectory (see below).

21 Our problems will be made simpler in that when two spheres collide the line connecting their centres will always be along the i-axis (or occasionally the j-axis but let s keep things simple), as in the first diagram in the previous slide. Because of this, the j-velocity of both objects will be unchanged by the collision, much like when the snooker ball hit the cushion obliquely when we discussed elasticity. The i velocities will be subject to both the principle of conservation of momentum and the coefficient of restitution equations. So essentially we treat the i velocities as a direct collisions question and let the j velocities be unchanged. Example: Two smooth spheres of mass M and 3M respectively are travelling at velocities of 3 i + 4 j and 4 i + 3 j respectively, where i is along their line of centres at the time of collision. The coefficient of restitution between the two spheres is e = 4/7. Find their velocities after impact.

22 The following is a visual idea of what is happening. Let v 1 = p i + r j and v = q i + s j be the final velocities of the M and 3M spheres respectively. As the forces the objects exert on each other upon impact are purely in the i direction, the j velocities are unchanged and so r = 4, s = 3. Also, by the principle of conservation of momentum, M(3) + 3M( 4) = Mp + 3Mq 9 = p + 3q. Also, from the equation for e we get that 4 7 = q p 4 3 q p = 4.

23 Solving this system of simultaneous equations gives us p = 1/4 and q = 5/4, and so the final velocities of the M and 3M spheres are 1 4 i + 4 j m/s and 5 4 i + 3 j m/s respectively. What is the angle through which the 3M mass is deflected? The following diagram shows the actual path of the ball (in black) and the trace of the original path (in grey). θ is the angle of deflection.

24 The easiest way to find θ is to find the slope of both arrows and use the formula for finding the angle between two lines. The original line has slope 3/4, and the slope of the actual path is 1/5. Then tan θ = ± = ± As θ is the acute angle (we can tell this from the diagram) we let ± = +, so that θ =

25 What is the percentage loss in kinetic energy? K.E. Before = M(5) K.E. After M(6.6) + 3M(5) + 3M(4.19) So the percentage loss in kinetic energy is (1.89M/50M) %. = 50M. = 48.11M.

26 Example: A smooth sphere of mass M collides with another smooth sphere of mass M which is at rest. Before impact the direction of motion of the lighter sphere is at an angle 45 with the line of centres at the instant of impact. The direction of motion of the lighter sphere is deflected through an angle θ. The coefficient of restitution between the two spheres is e. Show that tan θ = 1 e e +.

27 The initial velocity of the lighter sphere can be shown to be u i + u j for some u > 0. Let a i + b j and c i + d j be the final velocity of the lighter and heavier sphere respectively. Then b = u and d = 0. By the principle of conservation of momentum we have that um = am + cm u = a + c. By the equation for e, we have that e = c a u c a = eu. We can find that a = (1 e)u/3. As the new angle of direction of motion of the lighter sphere is θ + 45, we have that tan(θ + 45) = u/a = 3/(1 e). Using the formula for tan(a + B), we get that tan θ tan θ = 3/(1 e) tan θ = 1 + e e.