1.5 Vectors and Motion in Two Dimensions

Vectors are essential tools in physics, representing quantities with both magnitude and direction. Unlike scalar quantities (which have only magnitude), vectors allow us to describe physical phenomena that have directional properties, such as force, velocity, and acceleration.

They can be broken down into perpendicular components, making complex problems easier to solve and analyze. This decomposition is particularly useful when dealing with objects moving in two dimensions, like projectiles.

Understanding vector components is crucial for studying motion in two dimensions. By resolving vectors into their x and y components, we can apply trigonometric relationships to calculate resultant vectors and solve real-world physics problems involving multiple forces or velocities acting in different directions.

Components of Vectors

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Resultant of Perpendicular Components

Vectors can be mathematically modeled as the resultant of two perpendicular components. This approach allows us to break down complex vectors into simpler, more manageable parts.

When we combine horizontal (x) and vertical (y) components, we create a single resultant vector that represents the original vector's overall effect.

For example, a boat traveling northeast can be analyzed by breaking its motion into an eastward component and a northward component. This makes calculations much simpler than working with the diagonal vector directly.

The resultant vector's magnitude can be calculated using the Pythagorean theorem:

$R = \sqrt{A_x^2 + A_y^2}$

Where:

• $R$ is the resultant vector magnitude
• $A_x$ is the x-component magnitude
• $A_y$ is the y-component magnitude

The direction of the resultant vector can be found using the inverse tangent function:

$\theta = \tan^{-1}(\frac{A_y}{A_x})$

Where $\theta$ is the angle measured from the positive x-axis.

Resolution into Components

Resolving vectors into components involves breaking a vector down into its horizontal (x) and vertical (y) parts based on a chosen coordinate system. This technique is fundamental to vector analysis.

For instance, a force applied at an angle to the horizontal can be resolved into its x and y components. These components represent the vector's influence in each direction and allow for easier analysis when multiple vectors are involved.

Components are calculated using trigonometric functions:

$A_x = A \cos \theta$ $A_y = A \sin \theta$

Where:

• $A$ is the original vector magnitude
• $\theta$ is the angle between the vector and the positive x-axis
• $A_x$ is the x-component
• $A_y$ is the y-component

Trigonometric Relationships for Components

Trigonometric functions play a crucial role in resolving vectors into perpendicular components. These mathematical relationships connect the original vector to its components.

The three primary trigonometric functions used are:

$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$

$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$

In the context of vectors:

• The hypotenuse represents the original vector's magnitude
• The opposite side represents the y-component
• The adjacent side represents the x-component

The Pythagorean theorem relates the magnitudes of the components to the original vector:

$a^2 + b^2 = c^2$

Where:

• $a$ and $b$ are the perpendicular component magnitudes
• $c$ is the original vector magnitude

The angle between the vector and the positive x-axis ($\theta$) is a key factor in resolving components. For example, a force of 50 N applied at a 30° angle to the horizontal can be resolved into:

• x-component: $F_x = 50 \text{ N} \times \cos(30°) = 43.3 \text{ N}$
• y-component: $F_y = 50 \text{ N} \times \sin(30°) = 25 \text{ N}$

Motion in Two Dimensions

Two-dimensional motion can be analyzed by separating the motion into perpendicular x- and y-components and then applying one-dimensional kinematic ideas to each direction independently. The horizontal and vertical motions occur simultaneously but are analyzed separately. For example, horizontal position changes according to the horizontal velocity, while vertical position changes according to the vertical velocity and vertical acceleration. A change in one component does not directly change the other component.

This is particularly useful when analyzing projectile motion. In ideal projectile motion, the acceleration is zero in the horizontal direction ($a_x = 0$) and constant and nonzero in the vertical direction because gravity acts downward ($a_y = -g$, if upward is positive). Because of this, horizontal motion has constant velocity while vertical motion changes according to constant-acceleration kinematics.

Think of it this way: if you throw a ball horizontally off a cliff, gravity only pulls the ball downward — it doesn't slow down or speed up the ball's horizontal motion. The ball moves forward at a steady rate while simultaneously speeding up as it falls. These two motions combine to produce the familiar curved path of a projectile.

Example: A ball launched horizontally has $v_x$ constant because $a_x = 0$, while $v_y$ changes because $a_y = -g$. The ball keeps moving forward as it falls, producing a curved path. At any moment, the horizontal and vertical motions can be analyzed separately using one-dimensional kinematic equations for each direction.

Practice Problem 1: Vector Resolution

Solution

To solve this problem, we need to resolve the hiker's displacement vector into its eastward (x) and northward (y) components.

Given:

• Total displacement = 5.0 km
• Direction = 30° north of east

Step 1: Identify the angle relative to the x-axis. Since "30° north of east" means 30° above the positive x-axis, our angle θ = 30°.

Step 2: Calculate the eastward (x) component using the cosine function. $x = 5.0 \text{ km} \times \cos(30°)$ $x = 5.0 \text{ km} \times 0.866$ $x = 4.33 \text{ km}$ eastward

Step 3: Calculate the northward (y) component using the sine function. $y = 5.0 \text{ km} \times \sin(30°)$ $y = 5.0 \text{ km} \times 0.5$ $y = 2.5 \text{ km}$ northward

Therefore, the hiker's displacement can be represented as 4.33 km east and 2.5 km north.

Practice Problem 2: Projectile Motion Components

Solution

This problem involves resolving an initial velocity vector into components and then describing how each component behaves during projectile motion.

Given:

• Initial speed = 20.0 m/s
• Launch angle = 40° above the horizontal

Step 1: Calculate the horizontal component of the initial velocity. $v_x = 20.0 \text{ m/s} \times \cos(40°)$ $v_x = 20.0 \text{ m/s} \times 0.766$ $v_x = 15.3 \text{ m/s}$

Step 2: Calculate the vertical component of the initial velocity. $v_y = 20.0 \text{ m/s} \times \sin(40°)$ $v_y = 20.0 \text{ m/s} \times 0.643$ $v_y = 12.9 \text{ m/s}$

Step 3: Describe the motion of each component during flight.

• Horizontal: Because there is no horizontal acceleration ($a_x = 0$), the horizontal velocity stays constant at 15.3 m/s throughout the entire flight.
• Vertical: Because gravity acts downward ($a_y = -g = -9.8 \text{ m/s}^2$), the vertical velocity decreases on the way up, reaches zero at the peak, and then increases in the downward direction on the way back down.

The ball's curved path results from combining these two independent motions — steady horizontal travel with changing vertical motion due to gravity.