Janilo Santos I, 1 1 E-mail: janilo dfte. A good teacher knows the value of analogy and universality when explaining difficult concepts.
A hard problem can be much simpler to elucidate when students have been exposed to a similar problem. For instance, students understand electrical forces better after they are acquainted with gravitational forces.
Terms such as energy become gradually more familiar as it is encountered in a variety of contexts. In this paper we aim to introduce and enlarge the concept of impedance to undergraduates and advanced high school students by investigating energy transfer in mechanical collisions and tracing a parallel with the propagation of light in electromagnetic systems.
We define the characteristic impedance of a system as the ratio between a forcelike quantity and a velocity-like quantity [1]. From this definition we derive an expression for the mechanical impedance of a billiard ball, which tells how to enhance the energy transfer from one mass to another in elastic collisions. Most importantly, we investigate how impedance matching appears in mechanical systems and we compare our results with the well-known problem of impedance matching in optical systems.
Our mechanical system consists of the one-dimensional elastic non-relativistic collision between two or three particles with different masses. This is simply the popular textbook problem of one-dimensional elastically colliding billiard balls. We observe how much kinetic energy is transmitted from one ball to the other during the collision. Before we introduce the idea of impedance in mechanical systems, let us use the conservation laws of linear momentum and kinetic energy in elastic collisions to find the fraction of transmitted energy from one object to the other.
We consider three rigid billiard balls of masses m 1 , m 2 , and m 3. Let us assume that ball 1 has a finite speed and both balls 2 and 3 are at rest before the collision. After the first collision between m 1 and m part of the kinetic energy from ball 1 has been transmitted to ball 2, which now has a velocity in the same direction as the initial velocity of ball 1 see Fig.
Using momentum and kinetic energy conservation, one obtain, for the fraction of kinetic energy transmitted from the first to the second ball,. Analogously, in the second collision between balls 2 and 3, the fraction of kinetic energy that is transferred to the third ball is.
Equating this with Eq. The plot for this configuration is shown in Fig. In order to proceed further, we ask ourselves whether this special range of values can be enlarged such that a maximum amount of kinetic energy can be transferred from the first to the third ball. This answers our question about the value of the intermediate mass: when m 2 is equal to the geometric mean of m 1 and m 3 , the transmitted kinetic energy is a maximum.
A similar calculation for partially elastic collisions between n masses was carried out by J. Hart and R. Herrmann [2]. Is momentum conserved even though both cars stop? Yes, although there is no momentum after the collision, there was no total momentum before the collision. Momentum is a vector quantity. The positive momentum of one car must have been balanced out by the negative momentum of the other car.
Is kinetic energy conserved even though both cars stop? Kinetic energy is a scalar quantity so there is no balancing out of the different directions. Both cars were moving and had kinetic energy before the collision.
Now, let us turn to the second type of collision. An inelastic collision is one in which objects stick together after impact, and kinetic energy is not conserved.
This lack of conservation means that the forces between colliding objects may convert kinetic energy to other forms of energy, such as potential energy or thermal energy. The concepts of energy are discussed more thoroughly elsewhere.
For inelastic collisions, kinetic energy may be lost in the form of heat. Two objects that have equal masses head toward each other at equal speeds and then stick together. The two objects come to rest after sticking together, conserving momentum but not kinetic energy after they collide.
Some of the energy of motion gets converted to thermal energy, or heat. Since the two objects stick together after colliding, they move together at the same speed. This lets us simplify the conservation of momentum equation from. Ask students what they understand by the words elastic and inelastic. Ask students to give examples of elastic and inelastic collisions. This video reviews the definitions of momentum and impulse. It also covers an example of using conservation of momentum to solve a problem involving an inelastic collision between a car with constant velocity and a stationary truck.
How would the final velocity of the car-plus-truck system change if the truck had some initial velocity moving in the same direction as the car? What if the truck were moving in the opposite direction of the car initially? In this activity, you will observe an elastic collision by sliding an ice cube into another ice cube on a smooth surface, so that a negligible amount of energy is converted to heat.
The Khan Academy videos referenced in this section show examples of elastic and inelastic collisions in one dimension. In one-dimensional collisions, the incoming and outgoing velocities are all along the same line.
But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and just as we did with two-dimensional forces, we will solve these problems by first choosing a coordinate system and separating the motion into its x and y components.
One complication with two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass each other, they will spin in circles. We will not consider such rotation until later, and so for now, we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses —that is, structureless particles that cannot rotate or spin.
The simplest collision is one in which one of the particles is initially at rest. The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8. Because momentum is conserved, the components of momentum along the x - and y -axes, displayed as p x and p y , will also be conserved. With the chosen coordinate system, p y is initially zero and p x is the momentum of the incoming particle.
Along the x -axis, the equation for conservation of momentum is. But because particle 2 is initially at rest, this equation becomes. Conservation of momentum along the x -axis gives the equation. Along the y -axis, the equation for conservation of momentum is. But v 1 y is zero, because particle 1 initially moves along the x -axis.
Because particle 2 is initially at rest, v 2 y is also zero. The equation for conservation of momentum along the y -axis becomes. Therefore, conservation of momentum along the y -axis gives the following equation:. Review conservation of momentum and the equations derived in the previous sections of this chapter.
Say that in the problems of this section, all objects are assumed to be point masses. Explain point masses. In this simulation, you will investigate collisions on an air hockey table. Place checkmarks next to the momentum vectors and momenta diagram options. Experiment with changing the masses of the balls and the initial speed of ball 1.
How does this affect the momentum of each ball? What about the total momentum? Next, experiment with changing the elasticity of the collision.
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