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CP Physics Final Exam Review Flashcards

Scalar and Vector Quantities

Scalar Quantities: These are quantities that are fully described by a magnitude or numerical value. Examples include:
- Speed
- Distance
- Mass
- Temperature
- Time
Vector Quantities: These are quantities that are fully described by both a magnitude and a direction. Examples include:
- Velocity
- Displacement
- Force
- Acceleration
- Momentum

Speed vs. Velocity

Speed is a scalar quantity that refers to how fast an object is moving. It is the rate at which an object covers distance.
Velocity is a vector quantity that refers to the rate at which an object changes its position. It includes both the speed and the direction of motion.
The key difference is that velocity includes direction, while speed does not.

Comparing Motion of Two Cars

Car 1 travels east 250 m in 50 s.
Car 2 travels west 125 m in 25 s.
Speeds:
- Speed of Car 1: \frac{250 \text{ m}}{50 \text{ s}} = 5 \text{ m/s}
- Speed of Car 2: \frac{125 \text{ m}}{25 \text{ s}} = 5 \text{ m/s}
- Their speeds are the same.
Velocities:
- Velocity of Car 1: 5 m/s east
- Velocity of Car 2: 5 m/s west
- Their velocities are different because they are traveling in opposite directions.

Position-Time Graphs

Slope: The slope of a position-time graph represents the velocity of the object.
Interpretation of graphs:

Velocity-Time Graphs

Slope: The slope of a velocity-time graph represents the acceleration of the object.
Interpretation of graphs:

Analyzing a Position-Time Graph

Given a position-time graph (details of the graph are not fully described in the text):
- Type of motion: Determined by the shape of the graph (constant velocity, acceleration, etc.)
- Instantaneous velocity at t = 5.0 s: Find the slope of the tangent to the graph at t = 5.0 s.
- Average velocity between t = 5.0 s and t = 10.0 s: Calculate the displacement during this interval and divide by the time interval (5.0 s).

Further Analysis of Position-Time Graphs

Given another position-time graph:
- Velocity at t = 2.0 s, t = 5.0 s, and t = 9.0 s: Determine the slope of the graph at each of these times to find the instantaneous velocity at those points.

Displacement from Velocity-Time Graph

Finding Displacement: Displacement is found by calculating the area under the velocity-time graph.

Analyzing a Velocity-Time Graph

Given a velocity-time graph:
- Displacement in the first ten seconds: Calculate the area under the graph from t = 0 s to t = 10 s.
- Displacement between t = 15 s and t = 25 s: Calculate the area under the graph from t = 15 s to t = 25 s.

Kinematics Problems

Motorcyclist:
- Initial velocity, v_i = 12 \text{ m/s}
- Acceleration, a = 2.5 \text{ m/s}^2
- Time, t = 4.0 \text{ s}
- Distance, d = v_i t + \frac{1}{2} a t^2 = (12 \text{ m/s})(4.0 \text{ s}) + \frac{1}{2} (2.5 \text{ m/s}^2)(4.0 \text{ s})^2 = 48 \text{ m} + 20 \text{ m} = 68 \text{ m}
Driver accelerating from rest:
- Initial velocity, v_i = 0 \text{ m/s}
- Acceleration, a = 1.8 \text{ m/s}^2
- Time, t = 8.5 \text{ s}
- Final velocity, vf = vi + a t = 0 + (1.8 \text{ m/s}^2)(8.5 \text{ s}) = 15.3 \text{ m/s}
Semi-truck accelerating:
- Acceleration, a = 0.64 \text{ m/s}^2
- Distance, d = 125 \text{ m}
- Final velocity, v_f = 17 \text{ m/s}
- Initial velocity, vi = \sqrt{vf^2 - 2 a d} = \sqrt{(17 \text{ m/s})^2 - 2 (0.64 \text{ m/s}^2)(125 \text{ m})} = \sqrt{289 - 160} \approx 11.4 \text{ m/s}

Resultant Force Vector

Two forces acting on an object:
- 25 N, south
- 45 N, west
These forces are perpendicular, so we can use the Pythagorean theorem to find the magnitude of the resultant force:
- R = \sqrt{(25 \text{ N})^2 + (45 \text{ N})^2} = \sqrt{625 + 2025} = \sqrt{2650} \approx 51.5 \text{ N}
The direction can be found using the arctangent function:
- \theta = \arctan\left(\frac{25}{45}\right) \approx 29.1^\circ
Resultant force: Approximately 51.5 N at 29.1° south of west.

Acceleration Due to Gravity

The acceleration due to gravity on Earth is approximately 9.8 \text{ m/s}^2.
It is usually given a negative sign when considering upward motion as positive because gravity acts downwards, toward the center of the Earth.

Falling Objects and Air Resistance

Ignoring air resistance, objects fall at the same rate regardless of their mass. This is because the force of gravity is proportional to mass, so a heavier object experiences a greater gravitational force, but it also has greater inertia, which cancels out the effect.

Projectile Motion Problems

Ping pong ball rolling off a table:
- Horizontal speed, v_x = 0.85 \text{ m/s}
- Height of table, h = 0.95 \text{ m}
- Time to fall, t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(0.95 \text{ m})}{9.8 \text{ m/s}^2}} \approx 0.44 \text{ s}
- Horizontal distance, d = v_x t = (0.85 \text{ m/s})(0.44 \text{ s}) \approx 0.37 \text{ m}
Soccer ball rolling off a cliff:
- Horizontal velocity, v_x = 1.4 \text{ m/s}
- Height of cliff, h = 67 \text{ m}
- Time to fall, t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(67 \text{ m})}{9.8 \text{ m/s}^2}} \approx 3.7 \text{ s}
- Horizontal distance, d = v_x t = (1.4 \text{ m/s})(3.7 \text{ s}) \approx 5.2 \text{ m}

Projectile Launched at an Angle

Diagram of a projectile launched at an angle:
- vx and vy are the horizontal and vertical components of the initial velocity, respectively.
- Relationship between vx and vy: vx is constant throughout the motion (assuming no air resistance), while vy changes due to gravity.
- Value of vy at the highest point: At the highest point, vy = 0 \text{ m/s}.
- Value of gravity throughout the trajectory: Gravity is constant throughout the trajectory and equals 9.8 \text{ m/s}^2 downwards.

Newton's Laws of Motion

Newton's First Law (Law of Inertia): An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by a force.
Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on the object, is in the same direction as the net force, and is inversely proportional to the mass of the object. This is mathematically expressed as F = ma.
Newton's Third Law: For every action, there is an equal and opposite reaction. This means that if object A exerts a force on object B, then object B exerts an equal and opposite force on object A.

Inertia

The amount of inertia in an object is determined by its mass. The greater the mass, the greater the inertia.

Forces and Constant Motion

An object moving in a straight line at a constant speed does not necessarily have no forces acting on it. It means that the net force acting on it is zero. For example, friction and an applied force could be equal and opposite, resulting in no net force and constant velocity.

Motorcycle Crash

When you crash your motorcycle into a guard rail, you keep moving forward due to inertia (Newton's first law). Your body continues to move forward at the same velocity until an external force acts upon it to stop it.

Force, Mass, and Acceleration

According to Newton's second law (F = ma):
- If more force is applied to an object, the acceleration increases (assuming mass remains constant).
- If the mass of an object increases, the acceleration decreases (assuming force remains constant).

Net Force

Net Force: The net force is the vector sum of all the forces acting on an object. It determines the acceleration of the object.
An object will accelerate in the direction of the net force.

Calculating Net Force

Forces of 50 N in one direction and 30 N in the opposite direction:
- Net force = 50 \text{ N} - 30 \text{ N} = 20 \text{ N} in the direction of the 50 N force.
Forces of 10 N and 20 N in the same direction:
- Net force = 10 \text{ N} + 20 \text{ N} = 30 \text{ N} in that direction.

Applying Newton's Second Law

Net force acting on a 120 kg mass accelerated at 1.3 \text{ m/s}^2:
- F = ma = (120 \text{ kg})(1.3 \text{ m/s}^2) = 156 \text{ N}
Mass of an object experiencing a net force of 380 N while accelerating at 5.5 \text{ m/s}^2:
- m = \frac{F}{a} = \frac{380 \text{ N}}{5.5 \text{ m/s}^2} \approx 69.1 \text{ kg}

Mass and Weight

Relationship: Weight is the force of gravity acting on an object's mass.
The relationship is expressed as W = mg, where W is weight, m is mass, and g is the acceleration due to gravity (approximately 9.8 \text{ m/s}^2 on Earth).

Mass and Weight Calculation

A 65 kg man on Earth:
- Mass: 65 kg
- Weight: W = mg = (65 \text{ kg})(9.8 \text{ m/s}^2) = 637 \text{ N}

Comparing Weight and Gravity

Two boxes: 10 kg and 15 kg:
- The 15 kg box has the greater weight because weight is directly proportional to mass.
- The 15 kg box experiences a greater force of gravity because the force of gravity is proportional to the mass of the object.

Examples of Newton's Third Law

Examples:
- When you push against a wall, the wall pushes back on you with an equal and opposite force.
- When a swimmer pushes water backward, the water pushes the swimmer forward.
- When a rocket expels gas downward, the gas pushes the rocket upward.

Four Basic Forces in the Universe

The four basic forces in the universe are:
- Strong nuclear force (strongest)
- Electromagnetic force
- Weak nuclear force
- Gravitational force (weakest)

Equilibrant Force

Equilibrant Force: The equilibrant force is the force that, when applied to an object, produces equilibrium (net force of zero). It is equal in magnitude and opposite in direction to the resultant force.

Resultant and Equilibrant Forces

Forces: 31 N east and 56 N west:
- Resultant force: 56 \text{ N} - 31 \text{ N} = 25 \text{ N} west.
- Equilibrant force: 25 N east.
Forces: 20 N north and 8 N east:
- Resultant force: Magnitude = \sqrt{(20 \text{ N})^2 + (8 \text{ N})^2} \approx 21.5 \text{ N}. Direction = \arctan(\frac{20}{8}) \approx 68.2^\circ north of east.
- Equilibrant force: 21.5 N at 68.2° south of west.

Movers and Furniture

Forces: 250 N west and 335 N north:
- Resultant force: Magnitude = \sqrt{(250 \text{ N})^2 + (335 \text{ N})^2} \approx 418.1 \text{ N}. Direction = \arctan(\frac{335}{250}) \approx 53.3^\circ north of west.
- Equilibrant force: 418.1 N at 53.3° south of east to keep the furniture moving at a constant velocity (i.e., in equilibrium).

Free Body Diagram Analysis

Diagram 1:
- Forces:
  - F_N: Normal force
  - W_t: Weight (force of gravity)
  - F_f: Friction force
  - F_a: Applied force
- Relationships:
  - FN = Wt (in magnitude, if the object is on a horizontal surface)
  - Ff = Fa (in magnitude, if the object is moving at a constant velocity or is at rest)
- Newton's Laws: Newton's First Law applies if the object is at rest or moving at a constant velocity. Newton's Second Law applies if there is a net force and the object is accelerating.
- Motion: This object can be moving if Fa is enough to overcome Ff, or it can be static if Fa = Ff.
Diagram 2:
- Relationships:
  - The relationships between $FN$ and $Wt$, and $Ff$ and $Fa$ depend on whether the object is in equilibrium or accelerating.
- Newton's Laws: Similar to the previous case, Newton's First or Second Law could apply depending on the motion.
- Motion: How the object is moving cannot be determined without knowing the relative magnitudes of the forces.

Types of Friction

Static Friction: The force that opposes the start of motion between two surfaces in contact. It prevents an object from moving when a force is applied.
Kinetic (Sliding) Friction: The force that opposes the motion of two surfaces sliding against each other. It acts on a moving object.

Comparing Friction

Static friction is greater than sliding friction for two surfaces in contact. It takes more force to initially start an object moving than to keep it moving.

Direction of Friction

Friction acts in the opposite direction to an object's motion or intended motion.

Net Force on an Elevator

Elevator weight: 14,000 \text{ N}
Cable tension: 14,800 \text{ N}
Net force: 14,800 \text{ N} - 14,000 \text{ N} = 800 \text{ N} upwards.

Normal Force on a Steel Beam

Beam weight: 45,000 \text{ N}
Crane tension: 40,000 \text{ N}
Normal force: 45,000 \text{ N} - 40,000 \text{ N} = 5,000 \text{ N}

Object on an Inclined Plane

Forces:
- W_t: Weight (force of gravity)
- F_H: Horizontal component of the weight
- F_|: Component of weight parallel to the inclined plane
Normal Force Vector: The normal force (F_N) is perpendicular to the surface of the inclined plane.
Motion: When the object is given a push:
- If the push is large enough to overcome friction, the object will move down the inclined plane.
- The acceleration depends on the net force acting on the object (component of weight along the plane - friction).

Circular Motion

Centripetal Force Vector: The centripetal force vector points toward the center of the circle.
Acceleration: The object's acceleration (centripetal acceleration) also points toward the center of the circle.

Centripetal Force and Icy Roads

When the road is icy, the centripetal force is reduced because there is less friction between the tires and the road. The centripetal force is what keeps the vehicle moving in a circle.
To make it safely through the curve, the driver must reduce their speed to reduce the centripetal force required.

Centripetal Force Calculation

Car mass: 1700 \text{ kg}
Speed: 23 \text{ m/s}
Curve radius: 85.0 \text{ m}
Centripetal force: F_c = \frac{mv^2}{r} = \frac{(1700 \text{ kg})(23 \text{ m/s})^2}{85.0 \text{ m}} \approx 10540 \text{ N}

Centripetal Acceleration and Force on a Runner

Runner mass: 82.0 \text{ kg}
Speed: 1.3 \text{ m/s}
Curve radius: 22 \text{ m}
Centripetal acceleration: a_c = \frac{v^2}{r} = \frac{(1.3 \text{ m/s})^2}{22 \text{ m}} \approx 0.077 \text{ m/s}^2
Centripetal force: Fc = m ac = (82.0 \text{ kg})(0.077 \text{ m/s}^2) \approx 6.3 \text{ N}

Centrifugal Force

Centrifugal force is not a real force. It is a fictitious force that appears to act outward on an object moving in a circular path.
When you go around a sharp curve in a vehicle, what really causes you to be pushed toward the door is your inertia. Your body wants to continue moving in a straight line, but the car is turning, so you feel like you are being pushed outward.

Law of Universal Gravitation

The law of universal gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically, F = G \frac{m1 m2}{r^2}, where G is the gravitational constant (6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2), m1 and m2 are the masses of the two objects, and r is the distance between their centers.

Gravitational Force Changes

Using the rule of ones to determine how the gravitation force would change we can determine the effect on the gravitational force (F = G \frac{m1 m2}{r^2}):

One mass doubles:
- The gravitational force doubles.F' = G \frac{2m1 m2}{r^2} = 2F
Distance is doubled:
- The gravitational force decreases to one-quarter.F' = G \frac{m1 m2}{(2r)^2} = \frac{1}{4}F
Both masses are doubled:
- The gravitational force quadruples.F' = G \frac{2m1 2m2}{r^2} = 4F
Distance is tripled:
- The gravitational force decreases to one-ninth.F' = G \frac{m1 m2}{(3r)^2} = \frac{1}{9}F
One mass doubles and distance doubles:
- The gravitational force is halved.F' = G \frac{2m1 m2}{(2r)^2} = \frac{1}{2}F
Distance is reduced to half:
- The gravitational force quadruples.F' = G \frac{m1 m2}{(\frac{1}{2}r)^2} = 4F

Gravitational Force Calculation

Two objects 5.0 m apart with masses of 12.0 kg and 35.0 kg:
- F = G \frac{m1 m2}{r^2} = (6.674 \times 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}) \frac{(12.0 \text{ kg})(35.0 \text{ kg})}{(5.0 \text{ m})^2} \approx 1.12 \times 10^{-9} \text{ N}

Torque

Torque: Torque is a rotational force that tends to cause rotation. It is the measure of how much a force acting on an object causes that object to rotate.
What it does: Torque causes or changes rotational motion.

Factors Affecting Torque

The three factors that affect the amount of torque produced are:
- Magnitude of the force
- Distance from the axis of rotation (lever arm)
- Angle between the force and the lever arm

Torque Calculation

Perpendicular force of 4.2 N at a distance of 0.65 m from the door's hinges:
- \tau = rF \sin(\theta) = (0.65 \text{ m})(4.2 \text{ N}) \sin(90^\circ) = 2.73 \text{ N} \cdot \text{m}

Balanced Seesaw

Child 1: Weight = 215 N, Distance = 5.0 m left of center
Child 2: Weight = 285 N, Distance = r right of center
For a balanced seesaw, the torques must be equal:
- \tau1 = \tau2
- (215 \text{ N})(5.0 \text{ m}) = (285 \text{ N})r
- r = \frac{(215 \text{ N})(5.0 \text{ m})}{285 \text{ N}} \approx 3.77 \text{ m}

Object on a Board

Object 1: Weight = 345 N, Distance = 7.5 m right of center
Object 2: Distance = 6.2 m right of center
For equilibrium, the torques must balance. If we assume the board is supported at its center by a fulcrum then the torque from the two objects have to be equal
- \tau1 = \tau2
- (345 \text{ N})(7.5 \text{ m}) = W_2 (6.2 \text{ m})
- W_2 = \frac{(345 \text{ N})(7.5 \text{ m})}{6.2 \text{ m}} \approx 417.3 \text{ N}

Work

Work: Work is the transfer of energy that occurs when a force causes a displacement of an object.
Conditions to be met:
- A force must be applied to the object.
- The object must move a distance.
- The force must have a component in the direction of the displacement.
Examples:
- Lifting a box
- Pushing a car that moves
- Kicking a ball that travels a distance

Power

Power: Power is the rate at which work is done or energy is transferred. It is measured in watts (W), where 1 watt = 1 joule/second.

Machines

A machine makes work easier to do by changing the magnitude or direction of the force needed, but it does not reduce the amount of work required. Machines can multiply force (mechanical advantage) but also increase the distance over which the force must be applied.

Energy Transformation: Falling Object

As an object falls towards Earth:
- Kinetic energy increases because the object's speed increases.
- Potential energy decreases because the object's height decreases.

Energy Transformation: Object Thrown Upward

As an object is thrown up into the air:
- Kinetic energy decreases because the object's speed decreases as it rises.
- Potential energy increases because the object's height increases.

Potential and Kinetic Energy

If an object has 1250 J of potential energy at the top of a tall building, it will have 1250 J of kinetic energy just before it hits the ground (assuming no air resistance) because potential energy is converted into kinetic energy. By the conservation of energy we know that the sum of both potential and kinetic energy is constant.

Work Calculation

Force: 58 N
Distance: 23 m
Work done: \text{Work} = \text{Force} \times \text{Distance} = (58 \text{ N})(23 \text{ m}) = 1334 \text{ J}

Efficiency of an Inclined Plane

Input work: 2100 J
Output work: 1750 J
Efficiency: \text{Efficiency} = \frac{\text{Output Work}}{\text{Input Work}} \times 100\% = \frac{1750 \text{ J}}{2100 \text{ J}} \times 100\% \approx 83.3\%

Potential Energy Loss

Ball mass: 5.0 kg
Distance fallen: 32.0 m
Potential energy lost: \Delta PE = mg\Delta h = (5.0 \text{ kg})(9.8 \text{ m/s}^2)(32.0 \text{ m}) = 1568 \text{ J}

Momentum and Impulse

Momentum: Momentum is the product of an object's mass and its velocity. It is a measure of how difficult it is to stop a moving object. (p = mv).
Impulse: Impulse is the change in momentum of an object. It is equal to the force applied to the object multiplied by the time interval over which the force acts. (J = FΔt).

Car Crash and Collision Time

In a car crash, it is advantageous for an occupant to extend the time during which the collision is taking place because this reduces the force exerted on the occupant. Since impulse (change in momentum) is equal to force multiplied by time, increasing the time of the collision reduces the force experienced.
Cars are designed to extend the time of a collision through features like crumple zones, airbags, and seatbelts.

Bug and Windshield Collision

a. The forces of impact on the bug and the car are the same: True (Newton's Third Law).
b. The impulses on the bug and on the car are the same: True (Impulse is the change in momentum, and the change in momentum of the bug and the car are equal and opposite).
c. The changes in speed of the bug and of the car are the same: False (The bug experiences a much greater change in speed because it has a much smaller mass).
d. The changes in momentum of the bug and the car are the same: True (The magnitude of the change in momentum is the same, but the directions are opposite).

Momentum of a Car

Car mass: 815 kg
Velocity: 12 m/s
Momentum: p = mv = (815 \text{ kg})(12 \text{ m/s}) = 9780 \text{ kg} \cdot \text{m/s}

Freight Car Collision

Car 1: Mass = 6500 kg, Velocity = v_1
Car 2: Mass = 8500 kg, Velocity = 0 m/s
Combined mass after collision: 6500 kg + 8500 kg = 15000 kg
Final velocity of combined mass: 1.8 m/s
Using conservation of momentum:
- (6500 \text{ kg})v_1 + (8500 \text{ kg})(0 \text{ m/s}) = (15000 \text{ kg})(1.8 \text{ m/s})
- v_1 = \frac{(15000 \text{ kg})(1.8 \text{ m/s})}{6500 \text{ kg}} \approx 4.15 \text{ m/s}

Ball Collision

Ball 1: Mass = 0.75 kg, Velocity = 1.2 m/s
Ball 2: Mass = 1.6 kg, Velocity = 0 m/s
After collision, Ball 1 stops, and Ball 2 moves with velocity v_2
Using conservation of momentum:
- (0.75 \text{ kg})(1.2 \text{ m/s}) + (1.6 \text{ kg})(0 \text{ m/s}) = (0.75 \text{ kg})(0 \text{ m/s}) + (1.6 \text{ kg})v_2
- v_2 = \frac{(0.75 \text{ kg})(1.2 \text{ m/s})}{1.6 \text{ kg}} \approx 0.56 \text{ m/s}

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