Any problems on moving bodies in classical mechanics require knowledge of the concept of momentum. This article discusses this concept, gives an answer to the question of where the body momentum vector is directed, and also provides an example of solving the problem.
Amount of movement
To find out where the body momentum vector is directed, it is necessary, first of all, to understand its physical meaning. The term was first explained by Isaac Newton, but it is important to note that the Italian scientist Galileo Galilei already used a similar concept in his works. To characterize a moving object, he introduced a quantity called aspiration, onslaught, or impulse itself (impeto in Italian). The merit of Isaac Newton lies in the fact that he was able to connect this characteristic with the forces acting on the body.
So, initially and more correctly, what most understand by the momentum of the body is called the amount of movement. Indeed, the mathematical formula for the quantity in question is written as:
p¯ = m * v¯.
Here m is the mass of the body, v¯ is its speed. As can be seen from the formula, we are not talking about any impulse, there is only the speed of the body and its mass, that is, the momentum.
It is important to note that this formula does not follow from mathematical proofs or expressions. Its appearance in physics has an extremely intuitive, everyday character. So, any person is well aware that if the fly and the truck move at the same speed, it is much harder to stop the truck, since it has a lot more movement than an insect.
Where the concept of the body momentum vector came from is discussed below.
Impulse of force - the reason for the change in momentum
Newton was able to connect the intuitively introduced characteristic with the second law bearing his last name.
A momentum of force is a known physical quantity that is equal to the product of an external force applied to a certain body by the time of its action. Using the famous Newton's law and assuming that the force does not depend on time, we can come to the expression:
F¯ * Δt = m * a¯ * Δt.
Here Δt is the duration of the force F, a is the linear acceleration reported by the force F to a body of mass m. As you know, multiplying the acceleration of the body by the period of time that it acts gives an increment of speed. This fact allows us to rewrite the formula above in a slightly different form:
F¯ * Δt = m * Δv¯, where Δv¯ = a¯ * Δt.
The right-hand side of the equation is the change in the momentum (see the expression in the previous paragraph). Then it turns out:
F¯ * Δt = Δp¯, where Δp¯ = m * Δv¯.
Thus, using Newton's law and the concept of momentum of force, we can come to an important conclusion: the influence of an external force on an object for some time leads to a change in its momentum.
Now it becomes clear why the amount of motion is usually called an impulse, because its change coincides with the impulse of force (the word "force" is usually omitted).
The vector quantity p¯
Over some quantities (F¯, v¯, a¯, p¯) there is a bar. This means that this is a vector characteristic. That is, the momentum, as well as speed, force and acceleration, in addition to the absolute value (module), is also described by the direction.
Since each vector can be decomposed into separate components, using the Cartesian rectangular coordinate system, we can write the following equalities:
1) p¯ = m * v¯;
2) p x = m * v x ; p y = m * v y ; p z = m * v z ;
3) | p¯ | = √ (p x 2 + p y 2 + p z 2 ).
Here, the 1st expression is a vectorial representation of the momentum, the 2nd set of formulas allows us to calculate each of the components of the momentum p¯, knowing the corresponding velocity components (the indices x, y, z indicate the projection of the vector onto the corresponding coordinate axis). Finally, the 3rd formula allows you to calculate the length of the pulse vector (the absolute value of the quantity) through its components.
Where is the body momentum vector directed?
Having considered the concept of momentum p¯ and its basic properties, one can easily answer the question posed. The body momentum vector is directed in the same way as the linear velocity vector. Indeed, it is known from mathematics that multiplying the vector a¯ by the number k leads to the formation of a new vector b¯ with the following properties:
- its length is equal to the product of the number by the modulus of the original vector, that is, | b¯ | = k * | a¯ |;
- it is directed in the same way as the original vector if k> 0, otherwise it will be directed opposite to a¯.
In this case, the role of the vector a¯ is played by the velocity v¯, the momentum p¯ is the new vector b¯, and the number k is the body mass m. Since the latter is always positive (m> 0), then, answering the question: why is the body momentum vector p¯ co-directed, it should be said that it is co-directed with velocity v¯.
Movement change vector
It is interesting to consider another similar question: where is the vector of change in the momentum of the body directed, that is, Δp¯. To answer it, you should use the formula obtained above:
F¯ * Δt = m * Δv¯ = Δp¯.
Based on the reasoning in the previous paragraph, we can say that the direction of change in the momentum Δp¯ coincides with the direction of the force vector F¯ (Δt> 0) or with the direction of the velocity vector Δv¯ (m> 0).
It is important not to confuse here that it is a question of changing quantities. In the general case, the vectors p¯ and Δp¯ do not coincide, since they are in no way connected with each other. For example, if the force F¯ acts against the speed v¯ of movement of the object, then p¯ and Δp¯ will be directed in opposite directions.
Where is it important to consider the vector nature of the momentum?
The questions discussed above: where the vector of the body's momentum and the vector of its change are directed are not due to simple curiosity. The fact is that the momentum conservation law p¯ holds for each of its components. That is, in the most complete form, it is written like this:
p x = m * v x ; p y = m * v y ; p z = m * v z .
Each component of the vector p¯ retains its value in a system of interacting objects that are not affected by external forces (Δp¯ = 0).
How to use this law and vector representations of p¯ to solve the problems of interaction (collision) of bodies?
The task with two balls
The figure below shows two balls of different masses that fly at different angles to a horizontal line. Let the masses of the balls be equal to m 1 = 1 kg, m 2 = 0.5 kg, their velocities v 1 = 2 m / s, v 2 = 3 m / s. It is necessary to determine the direction of the momentum after hitting the balls, assuming the latter to be completely inelastic.
Starting to solve the problem, you should write down the law of invariability of the momentum in vector form, that is:
p 1 ¯ + p 2 ¯ = const.
Since each component of the momentum must be conserved, this expression needs to be rewritten, also taking into account that after the collision the two balls will begin to move as a single object (absolutely inelastic impact):
m 1 * v 1x + m 2 * v 2x = (m 1 + m 2 ) * u x ;
-m 1 * v 1y + m 2 * v 2y = (m 1 + m 2 ) * u y .
The minus sign for the projection of the momentum of the first body on the y axis appeared due to its direction against the selected ordinate axis vector (see. Fig.).
Now you need to express the unknown components of the velocity u, and then substitute the known values in the expressions (the corresponding projections of the velocities are determined by multiplying the moduli of the vectors v 1 ¯ and v 2 ¯ by trigonometric functions):
u x = (m 1 * v 1x + m 2 * v 2x ) / (m 1 + m 2 ), v 1x = v 1 * cos (45 o ); v 2x = v 2 * cos (30 o );
u x = (1 * 2 * 0.7071 + 0.5 * 3 * 0.866) / (1 + 0.5) = 1.8088 m / s;
u y = (-m 1 * v 1y + m 2 * v 2y ) / (m 1 + m 2 ), v 1y = v 1 * sin (45 o ); v 2y = v 2 * sin (30 o );
u y = (-1 * 2 * 0.7071 + 0.5 * 3 * 0.5) / (1 + 0.5) = -0.4428 m / s.
These are two components of the body’s speed after hitting and sticking balls. Since the direction of velocity coincides with the momentum vector p¯, it is possible to answer the question of the problem if u¯ is determined. Its angle relative to the horizontal axis will be equal to the arc tangent of the ratio of the components u y and u x :
α = arctan (-0.4428 / 1.8088) = -13.756 o .
A minus sign indicates that the impulse (speed) after the impact will be directed downward from the x axis.