The vector quantity in physics. Examples of vector quantities

Physics and mathematics can not do without the notion of "vector quantity". It must be known and recognized, and also be able to operate with it. It is necessary to learn this, so as not to be confused and not to make stupid mistakes.

How to distinguish a scalar value from a vector value?

The first always has only one characteristic. This is its numerical value. Most scalar quantities can take both positive and negative values. Their examples are electrical charge, work or temperature. But there are scalars that can not be negative, for example, length and mass.

A vector quantity, except for a numerical value, which is always taken in modulus, is also characterized by a direction. Therefore, it can be represented graphically, that is, in the form of an arrow, whose length is equal to the magnitude of the quantity directed to a certain side.

When writing, each vector value is indicated by the arrow sign on the letter. If we are talking about a numerical value, then the arrow is not written, or it is taken modulo.

What actions are most often performed with vectors?

First - a comparison. They can be equal or not. In the first case, their modules are the same. But this is not the only condition. They should have the same or opposite directions. In the first case, they should be called equal vectors. In the second they turn out to be opposite. If at least one of the above conditions is not met, then the vectors are not equal.

Then comes the addition. It can be made according to two rules: a triangle or a parallelogram. The first prescribes to postpone at first one vector, then from its end the second. The result of addition will be the one that must be drawn from the beginning of the first to the end of the second.

The parallelogram rule can be used when it is necessary to add the vector quantities in physics. Unlike the first rule, here they should be postponed from one point. Then finish them up to a parallelogram. The result of the action is the diagonal of the parallelogram drawn from the same point.

If the vector value is subtracted from the other, then they are again deposited from one point. Only the result will be a vector that coincides with what is postponed from the end of the second to the end of the first.

What vectors are studied in physics?

There are as many as scalars. One can simply remember what vector quantities exist in physics. Or know the signs by which they can be calculated. Those who prefer the first option, useful such a table. It contains the basic vector physical quantities.

The notation in formula	Name
V	speed
R	Move
a	acceleration
F	force
R	pulse
E	Electric field strength
AT	magnetic induction
M	moment of power

Now a little more about some of these quantities.

The first quantity is the speed

It is worth starting to give examples of vector quantities. This is due to the fact that it is studied among the first.

Speed is defined as a characteristic of the motion of a body in space. It is given a numerical value and direction. Therefore, the velocity is a vector quantity. In addition, it is customary to divide into species. The first is the linear velocity. It is introduced when considering the rectilinear uniform motion. In this case, it turns out to be equal to the ratio of the path traversed by the body to the time of motion.

This formula can be used for uneven motion. Only then it will be average. And the time interval, which must be chosen, must be as small as possible. When the time interval tends to zero, the speed is already instantaneous.

If arbitrary motion is considered, then always the velocity is a vector quantity. After all, it must be decomposed into components directed along each vector that directs the coordinate lines. In addition, it is defined as the derivative of the radius vector taken with respect to time.

The second quantity is the force

It determines the measure of the intensity of the impact that is on the body from the side of other bodies or fields. Since the force is a vector quantity, it necessarily has its value in modulus and direction. Since it acts on the body, then the point to which the force is applied is also important. To get a visual representation of the force vectors, you can refer to the following table.

Force	Application point	Direction
Gravity	Body center	To the center of the Earth
Of universal gravitation	Body center	To the center of another body
Elasticity	The place of contact of interacting bodies	Against external influence
Friction	Between contiguous surfaces	In the opposite direction to the movement

Also the vector quantity is the resultant force. It is defined as the sum of all the mechanical forces acting on the body. To determine it, you must perform the addition according to the rule of the triangle rule. Only to postpone the vectors need to take turns from the end of the previous one. The result will be the one that connects the beginning of the first with the end of the latter.

The third quantity is the displacement

During motion, the body describes a certain line. It is called a trajectory. This line can be completely different. More important is not its appearance, but the points of the beginning and the end of the movement. They are connected by a segment, which is called displacement. This is also a vector quantity. And it is always directed from the beginning of the movement to the point where the movement was stopped. It is denoted by the Latin letter r.

Here the following question may appear: "The path is a vector quantity?". In general, this statement is not true. The path is equal to the length of the trajectory and does not have a definite direction. An exception is the situation when straight-line traffic in one direction is considered. Then the modulus of the displacement vector coincides in value with the path, and the direction of them is the same. Therefore, when considering the motion along a straight line without changing the direction of displacement, the path can be included in examples of vector quantities.

The fourth quantity is the acceleration

It is a characteristic of the speed of change in speed. And acceleration can have both positive and negative value. With rectilinear motion, it is directed toward higher speed. If the displacement occurs along a curvilinear trajectory, then its acceleration vector is decomposed into two components, one of which is directed to the center of curvature along the radius.

The mean and instantaneous acceleration are selected. The former should be calculated as the ratio of the change in velocity over a certain period of time to this time. As the time interval tends to zero, we speak of instantaneous acceleration.

The fifth quantity is the momentum

In other words, it is also called the amount of movement. The impulse is a vector quantity because it is directly related to the speed and force applied to the body. Both of them have a direction and set its momentum.

By definition, the latter is equal to the product of the mass of the body by the velocity. Using the concept of momentum of a body, it is possible to write down in a different way the well-known Newton's law. It turns out that the change in momentum is equal to the product of the force over a time interval.

In physics, the momentum conservation law plays an important role, which asserts that in a closed system of bodies its total momentum is constant.

We very briefly listed which quantities (vector) are studied in the course of physics.

The problem of inelastic impact

Condition. On the rails is a fixed platform. The car approaches it at a speed of 4 m / s. The masses of the platform and the car are 10 and 40 tons, respectively. The car strikes against the platform, the autoscheme occurs. It is necessary to calculate the speed of the "platform car" system after the impact.

Decision. First, you need to enter the symbols: the speed of the car before the impact - v ₁ , the car with the platform after the coupling - v, the mass of the car m ₁ , the platform - m ₂ . By the condition of the problem, it is necessary to find the value of the velocity v.

Rules for solving such tasks require a schematic representation of the system before and after interaction. Axis OX is reasonable to direct along the rails in the direction where the car is moving.

Under these conditions, the system of cars can be considered closed. This is determined by the fact that external forces can be neglected. Gravity and reaction of the support are balanced, and friction on the rails is not taken into account.

According to the law of conservation of momentum, their vector sum before the interaction of the car and the platform is equal to the common for the coupling after the impact. At first the platform did not move, so its momentum was zero. Moved only the car, its momentum is the product of m ₁ and v ₁ .

Since the impact was inelastic, that is, the car clung to the platform, and then it began to roll together in the same direction, then the momentum of the system did not change direction. But its meaning has become different. Namely, the product of the sum of the mass of the car with the platform and the required speed.

One can write the following equality: m ₁ * v ₁ = (m ₁ + m ₂ ) * v. It will be true for the projection of the momentum vectors on the selected axis. From it, it is easy to derive an equality that will be required to calculate the required velocity: v = m ₁ * v ₁ / (m ₁ + m ₂ ).

According to the rules, values for the mass from tons to kilograms should be translated. Therefore, when you substitute them in the formula, you must first multiply the known values by a thousand. Simple calculations give a number of 0.75 m / s.

Answer. The speed of the car with the platform is 0.75 m / s.

The problem of dividing the body into parts

Condition . The speed of the flying grenade is 20 m / s. It breaks into two pieces. Weight of the first 1.8 kg. He continues to move in the direction in which the grenade flew, at a speed of 50 m / s. The second fragment has a mass of 1.2 kg. What is its speed?

Decision. Let the fragment masses be denoted by the letters m ₁ and m ₂ . Their velocities are, respectively, v ₁ and v ₂ . The initial speed of the grenade is v. In the task, you need to calculate the value of v ₂ .

In order for the larger fragment to continue moving in the same direction as the entire grenade, the second must fly in the opposite direction. If you select for the direction of the axis that which was at the initial pulse, then after the break, a large fragment flies along the axis, and a small one - against the axis.

In this problem it is allowed to use the law of conservation of momentum due to the fact that the grenade break occurs instantaneously. Therefore, in spite of the fact that gravity acts on the grenade and its part, it does not have time to act and change the direction of the momentum vector with its value modulo.

The sum of the momentum vector values after the grenade break is equal to the one that was before it. If we write the law of conservation of the momentum of the body in the projection onto the OX axis, it will look like this: (m ₁ + m ₂ ) * v = m ₁ * v ₁ - m ₂ * v ₂ . It simply expresses the required speed. It is determined by the formula: v ₂ = ((m ₁ + m ₂ ) * v - m ₁ * v ₁ ) / m ₂ . After the substitution of numerical values and calculations, 25 m / s is obtained.

Answer. The speed of the small fragment is 25 m / s.

The problem of a shot at an angle

Condition. A tool is mounted on the platform with a mass M. It is fired by a shell with a mass m. It flies at an angle α to the horizon at a speed v (given relative to the ground). It is required to know the value of the speed of the platform after the shot.

Decision. In this problem, we can use the law of conservation of momentum in the projection onto the OX axis. But only in the case when the projection of external resultant forces is zero.

For the direction of the OX axis, you need to choose the side where the projectile will fly, and parallel to the horizontal line. In this case, the projections of the forces of gravity and the reaction of the support on OX will be zero.

The problem will be solved in general form, since there is no specific data for known quantities. The answer is the formula.

The impulse of the system before the shot was zero, since the platform and the projectile were stationary. Let the required platform speed be denoted by the letter u. Then its momentum after the shot is determined as the product of the mass by the projection of the velocity. Since the platform will roll back (against the direction of the OX axis), the pulse value will be a minus sign.

The impulse of the projectile is the product of its mass by the projection of velocity on the OX axis. Because the velocity is directed at an angle to the horizon, its projection is equal to the speed multiplied by the cosine of the angle. In the letter equality this will look like this: 0 = - Mu + mv * cos α. From it by simple transformations we get the formula-answer: u = (mv * cos α) / M.

Answer. The speed of the platform is determined by the formula u = (mv * cos α) / M.

The problem of crossing the river

Condition. The width of the river along its entire length is the same and is equal to l, its banks are parallel. The speed of water flow in the river v ₁ and the speed of the boat v _{2 are} known. 1). When crossing the boat, the nose is directed strictly to the opposite shore. At what distance does s carry it downstream? 2). At what angle α should the nose of the boat be directed so that it reaches the opposite shore strictly perpendicular to the point of departure? How long does it take to cross such a ferry?

Decision. 1). The full speed of the boat is a vector sum of two quantities. The first of these is the current of the river, which is directed along the coast. The second is the speed of the boat perpendicular to the coast. In the drawing, two similar triangles are obtained. The first is formed by the width of the river and the distance the boat takes down. The second is the velocity vectors.

From them follows the following: s / l = v ₁ / v ₂ . After the transformation, we obtain the formula for the required quantity: s = l * (v ₁ / v ₂ ).

2). In this version of the problem, the total velocity vector is perpendicular to the shores. It is equal to the vector sum v ₁ and v ₂ . The sine of the angle to which the eigenvector must deviate is equal to the ratio of the moduli v ₁ and v ₂ . To calculate the time of motion, it is necessary to divide the width of the river into the calculated full speed. The value of the latter is calculated by the Pythagorean theorem.

V = √ (v ₂ ² - v ₁ ² ), then t = l / (√ (v ₂ ² - v ₁ ² )).

Answer. 1). S = l * (v ₁ / v ₂ ), 2). Sin α = v ₁ / v ₂ , t = l / (√ (v ₂ ² - v ₁ ² )).