The problems of physics, in which moving and hitting one another bodies are considered, require the knowledge of the laws of conservation of momentum and energy, as well as understanding the specifics of the interaction itself. This article provides theoretical information on elastic and inelastic impacts. Special cases of solving problems associated with these physical concepts are also given.
Amount of movement
Before considering an absolutely elastic and inelastic impact, it is necessary to determine the value, which is known as the momentum. It is customary to denote it with the Latin letter p. It is introduced into physics simply: this is the product of mass and the linear velocity of the body, that is, the formula holds:
p = m * v
This is a vector quantity, but for simplicity it is written in scalar form. In this understanding, the amount of movement was considered by Galileo and Newton in the XVII century.
This value is not displayed. Its appearance in physics is associated with an intuitive understanding of processes observed in nature. For example, everyone is well aware that stopping a horse that runs at a speed of 40 km / h is much harder than a fly flying at the same speed.
Impulse of power
Many call the amount of movement simply an impulse. This is not entirely true, since the latter is understood to mean the effect of force on an object for a certain period of time.
If the force (F) does not depend on the time of its action (t), then the momentum of the force (P) in classical mechanics is written as follows:
P = F * t
Using Newton's law, we rewrite this expression as follows:
P = m * a * t,
where F = m * a
Here a is the acceleration communicated to the body of mass m. Since the acting force does not depend on time, acceleration is a constant value, which is determined by the ratio of speed to time, that is:
P = m * a * t = m * v / t * t = m * v.
We got an interesting result: the impulse of force is equal to the amount of movement that it tells the body. That is why many physicists simply omit the word “force” and say momentum, bearing in mind the amount of motion.
The written formulas also lead to one important conclusion: in the absence of external forces, any internal interactions in the system retain its total momentum (the momentum of the force is zero). The last formulation is known as the law of conservation of momentum of an isolated system of bodies.
The concept of mechanical shock in physics
Now it's time to move on to considering absolutely elastic and inelastic impacts. A mechanical shock in physics is understood as the simultaneous interaction of two or more solids, as a result of which there is an exchange of energy and the momentum between them.
The main features of the impact are large effective forces and small time intervals of their application. Often the impact is characterized by the magnitude of the acceleration, expressed as g for the Earth. For example, the record 30 * g, says that as a result of the collision, the force informed the body the acceleration 30 * 9.81 = 294.3 m / s 2 .
Particular cases of collisions are absolute elastic and inelastic impacts (the latter is also called elastic or plastic). Consider what they are.
Ideal types of strokes
Elastic and inelastic impacts of bodies are idealized cases. The first of them (elastic) means that in the collision of two bodies no permanent deformation is created. When one body collides with another, then at some point in time, both objects deform in their contact area. This deformation serves as a mechanism for the transfer of energy (momentum) between objects. If it is absolutely elastic, then after an impact, no energy loss occurs. In this case, one speaks of the conservation of the kinetic energy of interacting bodies.
The second type of impact (plastic or absolutely inelastic) means that after one body hits another, they “stick together” with each other, so after the impact both objects begin to move as a whole. As a result of this shock, some of the kinetic energy is spent on the deformation of bodies, friction, and heat. In this type of collision, energy is not conserved, but the momentum remains unchanged.
Elastic and inelastic impacts are ideal particular cases of collisions between bodies. In real life, the characteristics of all collisions do not apply to either of these two types.
Totally elastic collision
We solve two problems of elastic and inelastic impact of balls. In this paragraph, we consider the first type of collision. Since the laws of energy and momentum are observed in this case, we write the corresponding system of two equations:
m 1 * v 1 2 + m 2 * v 2 2 = m 1 * u 1 2 + m 2 * u 2 2 ;
m 1 * v 1 + m 2 * v 2 = m 1 * u 1 + m 2 * u 2 .
This system is used to solve any problems with any initial conditions. In this example, we restrict ourselves to a special case: let the masses m 1 and m 2 of two balls be equal. In addition, the initial velocity of the second ball v 2 is equal to zero. It is necessary to determine the result of the central elastic collision of the considered bodies.
Given the conditions of the problem, we rewrite the system:
v 1 2 = u 1 2 + u 2 2 ;
v 1 = u 1 + u 2 .
We substitute the second expression into the first, we get:
(u 1 + u 2 ) 2 = u 1 2 + u 2 2
We open the brackets:
u 1 2 + u 2 2 + 2 * u 1 * u 2 = u 1 2 + u 2 2 => u 1 * u 2 = 0
The last equality is true if one of the velocities u 1 or u 2 is equal to zero. The second of them cannot be zero, because when the first ball hits the second, it will inevitably begin to move. This means that u 1 = 0, and u 2 > 0.
Thus, in an elastic collision of a moving ball with a stationary one, whose masses are the same, the first transfers its momentum and energy to the second.
Inelastic impact
In this case, the ball that rolls, in a collision with the second ball that is resting, sticks to it. Then both bodies begin to move, as one unit. Since the momentum of elastic and inelastic impacts is conserved, we can write the equation:
m 1 * v 1 + m 2 * v 2 = (m 1 + m 2 ) * u
Since v 2 = 0 in our problem, the final speed of a system of two balls is determined by the following expression:
u = m 1 * v 1 / (m 1 + m 2 )
In case of equality of masses of bodies, we obtain an even simpler expression:
u = v 1/2
The speed of two sticking balls will be half as much as this value for one ball until the moment of collision.
Recovery rate
This value is a characteristic of energy loss during a collision. That is, it describes how elastic (plastic) is the impact in question. It was introduced into physics by Isaac Newton.
Obtaining an expression for the recovery coefficient is not difficult. Suppose that two bodies collided with masses m 1 and m 2 . Let their initial velocities be v 1 and v 2 , and their final (after a collision) - u 1 and u 2 . Assuming that the impact is elastic (kinetic energy is conserved), we write two equations:
m 1 * v 1 2 + m 2 * v 2 2 = m 1 * u 1 2 + m 2 * u 2 2 ;
m 1 * v 1 + m 2 * v 2 = m 1 * u 1 + m 2 * u 2 .
The first expression is the kinetic energy conservation law, the second is the conservation of momentum.
After a number of simplifications, you can get the formula:
v 1 + u 1 = v 2 + u 2 .
It can be rewritten in the form of a ratio of the difference of speeds as follows:
1 = -1 * (v 1 -v 2 ) / (u 1 -u 2 ).
Thus, the ratio of the difference between the velocities of two bodies before the collision to the similar difference for them after the collision, taken with the opposite sign, is equal to unity if there is an absolutely elastic impact.
It can be shown that the last formula for an inelastic impact gives a value of 0. Since the conservation laws for elastic and inelastic impact are different for kinetic energy (it is preserved only in elastic collisions), the resulting formula is a convenient coefficient for characterizing the type of impact.
The recovery coefficient K has the form:
K = -1 * (v 1 -v 2 ) / (u 1 -u 2 ).
Calculation of the recovery coefficient for a jumping body
Depending on the nature of the impact, the K coefficient may vary significantly. Consider how it can be calculated for the case of a "jumping" body, such as a soccer ball.
First, the ball is held at a certain height h 0 above the ground. Then they let him go. It falls to the surface, bounces off it and rises to a certain height h, which is fixed. Since the speed of the surface of the earth before and after its collision with the ball was equal to zero, the formula for the coefficient will have the form:
K = v 1 / u 1
Here v 2 = 0 and u 2 = 0. The minus sign disappeared, because the speeds v 1 and u 1 are directed opposite. Since the fall and rise of the ball is a uniformly accelerated and equally slow motion, the formula is valid for it:
h = v 2 / (2 * g)
Expressing the speed, substituting the values of the initial height and after the ball bounces into the formula for the coefficient K, we obtain the final expression: K = √ (h / h 0 ).