Resolution of a Vector

Vectors are quantities that have both magnitude and direction. Sometimes, we encounter situations where a single vector can be thought of as the combination of two or more simpler vectors. This process is called the resolution of a vector. The most common type of resolution is breaking a vector into two perpendicular components.

Components of a Vector

Imagine a vector A in a 2D plane. We can resolve this vector into two perpendicular components, typically along the x-axis and y-axis. Let the angle the vector A makes with the positive x-axis be θ.

The component of A along the x-axis, denoted as A_x, is given by:

A_x = A cos θ

The component of A along the y-axis, denoted as A_y, is given by:

A_y = A sin θ

Here, A is the magnitude of the vector A. These components, A_x and A_y, are the projections of the vector A onto the x and y axes, respectively.

Vector Representation in Terms of Components

Using unit vectors i (along the x-axis) and j (along the y-axis), the vector A can be written in component form as:

A = A_x i + A_y j

The magnitude of A can be found using the Pythagorean theorem:

A = √{A_x² + A_y²}

The direction of A can be specified by the angle α it makes with the x-axis, where:

tan α = A_y / A_x

Example: Force Vector

Suppose a force of 10 N acts on an object at an angle of 30° with the horizontal. We can resolve this force into horizontal and vertical components.

Magnitude of force, F = 10 N

Angle, θ = 30°

Horizontal component, F_x = F cos θ = 10 cos 30° = 10 * (√{3}/2) = 5√{3} N

Vertical component, F_y = F sin θ = 10 sin 30° = 10 * (1/2) = 5 N

So, the force vector can be written as F = (5√{3}) i + 5 j N.

Relative Velocity

Concept of Relative Velocity

Velocity is a measure of the rate of change of position. However, velocity is always measured with respect to a frame of reference. When we talk about the velocity of an object, we implicitly mean its velocity with respect to the ground, or some other assumed stationary observer. Relative velocity is the velocity of an object as observed from a particular frame of reference, which may itself be in motion.

Relative Velocity in One Dimension

Consider two objects, A and B, moving along a straight line. Let their velocities with respect to a stationary observer (say, the ground) be v_A and v_B, respectively.

The velocity of object A relative to object B, denoted as v_AB, is given by:

v_AB = v_A - v_B

Similarly, the velocity of object B relative to object A, denoted as v_BA, is:

v_BA = v_B - v_A

Notice that v_AB = - v_BA, which means if A is moving towards B with a certain relative velocity, B is moving away from A with the same magnitude of relative velocity.

Example: Two Cars on a Highway

Car A is moving east at 60 km/h. Car B is moving east at 80 km/h.

Let the eastward direction be positive. So, v_A = +60 km/h and v_B = +80 km/h.

Velocity of A relative to B: v_AB = v_A - v_B = 60 - 80 = -20 km/h. This means from Car B's perspective, Car A is moving west at 20 km/h.

Velocity of B relative to A: v_BA = v_B - v_A = 80 - 60 = +20 km/h. This means from Car A's perspective, Car B is moving east at 20 km/h.

If Car B was moving west at 80 km/h (v_B = -80 km/h):

v_AB = v_A - v_B = 60 - (-80) = 60 + 80 = 140 km/h. From Car B's perspective, Car A is moving east at 140 km/h.

Relative Velocity in Two Dimensions

In two dimensions, we use vector subtraction. If v_A and v_B are the velocities of objects A and B, respectively, then the relative velocity of A with respect to B is:

v_AB = v_A - v_B

This subtraction is performed using vector components. For example, if v_A = v_Ax i + v_Ay j and v_B = v_Bx i + v_By j, then:

v_AB = (v_Ax - v_Bx) i + (v_Ay - v_By) j

Example: Boat Crossing a River

A boat can move with a speed of 10 m/s relative to the water. The river flows at 5 m/s.

Case 1: Boat moves directly across the river. Let the river flow be along the x-axis (v_river = 5 i m/s). Let the boat's velocity relative to water be along the y-axis (v_boat/water = 10 j m/s). The boat's actual velocity relative to the ground is v_boat = v_boat/water + v_river = 10 j + 5 i = (5 i + 10 j) m/s. The magnitude of the boat's velocity is √{5² + 10²} = √{25 + 100} = √{125} = 5√{5} m/s.

Case 2: Boat heads upstream at 10 m/s relative to water. v_boat/water = -10 j m/s (assuming upstream is negative y-direction). v_boat = -10 j + 5 i = (5 i - 10 j) m/s. Magnitude is again 5√{5} m/s.

Motion in a Plane

Introduction

Motion in a plane, also known as 2D motion, describes the movement of an object along a curved path in a two-dimensional space. This is an extension of one-dimensional motion, where we consider movement along a straight line. Examples include projectile motion, uniform circular motion, and the motion of a charged particle in a uniform magnetic field.

Decomposition of Motion

The key to analyzing motion in a plane is to decompose it into two independent motions along perpendicular axes (usually horizontal and vertical, or x and y axes). The motion along each axis can be analyzed using the one-dimensional kinematic equations.

Let the position of a particle at time t be given by the position vector r(t) = x(t) i + y(t) j.

The velocity vector is the time derivative of the position vector:

v(t) = dr/dt = (dx/dt) i + (dy/dt) j = v_x i + v_y j

The acceleration vector is the time derivative of the velocity vector:

a(t) = dv/dt = (dv_x/dt) i + (dv_y/dt) j = a_x i + a_y j

If the acceleration is constant, i.e., a_x = constant and a_y = constant, then we can use the following kinematic equations for each component:

• v_x = u_x + a_x t
• x = u_x t + (1/2) a_x t²
• v_x² = u_x² + 2 a_x x
• v_y = u_y + a_y t
• y = u_y t + (1/2) a_y t²
• v_y² = u_y² + 2 a_y y

Here, u_x and u_y are the initial velocity components, and x and y represent displacement along the respective axes.

Uniform Circular Motion

A special case of motion in a plane is uniform circular motion, where an object moves in a circular path at a constant speed. Although the speed is constant, the velocity is continuously changing because the direction of motion is changing. This change in velocity implies the presence of acceleration, known as centripetal acceleration, which is directed towards the center of the circle.

If the radius of the circle is r and the constant speed is v, the magnitude of the centripetal acceleration is:

a_c = v² / r

The direction of acceleration is always radial, pointing towards the center.

Example: Projectile Motion as Motion in a Plane

Projectile motion is a classic example of motion in a plane where the only force acting is gravity (neglecting air resistance). Gravity acts vertically downwards, so the acceleration is constant and directed downwards.

Let the initial velocity be u at an angle θ with the horizontal.

Initial velocity components: u_x = u cos θ, u_y = u sin θ.

Acceleration components: a_x = 0 (no horizontal force), a_y = -g (gravity acts downwards).

Using the kinematic equations for each component:

• Horizontal motion: v_x = u_x + 0 * t = u cos θ (constant horizontal velocity)
• Horizontal motion: x = (u cos θ) t
• Vertical motion: v_y = u_y - gt = u sin θ - gt
• Vertical motion: y = (u sin θ) t - (1/2) g t²

By combining these, we can find the trajectory (path) of the projectile.

Projectile Motion

Definition

Projectile motion is the motion of an object thrown or projected into the air, subject only to the acceleration of gravity (neglecting air resistance). The path followed by a projectile is called its trajectory.

Assumptions

For idealized projectile motion, we make the following assumptions:

• The only force acting on the projectile is gravity.
• Air resistance is negligible.
• The acceleration due to gravity, g, is constant in magnitude and direction (downwards).
• The Earth is considered flat for the range of motion.

Analysis of Projectile Motion

We analyze projectile motion by resolving the initial velocity and considering the motion along the horizontal (x) and vertical (y) axes independently.

Let the projectile be launched with an initial velocity u at an angle θ with the horizontal.

Initial velocity components:

• Horizontal component: u_x = u cos θ
• Vertical component: u_y = u sin θ

Acceleration components:

• Horizontal acceleration: a_x = 0
• Vertical acceleration: a_y = -g (negative because it's downwards)

Equations of Motion

Using the 1D kinematic equations:

Horizontal Motion (constant velocity):

• Velocity at time t: v_x(t) = u_x = u cos θ
• Position at time t: x(t) = u_x t = (u cos θ) t

Vertical Motion (constant acceleration):

• Velocity at time t: v_y(t) = u_y + a_y t = u sin θ - gt
• Position at time t: y(t) = u_y t + (1/2) a_y t² = (u sin θ) t - (1/2) g t²

Trajectory Equation

To find the shape of the path (trajectory), we eliminate time t from the position equations.

From the horizontal position equation, t = x / (u cos θ).

Substitute this into the vertical position equation:

y = (u sin θ) [x / (u cos θ)] - (1/2) g [x / (u cos θ)]²

y = x tan θ - (g x²) / (2 u² cos² θ)

This equation is of the form y = Ax - Bx², which is the equation of a parabola. Thus, the trajectory of a projectile is a parabola.

Key Parameters of Projectile Motion

1. Time of Flight (T)

The total time the projectile spends in the air. It is the time taken to reach the ground from the launch point. If launched and landing at the same height, the time to go up equals the time to come down. The vertical displacement y is zero.

0 = (u sin θ) T - (1/2) g T²

T [u sin θ - (1/2) g T] = 0

The non-zero solution gives the time of flight:

T = (2 u sin θ) / g

2. Maximum Height (H)

The highest vertical position reached by the projectile. At the maximum height, the vertical component of velocity v_y is zero.

Using v_y² = u_y² + 2 a_y y:

0² = (u sin θ)² + 2 (-g) H

0 = u² sin² θ - 2gH

H = (u² sin² θ) / (2g)

3. Horizontal Range (R)

The total horizontal distance covered by the projectile during its time of flight.

R = u_x * T

R = (u cos θ) * [(2 u sin θ) / g]

R = (u² * 2 sin θ cos θ) / g

Using the trigonometric identity sin(2θ) = 2 sin θ cos θ:

R = (u² sin(2θ)) / g

Condition for Maximum Range: The horizontal range R is maximum when sin(2θ) is maximum, which is 1. This occurs when 2θ = 90°, so θ = 45°.

Maximum Range, R_max = u² / g (at θ = 45°).

Example: A Shot Put Throw

A shot put is thrown with an initial velocity of 15 m/s at an angle of 40° with the horizontal. Calculate the time of flight, maximum height, and horizontal range (take g = 9.8 m/s²).

Given: u = 15 m/s, θ = 40°, g = 9.8 m/s².

sin 40° ≈ 0.6428

cos 40° ≈ 0.7660

sin(2 * 40°) = sin 80° ≈ 0.9848

Time of Flight (T): T = (2 * 15 * sin 40°) / 9.8 ≈ (30 * 0.6428) / 9.8 ≈ 19.284 / 9.8 ≈ 1.97 seconds.

Maximum Height (H): H = (u² sin² θ) / (2g) = (15² * (sin 40°)²) / (2 * 9.8) H ≈ (225 * (0.6428)²) / 19.6 ≈ (225 * 0.4132) / 19.6 ≈ 93.0 / 19.6 ≈ 4.74 meters.

Horizontal Range (R): R = (u² sin(2θ)) / g = (15² * sin 80°) / 9.8 R ≈ (225 * 0.9848) / 9.8 ≈ 221.58 / 9.8 ≈ 22.61 meters.