KISS Notes Moving About

Usage & copying is permitted according to the Site Licence Conditions only

1PPrreelliimmiinnaarryy PPhhyyssiiccss TTooppiicc 33 ““MMoovviinngg AAbboouutt””CCooppyyrriigghhtt ©© 22000055-22000099 kkeeeepp iitt ssiimmppllee sscciieenncceewwwwww..kkeeeeppiittssiimmpplleesscciieennccee..ccoomm..aauu

keep it simple science®

but first, let’s revise...

Preliminary Physics Topic 3

MOVING ABOUTWhat is this topic about?To keep it as simple as possible, (K.I.S.S.) this topic involves the study of:1. SPEED and VELOCITY

2. FORCE and ACCELERATION3. WORK and KINETIC ENERGY

4. MOMENTUM and IMPULSE5. SAFETY DEVICES in VEHICLES

...all in the context of moving vehicles.

WHAT IS SPEED?“Speed” refers to how fast you are going. You already know that mathematically:

SPEED = distance travelledtime taken

In this topic, you will extend your understandingof speed to include VELOCITY, which is just aspecial case of speed.

WHAT IS FORCE?A FORCE is a PUSH or a PULL.

Some forces, like gravity and electric/magneticfields, can exert forces without actuallytouching things. In this topic you will dealmainly with CONTACT FORCES, which push orpull objects by direct contact.

In the context of moving vehicles, the mostimportant force is FRICTION. Friction allows acar’s tyres to grip the road to get moving, andfor the brakes to stop it again. Without frictionthe car couldn’t get going, and couldn’t stop if itdid!

WHAT IS ENERGY?Energy is what causes changes....

change in temperature (Heat energy)change in speed (Kinetic energy)change in height

(gravitational Potential energy)change in chemical structure

(chemical P.E.)...and so on.

In this topic the most important energy form youwill study is the one associated with movingvehicles...

KINETIC ENERGY

WHAT MAKES A CAR GO?Overviewof Topic:

MOMENTUMchanges

KINETICENERGYchanges

ENGINE provides ENERGY(from chemical energy

in petrol)

VELOCITYchanges

FORCE causesACCELERATION

FORCE acts overa distance...

“WORK” done

Tyres PUSH on road...FORCE acts...




AddingVectors

Average &Instantaneous

Speed

Vectors & Scalars.Speed & Velocity

MeasuringMotion

Acceleration Mass&

Weight

Energy Transformations

Equivalence ofWork & Energy

Newton’s2nd Law

Centripetal Force

Momentum

Impulseof a Force

Newton’s3rd LawPhysics of

SafetyDevices

Inertia&

Newton’s1st Law Conservation

of Momentumin Collisions

Law ofConservation

of Energy

MOVINGABOUT

Speed&

Velocity Force&

Acceleration

WWoorrkk&&

KKiinneettiicc EEnneerrggyy

Momentum&

ImpulseSafety Devicesin

Vehicles

CONCEPT DIAGRAM (“Mind Map”) OF TOPICSome students find that memorising the OUTLINE of a topic helps them learn and remember

the concepts and important facts. As you proceed through the topic, come back to this page regularly to see how each bit fits the whole.

At the end of the notes you will find a blank version of this “Mind Map” to practise on.

MotionGraphs

Forces



PPrreelliimmiinnaarryy PPhhyyssiiccss TTooppiicc 33 ““MMoovviinngg AAbboouutt””CCooppyyrriigghhtt ©© 22000055-22000099 kkeeeepp iitt ssiimmppllee sscciieenncceewwwwww..kkeeeeppiittssiimmpplleesscciieennccee..ccoomm..aauu

Speed-Time GraphsThe same journey could also be represented bya different graph, showing the SPEED atdifferent times:

Study this graph carefully and compared it withthe other...

You must not confuse the 2 types of graph andhow to interpret them.

This graph is very unrealistic in one way. Itshows the speed changing INSTANTLY from(say) 100 km/hr to zero (stopped), without anytime to slow down. It also shows the cartravelling at exactly 100 km/hr for an hour at atime... very unlikely with hills, curves, traffic etc.

Changes of speed (ACCELERATION) will bedealt with in the next section. For now we’reKeeping It Simple!

3

1. SPEED & VELOCITY

00 11 22 33 44TTIIMMEE ((hhrr))

SSPPEEEE

DD((kk

mm//hh

rr))00

2200

44

00

6600

8800

110000 AA

BB

CC

DD

SSttooppppeedd.. SSppeeeedd ssccaallee rreeaaddss zzeerroo..

““FFllaatt”” ppaarrttssDDOO NNOOTT

mmeeaannssttooppppeedd,,bbuutt mmeeaannccoonnssttaanntt

ssppeeeedd

Distance-Time GraphsPerhaps your journey was similar to this graph.

Start at the bottom-left of the graph and considereach section A, B, C and D.

So although the average speed for the entirejourney was 75km/hr, in fact you never actuallymoved at that speed.

This raises the idea of INSTANTANEOUSSPEED: the speed at a particular instant of time.The speedometer in your car gives you amoment-by-moment reading of your currentspeed... this is your instantaneous speed.

On the graph, the GRADIENT at any given pointis equal to INSTANTANEOUS SPEED.

Graph section DTravelled 150km in 1.5 hr:Speed = 100 km/hr

Graph section CTravelled 50 km in 1.0 hr:Speed=50 km/hr

Graph section BZero distancemoved in 0.5 hr:Speed= zero.

Graph section ATravelled 100 kmin 1.0 hour:Speed =100 km/hr

00 11 22 33 44TTIIMMEE ((hhoouurrss))

DDIISS

TTAANN

CCEE TT

RRAAVVEE

LLLLEEDD

((kkmm

))00

5500

110000

11

5500

220 0

00

225500

330000

AA

BB

CC

DD

ggrraaddiieenntt == zzeerrooii..ee.. ssttooppppeedd

DDiissttaannccee-TTiimmee GGrraapphh

ggrraaddiieenntt == ddiissttaanncceettiimmee

== ssppeeeedd

These graphsrepresent thesame journey

DISTANCE-TIME GRAPHSshow the DISTANCE (from the starting point)

at each TIME.The GRADIENT at any point equals

INSTANTANEOUS SPEED.

A horizontal section means thatthe object was not moving

SPEED-TIME GRAPHS show the SPEED of a moving object

at each TIME.

The speed at any time can be read from the vertical scale of the graph.

A horizontal section means thatthe object was moving at constant speed.

Average Speed for a JourneyIf you travelled by car a distance of 300 km in exactly 4 hours, then your “average speed” was:

average speed = distance travelled = 300 = 75 km/hr (km.hr-1)time taken 4

However, this does not mean that you actually travelled at a speed of 75 km/hr the whole way. You probably went faster at times, slower at other times,

and may have stopped for a rest at some point.




A Scalar quantity is something that has a size(magnitude) but no particular direction.A Vector quantity has both size (magnitude) ANDDIRECTION.

So far we have dealt with only distances &speeds... these are Scalar quantities, since theydo not have any special direction associated.

Now you must learn the vector equivalents: “Displacement” = distance in a given direction,and“Velocity” = speed in a given direction.

Consider this journey:START drove 60 km EAST in 1 hour

thendrove 30 km WESTin 0.5 hour.

As a SCALAR journey:• travelled a total 90 km distance in 1.5 hours,• average speed = 90/1.5 = 60 km/hr

Scalars & VectorsBUT, consider the “NET journey”: at the end ofthe journey you end up 30 km EAST of thestarting point. So, your final displacement is “30km east”.

The VECTOR journey was:• travelled 30 km east displacement in 1.5 hours.• average velocity = 30/1.5 = 20 km/hr east.

Notice that both displacement and velocity havea direction (“east”) specified.... they areVECTORS!

To make better sense (mathematically) of thejourney, the directions east & west could have(+) or ( - ) signs attached. Let east be (+) andwest be ( - ).

Then the totaljourney

displacement was(+60) + (-30) = +30 km.

Note: The symbol “S” is used for Displacement

Average = DisplacementVelocity time

Vav = S t

MORE GRAPHS... Displacement - TimeRefer to the previous Distance-Time graph.

What if the 300km journey had been 150 km north(sections A, B, C) then 150 km south (section D)?The Displacement - Time Graph would be:

In vector terms; displacement north is positive (+)displacement south is negative ( - )

In section D:displacement = -150 km (south)

velocity = displacementtime

= -150 /1.5= -100 km/hr (i.e. 100km/hr southward)

00 11 22 33 44TTIIMMEE ((hhoouurrss))

DDiiss

ppllaacc

eemmeenn

tt NN

OORRTT

HH ((

kkmm))

00

55

00

1100

00

115500

AA

BB

CC

DD

GGrraaddiieenntt nneeggaattiivvee

Grrad

ient

poss

itive

DDoowwnn-ssllooppiinngg

lliinnee mmeeaannssttrraavveelllliinnggSSOOUUTTHH

BBaacckk aatt ssttaarrttiinngg ppooiinntt..((DDiissppllaacceemmeenntt == 00 ))

...and the corresponding Velocity - Time Graph:

The velocity values for each part of this graphare equal to the gradients of the correspondingparts of the Displacement - Time Graph.

VVeelloo

cciittyy

((kkmm

//hhrr))

ssoouutt

hhnnoo

rrtthh

-110000

-5500

00

5500

110000 AA

BB

CC

DD

11 22 33 44

TTIIMMEE ((hhrrss))

NNeeggaattiivvee vvaalluuee::ssoouutthh-bboouunndd

vveelloocciittyy

ZZeerroo vveelloocciittyy::mmeeaannss ssttooppppeedd

PPoossiittiivvee vvaalluueess mmeeaann nnoorrtthh-bboouunndd vveelloocciittyy

Note: Since the journey ends back at the starting point,

total displacement = zeroand average velocity = zero for the whole trip.

However, this simply points out how littleinformation the “average” gives you. The instant-by-instant Physics of the

journey is in the graph details.




Tape Measure & StopwatchThe simplest method of all: measure the distanceor displacement involved, and the time taken.Then use

speed (velocity) = distance (displacement)time

Typical ResultsDistance Time Velocity

(m) (s) (ms-1)

Car 87 6.2 14.0Bicycle 87 22.4 3.9

However, this can only give you the AVERAGEspeed or velocity. In Physics we often need toconsider INSTANTANEOUS velocity.

Electronic or Computer Timing You may use devices that use either “Light

Gates” or “SONAR” to record displacementsand times for you.

Once again, any velocities calculated areaverages, but the time intervals are so short

(e.g. as small as 0.001 s) that the velocitycalculated is essentially instantaneous.

Although this method is very out-dated, it isstill commonly used as a way for students to learn how to measure

instantaneous velocity.

A moving object drags a paper strip onwhich dots get printed (usually every 0.02

second) as it goes. The gap between dots isa record of displacement and time. Thisallows you to calculate the velocity over

every 0.02 s. It’s still an average, but oversuch small time intervals it approximates

the instantaneous velocity.

Prac Work: Measuring MotionYou will probably experience one or more of these commonly used ways

to measure motion in the laboratory.

Moving lab. trolleydrags a strip ofpaper behind it

““TTiicckkeerr-ttiimmeerr”” ddeevviiccee hhaass aa ssmmaallll hhaammmmeerrwwhhiicchh vviibbrraatteess uupp aanndd ddoowwnn eevveerryy 00..0022 sseecc..

EEvveerryy ttiimmee tthhee hhaammmmeerr hhiittss tthheemmoovviinngg ssttrriipp ooff ppaappeerr

iitt lleeaavveess aa ddoott..TThhee ssttrriinngg ooff ddoottss ccaann bbee

aannaallyysseedd ttoo ssttuuddyy tthhee mmoottiioonn ooff tthhee ttrroolllleeyy..

Landmark A

Landmark B

Time to travel from A to B measured by stopwatch

You might do some measurements assuggested by this diagram

Distance between landmarks measured with sports tape

The “Ticker-Timer”

Moving trolley equipped with asonar reflector.

(An aluminium pie dish will do)

Sonar “transponder” gives outpulses of ultra-ssound and picks

up any returning echoes

To computer foranalysis

6




Worksheet 1 Speed & VelocityFill in the blank spaces. Student Name...........................................

On a displacement-time graph, movement southwould result in the graph slopingm)................................... to the right and having anegative n)............................................

The vector equivalent of speed iso).................................... The average velocity isequal to p)............................... divided byq)....................... Instantaneous velocity refers tor)..................................................................

Laboratory methods for measuring motioninclude using a tape measure and stopwatch.This allows calculation of s)................................................ only. “Ticker-timers” recordboth t)............................... and .......................... ona paper tape. Average velocity can be calculatedfor short time intervals which are approximatelyequal to u)............................................ velocity.• Electronic or Computer-based devices oftenuse v)........................ or .........................................to gather displacement, time and velocity data atvery short time intervals.

The average speed of a moving object is equalto the a)............................ travelled, divided byb)....................... taken. On a Distance-Timegraph, the c)........................ of the graph is equalto speed. A horizontal graph meansd)................................. ...............................

On a Speed-Time graph, constant speed isshown by e).......................................... on thegraph. This does NOT mean stopped, unless thegraph section is lined up with f).............................

Speed and distance are both g)..............................quantities, because the direction doesn’t matter.Often in Physics we deal with h)............................quantities, which have both i)...............................and .......................................

The vector equivalent of distance is calledj)................................., and refers to distance in aparticular k).............................. For example, ifdisplacement was being measured in the northdirection, then a distance southward would beconsidered as l).............................. displacement.

Worksheet 2 Practice ProblemsMotion Graphs Student Name...........................................A car travelled 200 km north in 3.0 hours, thenstopped for 1.0 hr, and finally travelled south 100km in 1.0 hr.1. What was the total distance travelled?

2. What was the total displacement?

3. What was the total time for the whole journey?

4. Calculate the average speed for the whole journey.

5. Calculate the average velocity for the wholejourney.

6. Construct aDisplacement - TimeGraph for this trip.

TIME

DDiiss

ppllaacc

eemmeenn

tt

7. Use your graph to find:i) average velocity for the first 3 hours.

ii) velocity during the 4th hour.

iii) velocity during the last hour.

8. Construct a Velocity- Time Graph for this trip.

TTiimmee ((hhrr))1 2 3 4 5

VVeelloo

cciittyy

((kkmm

//hhrr))

SSoouutt

hh

NN

oorrtthh

-1100

-55

0

0

50

100




1. An aircraft tookoff from town P andflew due north totown Q where itstopped to re-fuel. Itthen flew due southto town R.

The trip issummarised by thegraph.

a) How far is it from towns P to Q?

b) How long did the flight P to Q take?

c) Calculate the average velocity for the flightfrom P to Q (include direction)

d) What is the value of the gradient of thegraph from t=3 hr, to t=6 hr.?

e) What part of the journey does thisrepresent?

f) Where is town R located compared to town P?

g) What was the aircraft’s position and velocity(including direction) at t=5 hr?

h) What was the:i) total distance

ii) average speed

iii) total displacement

iv) average velocity(for the entire 6 hr journey)

7

UNITS OF MEASUREMENTSo far all examples have used the familiar

km/hr for speed or velocity. The correct S.I.units are metres per second (ms-1). You needto be able to work in both, and convert from

one to the other..... here’s how:1 km/hr = 1,000 metres/hr

= 1,000m/(60x60) seconds= 1,000/3,600 m/s= 1/3.6

So, to convert km/hr ms-1 divide by 3.6to convert ms-1 km/hr multiply by 3.6

2. A car is travelling at 100 km/hr.a) What is this in ms-1?

b) The driver has a “micro-sleep” for 5.00 s. How farwill the car travel in this time?

c) At this velocity, how long does it take (in seconds)to travel 1.00km (1,000m)?

3.For this question consider north as (+), south as ( - ).

A truck is travelling at a velocity of +20.5 ms-1 as itpasses a car travelling at -24.5 ms-1.

a) What are these velocities in km/hr? (includingdirections?)

b) What is the displacement (in m) of each vehicle in30.0 s?

c) How long would it take each vehicle to travel 100 m?

4. Where does this aircraft end up in relation to itsstarting point?

Flight details:First flew west for 2.50 hr at 460 km/hr.

Next, flew east at 105 ms-1 for 50.0 minutes.

Next, flew west for 3.25 hours at 325 km/hr.

Finally flew east for 5.50 hours at velocity 125 ms-1.

Worksheet 3 Practice ProblemsMotion Graphs & Calculations Student Name...........................................

DDiiss

ppllaacc

eemmeenn

tt nnoorr

tthh ((kk

mm))

-4400

-220

0

P

200

400

60

0 8

00

QQ

RR

1 2 3 4 5 6

TTiimmee ((hhrr))

TTiimmee ((hhrr))1 2 3 4 5 6 VVee

llooccii

ttyy ((

kkmm//hh

rr))SSoo

uutthh

NN

oorrtthh

-3300

-

1100

0

100

2

00

300

4

00

i) Construct aVelocity- Time Graph for the flight.

Change of Velocity = AccelerationAny change in velocity is an acceleration.

Mathematically,acceleration = velocity change = final vel. - initial velocity

time taken time taken

ΔΔ (Greek letter “delta”) refersto a change in a quantity

v = final velocityu = initial velocityt = time involved

Units: if velocities are in ms-1, and time inseconds, then acceleration is measured inmetres/sec/sec (ms-2).

Explanation: Imagine a car that accelerates at 1 ms-2:

Start 1 sec. later 1 sec.later 1sec.laterv =0 v = 1 ms-1 v=2 ms-1 v=3ms-1

Every second, its velocity increases by 1 ms-1.Therefore, the rate at which velocity is changingis 1 ms-1 per second, or simply 1 ms-2.

Acceleration is a vector, so direction counts.

“Deceleration” (or negative acceleration) simplymeans that the direction of acceleration isopposite to the current motion... the vehicle willslow down rather than speed up.

Graphs of Accelerating VehiclesYou may have done laboratory work to study themotion of an accelerating trolley. If you used a“Ticker-timer”, the paper tape records wouldlook something like this:

The graphs that result from acceleration are asfollows:

8

2. FORCE & ACCELERATIONkeep it simple science

®



a = ΔΔvΔΔt

a = v - ut

VELOCITYVECTOR

ACCELERATIONVECTOR

+

-

THIS CAR IS SLOWINGDOWN... DECELERATING

TTaappee ooff ttrroolllleeyy mmoovviinngg aatt ccoonnssttaanntt vveelloocciittyy ((ffoorr ccoommppaarriissoonn))

TTaappee ooff ttrroolllleeyy aacccceelleerraattiinngg...... ddoottss ggeett ffuurrtthheerr aappaarrtt

TTrroolllleeyy ddeecceelleerraattiinngg ((nneeggaattiivvee aacccceelleerraattiioonn))...... ddoottss ggeett cclloosseerr

Time

Dis

plac

emen

t

DISPLACEMENT-TTIME GRAPH

GGrraaddiieennttss iinnccrreeaassiinngg((ccuurrvvee ggeettss sstteeeeppeerr))

GGrraaddiieenntt ccoonnssttaanntt((ssttrraaiigghhtt lliinnee))

GGrraaddiieennttss ddeeccrreeaassiinngg((ccuurrvvee ffllaatttteennss oouutt))

RReemmeemmbbeerr,,GGrraaddiieenntteeqquuaallss

VVeelloocciittyy

AAcccceellee

rraattiinngg

CCoonnss

ttaanntt

VVeelloo

cciittyy

DDeecceelleerraattiinngg

Time

Velo

city

VELOCITY-TTIME GRAPH

AAccccee

lleerraatt

iinngg

CCoonnssttaannttVVeelloocciittyy

DDeecceelleerraattiinnggVVeelloocciittyyiinnccrreeaassiinngg

VVeelloocciittyyddeeccrreeaassiinngg

Gradient positive Gradient negative

On a Velocity-TTime GraphGradient = Acceleration

AA ccoommmmoonn eerrrroorr iiss ttootthhiinnkk tthhaatt tthhiiss mmeeaannss tthhee

oobbjjeecctt iiss mmoovviinnggbbaacckkwwaarrddss.. WWrroonngg!! IItt iiss

mmoovviinngg ffoorrwwaarrdd,, bbuuttsslloowwiinngg ddoowwnn..

VVeelloocciittyy == 00

∴∴ SSttooppppeedd!!

Example Problem 1A motorcycle travelling at 10.0 ms-1, acceleratedfor 5.00s to a final velocity of 30.0 ms-1. What wasits acceleration rate?

Solution: a = v - u = 30.0-10.0/5.00 = 20.0/5.00t = 4.00 ms-2.

Example Problem 2A car moving at 25.0 ms-1 applied its brakesproducing an acceleration of -1.50 ms-2

(i.e. deceleration) lasting for 12.0 s. What was its final velocity?

Solution: a = v - u, so v = u + att = 25.0 + (-1.50) x 12.0

= 25.0 - 18.0= 7.00 ms-1.




Force Causes AccelerationA simple definition of “Force” is a push or a pull.However, in the context of moving vehicles,

Force is what causes velocity to change.

Note that a change of velocity could mean:• speeding up• slowing down• changing direction (velocity is a vector)

To actually result in a change of velocity, theforce must be

External and Unbalanced (Net) Force

For example, if you wereinside a moving car andkicked the dashboard,

this force would have NOEFFECT on the car’s

motion...

This is an “Internal Force”and cannot cause

acceleration.

BALANCED & UNBALANCED FORCES

WWEEIIGGHHTT FFOORRCCEECCaarr ppuusshheess oonn EEaarrtthh

RREEAACCTTIIOONN FFOORRCCEEEEaarrtthh ppuusshheess bbaacckk

The car above has a number of forces acting onit, but they are BALANCED... those acting in the

same line are equal and opposite, and cancel each other out.

This car will not alter its velocity or direction; itwill not accelerate. It is either travelling at a

constant velocity, or it is stationary.

FFrriiccttiioonnaanndd AAiirr

RReessiissttaannccee

TThhrruusstt

ffrroommEEnnggiinnee

PRESSING ON THE ACCELERATOR...WWeeiigghhtt

RReeaaccttiioonnFFoorrccee

TThhrruusstt FFoorrcceeIInnccrreeaasseedd FFrriiccttiioonn &&

AAiirr RReessiissttaanncceessmmaallll ffoorrcceess

VVeerrttiiccaall ffoorrcceess aarreebbaallaanncceedd,, aanndd ccaanncceell

TURNING THE STEERING WHEEL...

((TThhrruusstt FFoorrccee ffrroommEEnnggiinnee iiss eeqquuaall ttoo

FFrriiccttiioonn ffoorrcceess))

PRESSING ON THE BRAKES...

WWeeiigghhtt

RReeaaccttiioonnFFoorrccee

FFrriiccttiioonn IInnccrreeaasseess aassBBrraakkeess aarree

aapppplliieedd


HHoorriizzoonnttaall ffoorrcceess aarreeUUNNBBAALLAANNCCEEDD

This car will SLOW DOWN(Decelerate)

SSiiddeewwaayyssFFoorrcceess bbeeccoommee UUNNBBAALLAANNCCEEDD((TThheessee wwoouulldd

bbee eeqquuaall iiffwwhheeeell nnoottttuurrnneedd))

GOING UP A HILL(without increasing engine thrust)

WWeeiigghhtt ((ssttiillll vveerrttiiccaall))

RReeaaccttiioonn FFoorrccee iiss nnoottvveerrttiiccaall,, aanndd nnoo

lloonnggeerr ccaanncceellss tthheewweeiigghhtt ccoommpplleetteellyy......UUNNBBAALLAANNCCEEDD FFOORRCCEE

FFrriiccttiioonnssttiillll tthheessaammee

EEnnggiinnee TThhrruussttssttiillll tthhee ssaammee

Part of the Weight Force actsdownhill to cancelsome of the thrust

This bike will SLOW DOWN.(Going down a hill, it will speed up)

WWeeiigghhtt

RReeaaccttiioonn FFoorrcceeccaanncceellss WWeeiigghhtt

VViirrttuuaallllyy

nnoo FFrriiccttiioonnoonn IIccee

VViirrttuuaallllyy nnoo TThhrruusstt FFoorrcceebbeeccaauussee ttyyrreess

ccaann’’tt ggrriipp oonn iiccee

PASSING OVER AN ICY PATCH ON THE ROAD

OOppppoossiittee FFoorrcceess aarreeBBAALLAANNCCEEDD aanndd ccaanncceell

TThhiiss ccaarr wwiillll ccoonnttiinnuueeiinn aa ssttrraaiigghhtt lliinnee,, aatt aa

ccoonnssttaanntt vveelloocciittyy......wwhheetthheerr tthhee ddrriivveerrwwaannttss ttoo oorr nnoott......

CCaarr iiss oouutt ooff ccoonnttrrooll;;CCaann’’tt ssttoopp......CCaann’’tt ttuurrnn......

BALANCEDFORCE

SITUATION

This car will SPEED UP

HHoorriizzoonnttaall ffoorrcceessUUNNBBAALLAANNCCEEDD

FFoorrwwaarrdd && BBaacckkFFoorrcceess aarree bbaallaanncceedd

aanndd ccaanncceell


This car will turn a corner at constant speed

(but this is a changed velocitysince the direction changed)

EXAMPLES OFUNBALANCED

FORCES

TThhrruusstt FFoorrcceeddeeccrreeaasseess aass

aacccceelleerraattoorr iissrreelleeaasseedd




Newton’s 2nd Law of MotionWhoa! Why not start with his 1st Law?Newton’s 2nd Law is all about what happenswhen a force acts, and so is appropriate tostudy here. His 1st Law is all about whathappens when a force does NOT act... it will becovered later in the topic.

Sir Isaac Newton (1642-1727) figured out therole of forces in causing acceleration:

Units: Mass must be measured in kilograms (kg)Acceleration in metres/sec/sec (ms-2)Force will then be in “newtons” (N)1N of force would cause a 1kg mass

to accelerate at 1ms-2

Mass & Weight“Mass” is a measure of the amount of matter inan object. In terms of force and acceleration,mass is the stuff that tries to preventacceleration... the more mass there is, the lessacceleration an applied force will produce.

The mass of an object is the same wherever itmight be. “Weight” however, changes accordingto gravity conditions.

Weight is a Force (measured in N) due togravity. Gravity causes objects near the Earth toaccelerate at (approximately) g = 10ms-2

(actually 9.81ms-2, but K.I.S.S.).

F = m a, so Weight = mgExample: consider an astronaut with a mass of 80kg

Mass (kg) Weight (N)Astronaut on Earth 80 800Astronaut in orbit 80 zeroAstronaut on Moon 80 133

Study this information to get the idea.The confusion about mass and weight has beencaused by the unfortunate choice by society to talkabout the “weight” of things, but then measure it inkilograms... it should be in Newtons!!

The acceleration of an object is directly proportional to the

external, unbalanced (net) force acting on it,

and inversely proportional to its mass.

a = F or F = m am

Verifying 2nd LawYou may have done laboratory work similar to this:

PPoowweerr PPaacckk

TTiicckkeerr-ttiimmeerr

ddeevviiccee

PPaappeerr ttaappee

LLaabb..TTrroolllleeyy

EExxttrraa mmaasssseess oonn ttrroolllleeyy

WWeeiigghhtt oonn ssttrriinnggccaauusseess ttrroolllleeyy ttoo aacccceelleerraattee

The acceleration of the trolley isdetermined by analysing the

displacement & time data from theticker tape record.

This is repeated several times,tranferring some of the “extra”

masses from trolley to the hanging weight each time.

This means, for each trial the totalmass of the entire system stays

constant, but the force causing theacceleration (weight on the string)

is different each time.

The results are analysed by graphing the Force (weighton string) against the acceleration produced.

Acceleration (ms-22)

Forc

e (w

eigh

t on

strin

g) (N

)

Force v Acceleration Graph

LLIINNEE OOFF ““BBEE

SSTTFFIITT

””

Final Results and Conclusions• Within experimental error, the graph shows a straightline. This proves there is a direct relationship betweenthe force applied, and the acceleration produced.

• The gradient of the graph will be found to be equal tothe mass of the total system (i.e. trolley + masses) inkilograms:

Gradient = Force = MassAcceleration

F = ma

and therefore, F = m a

FFiinnddGGrraaddiieenntt ooff

lliinnee

Weight = mg




Adding VectorsForce is a vector quantity, the same as velocityand acceleration. To fully describe a force, youmust state the direction of the force.

Often, there are situations where 2 (or more) forcesact on the same object at the same time. To find theNET FORCE acting you need to add the vectorstogether to find their combined effect.

It’s very easy if their vector directions are in thesame line:

However, if the forces are acting in totallydifferent directions, the problem is morecomplicated.

Vectors in EquilibriumIt is often the case that 2 or more vectors mightall cancel each other out so the “resultant” iszero. In fact this is always the case when avehicle is moving in a straight line with aconstant velocity.

Since it is NOT accelerating, then the net forceacting must be zero. Since there are forcesacting, then it follows they must be cancellingeach other out.

Example: an aircraft flyingstraight and level at constantvelocity.

The vector diagram for thisplane must be:

The vectors all cancel out... the resultant is zero... no acceleration will occur.

Example:

Mathematically, you should assign (+ve) and (-ve) signs to the opposite directions, then

simply add the values:

e.g. let East be (+ve), and West (-ve)Then, Force A = +20 and Force B = -30

So the Resultant = +20 +(-30) = -10N (i.e. 10N west)

Force A20N east

Force B30N west

“Resultant”10N west

The sum of these 2 vectors is a single force:

Example:

Use Pythagorus’s Theorem to find the size of the“Resultant” force:

R2 = A2 + B2 = 202 + 302 = 400 + 900 = 1300∴∴R = Sq.Root(1300)

≅≅ 36N (approximately)

and find the angle ( φφ ) by Trigonometry:Tan φφ = opp/adj = 30/20 = 1.5

∴∴ φφ ≅≅ 56o

So, the resultant force R = 36N, direction 56o S of E(bearing from north=146o)

Force A20N east

Force B30N south

First, sketchthese vectors“head-to-tail”.

A = 20

B =

30

φφ Next, connect thebeginning to the

end, to from aright-angled

triangle.

The 3rd side is the“Resultant” vector.

RReessuullttaanntt

WeightForce

AirResistance

“Drag”

“Lift” Force(on wings)

Thrustfrom

engines

Lift Weight

Drag

Thrust

You may have done laboratory work to measuresome vectors and their sum. A commonexperiment is shown in the photo:

A

A

B

B

C

C

Three Force Vectorsin Equilibrium

F = m g

TTeennssiioonn FFoorrcceessiinn ssttrriinnggss AA && BB

mmeeaassuurreedd bbyySSpprriinngg BBaallaanncceess

TThhee aanngglleess bbeettwweeeennssttrriinnggss AA,, BB && CC nneeeedd

ttoo bbee mmeeaassuurreedd..

The 3 vectorscan then be

analysed

The vectors can be analysed either byaccurate scale drawing,

or by mathematics (e.g. Sine Rule in a triangle).

It will be found (within experimentalerror) that these vectos add to zero.

They are in equilibrium.




More Examples of VectorsSo far, all the examples given of vectors have been forces. Vector addition could involve any vector

quantity, of course... displacement, velocity, acceleration or force.

A Special Force: FrictionOften in Physics problems we ignore friction to keep things simple (KISS Principle). In reality, when

anything moves on or near the Earth, there is always friction... you need to know about it.

Friction (including air resistance) is a force which always acts in the opposite direction to themotion of any object. Generally, you may consider the force of friction as a negative value,

assuming that the direction of motion is considered positive.

An example of dealing with friction:

This 500kg car is accelerating at 2.5ms-2. The “thrust” force from the engine is 1,700N.

What is the force of friction acting against it?

An aircraft flies 200km east, then 100km south.

Where is it in relation to its starting point?

110000kk

mmResultant

Dispplacemment

R22 = 20022 + 10022

= 50,000∴∴ R = Sq.Root (50,000)

= 224 km

Tan φφ = 100/200= 0.500

∴∴ φφ ≅≅ 27oo

Final displacement = 224 km, direction 27oo S of E(bearing from north = 117oo)

A ship is travelling due east at velocity 5.0ms-1. The tide is flowing from the south at 1.8ms-1.

What is the ship’s actual velocity?

5.0

1.8RReessuulttaanntt

VVeeloocciitty

R22 = 5.022 + 1.822

= 28.24

∴∴ R = Sq.Root (28.24)= 5.3ms-11

φφ

Tan φφ = 1.8/5.0= 0.36

∴∴ φφ ≅≅ 20oo N of E(this angle is 70oo

clockwise from north,∴∴ bearing = 70oo)

Actual Velocity = 5.3ms-1, on bearing 70o

aacccceelleerraattiioonn

FFrriiccttiioonnFFoorrccee

220000kkmm

φφ

Displacement VectorsVelocity Vectors

Solution:The net, unbalanced force

causes acceleration.

This net force must be

F = m a= 500 x 2.5= 1,250N

This net force is the vector sum of all forces acting:

Net Force = Engine thrust + Friction1,250 = 1,700 + FF

FF = 1,250 - 1,700∴∴ Friction = -450N

(the negative value simply means that friction is in the opposite direction to the car’s motion)

TThhrruussttFFoorrccee




Object 1Weight Force

mg = 0.5 x 10= 5N

Object 2Weight Force

mg = 0.2 x 10= 2N

T11

T11

T22

T22

Acceleration due to gravity hasbeen taken as 10ms-2 for

simplicity (KISS Principle)

500g(0.5kg)

wweeiigghhtt == mmgg

wweeiigghhtt == mmgg

Two Masses Hanging on StringsTension force in top string must hold up both weights, so T1 = (5+2) = 7N

Tension force T1 simultaneously pulls down on the top support (assumed immovable) and

pulls the round object upwards.

Tension in the bottom string only holds up the 200g mass, so T2 = 2N

Tension force T2 simultaneously pulls down on the round object and

pulls the rectangular object upwards.

Nothing is moving, so all forces must be inequilibrium.

i.e. Net Force = zero, but can we prove it?

Consider all forces acting on each mass: (let up be (+ve), down (-ve)

Round Object Rectangular ObjectForce T1 is pulling it Force T2 pulls it up,upwards, while its while its weight pullsweight and T2 pull it downwards.downwards.

ΣΣF = T1 + T2 + mg ΣΣF = T2 + mg= (+7) + (-2) + (-5) = (+2) + (-2)= zero = zero

It all works! If you undertstand the tension forcesacting, you can explain that this system is not moving

because the net forces add up to zero.

Another Force to Know About: Tension“Tension” refers to the force which acts in a

rope, wire, chain or other coupling, whichattaches two objects together.

The tricky thing about tension is that it pullsin both directions at once.

Consider these 2 examples:

Tension forces acts in bothdirections in each string

200g(0.2kg)

This train engine produces 35,000N netthrust force.

Problema) What is the acceleration?b) What tension force acts in thecoupling between engine and carriage?c) What is the net force acting on theengine alone?

EEnnggiinnee CCaarrrriiaaggee2200,,000000kkgg 55,,000000kkgg

CCoouupplliinngg

Solution: The net force must accelerate the entiremass of 25,000kg.

a) F= m a b) Tension in coupling∴∴ a= F/m must cause the carriage

= 35,000/25,000 to accelerate.= 1.4 ms-22 F = m a

= 5,000 x 1.4= 7,000N

Example 2 Tension Under Acceleration

Does this make sense? Yes, it does, when youconsider the forces acting on the engine alone...

Engine thrust = 35,000N Tension in coupling = 7,000N

c) Since the engine is accelerating at 1.4 ms-22

the net force on the engine must be:F = ma

= 20,000 x 1.4 = 28,000 N




For any vehicle to turn a corner, it must changedirection and therefore, must accelerate. This meansit is being acted on by an unbalanced force.

To keep things as simple as possible (K.I.S.S.) let’sassume that all the corners being turned are circular.

So, where does the centripetal force come fromto push a moving vehicle, such as a car, aroundthe corner?

In the example of a car, the centripetal forcecomes from the frictional “grip” of the tyres onthe road. Turning the steering wheel creates newfriction forces which are directed to the centreof an imaginary circle.

So long as the frictional forces are strong enough, thevehicle will follow a circular path around the corner.

If the centripetal force required for a particular cornerexceeds the friction “grip” of the tyres, then thevehicle will not make it, and may “spin out” andcrash.

This can happen because:• speed is too high for the radius of the curve.

(i.e. the radius is too small compared to velocity)• loss of friction between tyres and road.

(e.g. road is wet, or tyres are worn smooth)

The force causing the turning isalways toward the centre of the

circle.This is called “Centripetal” force

The velocity vectors at any instant aretangents to the circle.

Path of avehicleturning acircular corner

VV

VV

F FF

F

Even though the speed may be constant, the vehicleis constantly accelerating because its direction isconstantly changing.

The force causing this acceleration is called“Centripetal Force” and is always directed to thecentre of the circle.

The acceleration vector is also pointed at the centre ofthe circle.

The velocity vector is constantly changing, but at anyinstant it is a tangent to the circle, and therefore, atright angles to the acceleration and force vectors.

Centripetal ac = v2Acceleration R

Centripetal Fc = mv2

Force RR = radius of the circle (in metres)V = instantaneous velocity (ms-1)

(also called “orbital speed”)m = mass of vehicle (in kg)

InstantaneousVelocity vectorstraight ahead

Wheel turned

Path car will take

CCeennttrriippeettaal

FFoorrccee

Example Problem 1A 900kg car turns a corner at 30ms-1. The radius ofthe curve is 50 metres. What is the centripetal force acting on the car?

Solution Fc = m v2 = 900 x 302 / 50R

= 16,200N

Example Problem 2The maximum frictional force possible from each tyre

of this 750kg car is 5,000N.What is the maximum speed that the car can go around

a curve with a radius of curvature of 40m?Solution:Max. Force possible from 4 tyres = 4x 5,000

=20,000NCentripetal Force cannot exceed this value.

Fc = m v2 / R, so v2 = FcR /m= 20,000 x 40 /750

v2 = 1067∴∴ v = Sq.Root(1067) ≅≅ 33ms-1

(This is almost 120 km/hr)

Turning Corners - Circular Motion


Acceleration is a change ina)................................ This could meanspeeding up, or b)............... ....................,or even changing c).......................... atconstant speed. Acceleration is ad).......................... (vector/scalar).“Deceleration” simply meanse).............................. acceleration. Theunit of measurement isf).................................

On a Displacement-Time graph,acceleration appears as ag)........................ On a Velocity-Timegraph, accelerations appear ash)....................................... (compared toconstant velocity, which shows as ai)................................ line). Thej).................................. of the line equalsthe rate of acceleration. A decelerationwould have a k).....................................gradient.

Acceleration is caused by the action of al)........................ The force must bem).................., and it is only ann).................................. (or “net”) forcewhich causes an acceleration.

Newton’s o)......... Law of Motion statesthat “the acceleration of an object isproportional to the p)..........................and q)......................................proportional to its mass”. The unit offorce is the r)........................., so long asmass is in s)................. and accelerationin t).........................

Mass is a measure of the amount ofu).......................... in an object, while“weight” is the v).................... due tow)................................... acting on themass.

15



COMPLETED WORKSHEETSBECOME SECTION SUMMARIES

Worksheet 4 Force & AccelerationFill in the blank spaces. Student Name...........................................

Vector quantities can only be addedtogether in a simple arithmetic way ifthey act x)........................ .........................If vectors are in different directions,they must be added using a vectordiagram (in which vectors are joinedy)...................... to .............................).This diagram can then be analysedmathematically using z)............................and/or trigonometry to find the“aa)................................” vector. Thecomplete answer must contain both themagnitude and ab)...................... of theresultant.

If 2 or more force vectors cancel eachother out, they are said to be inac)........................................ In such acase there is no ad)............................force and so the object or vehicle willcontinue to move ae)..................................................................... with noaf)......................................................

Friction is a force which alwaysag)......................... the motion beingconsidered. “ah)..........................” is theforce acting in a rope, chain or wireconnecting objects together. It acts inai)...................... directions within thecoupling.

When a vehicle turns a corner it isaccelerating, because theaj)........................... keeps changing. Theforce causing this is calledak)....................................... force, and isdirected at the al)................................ of acircle. The instantaneous velocity vectoris a am)...................................... to thecircle.


5. A truck with mass 8.00x103kg is travelling at22.5ms-1 when the brakes are applied. It comes to acomplete stop in 4.50s.a) What is its average rate of acceleration?

b) What net force is acting during this deceleration?

6. A 60kg cyclist exerts a net force of 100N pedallinghis 15kg bike for 10.0 seconds. Ignoring any friction;a) what acceleration will be produced?

b) From a standing start, what velocity will bike andrider reach in the 10s?

c) What is the final velocity in km/hr?

1. Starting from rest (i.e. u=0) a car reaches 22.5 ms-1

in 8.20 s. What is the rate of acceleration?

2. A truck decelerated at -2.60 ms-2. It came to astop (v = 0) in 4.80 s. How fast was it going whenthe brakes were applied?

3. A car was travelling at 12.0ms-1. How long would ittake for it to reach 22.5ms-1, if it accelerated at1.75ms-2?

4. A spacecraft was travelling in space at 850ms-1

when its “retro rockets” began to fire, producing aconstant deceleration of 50.0ms-2 (i.e. acceleration of-50.0ms-2) The engines fire for 20.0s. What is thespacecraft’s final velocity at the end of this time?Interpret the meaning of the mathematical answer.

5. The graph shows the motion of a “drag” race car.

16



Worksheet 5 Practice ProblemsAcceleration Student Name...........................................

Worksheet 6 Practice ProblemsNewton’s 2nd Law Student Name...........................................

a) Find the rate of acceleration of the racer.

b) Find the maximum speed it achieved in km/hr.

c) What distance (in metres) did it travel betweent=5.0s and t=8.0s?

d) At which TWO times was the car stationary?

e) Describe the car’s motion after t=8.0s.

f) Find the rate of acceleration at time t=10s.

g) Sketch a graph of displacement-time for thismotion. Values on the graph axes are NOT required.

00 22 44 66 88 1100 1122Time (s)

Velo

city

(m

s-11 )

0

20

4

0

60

1. What force is required to cause a 600kg car toaccelerate at 2.65ms-2?

2. A 120kg motorcycle and its 60kg rider areaccelerating at 4.50ms-2. What net force must beacting?

3. A 500N force acts on a truck with mass 3,500kg.What acceleration is produced?

4. What is the mass of a vehicle which accelerates at3.20ms-2 when a force of 1.25x103N acts on it?


1. A space capsule, ready for launch has a mass of25,000kg. Of this, 80% is fuel. By the time it reachesEarth orbit it has burned three-quarters of the fuel.Later, it proceeds to the Moon and lands, with fueltanks empty. (on Earth, assume gravity g=10ms-2.In orbit g=zero. On Moon, g=1.7ms-2)a) What is the capsule’s weight on Earth?

b) In orbit, what is its i) mass?ii) weight?

c) When it gets to the Moon, what is itsi) mass?

ii) weight?

2. In a laboratory experiment, a 500g trolley isattached by a string to a 250g mass hanging verticallyover the bench. (Take g=10ms-2 , assume no friction)

a) What is the size of the force which will causeacceleration?

17



Worksheet 7 Practice ProblemsMass & Weight Student Name...........................................

Worksheet 8 Practice ProblemsAdding Vectors Student Name...........................................

2. b) What is the total mass to be accelerated?

c) What acceleration will occur?

3. An extra-terrestrial has a weight of 1.80x104N onhis/her/its home planet where g=22.5ms-2.a) What is this creature’s mass?

b) What will he/she/it weigh on Earth, whereg=9.81ms-2?

c) The creature’s personal propulsion device canexert a net force of 5.00x103N. What acceleration canthe alien achieve while wearing the device? (Assume no friction, and that the device itself hasneglible mass)

1. Find the resultant force, if a 25N force pusheseastward, and a 40N force pushes northward. (Remember to find magnitude AND direction)

2. If a 10N force pushes westward, and a 20N forcepushes southward, and a 50N force pushesnorthward, what is the magnitude and direction of theresultant?

3. An aircraft is flying at a velocity of 200ms-1 pointeddue north, but there is a cross-wind blowing from theeast at 20ms-1. What is the plane’s true velocity?

4. A ship sailed 300km due east, then 200km duesouth, then 150km west. Where is it in relation to thestarting point?

5. An object is being simultaneously pushed by 3forces:Force A = 5.25N towards northForce B = 3.85N towards westForce C unknown.The object is NOT accelerating.Find the magnitude and direction of Force C.


3. A 3,000kg aircraft is flying at 300 km/hr in levelflight, and begins a circular turn with radius500m. What centripetal force is needed to effectthis turn? (Hint: first convert velocity to m/s)

4. a) The maximum “grip” force of each tyre ona 1,000kg car is 4,500N. What is the tightest turn(in terms of radius of curve) the car cannegotiate at 90 km/hr? (Hint: velocity units?)

b) The same car comes to a curve with doublethis radius, (ie a much gentler curve) but it istravelling at double the speed. Can it make it?

5. The tension force inthe coupling betweenthis 25,000kg engineand the 10,000kgcarriage is 1.5x103N.a) Calculate the acceleration of the whole train.

b) Find the force produced by the engine.

18



Worksheet 9 Practice ProblemsFriction, Tension & Turning Corners Student Name...........................................1. The engine of an 850kg car is producing a“thrust” force of 2.25x103N at the wheels. Thecar is accelerating at 2.15ms-2. What frictionalforce is acting?

2. A 1,200kg car is towing a 300kg caravan.a) If there was no friction, what force would theengine need to produce for the car and van toaccelerate at 3.50ms-2?

b) In this case, what tension force would act inthe tow-bar?

c) In fact, friction DOES act. Both car and vanare subjected to a frictional force of magnitude450N. (ie total 900N)What acceleration is achieved when the engineproduces the force calculated in (a)?

d) What tension force acts in the tow-bar?(Hint: Tension must overcome the friction on thevan AND cause acceleration... careful!)

Multiple Choice1. Which part of this graph(A, B, C or D) indicates anobject moving, but with alower velocity thanelsewhere?

2. On this grid, one unit onthe scales represents 1 metre & 1 sec.

The average speed overthe first 3 seconds (in ms-1) is:A. 0.75 B. 1.3 C. 2.0 D. 1.0

This Velocity-Timegraph refers to Q3, 4 and 5.

It shows the motionof an objecttravelling north.

3. In section D ofthis graph, theobject’s motion isbest described as:A. moving southward at constant velocity.B. moving southward, and decelerating.C. moving northward, and decelerating.D. moving northward at constant velocity.

4. In the first 3 seconds of this motion, the timewhen the object was stationary was:A. graph section AB. graph section BC. time = 2.5sD. time = zero

5. An instant of time when the acceleration is zero is:A. t = 1.25sB. t = 2.0sC. t = 4.0sD. never

6. The arrows represent 2 vectors.The numbers show the magnitudesof each vector.

12The “resultant” of these 2 vectors would be a single vector with a magnitude closest to: 4A. 16 B. 160 C. 8 D. 13

7. A aircraft taking off accelerated along the runwayfrom rest to 150ms-1 in 30s. The acceleration rate (in ms-2)isA. 4,500 B. 5.0 C. 50 D. 120

8. An astronaut, who on Earth ( g≅≅10ms-2 ) has aweight of 800N, lands on a moon of Jupiter where thegravity g=1.50ms-2. His weight on this moon would beA. 120N B. 1200N C. 80kg D. 800N

9. A laboratory trolley is found to have 5 differentforces acting on it. The trolley is motionless.Four of them are known:• 0.75N weight force, vertically down• 0.75N reaction force, vertically up• 3.2N east• 2.5N east

The 5th force must be:A. 7.2N in all directionsB. 0.7N westC. 5.7N westD. 0.7N east

10. In an experiment, a 700gram trolley was found toaccelerate at 1.70ms-2. What net force must have actedon it?A. 1190N B. 1.19N C. 412N D. 2.4N

11. A car is turning a clockwise, circular curve at aconstant speed. At a particular instant, its velocityvector is directed east. At that instant its accelerationvector is directed:A. north B. south C. east D. west

Longer Response QuestionsMark values shown are suggestions only, and are togive you an idea of how detailed an answer isappropriate. Answer on reverse if insufficient space.

12. (7 marks)A light aircraft flew 150km due north in 2.00 hours,then turned and flew 100km west in 1.00 hour.a) Calculate the average speed (in km/hr) for thewhole flight.

b) Find its final displacement from the starting point,including direction.

c) Calculate its average velocity for the whole flight.

13. (5 marks)An aircraft is being simultaneously affected by 4forces:• “Lift”, acting vertically upwards• “Weight”, acting vertically downwards• “Thrust”, acting horizontally forwards• “Drag”, acting horizontally backwards

Sketch the vector diagram of these forces to showany “resultant” net force acting when:a) the plane is in level flight at constant velocity.

b) the aircraft is speeding up AND gaining height.(No numerical values are required)

19




Worksheet 10 Test Questions sections 1 & 2 Student Name...........................................

TTIIMMEE

DDIISS

TTAANN

CCEE TT

RRAAVVEE

LLLLEEDD

BB

CC

DD

VVeelloo

cciittyy

NNoorr

tthh

00 11 22 33 44TTiimmee ((sseecc))

AA

BB

CC

DD

AA


14. (4 marks)The following Displacement-Time graph shows ajourney in a north-south line.

Sketch the corresponding Velocity-Time graph for thesame journey. There is no need to show anynumerical values on the axes, but sections A, B, C, Dshould be clearly labelled.

15. (4 marks)a) Calculate the net force acting on a 2.50kg trolleythat accelerates from rest to 3.50ms-1 in 5.00s.

b) The trolley is being pulled by a string.The tension in the string is found to be 2.20N.What force of friction is acting?

16. (7 marks)In a laboratory experiment, a trolley of fixed mass wasaccelerated by different forces. The acceleration wasmeasured in each case.

Results:Force Applied (N) Acceleration (ms-2)1.5 1.22.5 1.93.0 2.34.5 3.6

a) Graph these results appropriately.

b) State your interpretation of the graph.

c) Use your graph to find the mass of the trolley.

20



Worksheet 10 (Continued) Student Name...........................................

Dis

plac

emen

tso

uth

(-vve

) n

orth

(+ve

)

Time

AB C

D

17. (6 marks)A broken-down car is being towed as shown.

Both cars are accelerating at 1.50ms-2.Someone accidentally left the hand brake on in the carbeing towed, causing a friction force of 200N to act asshown. Other friction forces are minor and may beignored.a) What is the net force acting on the entire system?

b) What “thrust” force is being provided by the frontcar?

c) Calculate the tension force in the tow-cable.

18. (6 marks)This car is turning a corner to thedriver’s left, at constant speed.a) Mark clearly on the diagram(and label) vectors to represent

i) instantaneous velocityii) acceleration

iii) any net, unbalanced force

The radius of the curve is 25.0m.The car’s speed is 22.0ms-1, and mass is 500kg.b) Calculate the centripetal force acting between thetyres and the road.

c) The maximum “grip” possible from each tyre is2,500N. Explain what will happen, and why, if thecurve becomes tighter... e.g. radius decreases to23.0m.

19. (6 marks)An alien creature has a weight of 5.50x103N on

his/her/its home planet where g=15.3ms-2.a) What is this creature’s mass?

b) What will he/she/it weigh on Earth, whereg=9.81ms-2?

c) The creature’s personal propulsion device canexert a net force of 2.50x104N. What acceleration canthe alien achieve while wearing the device? (Assume no friction, and that the device itself hasneglible mass)

750kg 400kg

a = 1.50ms-22Friction = -2200N

TTIIMMEE

FFoorrcc

ee




Energy of a Moving VehicleYou will be already aware that any movingobject possesses “Kinetic Energy”. The“bigger” the object, and the faster it moves, themore energy it has.

In fact, the amount of energy due to an object’smotion is calculated as follows:

Note that Energy is a Scalar. Energy has nodirection associated with it. “Northbound”energy does NOT cancel “southbound” energy.If 2 vehicles collide head-on, their oppositedirections do not cancel their energies at all...that’s why so much damage can be done in acollision!

Effect of Mass & Velocity on Kinetic Energy

Some simple example calculations can make animportant point:

The Concept of “Work”In Physics, “work” doesn’t mean employmentfor money. “Work” has a very specificmathematical meaning.

If a force acts over a displacement, then Work is done.

From this equation you would expect that theunits of work would be “newton-metres” (Nm).You can use “newton-metres” as the unit, but itturns out that a “newton-metre” is equivalent toa joule of energy...

This means, for example, if a vehicle’s engineexerts a FORCE, we can now calculate theeffects of the force in various ways:

21

3. WORK & KINETIC ENERGY

Kinetic Ek = 1 mv2

Energy 2

Ek = Kinetic Energy, in joules ( J )m = mass of the object, in kgv = velocity, in ms-1

Mass1,000kg

Velocity10ms-11

Calculation 1How much Ekk does this vehicle have?

Ekk = 0.5mv22 = 0.5 x 1,000 x 1022

= 50,000 J (or 50 kJ)

Calculation 2What if you double themass? (same velocity)

Ekk = 0.5mv22

= 0.5 x 2,000x 1022

= 100,000 J (or 100 kJ)

So, 2X the massgives 2X the Kinetic Energy.

Calculation 3What if you double thevelocity? (same mass)

Ekk = 0.5mv22

= 0.5 x 1,000 x 2022

= 200,000 J (or 200 kJ)

So, 2X the velocitygives 4X the Kinetic Energy !!!

Work W = F.SF is Force in newtons (N)S is displacement (in metres)

Work & Energy are Equivalent

WORK = ENERGY

Force fromEngine acts

this way

Force causesacceleration

F = ma1,000 = 500 x a

∴∴ a = 2.0 ms-22

The accelerationgoes on for 10s

v = u + at= 0 + 2 x 10

v = 20 ms-11

Force acting over adistance does workwhich increases thecar’s Kinetic Energy

(and velocity)

Work, W= F.S= 1,000 x100= 100,000 J

Work = Gain of EkkDone

Ekk = 0.5 m v22

100,000 =0.5x500x v22

v22 = 400∴∴ v = 20 ms-11

m= 500kg

F = 1,000N

Force is applied over a distance of 100m.Time taken = 10 s.

Initial velocityu = 0

Notice how 2 totallydifferent calculationsgive the same result.....don’t you just love itwhen things work?!


Energy Transformation When Accelerating

When the car engine does “Work” to acceleratethe car, the energy transformation is:

Note: this transformation is really quiteinefficient, and only a fraction of the energy inthe petrol actually ends up as motion of the car.Most is “lost” as heat energy from the engine,gearbox, wheel bearings, etc.

Energy Transformation When Braking

When the brakes do“Work” to slow thetruck down, the main energytransformation is:

This heat seems “gone” because it dissipatesinto the surroundings... but the energy stillexists.



22

CHEMICALPOTENTIAL KINETICENERGY ENERGY(in petrol)

Initial Velocityu = 30 ms-11

DDiissppllaacceemmeenntt == 110000mmdduurriinngg bbrraakkiinngg

Interpretation: What do the negative

quantities mean?

Negative Ek means that the car hasLOST Kinetic Energy.

Negatve Force means that the forceof braking was in the opposite

direction to the motion.

You could also calculate the acceleration:

F = m a-22000 = 500 x a

∴∴ a = -44.0 ms-22

The negative shows that this is a deceleration.

mass=500kg

Work is Done to Slow Down, TooIn the previous example, the force applied by the car’s engine was used to increase the car’s

Kinetic Energy and velocity. What about when the car slows down?

What Change of Energy Occurred?

Change in = Final Ekk - Initial EkkKinetic Energy

ΔΔEkk = 0.5mv22 - 0.5mu22

= 0.5x500x1022-00.5x500x3022

= 25,000 - 225,000ΔΔEkk = -2200,000 J

This energy change must equal theWORK DONE by the brakes

to slow the car down.

Work W = F.S-2200,000 = F x 100

∴∴ F = -22,000NThe brakes applied a force of -22,000N

Final Velocity v = 10 ms-11

BRAKES APPLIED

Energy TransformationsEnergy can be changed from one form intoanother, and does so frequently.

Law of Conservation of EnergyThis is a very grand-sounding title for a verysimple concept:

Whenever you think energy has been “used up”and is “gone”, what has really happened is thatit has changed into another form which mightnot be obvious any longer.

Electricity Sound

Electricity Light

We find electricity very useful because it can be easily transformed into

many other types of energy.

Energy cannot be created nordestroyed, but only changed in form KINETIC HEAT

ENERGY ENERGY




Energy Transformations in a Collision

When a “bouncy” ball collides with a wall it willbounce off again. A lot of its original KineticEnergy is “conserved”, meaning that after thecollision, it is still in the form of Kinetic Energy.

A collision in which 100% of the Ek is conservedis said to be an “Elastic Collision”. True elasticcollisions occur only at the atomic level, suchas the particles in a gas bouncing off each other.

Even a really “bouncy” ball will lose some of its Ekwith each bounce, and so is not truly “elastic”.The energy itself is not lost, but transformed intoother energy types, such as heat.

When a moving vehicle has an accident, there israrely much “bounce” involved. The collision isalmost totally “Inelastic”, in that all of the Ek ofthe moving vehicle is rapidly transformed intoheat, sound and the damage done to vehiclesand people.

The Law of Conservation of Energy demandsthat the Kinetic Energy of a moving vehiclecannot just disappear when the vehicle collideswith something and stops suddenly.

There is some heat and sound energy producedat the instant of the collision, but this is only atiny fraction of the Ek to be accounted for.

Most of the energy is transformed as the “Workdone” on the vehicle and the people involved.

Remember, that “work” means a force acts overa distance. In a sudden collision, this oftenmeans a very large force acting over a shortdistance, to permanently distort, damage anddestroy the vehicle and the people.

And remember... double the speed means 4times as much energy to be converted intodeath and destruction!

As they say, “Speed Kills”.

Vehicle distorted by the energy transformation

Sound &Heat

produced

Example Calculation: Energy, Work & Force in a Collision

The driver of this820kg car lost

control at 140km/hrand hit a solid rock

embankment.

The vehicle’sstructure was badly

distorted. It isestimated that the“work done” on the

car was due to aforce which actedover a distance ofonly about 2.50m,in a fraction of a

second.

Calculate the force which acted on this car SolutionVelocity =140 km/hr = 140/3.6

v = 38.9 ms-11

Kinetic EnergyEkk = 0.5m v22

= 0.5 x 820 x 38.922

= 6.20 x 1055JDuring the collision, this energy

transformed into the work done on thecar, causing the damage.

Work W = F.S6.20x1055 = F x 2.50

∴∴ F = 6.20x1055/2.50= 2.48x1055N

i.e. a force of 248,000N This is equivalent to being underneath a25 Tonne weight !!PPhhoottoo:: DDaann MMiittcchheellll


Any moving object possessesa)............................ energy. The 2 factorswhich determine how much energy amoving object has, are its b)....................and its c)............................. Their effectsare not equal however; if the mass isdoubled, then the Ek is d).........................,but if velocity is doubled then the Ek ise)...........................

Energy is a f)..................... (vector/scalar)and the unit is the g)............................

“Work” is done when a h)........................acts over a i)..................................... If theeffect of the force is to speed up or slowdown a vehicle, then the work done isequal to the change in j).......................................................

6. The engine of a 900kg car provides a force of1,200N. If this force acts to accelerate the car fromrest (u=zero) over a 75.0m displacement,

a) how much work is done on the car?b) How much kinetic energy does it gain?c) What is the car’s final velocity?d) Find the acceleration of the car, using F=ma.e) How long did it accelerate for?

7. A fully laden truck with mass 10,000kg is travellingat 25.0ms-1 when the engine is switched off and it isallowed to “coast” on a level road. Over a distance of250m it gradually slows down to a new velocity of8.50ms-1.

a) How much kinetic energy does it lose?b) What is the average force acting on it as it slowsdown?c) What is the nature of the force acting?d) Use F=ma to find its average rate of deceleration,and hence find the time period involved.

8. The rider of a bicycle strapped a rocket engine onthe bike, in an attempt on the World Stupidity Record.The combined mass of bike+rocket+rider was 250kg.When fired, the rocket provided 8,000N of thrust forjust 5.20s.

a) Use F=ma to calculate the acceleration produced.b) From a=(v-u)/t, find the final velocity. (u=0)achieved, ignoring any air resistance or friction.c) Find the gain in Kinetic Energy.d) Since this equals the work done by the rocket,calculate the distance covered during theacceleration.

24



Worksheet 11 Work & Kinetic EnergyFill in the blank spaces. Student Name...........................................

1. Calculate the Ek possessed bya) a 200kg motorbike & rider, moving at 10ms-1.

b) the same bike and rider, travelling at 30ms-1.

c) Between parts (a) & (b) the velocity increased by afactor of 3. By what factor did the Ek increase?

2. A car with mass 800kg has 160,000J (160kJ) ofkinetic energy. What is its velocityi) in ms-1?

ii) in km/hr?

3. A 600kg vehicle accelerates from 12.5ms-1 to30.0ms-1. What is the change in its kinetic energy?

4. A 5,500kg truck was travelling at 20.0ms-1, but thenslowed down, losing 5.00x105J of kinetic energy as itdid so. What was its new velocity?

5. How much work is done in each case?

a) A 50N force acts on an object over a distance of4.5m.

b) A 4.0kg mass accelerates at 1.5ms-2, over adisplacement of 3.2m.

c) Over a 50m distance, a 30N force acts on a 6.0kgmass.

The Law of k)........................... of Energystates that “Energy cannot bel)............................ nor..............................,but can be m)............................................

The important energy transformation inan accelerating vehicle isn)........................................ energy (in thepetrol) is converted into o)................................................ energy. When braking,the p).................................. energy of thevehicle is mostly converted intoq)........................... energy in the brakes.

In a collision, most of the Ek possessedby the moving vehicle is used to “dowork” and cause r)....................................to the vehicle and its occupants.

Worksheet 12 Practice ProblemsWork & Kinetic Energy Student Name...........................................

MomentumMomentum is a vector quantity (i.e. directioncounts) which measures the combined effect ofa moving object’s mass and velocity.

Kinetic Energy also depends upon both massand velocity, but Momentum measures a totallydifferent property of a moving object.

Momentum is a vector, whereas Ek is scalar.

Ek is never conserved in a collision, butMomentum always is.

Conservation of Momentum in a Collision

Kinetic Energy can only be “conserved” in anelastic collision, which only happens at theatomic scale. In “real-life” vehicle collisionsmost of the kinetic energy is transformed intoheat and distortion to the vehicles.

Unlike kinetic energy, momentum is alwaysconserved.

When 2 vehicles collide:

25

4. MOMENTUM & IMPULSEkeep it simple science

®



Momentum = mass x velocity

ρρ = m v

The symbol used for momentum ( ρρ ) is theGreek letter “rho”.

Unit of momentum = kilogram-mmetre/sec (kgms-11)

ExampleCompare theMomentum of these Two Vehicles

Bicycle

ρρ = m v= 100 x 1.50= 150 kgms-11 east

Car

ρρ = m v= 600 x 25.0= 15,000 kgms-1

south

ComparisonThe car has 100 times more momentum thanthe bike, because• the car is much more massive, and• it is travelling at a higher velocity.

The momentum vectors are also in totally different directions.

MMAASSSS110000kkgg

Velocity1.50ms-11 east

v = 25.0ms-11

south

600kg

Total Momentum = Total Momentumbefore Collision after Collision

Total Momentumbefore collision ρρi = mA.uA + mB.uB

Note: Since momentum is a vector, you must assign(+ve) and ( -ve) signs to show that these cars are

travelling in opposite directions.

Now the vehicles collide. Let’s imagine that thewrecked cars re-bound from each other, eachwith a new, final velocity.

Total Momentumafter collision ρρf = mA.vA + mB.vB

(Again, the same (+ve) and ( -ve) signs as beforeneed to be assigned for opposite directions)

Conservation of Momentummeans that when you do the calculation

you will find that

Mass = mAA Mass = mBBInitial Velocity = uAA Initial Velocity = uBB

Car “A” Car “B”

FinalVelocity = vBB

FinalVelocity = vAA

Total Momentum = Total Momentumbefore Collision after Collision

ρρi = ρρf

mA.uA + mB.uB = mA.vA + mB.vB

Conservation of Momentum

STUDY THE EXAMPLES next page.

WORKSHEET at the end of this section




Example 1Collision with a Stationary Vehicle

Mass = mAA = 500kg Mass = mBB = 750kgInitial InitialVelocity = uAA Velocity = uBB = 0

= 20.0ms-11


FinalVelocity

vBB

Final Car A stops, Car B moves.Velocity = vA = 0 What is Car B’s velocity?

ρρi = ρρf mA.uA + mB.uB = mA.vA + mB.vB

500x20.0 + 750x0 = 500x0 + 750x vB10,000 + 0 = 0 + 750 vB

∴∴ vB = 10,000/750 = 13.3ms-1

Car B moves forward at 13.3 ms-1

Example 2Head-on Collision. Vehicles “lock” together.

Mass = mAA = 500kg Mass = mBB = 750kgInitial InitialVelocity = uAA Velocity = uBB

= 20.0ms-11 = (-))25.0ms-11

Car “A” Car “B”east-bbound (+ve) west-bbound ( -vve)

What isFinalVelocity?


Since the cars lock together, their finalvelocity is the same

500x20.0 + 750x (-25.0) = (500 + 750) x v10,000 - 18,750 = 1250 v

∴∴ v = -8,750/1250= -7.00ms-1

Both cars move at 7.00ms-1 west

Example 3Collision with an Immovable Objecte.g. Rock Cliff

Mass = mAA = 500kgInitialVelocity = uAA

= 20.0ms-11

Car’s Final Velocity = 0Car stops.Cliff does not move.Where has the momentum gone?

Momentum must be conserved, so the intitialmomentum (10,000kgms-1) still exists. It hasbeen absorbed by the Earth, so the Earth’srotation has been changed. However, the

immense mass of the Earth means that itsvelocity has been altered by such a tiny

amount that it is not measurable.

Cars locktogether

Example 4Collision with a Vehicle Moving in the Same Direction

Mass = mAA = 500kg Mass = mBB = 750kgInitial InitialVelocity = uAA Velocity = uBB

= 20.0ms-11 = 10.0ms-11


Car B is jolted forward at new velocity =15.0ms-11

What is Car A’s final velocity?


500x20.0 + 750x10.0 = 500.vA + 750x15.010,000 + 7,500 = 500vA + 11,250

17,500 - 11,250 = 500vA∴∴ vA = 6,250/500 = 12.5ms-1

Car A continues to moves forward,but at slower velocity of 12.5 ms-1

Examples ofConservation

ofMomentum

in Collisions

IInn eevveerryy eexxaammppllee,, tthhee MMoommeennttuumm iiss ccoonnsseerrvveedd..

IIff yyoouu ccaallccuullaattee tthhee TToottaall KKiinneettiicc EEnneerrggyy bbeeffoorree aannddaafftteerr eeaacchh ccoolllliissiioonn,, yyoouu wwiillll sseeee tthhaatt iitt iiss NNOOTT

ccoonnsseerrvveedd iinn aannyy ooff tthhee ccaasseess.. TThhee ““mmiissssiinngg”” eenneerrggyyiiss uusseedd ttoo ddaammaaggee aanndd ddeessttrrooyy tthhee vveehhiicclleess..

CCoonnsseerrvvaattiioonn ooff MMoommeennttuumm oofftteenn ggooeess aaggaaiinnsstt ccoommmmoonn sseennssee..AAfftteerr aa vveehhiiccllee ccoolllliissiioonn,, tthhiinnggss uussuuaallllyy ssttoopp mmoovviinngg aallmmoossttiimmmmeeddiiaatteellyy.. TThhiiss iiss bbeeccaauussee ooff ffrriiccttiioonn aaccttiinngg oonn ddaammaaggeedd

vveehhiicclleess wwiitthh bbrrookkeenn aaxxlleess ddrraaggggiinngg oonn tthhee ggrroouunndd,, eettcc..IInn tthhee iinnssttaanntt aafftteerr tthhee ccoolllliissiioonn hhoowweevveerr,,

tthhee mmoommeennttuumm HHAASS bbeeeenn ccoonnsseerrvveedd..

Stops

(+ve

) and

( -vv

e) s

igns

mus

t be

assi

gned

““AA”” kkeeeeppss mmoovviinngg,, bbuutt sslloowweerr ““BB”” iiss jjoolltteedd ffoorrwwaarrdd

Newton’s 3rd Law of MotionIt was Sir Isaac Newton who figured out WHYmomentum must be conserved in a collision. Itis because, when one object collides withanother, it exerts a force on the other object, andthat one pushes back!

Newton’s 3rd Law explains quite a few things...

... including Conservation of Momentum.

Why Momentum is ConservedIn a collision between moving Car A andstationary Car B

When A pushes on B, this force accelerates carB according to F=ma. This causes car B toaccelerate and gain momentum.

Meanwhile, car B’s reaction force pushes backon A, with an exactly equal, but opposite force.This causes A to decelerate and losemomentum.

and momentum = momentumlost by A gained by B

Since the momentum lost by one is equal to thatgained by the other, it follows that the totalamount of momentum has not changed, andtherefore that

Momentum has been Conserved!

Impulse of a ForceThe “Impulse” of a force is defined as theproduct of force and the time for which the forceacts.

So what? Well, study the maths...

Start with Newton’s 2nd Law, F = ma

Now a = v - u so F = m(v - u)t t

Multiply both sides by “t” F.t = m(v - u)F.t = mv - mu

This turns out to be a very useful relationship.



27 Usage & copying is permitted according to the Site Licence Conditions only

For Every “Action” Force there is an Equal,but Opposite “Reaction” Force

AAccttiioonn FFoorrccee RReeaaccttiioonn FFoorrcceeAA ppuusshheess oonn BB BB ppuusshheess bbaacckk oonn AA

wwiitthh eeqquuaall ffoorrccee

Impulse = Force x Time

I = F.t

If Force is in newtons (N), and time is in seconds (s)Then the units for Impulse

will be “newton-sseconds” (N.s)

Impulse = Change in Momentum

F.t = mv - mu

This means that the unit of Impulse (N.s)

must be the same as the unit of Momentum (kg.ms-11)

These units are inter-cchangeable

ExampleA car driver applied the brakes for 6.00s andslowed his 800kg vehicle from 25.0ms-11 downto 10.0ms-11.

What was the average force applied by thebrakes?

Solution Impulse = Change in MomentumF.t = mv - mu = m(v - u)

F x 6.00 = 800 (10.0 - 25.0)∴∴ F = -112,000 / 6.00

= -22,000N

The braking force was -22,000NNote that the answer is negative, indicating thatthe force is acting against the motion, causingdeceleration.

When a cannon fires, there isalways a “recoil” or kick-bback.

ActionReaction

AAccttiioonnffoorrccee

bbllaassttss tthheeeexxhhaauussttggaasseess

bbaacckkwwaarrddss..RReeaaccttiioonn

ffoorrcceetthhrruussttss tthhee

rroocckkeettffoorrwwaarrdd..

Action

Reaction

Walking would be impossible withoutNewton’s 3rd Law.

You push on the ground, and theground pushes back.

Action

ReactionImpulse

Change in Momentum


Momentum is the product of a).............................multiplied by b)............................... It is ac)............................. quantity (vector/scalar) withunits d)....................................

In any collision, momentum ise)............................................ This means that thetotal momentum before the collision is equal tothe f).........................................................................This is not always apparent and in agreementwith common sense. For example, after a carcollision everything g)............................ veryrapidly. It would seem that all momentum hasbeen h)......................... However, this is becauseof i).......................... acting on the wreckage. Infact, in the instant following the collision,momentum has been j).........................................

28



Worksheet 13 Momentum & ImpulseFill in the blank spaces. Student Name...........................................

1. Calculate the momentum of:a) a 120kg bicycle (including rider) travelling at5.25ms-1.

b) a 480kg car travelling at 22.5ms-1.

c) a 9,500kg truck travelling at 32.0ms-1.

2. A 750kg car has momentum of 1.15x104 kgms-1. What is its velocity?

3. A passenger bus is travelling at 80.0km/hr.Its momentum is 1.40x105kgms-1. What is its mass?

4. The bus in Q3 slowed down from 90.0km/hr to50.0km/hr. What was its change in momentum?

5. A motorcycle (total mass 180kg) is headingnorth at 35.0ms-1. Meanwhile a 630kg car isheading south at 10.0ms-1.Compare the momentum of these 2 vehicles.

Newton’s 3rd Law states that“k)..............................................................................................................................” This explainsrocket propulsion, and why gunsl)............................. when fired. It also explainsConservation of m).................................

The “n).................................” of a force is definedas force multiplied by the o)...................... forwhich the force acts. The units arep)....................................... The impulse of a forceis equal to the change inq)................................................. which it causes.

COMPLETED WORKSHEETSBECOME SECTION SUMMARIES

Worksheet 14 Practice ProblemsMomentum Student Name...........................................

6. A 600kg car is travelling at 27.0ms-1, when itcollides with a stationary 1,500kg utility. Thevehicles lock together on impact.Find the velocity of the wreckage immediatelyafter impact.

7. Two identical 700kg cars are travelling in thesame direction, but at different speeds. One ismoving with a velocity of 24.5ms-1 and fails tonotice the other in front doing just 8.50ms-1. The“rear-end” collision stops the back car instantly.Find the velocity of the front car immediatelyafter the collision.

8. A truck is heading north at 15.0ms-1 when ithas a head-on collision with a 900kg car, whichwas heading south at 35.0ms-1. On impact the 2vehicles lock together and move north at 6.25ms-1.Find the mass of the truck.

9. In a head-on collision, both vehicles arebrought to a stop. (i..e. final momentum = zero)a) Explain how this is possible.

b) If one vehicle was twice the mass of the other,what was true about their velocities?


1. Find the Impulse in each case:a) A 20N force acted for 4.0s.

b) 150N of force was applied for 1 minute.

c) For 22.5s a 900N force acted.

2. a) A force acted for 19.0s and resulted in 380Nsof Impulse. What was the size of the force?

b) To achieve 2,650Ns of impulse, for how longmust a 100N force be applied?

c) How much force is needed to achieve 1240Nsof impulse in a time of 32.5s?

3. A 400kg car accelerated from 10.0ms-1 to25.0ms-1 in 8.25s.a) Calculate its change in momentum.

b) What is the impulse?

c) What average net force caused theacceleration?

29



4. During braking, a car with mass 850kg slowedto a stop from a speed of 50km/hr (13.9ms-1).The average braking force had a magnitude of3,900N. How long did it take to stop?

5. In a rear-end collision, the stationary car isjolted forward with a new velocity of 8.50ms-1 inthe instant after collision. The car’s mass is750kg.a) How much momentum did the vehicle gain?

b) In the actual collision, the cars were incontact for just 0.350s.

What force acted on the struck vehicle?

c) How much momentum was lost by the othervehicle?

d) What force acted on it?

e) The moving vehicle had a mass of 1,450kgand was moving at 10.5ms-1 before the collision.What was its velocity immediately aftercollision?

Worksheet 15 Practice ProblemsImpulse Student Name...........................................

Multiple Choice1. A vehicle has mass “M” and velocity “V”.Another vehicle has mass “2M” and velocity“2V”. The ratio between their kinetic energieswould be:A. 8:1 B. 4:1 C. 2:1 D. 1:1

2. The work done on a vehicle is equivalent to theA. acceleration of the vehicleB. change in momentum of the vehicleC. change in kinetic energy of the vehicleD. force multiplied by time for which it acts

3. Which vehicle has the least momentum?A. 200kg motorcycle, at velocity 50ms-1.B. 800kg car, at velocity 3ms-1.C. 400kg mini-van, at velocity 2ms-1.D. 120kg bicycle and rider, at velocity 10ms-1.

4. Just before a “head-on” collision, themomentum vectors of 2 cars could berepresented as follows:

car P car Q5,000kgms-1 15,000kgms-1

In the instant after the collision, car Q’s velocityis zero. Which of the following shows car P’smomentum vector just after the collision?

A. C.20,000kgms-1 10,000kgms-1

B. D.10,000kgms-1 15,000kgms-1

5. The”Conservation of Momentum” in a collisionis a consequence of:A. Law of Conservation of EnergyB. Newton’s 1st Law of MotionC. Newton’s 2nd Law of MotionD. Newton’s 3rd Law of Motion

6. Which of the following shows a correctrelationship?A. Change in Momentum = ImpulseB. Change in Kinetic Energy = ImpulseC. Change in Momentum = Work doneD. Change in Kinetic Energy = Change in Momentum


7. (5 marks)A 600kg car braked from a velocity of 25.0ms-1 to8.50ms-1 over a distance of 50.0m.a) What force was applied by the brakes to achievethis?

b) What is meant by the “Law of Conservation ofEnergy”?

c) Considering your answer to (b), explain whathappened to this car’s Kinetic Energy as it sloweddown.

8. (4 marks)A 600kg car, heading north at 15.0ms-1 collided head-on with a 500kg car heading south at 10.0ms-1.The vehicles locked together in the collision.Find the velocity (including direction) of the wreckageimmediately after the collision.

9. ( 5 marks)For the same collision described in Q8:a) Calculate the change in Momentum of the north-bound car.

b) Given that the collision occurred in a time of0.200s, find the average force that acted on the north-bound car.

10. (6 marks)For the same collision described in Q8 (again!):a) Calculate the total Kinetic Energy of both carscombined before the collision.

(Ignore directions... energy is a scalar, remember)b) Calculate the Kinetic Energy of the combinedwreckage after the collision. (use your answer to Q8)

c) Explain any difference in the amount of energybefore and after the collision.

30




Worksheet 16 Test Questions sections 3&4 Student Name...........................................

Newton’s 1st Law of MotionFinally, we get to 1st Law!The 2nd and 3rd Laws are all about the thingsthat happen when forces act, but 1st Law isabout what happens when forces DON’T act.

Essentially 1st Law means that, if no net forceoccurs, then motion cannot change... noacceleration, no change in momentum ispossible.

Newton’s 1st Law is probably the most difficultto understand because it seems to conflict withcommon sense. For example, if a moving car isallowed to “coast” without engine power, on alevel road, it gradually slows down and stops.

Doesn’t 1st Law say that it should keep going atconstant velocity if no force is acting?

The explanation is, of course, FRICTION and airresistance. In all everyday situations there isalways some friction acting against the motion.

We are used to the fact that to maintain aconstant speed forward, the engine must supplya force.

InertiaInertia is defined as the

“tendency of any object to resist any change in its motion”.

This means that moving things have a tendencyto keep moving, and stationary things tend toremain at rest - unless a net force acts on them.

Newton’s 1st Law is often called the “Law ofInertia”.

Inertia is linked to the concept of mass...you could say that mass is the “stuff” thatpossesses inertia, or that inertia is a property ofmass. You know from 2nd Law that it is massthat “resists” accelerations... the bigger themass, the less acceleration occurs. Now we cansay that this is because of inertia.

In a moving vehicle, inertia causes many of thefamiliar things we observe:

The ultimate in inertia effectsoccur in collisions.

31

5. SAFETY DEVICES in VEHICLESkeep it simple science

®



A body will continue to travel in a straight line, at a constant velocity,

unless a net force acts upon it. If at rest, it will remain at rest

until a net force acts.

Engine off...car coasting

Weight forceW = mg

Reaction forceequals weight

Frictionretardsmotion

FORCES UNBALANCED.NET FORCE CAUSES

DECELERATION

Engine pushing car with force

equal toFriction

Weight forceW = mg

Reaction forceequals weight

Friction

FORCES BALANCED.NO NET FORCE,

VELOCITY CONSTANT

Bike stopssuddenly

The unfortunate rider has NOT really been “thrownforward”. His inertia has simply kept him in motion

after his bike stopped moving suddenly.

Sudden AccelerationForward

We feel pressed-bback in the seat

Net force

on car

Looseobjects

seem to flybackwards

In fact, our bodies,and the loose

objects, are simplytrying to stay wherethey were, while the

car acceleratesforward.

SuddenDeceleration

We feelthrownforward

Net force on car

Loose objects seemto fly forwards

In fact, ourbodies, and theloose objects,

are simplytrying to remainin motion, while

the cardecelerates around us.

The Physics of Safety DevicesIn a vehicle collision, most of the injuries topeople are caused by inertia.

Typically, when a car hits something there is arapid deceleration. The car comes to a suddenhalt, but the inertia of the driver and passengerscauses them to keep moving forward, withtragic results:

• can be thrown through the windscreen.• can sufferinjuries by hitting the dashboard.• drivers can be impaled on the steering wheel. • rear seat passengers can hit front seat

passengers with lethal force.

Energy & Momentum in CollisionsIn this topic you have learned that, in a collision:

Kinetic Energy is converted to distortion &destruction, and this is equal to the

WORK DONE = Force x distance

and

Momentum changes as the vehicle changes speed, and this is equal to

IMPULSE = Force x time

The effect of most safety devices is to maximisethe time and distance over which these changesoccur, because this will minimise the force.




Example Calculation 1In a collision at 50km/hr (approx 14ms-1), a 75kgpassenger is brought to rest (on the dashboard) ina time of 0.25sWhat force acts on the person’s body?

Solution Impulse = Change in MomentumF.t = m(v - u)

F x 0.25 = 75 (0 - 14)∴∴ F = - 4,200 N Lethal Force!

(The negative simply means the force was acting against the motion)

Example 2Same person, same collision, but because of a“crumple zone” in the car body, air bag and seatbelt, the time for them to stop moving is increasedto 1.25s. What force acts on the person this time?

Solution Impulse = Change in MomentumF.t = m(v - u)

F x 1.25 = 75 (0 - 14)∴∴ F = - 840 N Survivable!

CONCLUSION

CRUMPLE ZONE in Car BodyIn a collision, the car structure collapses,

one section after the other

This distortion absorbs the kinetic energyand increases the time to come to rest

SEAT BELTS

Seat Belts restrain people, and preventtheir inertia from throwing them intothe dash, or through the windscreen.

The belt has a little “give”, andstretches to increase the time of

momentum change... less force acts!

AIR BAGS

Air bags are “triggered” by inertia, andset off a chemical explosion thatreleases a gas to inflate the bag.

This cushions the person (especiallytheir head) and slows down their

change of momentum... less force acts!

Other Strategies... Reducing the Speed

Crumple Zones, Seatbelts and Air Bags all help toreduce the effects of a collision.

Another strategy is to reduce vehicle speed, so that vehicles generally have less Kinetic Energy

and less Momentum to lose in a collision. It alsogives drivers more time to react to danger and

perhaps avoid the collision.

How to force lower speeds, especially in residential areas:

50km/hr speed limits in residential streets.“Speed humps” and “chicanes”

force drivers to slow down

Safety DevicesIncrease the

Time & Distanceof Collision.

This Decreasesthe Forces

Acting on People

Newton’s 1st Law of motion is all about whathappens when forces a)........................................The Law states that a body in motion willb)........................................................ unlessc)...........................................If it is at rest, it willd)........................... until e).......................................

Observation of everyday events seems tocontradict 1st Law. For example, we observethat vehicles need to be powered to maintainf)....................................., and that they slow downand stop when no forces seem to be acting. Thisis because we don’t see g)....................................acting. To maintain a constant speed, a car’sengine must supply force equal toh)................................ Then, and only then, arethe forces i)....................................... and there isthere NO net force: 1st Law is obeyed.

j)............................. is the tendency of any objectto resist any k).......................................... Inertia isa property associated with l).........................., the“stuff” that resists m)........................................when a force acts.

When a car accelerates forward, it feels as if youare being n)..................................... In reality, youro)......................... is trying to keep you stationary,while the car p).................................. around you.

In a sudden stop you feel as if you areq).................... ..................................., but reallyyour inertia is trying to r)................................................... while the cars)................................... around you.

In a collision, most injuries are due tot)......................... When a car stops abruptly in acollision, the pasengers’ inertia keeps themmoving into the dash, or through theu)........................

Most safety devices such as v)............................and ..................................... work by increasingthe w).......................... over which the personstops moving. This helps by reducing thex)........................... acting on their body. Since“Impulse” equals change iny)................................, then for any given amountof momentum, the larger the z)........................involved, the aa)......................... the force acting.

Another strategy to minimise the effects ofvehicle accidents is to reduce driving speeds,because less speed means lessab)......................... energy and ac).........................to be lost in a collision. Strategies to slow trafficdown include lower speed limits inad)....................................... and the installation ofae).......................... and ....................................

33




Worksheet 17 Physics of SafetyFill in the blank spaces. Student Name...........................................

Worksheet 18 Test Questions section 5 Student Name...........................................

Multiple Choice1. Most safety devices in modern cars aredesigned to reduce the effects of a collision by:

A. reducing the time duration of the collision.B. increasing the change of momentum involved.C. decreasing the distance over which the forces act.D. increasing the time duration of the collision.

2. As the car accelerated when the traffic lightschanged, a book on the dashboard “jumpedback” into Sally’s lap. She immediately thoughtof several possible explanations for the motionof the book. Which one is correct?

A. The book was pushed by a backward, 3rd Lawreaction force.

B. The book stayed still as the car acceleratedforward.

C. The book was pushed by centripetal force.D. As the car moved forward, the book moved

back, to conserve momentum.


3. ( 3 marks)State Newton’s 1st Law of Motion, and use it toexplain why an unrestrained passenger may gothrough a car windscreen during a collision.

4. (3 marks)One of the important safety features of modern motorvehicles is the “crumple zone” built into the front andrear.a) Describe what happens to this “crumple zone” in acollision.b) Explain how this reduces the forces which act onpeople in the car during a collision.

5. (4 marks)a) Explain, with reference to how velocity contributesto kinetic energy, why government agencies mightseek ways to slow traffic down.b) List 2 strategies that local governments use toforce traffic to slow down.




CONCEPT DIAGRAM (“Mind Map”) OF TOPICSome students find that memorising the OUTLINE of a topic

helps them learn and remember the concepts and important facts. Practise on this blank version.

MOVINGABOUT

Answer SectionWorksheet 1a) distance b) timec) gradient d) stationary, not movinge) horizontal line f) zero on the speed scaleg) scalar h) vectori) magnitude & direction j) displacementk) direction l) negativem) down n) gradiento) velocity p) displacementq) timer) velocity at a particular instant of times) average velocity t) displacement and timeu) instantaneous v) sonar or “light gates”

Worksheet 21. 200+100 = 300km2. +200 + (-100) = 100km north3. 5hr4. Speed = dist/time = 300/5 = 60km/hr5. V = S/t

= 100/5 = 20km/hr6. graph

7. from graph:i) gradient = 200/3

≅≅ 67 km/hrii) gradient = zero

iii) gradient = -100/1= -100km/hr

(i.e. 100km/hr south)8. graph

Worksheet 31.a) 600km b) 1.5hrc) V = S/t = 600/1.5 = 400km/hr northd) gradient = -900/3 = -300e) Flight from Q to Rf) R is 300km south of Pg) Position = over town P. Velocity = 300km/hr southh) i) distance = 1,500km

ii) Speed = 1,500/6 = 250km/hriii) Final displacement = 300 km southiv) V = S/t = 300/6 = 50km/hr south

i) graph

2.a) 100/3.6 ≅≅ 27.8ms-1.b) V = S/t, so S = V.t = 27.8x5.00 ≅≅ 139m.c) V = S/t, so t = S/V = 1,000/27.8 ≅≅ 36.0s.

3.a) 20.5ms-1 = 73.8km/hr (north)

-24.5ms-1 = -88.2km/hr (south)b) S = V.t = 20.5x30,0 = 615m north

-24.5x30.0 = -735m ( 735m south)c) t = S/V = 100/20.5 = 4.88s

100/24.5 = 4.08s

4.1st leg: S = V.t = 460x2.50 = 1,150km west2nd leg: = 105x(50x60) =315,000m =315km east3rd leg: = 325x3.25 ≅≅ 1,056km west4th leg: = 125x(5.50x60x60) = 2,475,000m

= 2,475km eastLet east be (+ve), west be ( -ve)Final displacement = -1,150 + 315 -1,056 + 2,475

= +575 km (east) of starting point.

Worksheet 4a) velocity b) slowing downc) direction d) vectore) negative f) ms-2

g) curve h) sloping, straight linei) horizontal j) gradientk) negative l) forcem) external n) unbalancedo) 2nd p) net force appliedq) inversely r) newtons) kg t) ms-2

u) matter v) forcew) gravity x) in the same liney) head to tail z) Pythagorus’s Theoremaa) Resultant ab) directionac) equilibrium ad) net forceae) in a straight line at constant velocity af) acceleration ag) opposesah) Tension ai) bothaj) direction ak) centripetalal) centre am) tangent

Worksheet 51. a = (v - u)/t = (22.5-0)/8.20 = 2.74ms-2

2. u = v - at = 0-( -2.60x4.80) = 12.5ms-1

3. a = (v - u)/t \ t = (v - u)/a = (22.5 - 12.0)/1.75 = 6.00s

4. v = u + at = 850 + (-50.0)x20.0 = -150ms-1

The final negative velocity means it is movingbackwards, compared to its original direction.

continued...




0 1 2 3 4 5TTiimmee ((hhrr))

DDiiss

ppllaacc

eemmeenn

ttNN

oorrtthh

((kk

mm))

0

100

20

0

TTiimmee ((hhrr))1 2 3 4 5

VVeelloo

cciittyy

((kkmm

//hhrr))

SSoouutt

hh

NNoorr

tthh-1

00

0

100

TTiimmee ((hhrr))1 2 3 4 5 6

VVeelloo

cciittyy

((kkmm

//hhrr))

SSoouutt

hh

NN

oorrtthh

-300

-

100

0

100

2

00

300

4

00

Worksheet 5 (cont)5. a) in first 5.0 seconds, gradient = 70/5.0 = 14

∴∴ acceleration = 14ms-2.b) reached 70ms-1 70x3.6 = 252km/hrc) For these 3 seconds it was travelling at 70m/s

S = V.t = 70x3 = 210m.d) Stationary at t = zero and at t = 13s.e) It was decelerating to a stop.f) Acceleration = gradient = -70/5.0 = -14.0ms-2.g) rough sketch

Deceleration: curvesdown to horizontal

Constant Velocity:straight line

Acceleration: curves upfrom horizontal

Note: Although slowing down, the vehicle continuesto move away from the start, so the Displacement-Time graph never shows a negative gradient.

Worksheet 61. F = ma = 600x2.65 = 1,590 = 1.59x103N.

2. F = ma = (120+60)x4.50 = 810 = 8.10x102N.

3. F=ma, so a=F/m = 500/3,500 = 0.1428...= 1.43x10-1N.

4. m=F/a =1.25x103/3.20 =390.6... = 391kg (3.91x102kg)

5. a) a=(v-u)/t = (0 -22.5)/4.50 = -5.00ms-2 (deceleration

b) F=ma = 8.00x103x(-5.00) = -40,000N = -4.00x104N.(Negative force = opposing the motion)

6. a) a=F/m =100/(60+15) = 1.33ms-2.b) a=(v - u)/t, so v=u+at = 0 + 1.33x10.0 = 13.3ms-1.c) 13.3x3.6 = 47.9km/hr.

Worksheet 71. a) W=mg = 25,000x10 = 250,000 = 2.5x105N.

b) i) Take-off mass is 80% fuel=20,000kg of fuel +5,000kg capsule.3/4 is burned reaching orbit, so 5,000kg fuel + 5,000kgcapsule remain. Mass in orbit = 10,000kg.

ii) In orbit (free fall) weight = zero.

c) i) No fuel left, so mass = 5,000kg.ii) W=mg = 5,000x1.7 =8,500N = 8.5x103N.

2. a) W=mg = 0.250x10 = 2.5N.b) 750g = 0.750kg.c) a=F/m = 2.5/0.750 = 3.3ms-2.

3. a) W=mg, so m=W/g = 1.80x104/22.5 = 800kg.b) W=mg = 800x9.81 = 7,848 = 7.85x103N.c) a=F/m = 5.00x103/800 = 6.25ms-2.

Worksheet 81. R2 = 402 + 252 Tan φφ = 25/40

∴∴ R = sq.root(2,225) φφ = 32o

= 47NResultant = 47N at 32o bearing

2.R2 = 302 + 102 Tan φφ = 30/10R = sq.root(1,000) φφ = 72o

= 32

R = 32N, 72o north of west (bearing 342o).

3. R2 = 2002 + 202 Tanφφ = 20/200R = sq.root(40,400) φφ = 6o

= 201

R = 201ms-1, 6o W of N (bearing 354o)Note the directions in these last 2 problems.One angle was “N of W”, another “W of N”.Study the vector diagrams to see why.

“Bearings” (clockwise from north) are best.4. R2 = 1502 + 2002

R = sq.root(62,500)= 250

Tan φφ = 200/150φφ = 53o

R = 250km, 53o S of E (bearing 143o).

5. Not accelerating means there is NO net force, The 3 forcesmust be in equilibriumF2 = 5.252 + 3.852

F = sq.root(42.385) = 6.51

Tanφφ = 5.25/3.85 φφ = 54o

3rd Force = 6.51N, 54o S of E (bearing 144o).

Worksheet 91. Net Force: F= ma = 850x2.15 = 1.83x103N.

Net Force = “Thrust” + Friction1.83x103 = 2.25x103 + Friction

∴∴ Friction = -420N (-4.20x102N).(negative because it opposes the motion)

2. a) F=ma = (1,200+300)x3.50 = 5.25x103N.b) T=ma = 300x3.50 = 1.05x103N.c) Net force = “Thrust” + Friction

= 5.25x103 + (-900) = 4.35x103NF=ma, so a=F/m = 4.35x103/1,500 = 2.90ms-2.

d) Tension must overcome 450N of friction andaccelerate the van at 2.90ms-2. So T=ma +450 = 300x2.90 + 450 = 1.32x103N.

3. v = 300/3.6 = 83.3ms-1.F = mv2/R = 3,000x83.32/500 = 4.16x104N.




50

20

200

200

150

300

R

R

20

φ

φ

10

Rφ

5.25

F

3.85

φ

S

t

φ R


Worksheet 9 (cont)4.a) v=90/3.6 = 25ms-1.Total grip from 4 tyres = 4,500x4 = 18,000N.F=mv2/R, so R=mv2/F = 1,000x252/18,000 = 34.72...

= 35m.b) R=70m, v=50ms-1.Centripetal force needed: F=mv2/R = 1,000x502/70

= 35,714NSince the maximum grip of the tyres is only 18,000N,the tyres cannot provide the force needed to to turnthis corner... car will “spin out”.

5. a) Tension in coupling will accelerate carriage:T=ma, so a=T/m = 1.5x103/10,000 = 0.15ms-2.(and entire train must accelerate at the same rate)

b) Engine force must accelerate entire train:F=ma = (25,000+10,000)x0.15 = 5.3x103N.

Worksheet 101. C 2. B 3. C 4. D 5. A 6. D7. B 8. A 9. C 10. B 11. B

12. a) Total distance = 150 + 100 = 250km

Total time = 2+1 = 3.00hr.Av.Speed = distance/time = 250/3.00 = 83.3km/hr.b) vector diagram essential

R2 = 1002 + 1502

∴∴R = sq.root(32,500) = 180kmTan φφ = 100/150

φφ = 34o

Displacement = 180km, 34o W of N(bearing 326o)

c) v = S/t =180/3.00 = 60km/hr, bearing 326o.

13. a) Forces in equilibrium means the vector diagram must “close” so there is no resultant.

b) Since it is speeding up,then Thrust> Drag.

Since it isclimbing, then Lift > Weight.

14.



37

100

150Rφ

WWeeiigghhtt

TThhrruusstt

LLiifftt

DDrraagg

WWeeiigghhtt

TThhrruussttiinnccrreeaasseedd

RReessuullttaannttFFoorrcceeLLiifftt

iinnccrreeaasseedd

DDrraagg

VVeelloo

cciittyy

((-vvee

))00

((++

vvee))

AA.. ccoonnssttaannttnneeggaattiivvee vveelloocciittyy

CC.. aacccceelleerraattiinngg

time

BB.. ssttooppppeeddzzeerroo vveelloocciittyy

DD.. ccoonnssttaannttppoossiittiivvee vveelloocciittyy

15.a) F=ma and a = (v - u)/t, so F = m(v - u)/t

= 2.50x(3.50 -0)/5.00= 1.75N.

b) Visualise with a vector diagram.Tension 2.20N

Friction Net Force 1.75N

Net Force = Tension + Friction1.75 = 2.20 + F

Friction = -0.45N.

16.a)

b) Graph shows a directrelationship between forceand acceleration.

c) gradient =force/acceleration

= 3.0/2.5= 1.2

Trolley is approx. 1.2kg.

17.a) Since the net force causes acceleration:

F= ma = (750+400)x1.50 = 1,725 = 1.73x103N.b) Vector diagram:

ThrustNet force Friction

Net F = Thrust + Friction1.73x103 = Thrust + (-200)

Thrust = 1.73x103 + 200 = 1.93x103N.c) Tension must accelerate the towed car ANDovercome the friction. T = ma + 200

= 400x1.50 + 200 = 800N.

18.a) on diagram

b) F = mv2/R= 500x22.02/25.0= 9.68x103N.

19.a) W=mg, so m=W/g = 5.50x103/15.3 = 359kg.

b) W=mg = 359x9.81 = 3,522 = 3.52x103N.

c) a=F/m = 2.50x104/359 = 69.6ms-2.

0 1 2 3 4 5AAcccceelleerraattiioonn ((mmss-22))

FFoorrcc

ee ((NN

))0

1

2

3

4

5

22..55

33..00

iiiiii)) FFiiii)) aa

ttaannggeenntt ttoo cciirrccllee

Remember that for full marksin calculations, you need to show

FORMULA, NUMERICAL SUBSTITUTION,APPROPRIATE PRECISION and UNITS

ii)) VV

((ttoowwaarrddcceennttrree ooff

cciirrccllee))


8.a) F=ma, a = F/m = 8,000/250 = 32.0ms-2.b) a = (v - u)/t, v = u + at = 0 + 32.0x5.20 = 166ms-1.c) ΔΔEk = 0.5mv2 - 0.5mu2 = 0.5x250x1662 - 0

= 3.44x106 J.d) ΔΔEk = Work = F.S, so S = W/F =3.44x106/8,000

= 431 m.

Worksheet 13a) mass b) velocityc) vector d) kg.ms-1

e) conservedf) total momentum after collisiong) stops moving h) losti) friction j) conservedk) For every “action” force there is an equal, opposite“reaction” force.l) recoil (kick) m) momentumn) Impulse o) timep) newton-seconds (N.s) q) momentum

Worksheet 141. a) ρρ = mv = 120x5.25 = 630kgms-1.b) ρρ = mv = 480x22.5 = 10,800 = 1.08x104kgms-1.c)ρρ = mv = 9,500x32.0 = 304,000 = 3.04x105kgms-1.

2.ρρ = mv, so v = ρρ/m = 1.15x104/750 = 15.3ms-1.

3. v = 80.0/3.6 = 22.2ms-1

ρρ = mv, so m =ρρ/v = 1.4x105/22.2 = 6.31x103kg.

4. u = 90.0/3.6 = 25.0ms-1. v = 50.0/3.6 = 13.9ms-1

ΔΔρρ = mv - mu = 6.31x103x(13.9-25.0) = -700(negative, lost momentum) = -7.00x102kgms-1.

5.motorcycle: ρρ =mv =180x35.0 = 6.30x103kgms-1 north.car: ρρ = mv = 630x10.0 = 6.30x103kgms-1 south.Comparison: both vehicles have the same magnitudeof momentum, but in opposite directions. (Remember, momentum is a vector)

6. ρρi = ρρf mA.uA + mB.uB = mA.vA + mB.vB

Since the cars lock together, final velocity is the same.600x27.0 + 1,500x0 = (600+1,500)xV

2,100V = 16,200v = 7.71ms-1.

7. ρρi = ρρfmA.uA + mB.uB = mA.vA + mB.vB

700x24.5 + 700x8.50 = 0 + 700x VBVB = (17,150+5,950)/700

= 33.0ms-1.8. ρρi = ρρf

mA.uA + mB.uB = mA.vA + mB.vB(let north be +ve, south -ve)

mAx15.0 - 900x35.0 = mAx6.25 + 900x6.258.75xmA = 31,500 + 5,625

mA = 37,125/8.75 = 4.24x103kg.9.a) If they had equal magnitudes of momentum, butopposite directions, then the sum of theirmomentum = zero.b) To have equal magnitudes of momentum, theproduct MxV must be the same for each (ignoringdirection).If one has twice the mass, the other musthave twice the velocity.



38

Worksheet 11 a) kinetic b) massc) velocity d) doublede) quadrupled (4X) f) scalarg) joule ( J) h) forcei) distance j) kinetic energyk) Conservation l) created nor destroyedm) transformed (into other forms of energy)n) (chemical) potential o) kineticp) kinetic q) heatr) distortion/damage

Worksheet 121.a) Ek = 0.5mv2 = 0.5x200x102 =10,000 =1.0x104 J.b) = 0.5x200x302 =90,000 =9.0x104 J.c) increased 9 times (i.e. 32)

2. a)Ek = 0.5mv2 , so v2=2xEk/m = 2x160,000/800

v2 = 400, so v = 20ms-1.b) v=20x3.6 = 72km/hr.

3.ΔΔEk = 0.5m(v2 - u2) = 0.5x600x(30.02-12.52)

= 2.23x105 J.

4.ΔΔEk = 0.5mv2 - 0.5mu2

(-5.00x105) = 0.5x5,500xv2 - 0.5x5,500x20.02

Note: change in KE is negative, because energy was lost.

(-5.00x105) = 2,750v2 - 1.10x106

∴∴ v2 = (-5x105 + 1.1x106)/2,750v = sq.root(218.18...)

= 14.8ms-1.

5.a) W = F.S = 50x4.5 = 225 N.m (2.3x102 N.m)b) W = F.S and F = ma, so W = ma.S = 4.0x1.5x3.2 = 19 N.mc) W = F.S = 30x50 = 1500 = 1.5x103 N.m (mass not used)

6.a) W=F.S = 1,200x75.0 = 90,000 = 9.00x104 N.m.b) 9.00x104 N.m. (because Work = ΔΔEk)c) ΔΔEk = 0.5mv2 - 0.5mu2

9.00x104 = 0.5x900xV2 - 0∴∴ v2 = 9.00x104/450

v = sq.root(200) = 14.1ms-1.d) F=ma, a=F/m = 1,200/900 = 1.33 ms-1

e) a=(v-u)/t, t=(v-u)/a = (14.1-0)/1.33 = 10.6s.

7.a) ΔΔEk = 0.5mv2 - 0.5mu2 = 0.5x10,000x(8.502 - 25.02)

= -2.76x106J.(energy lost, so negative)

b) ΔΔEk = Work = F.S, so F = W/S = -2.76x106/250= -1.11x104N.

(Negative force, because it acts against the motion)

c) Frictiond) F=ma, a=F/m = -1.11x104/10,000 = -1.11ms-2

(deceleration)a = (v - u)/t, so t = (v - u)/a =(8.50-25.0/-1.11

= 14.9s.

Worksheet 151.a) I = F.t = 20x4.0 = 80Ns.b) I = F.t = 150x60 = 9x103Ns.c) I = F.t = 900x22.5 = 20,250 = 2.03x104Ns.

2. a) I=F.t, so F=I/t = 380/19.0 = 20.0N.b) I=F.t, so t = I/F = 2,650/100 = 26.5s.c) F=I/t = 1,240/32.5 = 38.2N.

3.a) ΔΔρρ = m(v - u) = 400x(25.0 - 10.0) = 6.00x103kgms-1.b) I = 6.00x103kgms-1.

(Impulse = change in momentum)c) I=F.t, so F = I/t = 6,000/8.25 = 727N.

4.ΔΔρρ = m(v - u) = 850(0 - 13.9)

= -11,815 = -1.18x104kgms-1.(negative because it lost momentum)

Change in momentum = Impulse = F.tt=I/F = -1.18x104/-3,900

(negative force, opposing motion)= 3.03s.

5.a) ΔΔρρ = m(v - u) = 750x(8.50 - 0)

= 6,375 = 6.38x103kgms-1.b) ΔΔρρ = Impulse = F.t, so F = I/t = 6.38x103/0.350

= 1.82x104N.c) Momentum is conserved, so momentum gained byone equals momentum lost by by the other.Momentum lost by the other vehicle =6.38x103kgms-1.

d) F = -1.82x104N (by Newton’s 3rd Law)e) ΔΔρρ = m(v - u) (momentum lost, so negative)

-6.38x103 = 1,450x(v - 10.5)v = -4.4 + 10.5 = +6.10ms-1.

(i.e. still moving forward, but slower)

Worksheet 161. A 2. C 3. C 4. A 5. D 6. A

7.a) Work = change in kinetic energyF.S = 0.5m(v2 - u2)=0.5x600x(8.502 - 25.02)= -165,825 J

F = -165,825/50.0 = -3.32x103N.(negative force, because opposing the motion)

b) It means that energy cannot be created ordestroyed... it never disappears or ceases to exist. Itsimply gets transformed from one type to another.

c) The car lost kinetic energy, but this energy didn’tdisappear... it was transformed, mainly into heat, bythe brakes.

8. ρρi = ρρf mA.uA + mB.uB = mA.vA + mB.vB

Let north be (+ve), south ( -ve)Since the cars lock together, their final velocity isthe same

600x15.0 + 500x(-10.0) = (600 + 500)xV9,000 - 5,000 = 1,100V

∴∴ V = 4,000/1,100 = 3.64ms-1 north(since answer is +ve)

9.a) ΔΔρρ = m(v - u) = 600x(3.64 - 15) = -6.82x103kgms-1.(negative, because the change in momentum wassouthward, or a loss of northward momentum)

b) ΔΔρρ = Impulse = F.t = -6.82x103

∴∴ F = -6.82x103/0.200 = -3.41x104N.

10.a) Ek = 0.5mv2

northbound car southbound carEk = 0.5x600x15.02 Ek = 0.5x500x10.02

= 67,500 J = 25,000 JTotal Ek = 92,500 = 9.25x104J.

b) After collision, velocity = 3.64ms-1

Ek = 0.5x(600+500)x3.642

= 7.29x103 J.

c) Over 90% of the original kinetic energy is gone.Some has been transformed into the sound and heatof the collision, but most has been used to distort anddamage the vehicles.

Worksheet 17a) do not actb) continue moving in a straight line, with constantvelocityc) acted upon by net force d) remain at reste) acted upon by net force f) constant speedg) friction/retarding forces h) frictioni) balanced/in equilibrium j) Inertiak) change of motion l) massm) acceleration n) pushed backwardso) inertia p) acceleratesq) flung forward r) keep you moving forwards) decelerates t) inertiau) windscreenv) seatbelts, airbags & crumple zonesw) time (and distance) x) Forcey) momentum z) timeaa) smaller ab) kinetic energyac) momentum ad) residential areasae) speed humps & chicanes





Worksheet 181. D 2. B

3.1st Law: A moving object will continue to move in astraight line at a constant velocity unless acted uponby a net force. If at rest it will remain at rest unless aforce acts on it.

In a collision in which a vehicle stops suddenly, anunrestrained passenger will continue to moveaccording to 1st Law, and may go through thewindscreen.

4. a) The car body is designed so that it collapses, onesection after another, and crumples in like aconcertina.b) This extends the time over which the car loses itsmomentum. Since change of momentum = Impulse =Force x time, then for any given amount ofmomentum, increasing the time involved reduces theforce acting on the people in the vehicle, anddecreases the risk of injury or death.

5.a) Kinetic energy depends upon both mass andvelocity, but velocity has the biggest contribution,since Ek = 0.5mv2. This means doubling the velocityincreases the energy by a factor of 4.Since velocity is so important, it means that reducingspeeds can greatly reduce the energy involved invehicle accidents, and reduce the incidence of deathand injury.

b) Build Speed humps & Chicanes in streets.Create low speed zones in residential areas, andaround schools.



40

NOTICE ANY ERRORS?

Our material is carefully proof-readbut we’re only human

If you notice any errors, please let us know

ABN 54 406 994 557

PO Box 2575PORT MACQUARIE NSW 2444

(02) 6583 4333 FAX (02) 6583 9467www.keepitsimplescience.com.au [email protected]

kkeeeepp iitt ssiimmppllee sscciieennccee

Need to contact us?

®

Topics Available

Year 11-12 Science CoursesBiologyPreliminary CoreLocal EcosystemPatterns in NatureLife on EarthEvolution Aust. BiotaHSC CoreMaintain. a BalanceBlueprint of LifeSearch for Better HealthOptionsCommunicationGenetics:Code Broken?

ChemistryPreliminary CoreChemical EarthMetalsWaterEnergyHSC CoreProduction of MaterialsAcidic EnvironmentChem.Monit.&MngmentOptionsShipwrecks, Corrosion...Industrial Chemistry

Earth & Envir.Science

Preliminary CorePlanet Earth...Local EnvironmentWater IssuesDynamic EarthHSC CoreTectonic ImpactsEnvirons thru TimeCaring for the CountryOptionIntroduced Species

PhysicsPreliminary CoreWorld CommunicatesElectrical Energy...Moving AboutCosmic EngineHSC CoreSpaceMotors & GeneratorsIdeas to ImplementationOptionsQuanta to QuarksAstrophysics

Year 7-8 General ScienceDisk Filename Topic Name01.Energy Energy02.Forces Forces03.Matter Solids, Liquids & Gases04.Mixtures Separating Mixtures05.Elements Elements & Compounds06.Cells Living Cells07.Life Living Things08.LifeSystems Plant & Animal Systems09.Astronomy Astronomy10.Earth The Earth11.Ecosystems Ecosystems

Year 9-10 General ScienceDisk Filename Topic Name12.Waves Wave Energy (inc. Light)13.Motion Forces & Motion14.Electricity Electricity15.Atoms Atoms & Elements16.Reactions Compounds & Reactions17.DNA Cell Division & DNA18.Evolution Evolution of Life19.Health Health & Reproduction20.Universe The Universe21.EarthScience Earth Science22.Resources Resources & Technology

Site Licence ConditionsA school (or other recognised educationalinstitution) may store the disk contents inmultiple computers (or other data retrievalsystems) to facilitate the following usages ofthe disk contents:

• School staff may print unlimited copies onpaper and/or make unlimited photocopies atone school and campus only, for use bystudents enrolled at that school and campusonly, for non-profit, educational use only.

• School staff may use the disk contents tomake audio-visual displays, such as viacomputer networks, or by using dataprojectors or overhead projectors, at oneschool and campus only, for viewing bystudents enrolled at that school and campusonly, for non-profit, educational use only.

• School staff may allow students enrolled atthat school and campus only to obtaincopies of the disk files and store them ineach student’s personal computer for non-profit, educational use only.

IN SUCH CASE, THE SCHOOLSHOULD MAKE PARTICIPATING

STUDENTS AWARE OF THESE SITELICENCE CONDITIONS AND ADVISE

THEM THAT COPYING OF DATAFILES BY STUDENTS MAY

CONSTITUTE AN ILLEGAL ACT.

• In every usage of the disk files, the KISSlogo and copyright declaration must beincluded on each page, slide or frame.

Please Respect Our Rights Under Copyright Law

All Topics Available as PHOTOCOPY MASTERS and/or KCiCPhotocopy Masters (PDF files)

Black & White, A4 portrait-orientationfor clear, economical photocopying.

KCiC = Key Concepts in ColourFull colour, formatted for on-screen studyand data projection. PDF + Powerpoint®

Powerpoint is a trademark of Microsoft Corp.

KISS Notes Moving About

Documents

Transcript of KISS Notes Moving About