AP Physics 1- 1.5 Vectors and Motion in Two Dimensions- Study Notes- New Syllabus
AP Physics 1-1.5 Vectors and Motion in Two Dimensions – Study Notes
AP Physics 1-1.5 Vectors and Motion in Two Dimensions – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Vector Resolution into Components
- Resolving Vectors Using Trigonometric Functions
- Motion in Two Dimensions Using Component Analysis
- Projectile Motion
Vector Resolution into Components
A vector can be mathematically modeled as the sum of two (or more) perpendicular components.This process is called resolving a vector.
- It is useful because motion and forces can be analyzed independently along each axis of a chosen coordinate system (usually x- and y-axes).
Method:
- Choose a coordinate system (x-axis, y-axis).
- For a vector \( \vec{A} \) of magnitude \( A \) making angle \( \theta \) with the x-axis:
- \( A_x = A \cos \theta \) (horizontal component)
- \( A_y = A \sin \theta \) (vertical component)
- The vector is expressed as → \( \vec{A} = A_x \hat{i} + A_y \hat{j} \).
The original vector can be reconstructed using the Pythagorean theorem:
- \( A = \sqrt{A_x^2 + A_y^2} \)
- \( \tan \theta = \dfrac{A_y}{A_x} \)
Example: Resolving a Force
A force of \( 50 \, \text{N} \) acts at an angle of \( 30^\circ \) above the horizontal. Find its horizontal and vertical components.
▶️Answer/Explanation
Step (1): Use formulas → \( F_x = F \cos \theta \), \( F_y = F \sin \theta \).
Step (2): \( F_x = 50 \cos 30^\circ = 50 \times 0.866 = 43.3 \, \text{N} \).
Step (3): \( F_y = 50 \sin 30^\circ = 50 \times 0.5 = 25 \, \text{N} \).
Final Answer: \( F_x = 43.3 \, \text{N}, \; F_y = 25 \, \text{N} \).
Resolving Vectors Using Trigonometric Functions
Vectors can be resolved into perpendicular components using basic trigonometric functions.
For a vector \( \vec{A} \) of magnitude \( A \) making an angle \( \theta \) with the x-axis:
- \( A_x = A \cos \theta \) (adjacent side → cosine rule)
- \( A_y = A \sin \theta \) (opposite side → sine rule)
Trigonometric relationships used:
- \( \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{A_y}{A} \)
- \( \cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{A_x}{A} \)
- \( \tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{A_y}{A_x} \)
- Pythagoras: \( A^2 = A_x^2 + A_y^2 \)
Thus, vector components are directly related to right triangle geometry.
Example:
A person walks \( 100 \, \text{m} \) at an angle of \( 37^\circ \) north of east. Find the eastward (x) and northward (y) components of displacement.
▶️Answer/Explanation
Step (1): Apply trigonometric relations.
\( A_x = A \cos \theta = 100 \cos 37^\circ \).
\( A_y = A \sin \theta = 100 \sin 37^\circ \).
Step (2): Use values \( \cos 37^\circ \approx 0.8, \; \sin 37^\circ \approx 0.6 \).
\( A_x = 100 \times 0.8 = 80 \, \text{m} \).
\( A_y = 100 \times 0.6 = 60 \, \text{m} \).
Final Answer: Eastward displacement = 80 m, Northward displacement = 60 m.
Motion in Two Dimensions Using Component Analysis
Any two-dimensional motion can be analyzed by breaking it into x- and y-components, each of which follows the same one-dimensional kinematic equations.Since the components are perpendicular, they are treated independently, with time \( t \) being the common variable.
Equations of motion for each component:
- For x-axis (horizontal): \( x = x_0 + v_{0x} t + \tfrac{1}{2} a_x t^2 \)
- For y-axis (vertical): \( y = y_0 + v_{0y} t + \tfrac{1}{2} a_y t^2 \)
Velocity components:
- \( v_x = v_{0x} + a_x t \)
- \( v_y = v_{0y} + a_y t \)
The resultant velocity is obtained by vector addition: \( v = \sqrt{v_x^2 + v_y^2} \), and direction \( \theta = \tan^{-1} \left( \dfrac{v_y}{v_x} \right) \).
Key idea: Treat 2D motion as two independent 1D motions, linked only by the same time \( t \).
Example:
A ball is kicked with speed \( 20 \, \text{m/s} \) at an angle of \( 30^\circ \) above the horizontal. Find the horizontal and vertical velocity components just after launch.
▶️Answer/Explanation
Step (1): Resolve velocity into components.
\( v_x = v \cos \theta = 20 \cos 30^\circ = 20 \times 0.866 = 17.3 \, \text{m/s} \).
\( v_y = v \sin \theta = 20 \sin 30^\circ = 20 \times 0.5 = 10 \, \text{m/s} \).
Step (2): These components evolve separately using 1D kinematics:
- Horizontal → constant velocity motion (if \( a_x = 0 \)).
- Vertical → uniformly accelerated motion (with \( a_y = -9.8 \, \text{m/s}^2 \)).
Final Answer: \( v_x = 17.3 \, \text{m/s}, \; v_y = 10 \, \text{m/s} \).
Projectile Motion
Projectile motion is a special case of two-dimensional motion.
- Horizontal motion → constant velocity (\( a_x = 0 \)).
- Vertical motion → uniformly accelerated motion under gravity (\( a_y = -g \)).
- The path of a projectile is a parabola, due to the combination of linear horizontal motion and accelerated vertical motion.
Assumptions:
- Air resistance is negligible.
- Acceleration due to gravity is constant and downward (\( g \approx 9.8 \, \text{m/s}^2 \)).
Equations of Motion in Components
Initial velocity: \( v_0 \) at angle \( \theta \).
- Horizontal: \( v_{0x} = v_0 \cos \theta \)
- Vertical: \( v_{0y} = v_0 \sin \theta \)
Horizontal position: \( x(t) = v_{0x} t \)
Vertical position: \( y(t) = v_{0y} t – \tfrac{1}{2} g t^2 \)
Velocity at time \( t \):
- Horizontal: \( v_x = v_{0x} \) (constant)
- Vertical: \( v_y = v_{0y} – g t \)
Key Results in Projectile Motion
- Time of Flight (T): \( T = \dfrac{2 v_0 \sin \theta}{g} \)
- Maximum Height (H): \( H = \dfrac{(v_0 \sin \theta)^2}{2g} \)
- Horizontal Range (R): \( R = \dfrac{v_0^2 \sin 2\theta}{g} \)
- Trajectory Equation (parabola): \( y = x \tan \theta – \dfrac{g x^2}{2 v_0^2 \cos^2 \theta} \)
Example:
A ball is launched with initial velocity \( 20 \, \text{m/s} \) at an angle of \( 45^\circ \). Find its time of flight, maximum height, and range.
▶️Answer/Explanation
Step (1): Break velocity into components.
\( v_{0x} = 20 \cos 45^\circ = 14.14 \, \text{m/s} \).
\( v_{0y} = 20 \sin 45^\circ = 14.14 \, \text{m/s} \).
Step (2): Time of flight:
\( T = \dfrac{2 v_{0y}}{g} = \dfrac{2 \times 14.14}{9.8} \approx 2.89 \, \text{s} \).
Step (3): Maximum height:
\( H = \dfrac{(v_{0y})^2}{2g} = \dfrac{14.14^2}{19.6} \approx 10.2 \, \text{m} \).
Step (4): Range:
\( R = v_{0x} \times T = 14.14 \times 2.89 \approx 40.9 \, \text{m} \).
Final Answer: Time of flight ≈ 2.9 s, Maximum height ≈ 10.2 m, Range ≈ 40.9 m.