Physics 2A Chapter 8 - Impulse & Momentum Fall 2017

Transcription

1 These notes are six pages. A quick summary: Impulse is the product of force and time; momentum is the product of mass and velocity. The impulse given to an object is equal to its change in momentum. When two objects collide, they impart equal and opposite impulses to each other. So the total momentum of the two objects is the same before and after the collision. Momentum & Impulse We have discovered in Chapter 6 that taking the product of the force acting on an object and the displacement of the object results in a useful definition: work done by the force, which is the kinetic energy the force gives to the object. In Chapter 8 we make a similar discovery : we will consider the product of the force acting on an object and the time for which it acts. We can start with Newton s Second law: F = ma (For clarity, I have boldfaced the vector quantities.) and mutiply both sides by the time for which the force acts F t = ma t Recall that from our Chapter 2 equations, v f = v i + at, or v f - v i = at so: F t = m (v f - v i ) We can now distribute the mass on the right side and we get: F t = mv f - mv i Notice that on the right side we have a combination of mass and velocity... similar to how we defined kinetic energy in Chapter 6. In this case, we make a new definition: momentum = mass x velocity We use a p for momentum (because we can t use m ). Momentum is a vector quantity; it is in the same direction as the velocity of the object. The SI unit of momentum is simply kg-m/s (i.e. we don t name this after anyone...) Page 1 of 6

2 We also define the impulse imparted to an object as the product of the force acting on the object and the time for which the force acts. That is: impulse = force x time Now our equation above: F t = mv f - mv i becomes: J = p f - p i where we use a J for impulse (because we can t use I... I ll explain later.) Notice that this expresses the same kind of relationship between impulse and momentum that we had between work and kinetic energy: the work done on an object is equal to the change in kinetic energy; the impulse imparted to an object is equal to its change in momentum. Just as we found when finding the work done by a force, we have to discern between situations where the force is constant or the force is changing. If the force is constant: F t = p i.e. the impulse is simply the product of force and time. If the force is changing: Favg t = p i.e. the impulse is the product of the average force and time. Collisions Two objects collide when they apply a force to each other that typically lasts for a relatively short time and is significantly greater than other forces acting on the objects during that time. These conditions allow us to define a point immediately before the collision and a point immediately after the collision. Since the objects apply equal and opposite forces to each other for the same amount of time, they apply equal and opposite impulses to each other, and we can write (using subscripts 1 and 2 for the two objects): J 1 = -J 2 p 1 = - p 2 p 1f - p 1i = - (p 2f - p 2i ) p 1i + p 2i = p 1f + p 2f Page 2 of 6

3 The last equation is simply another way to state the first equation: since the total impulse acting on the two objects is zero, the total change in momentum is zero, so the momentum of the two objects initially (i.e. before the collision) must equal the final momentum (i.e. after the collision) of the two objects. We use this statement of conservation of momentum to analyze the motion of objects prior to and following a collision. Very important: we only use conservation of momentum for collisions. A common pitfall for students using this concept for the first time is to confuse it with conservation of energy that was introduced in Chapter 7. Conservation of Energy: one object that changes speed and height, possibly compresses a spring Conservation of Momentum: two objects that are involved in a collision Elastic Collisions During most collisions we find that energy is lost in one or more ways, i.e. the combined kinetic energy of both objects after the collision is less than the combined kinetic energy before the collision. This principle can actually be used to check your calculations when using the principle of conservation of momentum: the combined kinetic energy after the collision cannot possibly be greater than the combined kinetic energy prior to the collision. However, there is a special class of collisions for which combined kinetic energy before the collision is equal to the combined kinetic energy after the collision. We call these elastic collisions. In general, collisions are not elastic. But if you encounter an elastic collision within a problem, you can write down one more equation: initial kinetic energy of both objects = final kinetic energy of both objects K 1i + K 2i = K if + K 2f ½ m 1 v 1i 2 + ½ m 2 v 2i 2 = ½ m 1 v 1f 2 + ½ m 2 v 2f 2 Note that when we multiply both sides by 2 and divide both sides by m 1 this equation reduces to: v 1i 2 + α v 2i 2 = v 1f 2 + α v 2f 2 where α is the ratio m 2 / m 1 If you are told the collision is elastic, you can write one additional equation, so the problem can leave you with one more unknown (than with a regular collision problem). There is nothing more to it. Note that the equation above contains the same information as the conservation of momentum equation: masses, initial and final velocities of both objects. Page 3 of 6

4 If you have an elastic collision, the rules for a collision problem still apply. Follow the rules (draw two pictures, label information, write conservation of momentum) and also write the second equation for the collision. Rules for Collision Problems 1. Draw two pictures: one immediately before the collision, one immediately after the collision. 2. Define directions: label the positive direction if the collision is in one dimension, label x and y axes if the collision is in two dimensions 3. Label mass and velocity for both objects in both pictures. 4. Write the conservation of momentum equation. If the collision is in one dimension, write the equation once; if in two dimensions, write the equation for each dimension. 5. If the collision is elastic, also write the kinetic energy equation. You only have one kinetic energy equation, whether the problem is one or two dimensions (because energy is not a vector.) 6. Count your equations and unknowns... and solve. Center of Mass Center of mass is a relatively simple and intuitive concept: it is the point at which we can balance an object. Technically, it is the average position of the mass distribution of an object. For any symmetrical object, the center of mass is at the center of symmetry. The center of mass of the Earth is at the center of the Earth. The center of mass of a wheel is in the middle of its axle. For a dinner plate? At its center also. We can define the position of the center of mass of a collection of point masses, i.e. objects that can be treated as if all of their mass is at a single point (more on this in a moment). We can define our point masses as m 1, m 2, m 3, etc and their positions along the x-axis as x 1, x 2, x 3, etc (we an also define their positions along the y-axis and z-axis, i.e. in two or three dimensions... the principle below is exactly the same). The position of the center of mass is then calculated by finding the weighted average of the position of the individual point masses. The calculation is: Page 4 of 6

5 x cm 1 = ( m x1 + m2x2 + m3x m tot ) For the simple case that each mass is 1 and there are ten masses, note that we would add the x positions and divide by 10 (the total mass). In other words, if the masses are equal the expression gives us just the simple average of the positions. If the masses are not equal, the expression gives the weighted average of the positions where the masses are the weight factor. The y and z coordinates of the center of mass can be found by simply replacing x in the above expression with y or z. Of course the above expression assumes that you are working with a collection of point masses, which is a very strange concept: an object with all of its mass at one point! These do not exist in reality. But a simple principle makes the above expression very useful: For the purpose of center of mass calculation, any symmetrical object can be treated as if it is a point mass at its center of mass. This principle allows us to take any non-symmetrical object, divide it into symmetrical pieces, and treat each piece as if its mass was at its center of symmetry. In this way, any non-symmetrical object can be represented as a collection of point masses and the position of the center of mass of the nonsymmetrical object can be calculated. So what does center of mass have to do with momentum? There is a reason that the concept of center of mass is included in the momentum chapter. Imagine we have a collection of objects that are interacting. We can call this collection a system. Assume that the system is isolated, i.e. the net force acting on the system from objects external to the system is zero. Within the system the objects might be applying forces to each other, changing each other s momentum, but because any two objects apply equal and opposite changes in momentum to each other, we know that the total change in the momentum of the system must be zero. Now consider the position of the center of mass of the system, which can be calculated using the expression above. And imagine the rate of change with respect to time of the position of the center of mass: we can do this by mathematically taking the time derivative of each side of the above equation. Since the masses are constants, all of the x terms in the expression become dx/dt. But dx/dt is simply the velocity of the center of mass (on the left side) and the velocity of the individual objects on the right side. Like this: Page 5 of 6

6 v cm 1 = ( m v1 + m2v2 + m3v m tot Notice that the right side of the equation now includes the momentum of each object in the system, specifically the sum of the individual momenta of the objects in the system. Again, if we assume there is no net external force acting on the objects in the system, the total momentum remains constant (i.e. there is no change in the total momentum of the objects in the system). Since the total mass also remains constant, we arrive at a very simple result:...) If the net external force acting on a system of objects is zero, the velocity of the center of mass of that system remains constant. This means that if the center of mass is at rest relative to some external reference point, the center of mass will remain in the same place if there is no net external force acting on the system and regardless of the motion of the individual objects. The objects of the system can move around relative to each other, but their center of mass will remain in the same place. This principle is often put to use in problems that include people in a small boat that move inside the boat. The idea is that the net external force acting on the boat (gravity down, buoyancy up) is zero, so the boat will move around when the people inside move around, but the center of mass of the system (i.e. boat and people) will remain at the same position relative to a fixed point (usually the shore, or a tree, or something similar). Problem 39 in Chapter 8 (4 th Edition), which presents Romeo and Juliet together in a small boat, is this kind of problem. Page 6 of 6