Physics Exam 1 Jeopardy: Kinematics and Newton's Second Law

Projectiles

Projectile Motion: Basic Concepts

Projectile motion describes the path of an object that is launched into the air and is subject only to gravity and its initial velocity. The motion can be analyzed in horizontal and vertical components independently.

• Key Equations: Vertical displacement: $\Delta y = v_{iy} t + \frac{1}{2} a_y t^2$ Horizontal displacement: $\Delta x = v_{ix} t + \frac{1}{2} a_x t^2$ Time of flight (vertical drop): $t = \sqrt{\frac{2\Delta y}{a_y}}$

• Example: A ball thrown horizontally from a 20-m-high building with a speed of 5 m/s lands 10 m from the base after 2 seconds. Calculation: $\Delta y = -20\,m$, $v_{ix} = 5\,\frac{m}{s}$, $a_y = -9.8\,\frac{m}{s^2}$ $t = \sqrt{\frac{2(-20)}{-9.8}} = 2\,s$ $\Delta x = 5 \times 2 = 10\,m$

Projectile Motion with Angled Launch

When a projectile is launched at an angle, its initial velocity is split into horizontal and vertical components using trigonometric functions.

• Key Equations: $v_{ix} = v_i \cos \theta$ $v_{iy} = v_i \sin \theta$

• Example: An archerfish shoots a water stream at 3.7 m/s at a 39° angle to hit an insect 26 cm away and 17 cm above the water. Calculation: $v_{ix} = 3.7 \cos 39^\circ$ $v_{iy} = 3.7 \sin 39^\circ$ $t = \frac{0.26}{3.7 \cos 39^\circ} = 0.09\,s$ $y = 3.7 \sin 39^\circ (0.09) + \frac{1}{2}(-9.8)(0.09)^2$ The stream reaches the insect: "It's lunch time for Fishy!"

Projectile Motion with Wind or Additional Acceleration

External forces such as wind can add horizontal acceleration to projectile motion, affecting both time of flight and horizontal displacement.

• Key Equations: $a_x$ (horizontal acceleration) is included in $\Delta x = v_{ix} t + \frac{1}{2} a_x t^2$

• Example: A person walks off a 15.0 m cliff at 3.00 m/s horizontally, with a wind causing $a_x = -3.5\,\frac{m}{s^2}$. $t = \sqrt{\frac{2(-15)}{-9.8}} = 1.75\,s$ $\Delta x = 3.00 \times 1.75 + \frac{1}{2}(-3.5)(1.75)^2 = 12\,m$

Newton's Second Law

Newton's Second Law: Definition and Application

Newton's Second Law states that the net force acting on an object is equal to the product of its mass and acceleration.

• Key Equation: $F_{net} = m a$

• Example: A car with a mass of 1200 kg has a maximum acceleration of 4.0 m/s2. After picking up passengers and luggage, the mass increases to 1600 kg. The new maximum acceleration is: $a = \frac{F_{net}}{m} = \frac{4800}{1600} = 3.0\,\frac{m}{s^2}$

Force, Mass, and Acceleration Relationships

For a given force, increasing mass decreases acceleration. This relationship is often visualized in acceleration vs. force graphs.

• Key Point: The slope of an acceleration vs. force graph is $\frac{1}{m}$, so a steeper slope means a smaller mass.

• Example Table:

  Object	Force (N)	Acceleration (m/s2)	Mass (kg)
  1	2	4	0.5
  2	2	1	2.0
  3	10	2	5.0

  Additional info: Table values inferred from context and sample calculations.

Applications: Braking and Reaction Time

When stopping a car, both reaction time and braking acceleration must be considered to determine if a collision can be avoided.

• Key Equations: Distance during reaction: $\Delta x_1 = v t_{reaction}$ Distance during braking: $\Delta x_2 = \frac{v^2}{2a}$

• Example: A driver at 20 m/s with a reaction time of 0.5 s and maximum braking acceleration of 6.0 m/s2 can stop in $\Delta x_1 + \Delta x_2 = 43.3\,m$ before hitting an obstacle 50 m away.

Impulse and Force in Sports

Impulse is the change in momentum resulting from a force applied over a time interval. In sports, this is seen when a bat hits a ball.

• Key Equations: $a = \frac{v_f - v_i}{\Delta t}$ $F = m a$

• Example: A ball of mass 0.14 kg is hit to 12 m/s in 2 ms: $a = \frac{12}{0.002} = 6000\,\frac{m}{s^2}$ $F = 0.14 \times 6000 = 840\,N$

Friction and Energy in Curling

Friction opposes motion and causes objects to slow down. In curling, the frictional force determines how far the stone travels.

• Key Equations: $v_f^2 = v_i^2 + 2 a \Delta x$ $a = \frac{F_{friction}}{m}$

• Example: A 20 kg stone with friction force 2 N travels 27.9 m before stopping: $a = \frac{-2}{20} = -0.1\,\frac{m}{s^2}$ $v_i = \sqrt{-2(-0.1)(27.9)} = 2.36\,m/s$

Summary Table: Key Equations

Concept	Equation (LaTeX)
Projectile vertical displacement	$\Delta y = v_{iy} t + \frac{1}{2} a_y t^2$
Projectile horizontal displacement	$\Delta x = v_{ix} t + \frac{1}{2} a_x t^2$
Newton's Second Law	$F_{net} = m a$
Impulse	$F = m \frac{\Delta v}{\Delta t}$
Friction acceleration	$a = \frac{F_{friction}}{m}$

Additional info:

• All problems are based on introductory college physics topics: kinematics, projectile motion, Newton's laws, and basic dynamics.

• Some values and table entries were inferred for completeness and clarity.