Problems of physics, in which bodies moving and hitting each other, require knowledge of the laws of conservation of momentum and energy, as well as an understanding of the specifics of the interaction itself. This article provides theoretical information about elastic and inelastic impacts. Particular cases of solving problems related to these physical concepts are also given.
Amount of movement
Before considering perfectly elastic and inelastic impact, it is necessary to define the quantity known as momentum. It is usually denoted by the Latin letter p. It is introduced into physics simply: this is the product of the mass by the linear speed of the body, that is, the formula takes place:
p=mv
This is a vector quantity, but for simplicity it is written in scalar form. In this sense, the momentum was considered by Galileo and Newton in the 17th century.
This value is not displayed. Its appearance in physics is associated with an intuitive understanding of the processes observed in nature. For example, everyone is well aware that it is much harder to stop a horse running at a speed of 40 km/h than a fly flying at the same speed.
Impulse of power
The amount of movement is simply referred to by many as momentum. This is not entirely true, since the latter is understood as the effect of force on an object over a certain period of time.
If the force (F) does not depend on the time of its action (t), then the impulse of the force (P) in classical mechanics is written by the following formula:
P=Ft
Using Newton's law, we can rewrite this expression as follows:
P=mat, where F=ma
Here a is the acceleration imparted to a body of mass m. Since the acting force does not depend on time, the acceleration is a constant value, which is determined by the ratio of speed to time, that is:
P=mat=mv/tt=mv.
We got an interesting result: the momentum of the force is equal to the amount of motion that it tells the body. That is why many physicists simply omit the word "force" and say momentum, referring to 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 preserve its total momentum (the momentum of the force is zero). The last formulation is known as the law of conservation of momentum for an isolated system of bodies.
The concept of mechanical impact in physics
Now it's time to move on to considering absolutely elastic and inelastic impacts. In physics, mechanical impact is understood as the simultaneous interaction of two or more solid bodies, as a result of which there is an exchange of energy and momentum between them.
The main features of the impact are large acting forces and short periods of time of their application. Often the impact is characterized by the magnitude of the acceleration, expressed as g for the Earth. For example, the entry 30g says that as a result of the collision, the force imparted to the body an acceleration of 309, 81=294.3 m/s2.
Special cases of collision are absolute elastic and inelastic impacts (the latter is also called elastic or plastic). Consider what they are.
Ideal shots
Elastic and inelastic impacts of bodies are idealized cases. The first one (elastic) means that when two bodies collide, no permanent deformation is created. When one body collides with another, at some point in time both objects are deformed in the area of their contact. This deformation serves as a mechanism for transferring energy (momentum) between objects. If it is perfectly elastic, then no energy loss occurs after the impact. In this case, one speaks of the conservation of the kinetic energy of the interacting bodies.
The second type of impacts (plastic or absolutely inelastic) means that after the collision of one body against another, they"stick together" with each other, so after the impact, both objects begin to move as a whole. As a result of this impact, some part of the kinetic energy is spent on the deformation of bodies, friction, and heat release. In this type of impact, energy is not conserved, but momentum remains unchanged.
Elastic and inelastic impacts are ideal special cases of collision of bodies. In real life, the characteristics of all collisions do not belong to either of these two types.
Perfectly elastic collision
Let's solve two problems for elastic and inelastic impact of balls. In this subsection, 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:
m1v12+m2 v22 =m1u1 2+m2u22;
m1v1+m2v 2=m1u1+m2u 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 m1 and m2 of two balls be equal. In addition, the initial speed of the second ball v2 is zero. It is necessary to determine the result of the central elastic collision of the considered bodies.
Taking into account the condition of the problem, let's rewrite the system:
v12=u12+ u22;
v1=u1+ u2.
Substitute the second expression into the first, we get:
(u1+ u2)2=u 12+u22
Open brackets:
u12+ u22+ 2u1u2=u12+ u22=> u1u2 =0
The last equality is true if one of the speeds u1 or u2 equals zero. The second of them cannot be zero, because when the first ball hits the second, it will inevitably start moving. This means that u1 =0 and u2 > 0.
Thus, in an elastic collision of a moving ball with a ball at rest, the masses of which are the same, the first one transfers its momentum and energy to the second one.
Inelastic impact
In this case, the ball that is rolling, when colliding with the second ball that is at rest, sticks to it. Further, both bodies begin to move as one. Since the momentum of elastic and inelastic impacts is conserved, we can write the equation:
m1v1+ m2v 2=(m1 + m2)u
Since in our problem v2=0, the final speed of the system of two balls is determined by the following expression:
u=m1v1 / (m1 + m 2)
In the case of equality of body masses, we get an even simplerexpression:
u=v1/2
The speed of two balls stuck together will be half as much as this value for one ball before the collision.
Recovery Rate
This value is a characteristic of energy losses during a collision. That is, it describes how elastic (plastic) the impact in question is. It was introduced into physics by Isaac Newton.
Getting an expression for the recovery factor is not difficult. Suppose that two bodies of masses m1 and m2 have collided. Let their initial velocities be v1and v2, and their final velocities (after collision) be u1 and u2. Assuming that the impact is elastic (kinetic energy is conserved), we write two equations:
m1v12 + m2 v22 =m1u1 2 + m2u22;
m1v1+ m2v 2=m1u1+ m2u 2.
The first expression is the law of conservation of kinetic energy, the second is the conservation of momentum.
After a number of simplifications, we can get the formula:
v1 + u1=v2 + u 2.
It can be rewritten as the ratio of the speed difference as follows:
1=-1(v1-v2) / (u1 -u2).
SoThus, taken with the opposite sign, the ratio of the difference in the velocities of two bodies before the collision to the similar difference for them after the collision is equal to one if there is an absolutely elastic impact.
It can be shown that the last formula for an inelastic impact will give a value of 0. Since the conservation laws for elastic and inelastic impact are different for kinetic energy (it is conserved only for an elastic collision), the resulting formula is a convenient coefficient for characterizing the type of impact.
The recovery factor K is:
K=-1(v1-v2) / (u1 -u2).
Calculation of the recovery factor for a "jumping" body
Depending on the nature of the impact, the K factor may vary significantly. Let's consider how it can be calculated for the case of a "jumping" body, for example, a soccer ball.
First, the ball is held at a certain height h0above the ground. Then he is released. It falls on the surface, bounces off it and rises to a certain height h, which is fixed. Since the speed of the ground surface before and after its collision with the ball was equal to zero, the formula for the coefficient will look like:
K=v1/u1
Here v2=0 and u2=0. The minus sign has disappeared because the speeds v1 and u1 are opposite. Since the fall and rise of the ball is a movement of uniformly accelerated and uniformly slowed down, then for himthe formula is valid:
h=v2/(2g)
Expressing the speed, substituting the values of the initial height and after the ball bounces into the formula for the coefficient K, we get the final expression: K=√(h/h0).
