Motion

Chapter 11

Standards

• Students will:• SPS8. Determine the relationship between

force, mass and motion• SPS8a Calculate velocity and acceleration• SPS8c Relate falling objects to gravitational

force• SPS8d Explain difference in mass and weight

Observing Motion

• Motion- change in position in relation to a reference point.

Measuring Motion: Distance

• Distance- how far an object moves on a path• Displacement- how far between starting and

ending points on a path

Measuring Motion: Speed

• Speed– rate of motion – distance traveled

per unit time speed = distance time

Measuring Motion: Speed (cont’d)

• Instantaneous Speed– speed at a given

instant

• Average Speed-

= total distance

total time

Measuring Motion: Velocity

• Velocity– speed in a given

direction – can change even

when the speed is constant!

Calculating Velocity:Distance/Speed/Time Triangle

Graphing Speed/Velocity

• X axis- usually independent variable (time)• Y axis- usually dependent variable (distance)• Slope of straight line= vertical change horizontal change

Graphing (cont’d)

• slope = velocity

• steeper slope = faster velocity

• straight line = constant velocity

• flat line = 0 velocity

(no motion)

Calculating Slope

1. Choose two points on graph to calculate slope.

2. Calculate the vertical and horizontal change.

3. Divide the vertical change by the horizontal change.

Distance Vs Time

Time (s)0 1 2 3 4 5

2

4

6

8

10

12

14

16

.

.

Dis

tanc

e (m

)

Calculating Slope(cont’d)

1. point 1: d= 6m t= 1s point 2: d= 12m t= 4s 2. vert ∆- 12m-6m= 6m horiz ∆- 4s-1s= 3s

3. slope= 6m = 2m/s 3s

Distance Vs Time

Time (s)0 1 2 3 4 5

2

4

6

8

10

12

14

16

.

.

Dis

tanc

e (m

)HorizontalChange

Verticalchange

Practice Problem

• Who started out faster?– A (steeper slope)

• Who had a constant speed?– A

• Describe B from 10-20 min.– B stopped moving

• Find their average speeds.– A = (2400m) ÷ (30min)

A = 80 m/min– B = (1200m) ÷ (30min)

B = 40 m/min

A

B

Measuring Motion: Acceleration

• Acceleration– the rate of change of velocity– change in speed or direction• Centripetal acceleration– circular motion (even if speed is constant,

direction is always changing) ex. Moon accelerates around Earth

Measuring Motion: Acceleration

• Positive acceleration -“speeding up”

• Negative acceleration

-“slowing down”

Calculating Acceleration

• a= vf – vi t • a= ∆ v t

• a- acceleration• vf- final velocity • vi- initial velocity• t- time

Calculating Acceleration (cont’d)

1. List given, then unknown values.

2. Write equation for acceleration.

3. Insert known values into equation and solve.

t

Vf-Vi

a t

Practice Problem 1 A flowerpot falls off a second-story windowsill. The flowerpot starts from rest and hits Mr. Mertz 1.5s later with a velocity of 14.7m/s. Find the average acceleration of the flowerpot. Given: Remember: Solve: t= 1.5s a= vf – vi a= 14.7m/s-

0m/s Vi= 0m/s t 1.5s Vf= 14.7m/s a= 14.7m/s a= ? 1.5s a= 9.8m/s2

Practice Problem 2Joseph’s car accelerates at an average rate of2.6m/s2. How long will it take his car to speed up from 24.6m/s to 26.8m/s2?Given: Remember: Solve:a= 2.6m/s2 t= (vf-vi) ÷ avf= 26.8m/s2 vf-vi t= 26.8m/s2-24.6m/s2Vi= 24.6m/s2 a t 2.6m/s2t= ? t= 2.2m/s 2.6m/s2 t= 0.85s

Practice Problem 3A cyclist travels at a constant velocity of 4.5m/swestward and then speeds up with a steadyacceleration of 2.3m/s2. Calculate the cyclist’sspeed after accelerating for 5.0s.Given: Remember: Solve:vi= 4.5m/s vf= vi + a x tvf= ? vf-vi vf= 4.5m/s + (2.3m/s2 x 5.0s)

a= 2.3m/s2 a t vf= 4.5m/s + 11.5m/st= 5.0s vf= 16m/s

Graphing Acceleration

On Distance-Timegraph:• Acceleration is shown as

a curved line

0

100

200

300

400

0 5 10 15 20

Time (s)

Dis

tan

ce (

m)

Distance/Time

Graphing Acceleration

On a Speed-Time graph:• Slope of straight line=

acceleration• Positive slope- speeding

up• Negative slope- slowing

down• Flat line- constant

velocity (no acceleration)

0

1

2

3

0 2 4 6 8 10

Time (s)

Sp

ee

d (

m/s

)

Speed/Time

Newton’s Laws of Motion

• 2nd Law of Motion: The acceleration of an object is directly

proportional to the net force acting on it and inversely proportional to its mass.

Force = mass x acceleration oror

F= ma

Force

Force- push or pull that one body exerts on

another

Fundamental Forces

• 4 types:1. gravity2. electromagnetic3. weak nuclear4. strong nuclear• Vary in strength• Act through contact

or at a distance

Forces• Force Pairs: forces moving in opposite directions• Balanced forces: do not move; push equally on

each other• Unbalanced forces: acceleration (movement)

in the direction of larger force

Force Pair Examples

FrictionFriction: force that opposes

motion between 2 surfaces

• Static Friction: nonmoving surfaces

• Kinetic Friction: moving surfaces- sliding or rolling

(sliding friction is greater than rolling friction)

Friction Facts• Necessary for all

motion• Rougher surfaces

create greater friction

• Greater mass creates greater friction

• Lubricants reduce friction

Newton’s First and Second Laws

Chapter 12

Newton’s Laws of Motion

• 1st Law of Motion: Law of Inertia• An object at rest will remain at rest;• An object in motion will continue moving at

a constant velocity unless acted upon by a net force.

Inertia• Objects move only when

net force is applied.• Objects maintain state of

motion.• Inertia is related to mass. (small mass can be

accelerated by small force large mass can be

accelerated by large force)

Newton’s Laws of Motion

• 2nd Law of Motion: The acceleration of an object is directly

proportional to the net force acting on it and inversely proportional to its mass.

Force = mass x acceleration oror

F= ma

Newton’s Second Law

• For equal forces, large masses accelerate less

• Force is measured in newtons (N)

• 1N= 1kg x 1m/s2

F

m a

Practice Problem 1 Zoo keepers lift a stretcher that holds a sedate lion. The lion’s mass is 175 kg and the upward acceleration of the lion and stretcher is 0.657m/s2. What force is needed to produce acceleration of the lion and stretcher?Given: Remember: Solve:m= 175 kg F = m x a

a= 0.657m/s2 F = 175 kg x 0.657m/s2

F= ? m a F = 11.49 N

F

F

Practice Problem 2

A baseball accelerates downward at 9.8 m/s2. If gravity is the only force acting on the baseball and is 1.4N, what is the baseball’s mass?

Given: Remember: Solve:F= 1.4 kg/m/s2 m = F a= 9.8 m/s2 a m= ? m = 1.4 N 9.8 m/s2 m = .14 kg

Fm a

Weight and Mass

• Weight- measure of gravity on an object• Not equal to mass (constant everywhere)• Measured in Newtons weight = mass x free-fall acceleration

(9.8m/s2)

free-fall m g acceleration

w

Gravity

• Force of attraction between 2 objects in the universe

• Increases as: - mass increases - distance decreases• Affects all matter

Gravity Quiz 1Who experiences more gravity - the astronaut orthe politician?

Less distance

More distance

Gravity Quiz 2

Which exerts more gravity, your hand or yourpencil ?

Gravity Quiz 3

• Would you weigh more on Earth or Jupiter?

(Hint: Which planet has the greater mass?)

Free Fall Acceleration

• Occurs when Earth’s gravity is only force acting on an object

• In absence of air resistance, all objects accelerate at same rate

• g = 9.8 m/s2• g = 9.8 m/s2 g = 9.8

m/s2 g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2

Air Resistance• Force of air on a

moving object which opposes its motion

• Aka fluid friction or drag

• Depends on objects: - speed - surface area - shape - density

Air Resistance (cont’d)

• Terminal velocity= maximum velocity

reached by a falling object

• Reached when… F gravity = F air resistance

(no net force)

Projectile MotionProjectile-• Any object thrown in

air• Only acted on by

gravity • Follows parabolic

path- trajectory• Has horizontal and vertical velocities

V oy

V ox

Projectile Motion (cont’d)

• Horizontal Velocity– depends on inertia– remains constant

• Vertical Velocity– depends on gravity– accelerates

downward at 9.8 m/s2

Projectile Motion Quiz

• A moving truck launches a ball vertically (relative to the truck). If the truck maintains a constant horizontal velocity after the launch, where will the ball land (ignore air resistance)?

A) In front of the truck B) Behind the truck C) In the truck

Answer: C because horizontal and vertical velocities are independent of each other

Newton’s Laws of Motion

• 3rd Law of Motion: When one object exerts a force on a second

object, the second object exerts an equal but opposite force on the first.

Newton’s Third Law

• Forces always occur in pairs

• Forces in a pair do not act on the same object

• Equal forces don’t always have equal effects

Newton’s Third Law

Problem: How can a horse pull a cart if the cartis pulling with equal force back on the horse?

Newton’s Third Law

Answer:1. Forces are equal and opposite but are acting

on different objects2. Forces are not balanced3. The movement of the horse depend on the

forces working on the horse.

Momentum

Momentum- • quantity of motion• = mass x velocity• units: kg x m/s

p

m v

Practice Problem 1Find the momentum of a bumper car if it has atotal mass of 280 kg and a velocity of 3.2 m/s.

Given Remember Solvem = 280 kg p = mvv = 3.2 m/s p =(280kg)(3.2m/s)P = ? P = 896 kg m/s

vv

p

vm

Practice Problem 2

The momentum of a second bumper car is 675kg·m/s. What is its velocity if its total mass is300 kg?

Given Remember Solvep = 675 kg·m/s v = p ÷ m m = 300kg v =(675kg·m/s)÷(300kg)

v = ? v = 2.25m/s

p

m v

Conservation of Momentum

Law of Conservation of Momentum:• Total momentum of two or more objects

before a collision is the same as it was before the collision (momentum is conserved).

Jet Car Challenge

CHALLENGE:Construct a car that will travel a least 3 meters using only the following materials:• scissors • tape• 4 plastic lids• 2 plastic straws

Use your knowledge of Newton’s Laws to make this thing go!

• 2 skewers• 1 balloon• 1 cardboard base