Ch. 5Ch. 5Motion & ForcesMotion & Forces

II. Describing Motion Using Newton’s Laws of Motion

A. MotionA. Motion

Problem: Is your desk moving?

We need a reference point...nonmoving point from which

motion is measured

A. MotionA. Motion

MotionChange in position in relation to

a reference point.

Reference point

Motion

A. MotionA. Motion

Problem:You are a passenger in a car

stopped at a stop sign. Out of the corner of your eye, you notice a tree on the side of the road begin to move forward.

You have mistakenly set yourself as the reference point.

A. ForceA. Force

Force a push or pull that one body exerts on another What forces are being

exerted on the football?

Fkick

Fgrav

A. ForceA. Force

Balanced Forces

forces acting on an object that are opposite in direction and equal in size

no change in velocity

A. ForceA. Force

Net Force unbalanced forces that are not

opposite and equal velocity changes (object accelerates)

Ffriction

W

Fpull

Fnet

NN

B. GravityB. Gravity

Gravity force of attraction between any two

objects in the universe

increases as...

•mass increases

•distance decreases

B. GravityB. Gravity

Who experiences more gravity - the astronaut or the politician?

less distance

more mass

Which exerts more gravity - the Earth or the moon?

B. GravityB. Gravity

Weight the force of gravity on an object

MASSalways the same

(kg)

WEIGHTdepends on gravity

(N)

W = mgW: weight (N)m: mass (kg)g: acceleration due

to gravity (m/s2)

B. GravityB. Gravity

Would you weigh more on Earth or Jupiter?

greater gravity

greater weight

greater mass

Jupiter because...

B. GravityB. Gravity

Accel. due to gravity (g) In the absence of air

resistance, all falling objects have the same acceleration!

On Earth: g = 9.8 m/s2

mW

g

elephant

m

Wg

featherAnimation from “Multimedia Physics Studios.”

Newton’s First LawNewton’s First Law

Newton’s First Law of MotionAn object at rest will remain at

rest and an object in motion will continue moving at a constant velocity unless acted upon by a net force.

motion

constant velocitynet force

B. Newton’s First LawB. Newton’s First Law

Newton’s First Law of Motion “Law of Inertia”

Inertia (Greek word meaning Lazyness)

tendency of an object to resist any change in its motion

increases as mass increases

ConcepTest 2ConcepTest 2

You are a passenger in a car and not wearing your seat belt.

Without increasing or decreasing its speed, the car makes a sharp left turn, and you find yourself colliding with the right-hand door.

Which is the correct analysis of the situation? ...

ConcepTest 2ConcepTest 2

1. Before and after the collision, there is a rightward force pushing you into the door.

2. Starting at the time of collision, the door exerts a leftward force on you.

3. both of the above

4. neither of the above

2. Starting at the time of collision, the door exerts a leftward force on you.

What is a Force?What is a Force?

Push or pullMeasured in (N) Newton

Net forcesNet forces

Net force: all forces exerted on an object

Unbalanced and Balanced Unbalanced and Balanced ForcesForces

Unbalanced forces produce a change in motion(acceleration)

Balanced Forces produce no change in motion

C. FrictionC. Friction

Friction force that opposes motion between

2 surfaces depends on the:

• types of surfaces• force between the surfaces

C. FrictionC. Friction

Friction is greater... between rough surfaces when there’s a greater

force between the surfaces (e.g. more weight)

Pros and Cons?

Types of FrictionTypes of Friction

Sliding Friction

Rolling Friction

Fluid Friction

Static Friction

Newton’s Second LawNewton’s Second Law

Newton’s Second Law of MotionThe acceleration of an object is

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

F = ma

B. Speed & VelocityB. Speed & Velocity

Speed rate of motion distance traveled per unit time

time

distancespeed

vd

t

B. Speed & VelocityB. Speed & Velocity

Instantaneous Speedspeed at a given instant

Average Speed

time total

distance totalspeed avg.

B. Speed & VelocityB. Speed & Velocity

Problem:A storm is 10 km away and is

moving at a speed of 60 km/h. Should you be worried?

It depends on the storm’s direction!

B. Speed & VelocityB. Speed & Velocity

Velocityspeed in a given directioncan change even when the

speed is constant!

C. AccelerationC. Acceleration

Acceleration the rate of change of velocitychange in speed or direction

t

vva if

a: acceleration

vf: final velocity

vi: initial velocity

t: time

a

vf - vi

t

D. CalculationsD. CalculationsYour neighbor skates at a speed of 4 m/s.

You can skate 100 m in 20 s. Who skates faster?

GIVEN:

d = 100 m

t = 20 s

v = ?

WORK:

v = d ÷ t

v = (100 m) ÷ (20 s)

v = 5 m/s

You skate faster!vd

t

D. CalculationsD. CalculationsA roller coaster starts down a hill at 10 m/s.

Three seconds later, its speed is 32 m/s. What is the roller coaster’s acceleration?

GIVEN:

vi = 10 m/s

t = 3 s

vf = 32 m/s

a = ?

WORK:

a = (vf - vi) ÷ t

a = (32m/s - 10m/s) ÷ (3s)

a = 22 m/s ÷ 3 s

a = 7.3 m/s2a

vf - vi

t

D. CalculationsD. CalculationsSound travels 330 m/s. If a lightning bolt

strikes the ground 1 km away from you, how long will it take for you to hear it?

GIVEN:

v = 330 m/s

d = 1km = 1000m

t = ?

WORK:

t = d ÷ v

t = (1000 m) ÷ (330 m/s)

t = 3.03 s

vd

t

D. CalculationsD. CalculationsHow long will it take a car traveling 30 m/s

to come to a stop if its acceleration is -3 m/s2?

GIVEN:

t = ?

vi = 30 m/s

vf = 0 m/s

a = -3 m/s2

WORK:

t = (vf - vi) ÷ a

t = (0m/s-30m/s)÷(-3m/s2)

t = -30 m/s ÷ -3m/s2

t = 10 sa

vf - vi

t

D. CalculationsD. CalculationsWhat force would be required to

accelerate a 40 kg mass by 4 m/s2?

GIVEN:

F = ?

m = 40 kg

a = 4 m/s2

WORK:

F = ma

F = (40 kg)(4 m/s2)

F = 160 N

m

F

a

D. CalculationsD. CalculationsA 4.0 kg shotput is thrown with 30 N of

force. What is its acceleration?

GIVEN:

m = 4.0 kg

F = 30 N

a = ?

WORK:

a = F ÷ m

a = (30 N) ÷ (4.0 kg)

a = 7.5 m/s2

m

F

a

D. CalculationsD. CalculationsMs. Wills weighs 557 N. What is her

mass?

GIVEN:

F(W) = 557 N

m = ?

a(g) = 9.8 m/s2

WORK:

m = F ÷ a

m = (557 N) ÷ (9.8 m/s2)

m = 56.8 kg

m

F

a

ConcepTestConcepTest

Is the following statement true or false? An astronaut has less mass on the

moon since the moon exerts a weaker gravitational force.

False! Mass does not depend on gravity, weight does. The astronaut has less weight on the moon.

A. Newton’s Third LawA. Newton’s Third Law

Newton’s Third Law of MotionWhen one object exerts a force

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

A. Newton’s Third LawA. Newton’s Third Law

Problem:

How can a horse pull a cart if the cart is pulling back on the horse with an equal but opposite force?

NO!!!

Aren’t these “balanced forces” resulting in no acceleration?

A. Newton’s Third LawA. Newton’s Third Law

forces are equal and opposite but act on different objects

they are not “balanced forces” the movement of the horse

depends on the forces acting on the horse

Explanation:

A. Newton’s Third LawA. Newton’s Third Law

Action-Reaction Pairs

The hammer exerts a force on the nail to the right.

The nail exerts an equal but opposite force on the hammer to the left.

A. Newton’s Third LawA. Newton’s Third Law

Action-Reaction Pairs

The rocket exerts a downward force on the exhaust gases.

The gases exert an equal but opposite upward force on the rocket.

FG

FR

A. Newton’s Third LawA. Newton’s Third Law

Action-Reaction PairsBoth objects accelerate.The amount of acceleration

depends on the mass of the object.

a Fm

Small mass more accelerationLarge mass less acceleration

E. Graphing MotionE. Graphing Motion

slope =

steeper slope =

straight line =

flat line =

Distance-Time Graph

A

B

faster speed

constant speed

no motion

speed

E. Graphing MotionE. Graphing Motion

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

Distance-Time Graph

A

B

0

100

200

300

400

0 5 10 15 20

Time (s)

Dis

tan

ce (

m)

Distance-Time Graph

E. Graphing MotionE. Graphing Motion

Acceleration is indicated by a curve on a Distance-Time graph.

Changing slope = changing velocity

E. Graphing MotionE. Graphing Motion

0

1

2

3

0 2 4 6 8 10

Time (s)

Sp

ee

d (

m/s

)

Speed-Time Graph

slope =

straight line =

flat line =

acceleration +ve = speeds up -ve = slows down

constant accel.

no accel. (constant velocity)

E. Graphing MotionE. Graphing Motion

0

1

2

3

0 2 4 6 8 10

Time (s)

Sp

ee

d (

m/s

)

Speed-Time GraphSpecify the time period

when the object was... slowing down

5 to 10 seconds speeding up

0 to 3 seconds

moving at a constant speed 3 to 5 seconds

not moving 0 & 10 seconds

B. MomentumB. Momentum

Momentum quantity of motion

p = mvp: momentum (kg ·m/s)m: mass (kg)v: velocity (m/s)m

p

v

B. MomentumB. MomentumFind the momentum of a bumper car if it

has a total mass of 280 kg and a velocity of 3.2 m/s.

GIVEN:

p = ?

m = 280 kg

v = 3.2 m/s

WORK:

p = mv

p = (280 kg)(3.2 m/s)

p = 896 kg·m/s

m

p

v

B. MomentumB. MomentumThe momentum of a second bumper car

is 675 kg·m/s. What is its velocity if its total mass is 300 kg?

GIVEN:

p = 675 kg·m/s

m = 300 kg

v = ?

WORK:

v = p ÷ m

v = (675 kg·m/s)÷(300 kg)

v = 2.25 m/s

m

p

v

C. Conservation of MomentumC. Conservation of Momentum

Law of Conservation of Momentum The total momentum in a group of

objects doesn’t change unless outside forces act on the objects.

pbefore = pafter

C. Conservation of MomentumC. Conservation of Momentum

A 5-kg cart traveling at 4.2 m/s strikes a stationary 2-kg cart and they connect. Find their speed after the collision.

BEFORECart 1:m = 5 kgv = 4.2 m/s

Cart 2 :m = 2 kgv = 0 m/s

AFTERCart 1 + 2:m = 7 kgv = ?

p = 21 kg·m/s

p = 0

pbefore = 21 kg·m/s pafter = 21 kg·m/s

m

p

vv = p ÷ mv = (21 kg·m/s) ÷ (7 kg)v = 3 m/s

C. Conservation of MomentumC. Conservation of Momentum

A 50-kg clown is shot out of a 250-kg cannon at a speed of 20 m/s. What is the recoil speed of the cannon?

BEFOREClown:m = 50 kgv = 0 m/s

Cannon:m = 250 kgv = 0 m/s

AFTERClown:m = 50 kgv = 20 m/s

Cannon:m = 250 kgv = ? m/s

p = 0

p = 0

pbefore = 0

p = 1000 kg·m/s

pafter = 0

p = -1000 kg·m/s

C. Conservation of MomentumC. Conservation of Momentum

So…now we can solve for velocity.

GIVEN:

p = -1000 kg·m/s

m = 250 kg

v = ?

WORK:

v = p ÷ m

v = (-1000 kg·m/s)÷(250 kg)

v = - 4 m/s (4 m/s backwards)

m

p

v