Physics CN03 Kinematics in 2D

Kinematics in Two Dimensions

Change in Velocity

• The change in velocity ($\Delta v$) between any two times is equal to the area under the velocity-time graph between those two times.

Position-Time Graph for Two Trains

• Scenario: Graph showing position as a function of time for two trains running on parallel tracks.

• Statements to analyze:

  • A) At time $t_B$, both trains have the same velocity.

  • B) Both trains speed up all the time.

  • C) Both trains have the same velocity at some time before $t_B$.

  • D) Somewhere on the graph, both trains have the same acceleration.

Kinematics in Three Dimensions

• The x, y, and z components of motion are independent of each other.

• This independence allows us to break down three-dimensional motion into three separate one-dimensional problems.

Projectile Motion

Definitions and Key Concepts

• Projectile Motion: A special case of two-dimensional motion defined as the motion of an object thrown or launched with an initial velocity, which, after release, moves in a vertical plane and is solely affected by the Earth’s gravitational field.

  • Vertical acceleration at all points in the trajectory is $9.8 \text{ m/s}^2$ downward.

  • The velocity in the x-direction remains constant ($a_x = 0$).

  • All projectiles follow a parabolic trajectory.

Independence of Motion

• Horizontal and vertical motions are independent:

  • Neither motion affects the other.

  • Problems can be decomposed into two one-dimensional problems - one for horizontal motion and one for vertical motion.

Projectile Motion Examples

• CQ_3K7: A projectile is launched from the ground at a $30^{\circ}$ angle. The point of least speed in its trajectory is at:

  • C) at the highest point in its flight.

• CQ_3K1: A velocity $\mathbf{v} = (25, -5)$ indicates that:

  • A) yes, the ball has passed its highest point.

Range and Angle of Projection

• Maximum Range: Occurs at a launch angle of $45^{\circ}$ on a horizontal surface.

• Symmetry exists in the range function relative to the firing angle.

• Time of Flight: Determined by maximum vertical height of the projectile.

• CQ: Determine which angle gives the same range as $25^{\circ}$:

  • B) $35^{\circ}$ is the complementary angle to achieve the same range.

• CQ_3K8: Which of three punts has the longest hang time?

  • (D) All have the same hang time due to symmetrical nature of projectile motion.

Projectile Motion Scenarios

• Pirate Ship Example: Two shells are launched at the same initial speed towards two enemy ships. The question is which ship gets hit first. Potential answers include:

  • A) Ship A

  • B) Ship B

  • C) They are hit at the same time dependent on the trajectory paths.

Problem Solving Strategies for Projectile Motion

1. Read the Problem Carefully: Understand every word as each is important for solving the problem.

2. Draw a Picture: A diagram helps visualize the problem.

3. Label Everything: Note what is known and what needs to be found. Translate words into physical equations.

4. Choose a Coordinate System: Define the origin and axes for calculations.

5. Break into Components: Analyze the x and y components separately.

6. List Initial Conditions and Desired Outcome: Make sure to align signs appropriately.

7. Write Relevant Equations: Ensure that the equations used fit the scenario (e.g., equations of motion must be valid for constant acceleration).

8. Solve in Terms of Symbols First: This is important before substituting numbers.

9. Insert Known Values into Equations: Write everything with correct units.

10. Calculate While Keeping Extra Digits: Round to the correct significant figures at the end of calculations.

11. Report Your Answer with Correct Units: Verify units are included and consistent.

12. Check Physicality of Answer: A rough estimate can indicate if results are reasonable.

Example Problems

• Example: A ball is kicked with an initial speed of $v<em>0 = 20.0 \text{ m/s}$ at an angle $\theta</em>0 = 30.0^{\circ}$. Tasks include:

  • (a) Find the time to reach the highest point: $t_1 = ?$

  • (b) Find the maximum height.

  • (c) Find total time before the ball hits the ground.

  • (d) Find horizontal range and velocity just before landing.

Comparative Analysis of Dropped vs. Thrown Ball

• CQ_3K5: One ball is dropped from a height while another is fired horizontally:

  • A) The dropped ball and the fired ball will hit the ground at the same time ignoring air resistance.

Cliff Scenario with a Car

• Example: A car drives horizontally off a $50.0 \text{ m}$ high cliff and land $90.0 \text{ m}$ from the base:

  • Find initial speed of the car and its velocity before hitting the ground.

• Question of Water Balloon: When a water balloon is shot at a friend who jumps off the roof, it may or may not hit him:

  • Assumed the speed is sufficient to reach him creates ambiguity depending on timing and trajectory.

Summary of Kinematic Formulas

These formulas describe motion with constant acceleration.

General Equations for Constant Acceleration (1D)

For motion along a single dimension (e.g., x-axis) with constant acceleration $a$:

1. Velocity as a function of time:
   $v = v<em>0 + at$ Where $v$ is final velocity, $v</em>0$ is initial velocity, $a$ is constant acceleration, and $t$ is time.

2. Position as a function of time:
   $x = x<em>0 + v</em>0 t + \frac{1}{2} a t^2$
   Where $x$ is final position, $x_0$ is initial position.

3. Velocity as a function of displacement:
   $v^2 = v<em>0^2 + 2a(x - x</em>0)$

4. Position using average velocity:
   $x = x<em>0 + \frac{1}{2} (v</em>0 + v) t$

Projectile Motion Component Equations (2D)

Projectile motion is a special case of two-dimensional motion where horizontal acceleration is zero ($a<em>x = 0$) and vertical acceleration is constant ($a</em>y = -g = -9.8 \text{ m/s}^2$, assuming positive y is upwards).

Horizontal Motion (x-direction)

Acceleration is zero, so velocity is constant.

1. Horizontal velocity:
   $v<em>x = v</em>{0x} = v<em>0 \cos(\theta</em>0)$
   Where $v<em>0$ is initial speed and $\theta</em>0$ is the launch angle.

2. Horizontal position:
   $x = x<em>0 + v</em>{0x} t = x<em>0 + (v</em>0 \cos(\theta_0)) t$

Vertical Motion (y-direction)

Acceleration is constant, equal to $-g$ (magnitude $g = 9.8 \text{ m/s}^2$).

1. Vertical velocity:
   $v<em>y = v</em>{0y} + a<em>y t = v</em>0 \sin(\theta_0) - g t$

2. Vertical position:
   $y = y<em>0 + v</em>{0y} t + \frac{1}{2} a<em>y t^2 = y</em>0 + (v<em>0 \sin(\theta</em>0)) t - \frac{1}{2} g t^2$

3. Vertical velocity as a function of vertical displacement:
   $v<em>y^2 = v</em>{0y}^2 + 2 a<em>y (y - y</em>0) = (v<em>0 \sin(\theta</em>0))^2 - 2 g (y - y_0)$

Kinematics in Two Dimensions

Change in Velocity

• The change in velocity is represented by: $\Delta v$

Projectile Motion

Definitions and Key Concepts

• Vertical acceleration at all points in the trajectory is $9.8 \text{ m/s}^2$ downward.

• The velocity in the x-direction remains constant ($a_x = 0$).

Summary of Kinematic Formulas

These formulas describe motion with constant acceleration.

General Equations for Constant Acceleration (1D)

For motion along a single dimension (e.g., x-axis) with constant acceleration $a$:

1. Velocity as a function of time:

   $v = v_0 + at$

   Where $v$ is final velocity, $v_0$ is initial velocity, $a$ is constant acceleration, and $t$ is time.

2. Position as a function of time:

   $x = x<em>0 + v</em>0 t + \frac{1}{2} a t^2$

   Where $x$ is final position, $x_0$ is initial position.

3. Velocity as a function of displacement:

   $v^2 = v<em>0^2 + 2a(x - x</em>0)$

4. Position using average velocity:

   $x = x<em>0 + \frac{1}{2} (v</em>0 + v) t$

Projectile Motion Component Equations (2D)

Projectile motion is a special case of two-dimensional motion where horizontal acceleration is zero ($a<em>x = 0$) and vertical acceleration is constant ($a</em>y = -g = -9.8 \text{ m/s}^2$, assuming positive y is upwards).

Horizontal Motion (x-direction)

Acceleration is zero, so velocity is constant.

1. Horizontal velocity:

   $v<em>x = v</em>{0x} = v<em>0 \cos(\theta</em>0)$

   Where $v<em>0$ is initial speed and $\theta</em>0$ is the launch angle.

2. Horizontal position:

   $x = x<em>0 + v</em>{0x} t = x<em>0 + (v</em>0 \cos(\theta_0)) t$

Vertical Motion (y-direction)

Acceleration is constant, equal to $-g$ (magnitude $g = 9.8 \text{ m/s}^2$).

1. Vertical velocity:

   $v<em>y = v</em>{0y} + a<em>y t = v</em>0 \sin(\theta_0) - g t$

2. Vertical position:

   $y = y<em>0 + v</em>{0y} t + \frac{1}{2} a<em>y t^2 = y</em>0 + (v<em>0 \sin(\theta</em>0)) t - \frac{1}{2} g t^2$

3. Vertical velocity as a function of vertical displacement:

   $v<em>y^2 = v</em>{0y}^2 + 2 a<em>y (y - y</em>0) = (v<em>0 \sin(\theta</em>0))^2 - 2 g (y - y_0)$