Phys3.MovingAbout

keep it simple science TM

www.keepitsimplescience.com.au

St Johns Park High School SL#802444

1

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 will alreadyknow that mathematically:

SPEED = distance travelledtime taken

In this topic, you will extend your understanding of speed toinclude VELOCITY, which is just a special case of speed.

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

Some forces, like gravity and electric/magnetic fields, canexert forces without actually touching things. In this topicyou will deal mainly with CONTACT FORCES, whichpush or pull objects by direct contact.

In the context of moving vehicles, the most importantforce is FRICTION. Friction allows a car’s tyres to grip theroad to get moving, and for the brakes to stop it again.Without friction the car couldn’t get going, and couldn’tstop if it did!

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 you will studyis the one associated with moving vehicles...

KINETIC ENERGY

WHAT MAKES A CAR GO?

Overviewof Topic:

MOMENTUMchanges

KINETICENERGY

changes

ENGINE provides ENERGY(from chemical energy in petrol)

VELOCITY changes

FORCE causesACCELERATION

FORCE acts overa distance...

“WORK” done

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

Preliminary Physics Topic 3copyright © 2005-2007 keep it simple science


www.keepitsimplescience.com.auPreliminary Physics Topic 3copyright © 2005-2007 keep it simple science


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Velocity-TTime Displacement-TTime

InstantaneousVelocity

VectorDiagrams

Friction

Net Force

•safety belts•crumple zones•air bags

AverageSpeed orVelocity

v = St

Vectors&

Scalars

MeasuringMotion

a = v - ut

F = ma

F = m v22

r

Ekk = 1 mv222

W = F.s

Acceleration Mass&

Weight

Energy Transformations

Equivalence ofWork & Energy

Newton’s2nd Law

Centripital Force

p = mv

I = F.t

Momentum

Impulseof a Force

Newton’s3rd Law

Physics of Safety devices

Inertia&

Newton’s1st Law

Conservationof Momentumin Collisions

Law ofConservation

of Energy

MMOOVVIINNGGAABBOOUUTT

Speed&

Velocity

Force&

Acceleration

Work&

Kinetic Energy

Momentum&

Impulse

Safety Devicesin

Vehicles

CONCEPT DIAGRAM (“Mind Map”) OF TOPICSome students find that memorizing the OUTLINE of a topic helps them learn and remember the concepts andimportant facts. As you proceed through the topic, come back to this page regularly to see how each bit fits thewhole. At the end of the notes you will find a blank version of this “Mind Map” to practise on.

MotionGraphs

Concept of

Force




Speed-Time GraphsThe same journey could also be represented by a differentgraph, showing the SPEED at different times:

Study this graph carefully and compared it with theother...

You must not confuse the 2 types of graph and how tointerpret them.

This graph is very unrealistic in one way. It shows thespeed changing INSTANTLY from (say) 100 km/hr tozero (stopped), without any time to slow down. It alsoshows the car travelling at exactly 100 km/hr for an hourat a time... very unlikely with hills, curves, traffic etc.

Changes of speed (ACCELERATION) will be dealt within the next section. For now we’re Keeping It Simple S...

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1. SPEED & VELOCITY

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Stopped. Speed scale reads zero.

“Flat” partsDO NOT

meanstopped,but meanconstant

speed

Distance-Time GraphsPerhaps your journey was as shown by this graph.

Start at the bottom-left of the graph and consider eachsection A, B, C and D.

So although the average speed for the entire journey was75km/hr, in fact you never actually moved at that speed.

This raises the idea of INSTANTANEOUS SPEED: thespeed at a particular instant of time. The speedometer inyour car gives you a moment-by-moment reading of yourcurrent speed... this is your instantaneous speed.

On the graph, the GRADIENT at any given point is equalto 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 distance movedin 0.5 hr:Speed= zero.

Graph section ATravelled 100 km in1.0 hour:Speed =100 km/hr

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Distance-Time Graph

gradient = distancetime

= speed

These graphsrepresent thesame journey

DISTANCE-TIME GRAPHS show 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 in a 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.




4

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... theseare Scalar quantities, since they do not have any specialdirection 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 then

drove 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 & Vectors

BUT, consider the “NET journey”: at the end of thejourney you end up 30 km EAST of the starting point.Your final displacement is “30 km east”.

So 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 have a direction(“east”) specified.... they are VECTORS!

To make better sense (mathematically) of the journey, thedirections east & west could have (+) or ( - ) signs attached.Let east be (+) and west be ( - ). Then the total journeydisplacement was (+60) + (-30) = +30 km.

Note:The symbol “S” isused for Displacement

Average = DisplacementVelocity time

Vav = S t

TRY THE WORKSHEET at the end of this section

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

What if the 300km journey had been 150 km north (graphsections 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)

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Grad

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Down-sloping

line meanstravellingSOUTH

Back at starting point.(Displacement = 0 )

...and the corresponding Velocity - Time Graph:

The velocity values for each part of this graph are equal tothe gradients of the corresponding parts of theDisplacement - Time Graph.

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Negative value:south-bound

velocity

Zero velocity:means stopped

Positive values mean north-bound velocity

Note: Since the journey ends back at the starting point,total displacement = zero and average velocity = zero

for the whole trip.

However, this simply points out how little informationthe “average” gives you. The instant-by-instant Physics

of the journey is in the graph details.




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• Tape Measure & Stopwatch.The simplest method of all: measure the distance ordisplacement involved, and the time taken. Then use

speed (velocity) = distance (displacement)time

However, this can only give you the AVERAGE speedor velocity. In Physics we often need to considerINSTANTANEOUS velocity.

• Electronic or Computer Timing devicesuse either “Light Gates” or a “SONAR”device to record displacements and times foryou. 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.

• “Ticker-timer”: a moving object drags a paperstrip on which dots get printed (usually every 0.02second) as it goes. The gap between dots is arecord of displacement and time. This allows youto calculate the velocity over every 0.02 s. It’s stillan average, but over such small time intervals itapproximates 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

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Worksheet 1Part AFill in the blanks. Check answers at the back

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

On a Speed-Time graph, constant speed is shownby e)................................................. on the graph.This does NOT mean stopped, unless the graphsection is lined up withf).................................................................................

Speed and distance are bothg)................................... quantities, because thedirection doesn’t matter. Often in Physics we dealwith h)................................................... quantities,which have both i).................................. and.......................................

The vector equivalent of distance is calledj)......................................, and refers to distance in aparticular k).......................................... For example,if displacement was being measured in the northdirection, then a distance southward would beconsidered as l)............................... displacement.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 velocityis equal to p)........................................ divided byq)...................................... Instantaneous velocityrefers to r)..................................................................

Laboratory methods for measuring motioninclude:• using a tape measure and stopwatch. This allowscalculation of s)................................................ only.• “Ticker-timers” record both t)...............................and .......................... on a paper tape. Averagevelocity can be calculated for short time intervalswhich are approximately equal tou)..................................................... velocity.• Electronic or Computer-based devices often usev)........................ or .....................................................to gather displacement, time and velocity data atvery short time intervals.

COMPLETED WORKSHEETSBECOME SECTION SUMMARIES

Preliminary Physics Topic 3copyright © 2005-2007 keep it simple science www.keepitsimplescience.com.au

Worksheet 1Part B Practice Problems

1.A car travelled 200 km north in 3.0 hours, then stopped for 1.0 hr,and finally travelled south 100 km in 1.0 hr.

a) What was the total distance travelled? b) What was the total displacement?c) What was the total time for the whole journey?d) Calculate the average speed for the whole journey.e) Calculate the average velocity for the whole journey.f) Construct a Displacement - Time Graph for this trip.g) 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.

h) Construct a Velocity - Time Graph for the journey.

2.An aircraft took off from town P and flew due north to town Qwhere it stopped to re-fuel. It then flew due south to town R.

The trip is summarized by the following graph.

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 flight from P to Q(include direction)d) What is the value of the gradient of the graph fromt=3 hr, to t=6 hr.?

e) What part of the journey does this represent?f) Where is town R located compared to town P?g) What was the aircraft’s position and velocity (includingdirection) at t=5 hr?h) What was the:

i) total distance )ii) average speed ) over the wholeiii) total displacement ) 6 hours?iv) average velocity )

i) Construct a Velocity- Time Graph for this flight.

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200

400

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1 2 3 4 5 6

Time (hr)

UNITS OF MEASUREMENTSo far all examples have used the familiar km/hr for speed orvelocity. The correct S.I. units are metres per second (ms-1).You need to be able to work in both, and convert from one to theother..... 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

3.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 far will the cartravel in this time?

c) At this velocity, how long does it take (in seconds) to travel1.00km (1,000m)?

4.The Velocity - Time Graph shows a journey in an east-west line,by motorcycle.

a) For each part of the graph (A, B, C & D):i) Calculate the velocity in ms-1.ii) Calculate the displacement (in km) during each part.

b) Use the displacement data to construct a Displacement-TimeGraph (use km & hour scales) for this journey.

c) Find i) total distance coveredii) average speediii) total displacementiv) average velocity for the 2 hour trip.

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

A truck is travelling at a velocity of +20.5 ms-1 as it passes a cartravelling at -24.5 ms-1.

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

b) What is the displacement (in m) of each vehicle in 30.0 s?

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

6.Where does this aircraft end up in relation to its starting 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.

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C

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1.0 2.0

Time(hr)

FULLY WORKED SOLUTIONSIN THE ANSWERS SECTION


Change of Velocity = AccelerationVehicles increase and decrease their speed all the time.Any 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 in seconds, thenacceleration is measured in metres/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 changing is 1 ms-1 per second,or simply 1 ms-2.

Acceleration is a vector, so direction counts.

“Deceleration” (or negative acceleration) simply meansthat the direction of acceleration is opposite to the currentmotion... the vehicle will slow down rather than speed up.


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

The graphs that result from acceleration are as follows:

7


2. FORCE & ACCELERATION




a = ΔΔvΔΔt

a = v - ut

VELOCITY VECTOR

ACCELERATIONVECTOR

+

-

THIS CAR IS SLOWINGDOWN... DECELERATING

Tape of trolley moving at constant velocity (for comparison)

Tape of trolley accelerating... dots get further apart

Trolley decelerating (negative acceleration)... dots get closer

Time

Dis

plac

emen

t

DISPLACEMENT-TTIME GRAPH

Gradients increasing(curve gets steeper)

Gradient constant(straight line)

Gradients decreasing(curve flattens out)

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Gradient positive Gradient negative

On a Velocity-TTime GraphGradient = Acceleration

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Example Problem 1A motorcycle travelling at 10.0 ms-1, accelerated for 5.00s toa final velocity of 30.0 ms-1. What was its acceleration?

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 brakes producing anacceleration 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.




8

Force Causes AccelerationA simple definition of “Force” is that it is when somethingpushes or pulls on something else. However, in the contextof moving vehicles,

Force is what causes velocity to change.

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

To actually result in a change of velocity, the force 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

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The car above has a number of forces acting onit, but they are BALANCED... those acting in thesame line are equal and opposite, and canceleach other out.

This car will not alter its velocity or direction; itwill not accelerate. It is either travelling at aconstant velocity, or it is stationary.

FrictionandAirResistance

Thrust

fromEngine

PRESSING ON THE ACCELERATOR...

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Vertical forces arebalanced, and cancel

Horizontal forces areUNBALANCED

This car will SPEED UP

TURNING THE STEERING WHEEL...

WWeeiigghhtt


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Forward & BackForces are balanced

and cancel

This car will turn a corner

at constant speed(but new velocity

since new direction)

PRESSING ON THE BRAKES...

WWeeiigghhtt


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Horizontal forces areUNBALANCED

This car will SLOW DOWN(Decelerate)

EXAMPLES OFUNBALANCED

FORCESBALANCED

FORCESITUATION

SidewaysForces become UNBALANCED(These would

be equal ifwheel notturned)

GOING UP A HILL(without pushing harder on accelerator)

WWeeiigghhtt (still vertical)

RReeaaccttiioonn FFoorrccee is notvertical, and nolonger cancels theweight completely...UNBALANCED

Frictionstill thesame

Engine Thruststill the same

Part of theWeight Force actsto cancel someof the thrust

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

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PASSING OVER AN ICY PATCH ON THE ROAD

Opposite Forces areBALANCED and cancel

This car will continuein a straight line, at a

constant velocity...whether the driverwants to or not...

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9

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

Sir Isaac Newton (1642-1727) figured out the role of forcesin 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 in an object.In terms of force and acceleration, mass is the stuff thattries to prevent acceleration... the more mass there is, theless acceleration an applied force will produce.

The mass of an object is the same wherever it might be.

“Weight” however, changes according to gravity conditions.Weight is a Force (measured in N) due to gravity.Gravity causes objects near the Earth to accelerate at(approximately) g= 10ms-2 (actually 9.81ms-2, but K.I.S.S.).

F = m a, so Weight = mgFor example, 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 been caused bythe unfortunate choice by society to talk about the “weight”of things, but then measure it in kg... it should be in N!!

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:

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ddeevviiccee

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Extra masses on trolley

Weight on stringcauses trolley to accelerate

The acceleration of the trolley is determined byanalysing the displacement & time data fromthe ticker tape record.

This is repeated several times, tranferring someof the “extra” masses from trolley to thehanging weight each time.

This means, for each trial:• total mass of the entire system stays

constant• the force causing the acceleration (weight

on the string) is different each time

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

Acceleration (ms-22)

Forc

e(w

eigh

t on

str

ing)

(N)

Force v Acceleration Graph

LINE OF “BE

STFIT

”

Final Results and Conclusions• Within experimental error, the graph shows a straight line.This proves there is a direct relationship between the forceapplied, and the acceleration produced.

• The gradient of the graph will be found to be equal to the massof the total system (i.e. trolley + masses) in kilograms:

Gradient = Force = MassAcceleration

F = ma

and therefore, F = m a

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Weight = mg

UNCHANGED CHANGES




10

Vector AnalysisForce is a vector quantity, the same as velocity andacceleration. To fully describe a force, you must state thedirection of the force.

Often, there are situations where 2 (or more) forces act onthe same object at the same time. To find the NETFORCE acting you need to add the vectors together to findtheir combined effect.

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

However, if the forces are acting in totally differentdirections, the problem is more complicated.

Vectors in EquilibriumIt is often the case that 2 or more vectors might all canceleach other out so the “resultant” is zero. In fact this isalways the case when a vehicle is moving in a straight linewith a constant velocity.

Since it is NOT accelerating, then the net force acting mustbe zero. Since there are forces acting, then it follows theymust be cancelling each other out.

Example:an aircraft flying straight and level at constant velocity.

The vector diagram for this plane must be:

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

Example:

Mathematically, you should assign (+ve) and (-ve) signs tothe 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 ( φ ) by Trigonometry:Tan φ = opp/adj = 30/20 = 1.5∴ φ ≅ 56o

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

Force A20N east

Force B30N south

First, sketch thesevectors “head-to-tail”.

A = 20

B =

30

φNext, connect thebeginning to theend, to from a right-angled triangle.

The 3rd side is the“Resultant” vector.

RReessuullttaanntt

WeightForce

AirResistance“Drag”

“Lift” Force(on wings)

Thrustfromengine

Lift Weight

Drag

Thrust

You may have done laboratory work to measure somevectors and their sum. A common experiment is shown inthe photo:

A

A

B

B

C

C

Three Force Vectorsin Equilibrium

F = m g

TTeennssiioonn FFoorrcceessiinn ssttrriinnggss AA && BBmmeeaassuurreedd bbyySSpprriinngg BBaallaanncceess

TThhee aanngglleess bbeettwweeeennssttrriinnggss AA,, BB && CC nneeeeddttoo bbee mmeeaassuurreedd..

The 3 vectorscan then beanalysed

The vectors can be analysed either byaccurate scale drawing, or bymathematics (e.g. Sine Rule in triangle).

It will be found (within experimentalerror) that these vectos add to zero.They are in equilibrium.




11

More Examples of VectorsSo far, all the examples given of vectors have been forces.

Vector analysis 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 the motion 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 flies 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 27o S of E(bearing from north = 117o)

A ship is travelling due east at a velocity of 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

(this angle is 70oo

clockwise from north,∴∴ bearing = 70oo)

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

aacccceelleerraattiioonn

TThhrruussttFFoorrccee

FFrriiccttiioonnFFoorrccee

220000kkmm

φφ

Displacement Vectors

Velocity 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 theopposite direction to the car’s motion)




12

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 gravityhas been taken as 10ms-2 forsimplicity (KISS Principle)

500g(0.5kg)

weight = mg

weight = mg

Two Masses Hanging on Strings

Tension force in top string must hold up both weights, soT1 = (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 in equilibrium.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 = T11 + T22 + mg ΣΣF = T22 + mg= (+7) + (-22) + (-55) = (+2) + (-22)= zero = zero

It all works! If you undertstand the tension forcesacting, you can explain that this system is notmoving 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, which attaches two

objects together. The tricky thing about tension is thatit pulls in 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 net thrustforce.

Problema) What is the acceleration?b) What tension force acts in the couplingbetween engine and carriage?c) What is the net force acting on the enginealone?

Engine Carriage20,000kg 5,000kg

Coupling

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




13

Turning Corners - Circular MotionFor any vehicle to turn a corner, it must change directionand therefore, must accelerate. This means it is being actedon by an unbalanced force.

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

So, where does the centripital force come from to push amoving vehicle, such as a car, around the corner?

In the example of a car, the centripital force comes fromthe frictional “grip” of the tyres on the road. Turning thesteering wheel creates new friction forces which aredirected to the centre of an imaginary circle.

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

If the centripital force required for a particular cornerexceeds the friction “grip” of the tyres, then the vehicle willnot make it, and may “spin out” and crash.

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 turningis always toward the centre ofthe circle.This is called “Centripital” 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 vehicle isconstantly accelerating because its direction is constantlychanging.

The force causing this acceleration is called “CentripitalForce” and is always directed to the centre of the circle.

The acceleration vector is also pointed at the centre of thecircle.

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

Centripital ac = v2

Acceleration R

Centripital Fc = m v2

Force R

Where R = radius of the circle (in metres)V = instantaneous velocity (or “orbital speed”)(ms-1)m = mass of vehicle (in kg)

InstantaneousVelocity vector

Wheel turned

Path car willtake

CentripitalForce

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

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

= 16,200N

Example Problem 2The maximum frictional forcepossible from each tyre of this 750kgcar is 5,000N.

What is the maximum speed that thecar can go around a curve with aradius of curvature of 40m?

Solution:Max. Force possible from 4 tyres = 4x5,000

=20,000NCentripital 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)

Worksheet 2

Part A Fill in the blanks. Check your answers at the back.Acceleration is a change in a)............................................. This couldmean speeding up, or b)...................................................., or evenchanging c)........................................ at constant speed. Accelerationis a d)........................................ (vector/scalar). “Deceleration”simply means e)..................................... acceleration. The unit ofmeasurement is f)................................. On a Displacement-Timegraph, acceleration appears as a g)............................................... Ona Velocity-Time graph, accelerations appear as h)................................................................ (compared to constant velocity, which showsas a i)................................ line). The j)..................................... of theline equals the rate of acceleration. A deceleration would have ak)..................................... gradient.

Acceleration is caused by the action of a l)......................................The force must be m)..................................., and it is only ann)............................................. (or “net”) force which causes anacceleration.

Newton’s o)................. Law of Motion states that “the accelerationof an object is proportional to the p)..................................................and q)...................................... proportional to its mass”. The unit offorce is the r)........................., so long as mass is in s)................. andacceleration in t).........................

Mass is a measure of the amount of u).......................... in an object,while “weight” is the v)....................... due to w)...................................acting on the mass.

Vector quantities can only be added together in a simple arithmeticway if they act x)...................................................................................If vectors are in different directions, they must be added using avector diagram (in which vectors are joined y)...................... to...................................). This diagram can then be analysedmathematically using z)....................................... and/ortrigonometry to find the “aa)........................................” vector. Thecomplete answer must contain both the magnitude andab)............................................ of the resultant.

If 2 or more force vectors cancel each other out, they are said tobe in ac).......................................... In such a case there is noad)............................ force and so the object or vehicle will continueto move ae)................................................................................................with no af)......................................................

Friction is a force which always ag)..................................... themotion being considered. “ah).....................................” is the forceacting in a rope, chain or wire connecting objects together. It actsin ai)............................ directions within the coupling.

When a vehicle turns a corner it is accelerating, because theaj)............................................ keeps changing. The force causing thisis called ak)............................................ force, and is directed at theal)..................................... of a circle. The instantaneous velocityvector is a am)...................................... to the circle.

Part B Practice Problems

Acceleration Problems

1.Starting from rest (i.e. u=0) a car reaches 22.5 ms-1 in 8.20 s. Whatis the rate of acceleration?

2.A truck decelerated at -2.60 ms-2. It came to a stop (v = 0) in 4.80 s. How fast was it going when the brakes wereapplied?

3.A car was travelling at 12.0ms-1. How long would it take for it toreach 22.5ms-1, if it accelerated at 1.75ms-2?

4.A spacecraft was travelling in space at 850ms-1 when its “retrorockets” began to fire, producing a constant deceleration of50.0ms-2 (i.e. acceleration of -50.0ms-2) The engines fire for 20.0s.What is the spacecraft’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.

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 between t=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 this motion.Values on the graph axes are NOT required.

Worksheet continued next page...

14






00 22 44 66 88 1100 1122

Time (s)

Velo

city

(m

s-11 )

0

20

4

0

60

Worksheet 2 Part B Practice Problems (continued)

Newton’s 2nd Law6. What force is required to cause a 600kg car to accelerate at2.65ms-2?

7. A 120kg motorcycle and its 60kg rider are accelerating at4.50ms-2. What net force must be acting?

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

9. What is the mass of a vehicle which accelerates at 3.20ms-2

when a force of 1.25x103N acts on it?

10. A truck with mass 8.00x103kg is travelling at 22.5ms-1 whenthe brakes are applied. It comes to a complete stop in 4.50s.a) What is its average rate of acceleration?b) What net force is acting during this deceleration?

11. A 60kg cyclist exerts a net force of 100N pedalling his 15kgbike for 10.0 seconds. Ignoring any friction;a) what acceleration will be produced?b) From a standing start, what velocity will bike and rider reach inthe 10s?c) What is the final velocity in km/hr?

Mass and Weight 12. A space capsule, ready for launch has a mass of 25,000kg. Ofthis, 80% is fuel. By the time it reaches Earth orbit it has burnedthree-quarters of the fuel. Later, it proceeds to the Moon andlands, with fuel tanks 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 its

i) mass?ii) weight?

13. In a laboratory experiment, a 500g trolley is attached by a stringto a 250g mass hanging vertically over the bench.(Take g=10ms-2 , and assume no friction))a) What is the size of the force which will cause acceleration?b) What is the total mass to be accelerated?c) What acceleration will occur?

14. An extra-terrestrial has a weight of 1.80x104N on his/her/itshome planet where g=22.5ms-2.a) What is this creature’s mass?b) What will he/she/it weigh on Earth, where g=9.81ms-2?c) The creature’s personal propulsion device can exert a net forceof 5.00x103N. What acceleration can the alien achieve whilewearing the device? (Assume no friction, and that the device itself has neglible mass)

Vector Analysis15. Find the resultant force, if a 25N force pushes eastward, anda 40N force pushes northward.(Remember to find magnitude AND direction)

16. If a 10N force pushes westward, and a 20N force pushessouthward, and a 50N force pushes northward, what is themagnitude and direction of the resultant?

17. An aircraft is flying at a velocity of 200ms-1 pointed due north,but there is a cross-wind blowing from the east at 20ms-1. What isthe plane’s true velocity?

18. A ship sailed 300km due east, then 200km due south, then150km west. Where is it in relation to the starting point?

19. An object is being simultaneously pushed by 3 forces: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.

Friction, Tension and Turning Corners20. The engine of an 850kg car is producing a “thrust” force of2.25x103N. The car is accelerating at 2.15ms-2.What frictional force is acting?

21. A 1,200kg car is towing a 300kg caravan.a) If there was no friction, what force would the engine need toproduce for the car and van to accelerate at 3.50ms-2?b) In this case, what tension force would act in the tow-bar?c) In fact, friction DOES act. Both car and van are subjected to africtional force of magnitude 450N. (ie total 900N)What acceleration is achieved when the engine produces the forcecalculated in (a)?d) What tension force acts in the tow-bar?(Hint: Tension must overcome the friction on the van AND causeacceleration... careful!)

22. An 3,000kg aircraft is flying at 300 km/hr in level flight, andbegins a circular turn with radius 500m. What centripital force isneeded to effect this turn? (Hint: first convert velocity to m/s)

23. a) The maximum “grip” force of each tyre on a 1,000kg car is4,500N. What is the tightest turn (in terms of radius of curve) thecar can negotiate at 90 km/hr? (Hint: velocity units?)b) The same car comes to a curve with double this radius, (ie amuch gentler curve) but it is travelling at double the speed. Can itmake it?

24. The tension force in thecoupling between this 25,000kgengine and the 10,000kg carriageis 1.5x103N.a) Calculate the acceleration of the whole train.b) Find the force produced by the engine.

15








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

In fact, the amount of energy due to an object’s motion iscalculated as follows:

Note that Energy is a Scalar. Energy has no directionassociated with it. “Northbound” energy does NOT cancel“southbound” energy. If 2 vehicles collide head-on, theiropposite directions do not cancel their energies at all...that’s why so much damage can be done in a collision!

Effect of Mass & Velocity on Kinetic EnergySome simple example calculations can make an importantpoint:

The Concept of “Work”In Physics, “work” doesn’t mean employment for money.“Work” has a very specific mathematical meaning.

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

From this equation you would expect that the units ofwork would be “newton-metres” (Nm). You can use“newton-metres” as the unit, but it turns out that a“newton-metre” is equivalent to a joule of energy...

This means, for example, if a vehicle’s engine exerts aFORCE, we can now calculate the effects of the force invarious ways:

16


3. WORK & KINETIC ENERGY

Kinetic Ek = 1 mv2

Energy 2

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

Mass1,000kg

Velocity10ms-11

Calculation 1How much Ek does this vehicle have?

Ek = 0.5mv2 = 0.5 x 1,000 x 102

= 50,000 J (or 50 kJ)

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

Ek = 0.5mv2

= 0.5 x 2,000x 102

= 100,000 J (or 100 kJ)

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

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

Ek = 0.5mv2

= 0.5 x 1,000 x 202

= 200,000 J (or 200 kJ)

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

Work W = F.S

F is Force in newtons (N)S is displacement (in metres)

Work & Energy are Equivalent

WORK = ENERGY

Force fromEngine actsthis 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= 500kgF = 1,000N

Force is applied over a distance of 100m.It takes 10 sec to movethis 100m distance.

Initial velocityu = 0

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



17

Work is Done to Slow Down, TooIn the previous example, the force applied by the car’sengine, via the gearbox, axles and wheels, was used toincrease the car’s Kinetic Energy and velocity.

What about when the car slows down?


Energy TransformationsEnergy can be changed from one form into another, anddoes so frequently.

The Law of Conservation of EnergyThis is a very grand-sounding title for a very simpleconcept:

Whenever you think energy has been “used up” and is“gone”, what has really happened is that it has changed intoanother form which might not be obvious any longer.

Energy Transformation When AcceleratingWhen the car engine does “Work” to accelerate the car, theenergy transformation is:

Note: this transformation is really quite inefficient, andonly a fraction of the energy in the petrol actually ends upas motion of the car. Most is “lost” as heat energy from theengine, gearbox, wheel bearings, etc.

Energy Transformation When BrakingWhen the brakes do “Work” to slow the car down, the mainenergy transformation is:

This heat seems “gone” because it dissipates into thesurroundings... but the energy still exists.



Electricity Sound

Electricity Light

CChheemmiiccaall HHeeaattPPootteennttiiaallEEnneerrggyy LLiigghhtt

We find electricityvery useful because

it can be easilytransformed into

many other types ofenergy.

Energy cannot be created nor destroyed,but only changed in form

CHEMICALPOTENTIAL KINETICENERGY ENERGY(in petrol)

KINETIC HEATENERGY ENERGY

Initial BRAKES APPLIED FinalVelocity Velocityu = 30 ms-11 v = 10 ms-11

DDiissppllaacceemmeenntt == 110000mmdduurriinngg bbrraakkiinngg

WHAT FORCE ACTED IN THE BRAKES?

Change in = Final Ek - Initial EkKinetic Energy

ΔΔEkk = 0.5mv22 - 0.5mu22

= 0.5x500x1022-00.5x500x3022

= 25,000 - 225,000ΔΔEkk = -2200,000 J

This energy change must equal the WORKDONE 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 -2,000N

Interpretation: What do the negative quantities mean?•Negative Ek means that the car has LOST

Kinetic Energy•Negatve Force means that the force of

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




18

Energy Transformations in a CollisionWhen a “bouncy” ball collides with a wall it will bounce offagain. A lot of its original Kinetic Energy is “conserved”,meaning that after the collision, it is still in the form ofKinetic Energy.

A collision in which 100% of the Ek is conserved is said tobe an “Elastic Collision”. True elastic collisions occur onlyat the atomic level, such as the particles in a gas bouncingoff each other.

Even a really “bouncy” ball will lose some of its Ek with eachbounce, and so is not truly “elastic”. The energy itself is notlost, but transformed into other energy types, such as heat.

When a moving vehicle has an accident, there is rarelymuch “bounce” involved. The collision is almost totally“Inelastic”, in that all of the Ek of the moving vehicle israpidly transformed into heat, sound and the damage doneto vehicles and people.

The Law of Conservation of Energy demands that theKinetic Energy of a moving vehicle cannot just disappearwhen the vehicle collides with something and stopssuddenly.

There is some heat and sound energy produced at theinstant of the collision, but this is only a tiny fraction of theEk to be accounted for.

Most of the energy is transformed as the “Work done” onthe vehicle and the people involved. Remember, that“work” means a force acts over a distance. In a suddencollision, this often means a very large force acting over ashort distance, to permanently distort / damage / destroythe vehicle and the people.

And remember... double the speed means 4 times asmuch energy to be converted into death anddestruction!

As they say, “Speed Kills”.

Vehicles distortedby the energy

absorbed

There is still some Kinetic Energyhere, but it be converted into

“damage” when he lands!

Sound & Heat produced

Example Calculation: Energy, Work & Force in a Collision

The driver of this820kg car lost controlat 140km/hr and hit asolid rockembankment.

The vehicle’sstructure was badlydistorted. It isestimated that the“work done” on thecar was due to a forcewhich acted over adistance of only about2.50m, in a fractionof a second.

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

v = 38.9 ms-1

Kinetic EnergyEk = 0.5m v2

= 0.5 x 820 x 38.92

= 6.20 x 105JThis energy is equivalent to the workdone on the car, causing damage.

Work W = F.S6.20x105 = F x 2.50

∴ F = 6.20x105/2.50= 2.48x105N

i.e. a force of 248,000N This is equivalent to being underneatha 25 Tonne weight !!Photo: Dan Mitchell

Worksheet 3 Part AFill in the blanks. Check answers at the back.

Any moving object possesses a)..............................energy. The 2 factors which determine how muchenergy a moving object has, are itsb)............................ and its c)......................................Their effects are not equal however; if the mass isdoubled, then the Ek is d)........................................,but if velocity is doubled then the Ek ise)........................................... Energy is af)......................................... (vector/scalar) and theunit is the g)................................

“Work” is done when a h).....................................acts over a i).............................................. If theeffect of the force is to speed up or slow down avehicle, then the work done is equal to the changein j).......................................................

The Law of k)..................................... of Energystates that “Energy cannot be l)................................nor.............................., but can bem)..................................................................................The important energy transformation in anaccelerating vehicle is n)..................................................................... energy (in the petrol) isconverted into o)..................................................energy. When braking, thep)........................................ energy of the vehicle ismostly converted into q)......................................energy in the brakes. In a collision, most of theEk possessed by the moving vehicle is used to “dowork” and cause r)............................................ tothe vehicle and its occupants.

Part BPractice Problems: Kinetic Energy & Work1.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 a factorof 3. By what factor did the Ek increase?

2.A car with mass 800kg has 160,000J (160kJ) of kineticenergy. What is its velocity i) in ms-1?

ii) in km/hr?

3.A 600kg vehicle accelerates from 12.5ms-1 to 30.0ms-1.What is the change in its kinetic energy?

4.A 5,500kg truck was travelling at 20.0ms-1, but then sloweddown, losing 5.00x105J of kinetic energy as it did so. Whatwas its new velocity?

5.How much work is done in each case?

a) A 50N force acts on an object over a distance of 4.5m.b) A 4.0kg mass accelerates at 1.5ms-2, over a displacementof 3.2m.c) Over a 50m distance, a 30N force acts on a 6.0kg mass.

6.The engine of a 900kg car provides a force of 1,200N. Ifthis force acts to accelerate the car from rest (u=zero) overa 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 travelling at25.0ms-1 when the engine is switched off and it is allowedto “coast” on a level road. Over a distance of 250m itgradually slows down to a new velocity of 8.50ms-1.

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

8.The rider of a bicycle (combined mass of bike+rider =95.0kg) strapped a rocket engine on the bike, in an attempton the World Stupidity Record. When fired, the rocketprovided 4,000N of thrust for just 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) Hence find the gain in Kinetic Energy.d) Since this equals the work done by the rocket, calculatethe distance covered during the acceleration.

Just after attaining maximum velocity, the bicycle and riderstruck a large tree. The tree was “dented” inwards by 5cm(0.05m) as it absorbed the energy of the collision.e) Calculate the average force which “did work” upon bikeand rider over this distance.

19






MomentumMomentum is a vector quantity (i.e. direction counts)which measures the combined effect of a moving object’smass and velocity.

Conservation of Momentum in a CollisionKinetic Energy can only be “conserved” in an elasticcollision, which only happens at the atomic scale. In “real-life” vehicle collisions most of the kinetic energy istransformed into heat and distortion to the vehicles.

Unlike kinetic energy, momentum is always conserved.

When 2 vehicles collide:

20


4. MOMENTUM & IMPULSE




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 the Momentum of theseTwo Vehicles

Bicycle

ρρ = m v= 100 x 1.50= 150 kgms-1 east

Car

ρρ = m v= 600 x 25.0= 15,000 kgms-1

south

ComparisonThe car has 100 times more momentum than the 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.

car15,000kgms-1

bike south150kgms-1 east

MMAASSSS110000kkgg

Velocity1.50ms-11 east

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 inopposite directions.

Now the vehicles collide. Let’s imagine that the wreckedcars re-bound from each other, each with a new, finalvelocity.

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

(Again, the same (+ve) and ( -ve) signs as before need tobe assigned for opposite directions)

Mass = mA Mass = mBInitial InitialVelocity = uA Velocity = uB

Car “A” Car “B”

FinalVelocity = vB

FinalVelocity = vA

Conservation of Momentum

Total Momentum = Total Momentumbefore Collision after Collision

ρρi = ρρf

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

STUDY THE EXAMPLES next page.





21

Example 1Collision with a Stationary Vehicle

Mass = mA = 500kg Mass = mB = 750kgInitial InitialVelocity = uA Velocity = uB = 0

= 20.0ms-1


FinalVelocity = vB

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

ρρi = ρρfmA.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 = mA = 500kg Mass = mB = 750kgInitial InitialVelocity = uA Velocity = uB

= 20.0ms-1 = (-)25.0ms-1

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

What isFinalVelocity?


Since the cars lock together, their final velocity isthe 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 Object e.g. Rock Cliff

Mass = mA = 500kgInitialVelocity = uA

= 20.0ms-1

Car “A”

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 has been

absorbed by the Earth, so the Earth’s rotation has beenchanged. However, the immense mass of the Earth

means that its velocity has been altered by such a tinyamount that it is not noticeable.

Cars locktogether

Example 4Collision with a Vehicle Moving in theSame Direction

Mass = mA = 500kg Mass = mB = 750kgInitial InitialVelocity = uA Velocity = uB

= 20.0ms-1 = 10.0ms-1


FinalVelocity vB = 15.0ms-1

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 + 11,25017,500 - 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

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22

Newton’s 3rd Law of MotionIt was Sir Isaac Newton who figured out WHY momentummust be conserved in a collision. It is because, when oneobject collides with another, it exerts a force on the otherobject, and that one pushes back!

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

... including Conservation of Momentum.

In a collision between moving Car A and stationary Car B

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

Meanwhile, car B’s reaction force pushes back on A, withan exactly equal, but opposite force.This causes A to decelerate and lose momentum.

andmomentum = momentumlost by A gained by B

Since the momentum lost by one is equal to that gained bythe other, it follows that the total amount of momentumhas not changed, and therefore that

Momentum has been Conserved!

Impulse of a ForceThe “Impulse” of a force is defined as the product of forceand the time for which the force acts.

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.



For Every “Action” ForceThere is an Equal, but Opposite

“Reaction” Force

Action Forcepushes onexhaust gasses,acceleratingthembackwards

Reaction force pushesrocket forward

Action force pushes bullet

Reaction force kicks back

Weight forcepushes onEarth,

Reactionforcepushes back

AAccttiioonn FFoorrccee RReeaaccttiioonn FFoorrcceeA pushes on B B pushes back on A

with equal force

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

Impulse

Change in Momentum

ExampleA car driver applied the brakes for 6.00s andslowed his 800kg vehicle from 25.0ms-1 downto 10.0ms-1.

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.

Worksheet 4

Part A Fill in the blanks. Check your answers at the back.

Momentum is the product of a)................................... multiplied byb)............................................. It is a c).........................................quantity (vector/scalar) with units d)....................................

In any collision, momentum is e)........................................................This means that the total momentum before the collision is equalto the f)......................................................................................................This is not always apparent and in agreement with common sense.For example, after a car collision everything g)..................................very rapidly. It would seem that all momentum has beenh).................................. However, this is because of i)..........................acting on the wreckage. In fact, in the instant following thecollision, momentum has been j).........................................................

Newton’s 3rd Law states that “k)............................................................................................................................................................................”This explains rocket propulsion, and why guns l).............................when fired. It also explains Conservation of m).................................

The “n)......................................” of a force is defined as forcemultiplied by the o)................................. for which the force acts.The units are p)......................................................... The impulse of aforce is equal to the change in q)........................................................which it causes.

Part B Practice Problems. Answers at the back.Momentum

1. Calculate the momentum of:a) a 120kg bicycle (including rider) travelling at 5.25ms-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 to 50.0km/hr.What was its change in momentum?

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

Conservation of Momentum6. A 600kg car is travelling at 27.0ms-1, when it collides with astationary 1,500kg utility. The vehicles lock together on impact.Find the velocity of the wreckage immediately after impact.

7. Two identical 700kg cars are travelling in the same direction, butat different speeds. One is moving with a velocity of 24.5ms-1 andfails to notice 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 immediately after the collision.

8. A truck is heading north at 15.0ms-1 when it has a head-oncollision with a 900kg car, which was heading south at 35.0ms-1.On impact the 2 vehicles lock together and move north at 6.25ms-1.Find the mass of the truck.

9. In a head-on collision, both vehicles are brought 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 trueabout their velocities?

Impulse & Momentum10. 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.

11.a) A force acted for 19.0s and resulted in 380Ns of Impulse.

What was the size of the force?b) To achieve 2,650Ns of impulse, for how long must a 100Nforce be applied?c) How much force is needed to achieve 1240Ns of impulse in atime of 32.5s?

12. A 400kg car accelerated from 10.0ms-1 to 25.0ms-1 in 8.25s.a) Calculate its change in momentum.b) What is the impulse?c) What average net force caused the acceleration?

13. During braking, a car with mass 850kg slowed to a stop froma speed of 50km/hr (13.9ms-1). The average braking force had amagnitude of 3,900N. How long did it take to stop?

14. In a rear-end collision, the stationary car is jolted forward witha new velocity of 8.50ms-1 in the instant after collision. The car’smass is 750kg.a) How much momentum did the vehicle gain?b) In the actual collision, the cars were in contact for just 0.350s.

What force acted on the struck vehicle?c) How much momentum was lost by the other vehicle?d) What force acted on it?e) The moving vehicle had a mass of 1,450kg and was moving at10.5ms-1 before the collision. What was its velocity immediatelyafter collision?

23






Newton’s 1st Law of MotionFinally, we get to 1st Law!The 2nd and 3rd Laws are all about the things that happenwhen forces act, but 1st Law is about what happens whenforces DON’T act.

Essentially 1st Law means that, if no net force occurs, thenmotion cannot change... no acceleration, no change inmomentum is possible.

Newton’s 1st Law is probably the most difficult tounderstand because it seems to conflict with commonsense. For example, if a moving car is allowed to “coast”without engine power, on a level road, it gradually slowsdown and stops.

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

The explanation is, of course, FRICTION and airresistance. In all everyday situations there is always somefriction acting against the motion.

We are used to the fact that to maintain a constant speedforward, the engine must supply a force.

InertiaInertia is defined as the“tendency of any object to resist any change in its motion”.

This means that moving things have a tendency to keepmoving, and stationary things tend to remain at rest -unless a force acts on them.

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

Inertia is linked to the concept of mass...you could say that mass is the “stuff ” that possesses inertia,or that inertia is a property of mass. You know from 2ndLaw that it is mass that “resists” accelerations... the biggerthe mass, the less acceleration occurs. Now we can say thatthis is because of inertia.

In a moving vehicle, inertia causes many of the familiarthings we observe:

The ultimate in inertia effects occur in collisions.

24


5. SAFETY DEVICES in VEHICLES




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 force

W = mg

Reaction forceequals weight

FrictionretardsmotionFORCES UNBALANCED

NET FORCE CAUSESDECELERATION

Engine pushing car with forceequal toFriction

Weight forceW = mg

Reaction forceequals weight

FrictionretardsmotionFORCES BALANCED

NO NET FORCE VELOCITY CONSTANT

Sudden AccelerationForward

We feel pressed-back in the seat

acceleration

of car

Loose objectsseem to flybackwards

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IInneerrttiiaa!!

Sudden Deceleration

We feel thrownforward

deceleration

of car

Loose objectsseem to flyforwards

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IInneerrttiiaa!!

Bike stopssuddenly

The unfortunaterider has NOTbeen “thrownforward”.His inertiahas simply kept him inmotion afterhis bike stopped.



25

The Physics of Safety DevicesIn a vehicle collision, most of the injuries to people arecaused by inertia.

Typically, when a car hits something there is a rapiddeceleration. The car comes to a sudden halt, but theinertia of the driver and passengers causes them to keepmoving forward, with tragic results:

• can be thrown through the windscreen• can suffer terrible injuries on impact with the dashboard• drivers can be impaled on the steering wheel • rear seat passengers can impact on 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 maximise the timeand distance over which these changes occur, because thiswill minimise the force.



Example Calculation 1In a collision at 50km/hr (approx 14ms-1), a 75kgpassenger is brought to rest (on the dashboard) in a timeof 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 “crumplezone” in the car body, air bag and seat belt, the time forthem to stop moving is increased to 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 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 ofmomentum 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 theirchange of momentum... less force acts!

Other Strategies... Reducing Speed

Crumple Zones, Seatbelts and Air Bags all help to reduce theeffects of a collision.

Another strategy is to reduce vehicle speed, so that vehiclesgenerally have less Kinetic Energy and less Momentum tolose in a collision. It also gives drivers more time to react todanger 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

Worksheet 5

Fill in the blanks. Check your answers at the back.

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

Observation of everyday events seems tocontradict 1st Law. For example, we observe thatvehicles need to be powered to maintainf)................................................., and that they slowdown and stop when no forces seem to be acting.This is because we don’t seeg)............................................... acting. To maintaina constant speed, a car’s engine must supply forceequal to h)....................................... Then, and onlythen, are the forces i).............................................and there is there NO net force: 1st Law isobeyed.

j)..................................... is the tendency of anyobject to resist any k)............................................................................. Inertia is aproperty 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,your o).................................... is trying to keep youstationary, while the car p)..................................around you. In a sudden stop you feel as if youare q).................... ..............................................., butreally your inertia is trying tor)....................................................................................while the car s)............................................... aroundyou.

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

Most safety devices such asv)........................................, ........................................and .................................................... work byincreasing the w)................................... over whichthe person stops moving. This helps by reducingthe x)............................................ acting on theirbody. Since “Impulse” equals change iny)..............................................., then for any givenamount of momentum, the larger thez)................................... involved, theaa)...................................... the force acting.

Another strategy to minimise the effects ofvehicle accidents is to reduce driving speeds,because less speed means lessab)..................................... energy andac).............................................. to be lost in acollision. Strategies to slow traffic down includelower speed limits in ad).............................................................. and the installation ofae)................................................. and............................................................

26









27

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

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

MMOOVVIINNGGAABBOOUUTT

Practice QuestionsThese are not intended to be "HSC style" questions, but tochallenge your basic knowledge and understanding of thetopic, and remind you of what you NEED to know at theK.I.S.S. principle level.

When you have confidently mastered this level, it is stronglyrecommended you work on questions from past exampapers.

Part A Multiple Choice1. Which part of thisgraph (A, B, C or D)indicates an objectmoving, but with alower velocity thanelsewhere?

2. On this grid, oneunit on the scalesrepresents 1 metre & 1 sec.

The average speedover the 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 is bestdescribed 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 time when theobject 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 runway fromrest to 150ms-1 in 30s. The acceleration rate (in ms-2)isA. 4,500 B. 5.0 C. 50 D. 120

8. A laboratory trolley is found to have 5 different forcesacting on it.Four of them are known:• 0.75N weight force, vertically down• 0.75N reaction force, vertically up• 3.2N east• 2.5N eastThe trolley is motionless.The 5th force must be:A. 7.2N in all directionsB. 0.7N westC. 5.7N westD. 0.7N east

9. An astronaut, who on Earth ( g≅10ms-2 ) has a weight of800N, lands on a moon of Jupiter where the gravityg=1.50ms-2.His weight on the moon would beA. 120N B. 1200N C. 80kg D. 800N

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

11.

A 400kg broken-down car is being towed by another carwith mass of 600kg. The net force being provided by thefront car is 1,500N.The tension force in the tow-rope is:A. 3.75N B. 1,500N C. 500N D. 600N

12. A car is turning a clockwise, circular curve at a constantspeed. At a particular instant, its velocity vector is directedeast. At that instant its acceleration vector is directed:A. north B. south C. east D. west

13. A vehicle has mass “M” and velocity “V”. Anothervehicle has mass “2M” and velocity “2V”. The ratiobetween their kinetic energies would be:A. 8:1 B. 4:1 C. 2:1 D. 1:1

28





TIME

DIS

TAN

CE T

RAVE

LLED

BB

CC

DD

Velo

city

Nor

th

00 11 22 33 44TTiimmee ((sseecc))

AA

BB

CC

DD

AA

14. 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

15. 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.

16. Just before a “head-on” collision, the momentum vectors of 2cars could be represented as follows:

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

In the instant after the collision, car Q’s velocity is zero.Which of the following shows car P’s momentum vector just afterthe collision?

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

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

17. The”Conservation of Momentum” in a collision is aconsequence 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

18. Most safety devices in modern cars are designed to reduce theeffects 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.

19. As the car accelerated when the traffic lights changed, a bookon the dashboard “jumped back” into Sally’s lap. She immediatelythought of several possible explanations for the motion of thebook. Which one is correct?A. The book was pushed by a backward, 3rd Law reaction force.B. The book stayed still as the car accelerated forward.C. The book was pushed by centripital force.D. As the car moved forward, the book moved back, to conserve

momentum.

20. Which of the following shows a correct relationship?A. Change in Momentum = ImpulseB. Change in Kinetic Energy = ImpulseC. Change in Momentum = Work doneD. Change in Kinetic Energy = Change in Momentum

Longer Response QuestionsMark values shown are suggestions only, and are to give you anidea of how detailed an answer is appropriate.

21. (7 marks)A light aircraft flew 150km due north in 2.00 hours, then turnedand flew 100km west in 1.00 hour.a) Calculate the average speed (in km/hr) for the whole flight.b) Find its final displacement from the starting point, includingdirection.c) Calculate its average velocity for the whole flight.

22. (4 marks)The following Displacement-Time graph shows a journey in anorth-south line.

Sketch the corresponding Velocity-Time graph for the samejourney. There is no need to show any numerical values on theaxes, but sections A, B, C, D should be clearly labelled.

23. (5 marks)An aircraft is being simultaneously affected by 4 forces:• “Lift”, acting vertically upwards• “Weight”, acting vertically downwards• “Thrust”, acting horizontally forwards• “Drag”, acting horizontally backwards

Sketch the vector diagram of these forces to show any “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)

24. (4 marks)a) Calculate the net force acting on a 2.50kg trolley that acceleratesfrom 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?

29





Remember that for full marksin calculations, you need to show

FORMULA, NUMERICAL SUBSTITUTION,APPROPRIATE PRECISION and UNITS

Dis

plac

emen

tso

uth

(-ve

) n

orth

(+ve

)Time

AB C

D

25. (7 marks)In a laboratory experiment, a trolley of fixed mass was acceleratedby different forces. The acceleration was measured 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.

26. (6 marks)A broken-down car is being towed as shown in the diagram.

Both cars are accelerating at 1.50ms-2.Someone accidentally left the hand brake on in the car beingtowed, causing a friction force of 200N to act as shown. Otherfriction forces are minor and may be ignored.a) What is the net force acting on the entire system?b) What “thrust” force is being provided by the front car?c) Calculate the tension force in the tow-cable.

27. (6 marks)This car is turning a corner to thedriver’s left, at constant speed.a) Mark clearly on the diagram (andlabel) vectors to represent

i) instantaneous velocityii) accelerationiii) 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 centripital force acting between the tyres and theroad.

The maximum “grip” possible from each tyre is 2,500N.c) Explain what will happen, and why, if the curve becomestighter... e.g. radius decreases to 23.0m.

28. (6 marks)An alien creature has a weight of 5.50x103N on his/her/its homeplanet where g=15.3ms-2.a) What is this creature’s mass?b) What will he/she/it weigh on Earth, where g=9.81ms-2?c) The creature’s personal propulsion device can exert a net forceof 2.50x104N. What acceleration can the alien achieve whilewearing the device? (Assume no friction, and that the device itself has neglible mass)

29. (5 marks)A 600kg car braked from a velocity of 25.0ms-1 to 8.50ms-1 overa distance of 50.0m.a) What force was applied by the brakes to achieve this?b) What is meant by the “Law of Conservation of Energy”?c) Considering your answer to (b), explain what happened to thiscar’s Kinetic Energy as it slowed down.

30. (4 marks)A 600kg car, heading north at 15.0ms-1 collided head-on with a500kg 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.

31. ( 5 marks)For the same collision described in Q30:a) Calculate the change in Momentum of the north-bound car.b) Given that the collision occurred in a time of 0.200s, find theaverage force that acted on the north-bound car.

32. (6 marks)For the same collision described in Q30 (again!):a) Calculate the total Kinetic Energy of both cars combinedbefore the collision.(Ignore directions... energy is a scalar, remember)b) Calculate the Kinetic Energy of the combined wreckage afterthe collision. (use your answer to Q30)c) Explain any difference in the amount of energy before andafter the collision.

33. ( 3 marks)State Newton’s 1st Law of Motion, and use it to explain why anunrestrained passenger may go through a car windscreen during acollision.

34. (3 marks)One of the important safety features of modern motor vehicles isthe “crumple zone” built into the front and rear.a) Describe what happens to this “crumple zone” in a collision.b) Explain how this reduces the forces which act on people in thecar during a collision.

35. (4 marks)a) Explain, with reference to how velocity contributes to kineticenergy, why government agencies might seek ways to slow trafficdown.b) List 2 strategies that local governments use to force traffic toslow down.

30





750kg 400kg

a = 1.50ms-22 Friction = -2200N



31

Answer Section

Worksheet 1 Part Aa) distance b) timec) gradient d) stationary, not movinge) horizontal line f) zero on the speed scaleg) scalar h) vectori) magnitude and 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”

Part B1.a) 200+100 = 300kmb) +200 + (-100) = 100km northc) 5hrd) Speed = dist/time = 300/5 = 60km/hre) V = S/t

= 100/5 = 20km/hrf) graph

g) from graph:i) gradient = 200/3

≅ 67 km/hrii) gradient = zero

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

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

2.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

3.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.

4.a) i) A 40km/hr B zero C 60km/hr D -50km/hr.

ii) Using S = V.t in each case,A = 40x0.4 = 16km B zeroC = 60x0.2 = 12km D = -50x0.8 = -40km

b) graphc) i) 16+12+40

= 68kmii) Sp = 68/2

= 34km/hriii) S = 16+12-40

= -12kmiv) V = S/t

= -12/2= -6km/hr

(i.e. 6km/hr west)

5.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

6.1st 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 2Part Aa) 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) centripitalal) centre am) tangent



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Worksheet 2 (cont)Part B Practice Problems1. 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 = 6.00s4. v = u + at = 850 + (-50.0)x20.0 = -150ms-1

The final negative velocity means it is moving backwards,compared to its original direction.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 continues to moveaway from the start, so the Displacement-Time graph nevershows a negative gradient.

Newtons 2nd Law6. F = ma = 600x2.65 = 1,590 = 1.59x103N.7. F = ma = (120+60)x4.50 = 810 = 8.10x102N.8. F=ma, so a=F/m = 500/3,500 = 0.1428...= 1.43x10-1N.9. m = F/a = 1.25x103/3.20 = 390.6... = 391kg (3.91x102kg)10. 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)11. 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.Mass & Weight12. 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 + 5,000kg capsuleremain. 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.13. a) W=mg = 0.250x10 = 2.5N.b) 750g = 0.750kg.c) a=F/m = 2.5/0.750 = 3.3ms-2.14. 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.Vector Analysis15. R2 = 402 + 252 Tan φ = 25/40

∴ R = sq.root(2,225) φ = 32o

= 47NResultant = 47N at 32o bearing

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

= 32

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

17. R2 = 2002 + 202 Tan φ = 20/200R = sq.root(40,400) φ = 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.

18.R2 = 1502 + 2002

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

Tan φ = 200/150φ = 53o

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

19. Not accelerating means there is NO net force, The 3 forcesmust be in equilibriumF2 = 5.252 + 3.852 Tan φ = 5.25/3.85F = sq.root(42.385) φ = 54o

= 6.513rd Force = 6.51N, 54o S of E (bearing 144o).

Friction, Tension & Turning Corners20. 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)

21. 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 and accelerate thevan at 2.90ms-2. So T=ma +450 = 300x2.90 + 450 = 1.32x103N.

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

23.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.Centripital force needed: F=mv2/R = 1,000x502/70

= 35,714NSince the maximum grip of the tyres is only 18,000N, the tyrescannot provide the force needed to to turn this corner... car will“spin out”.



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Worksheet 2 Part B (cont.)24.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 3Part Aa) 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

Part B Practice Problems1.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.ΔEk = 0.5m(v2 - u2) = 0.5x600x(30.02-12.52)

= 2.23x105 J.4.

ΔEk = 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 = ΔEk)c) ΔEk = 0.5mv2 - 0.5mu2

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

v = sq.root(200) = 14.1ms-1.7.a) ΔEk = 0.5mv2 - 0.5mu2 = 0.5x10,000x(8.502 - 25.02)

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

8.a) F=ma, a = F/m = 4,000/95 = 42.1ms-2.b) a = (v - u)/t, v = u + at = 0 + 42.1x5.20 = 219ms-1.c) ΔEk = 0.5mv2 - 0.5mu2 = 0.5x95x2192 - 0 = 2.28x106 J.d) ΔEk = Work = F.S, so F = W/S = 2.28x106/.0.05

= 4.6x107N.

Worksheet 4Part Aa) mass b) velocityc) vector d) kg.ms-1

e) conserved f) 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) Impluse o) timep) newton-seconds (N.s)Part B Practice Problems1.a) ρ = mv = 120x5.25 = 630kgms-1.b) ρ = mv = 480x22.5 = 10,800 = 1.08x104kgms-1.c) ρ = mv = 9,500x32.0 = 304,000 = 3.04x105kgms-1.2.ρ = mv, so v = ρ/m = 1.15x104/750 = 15.3ms-1.3. v = 80.0/3.6 = 22.2ms-1

ρ = mv, so m = ρ/v = 1.4x105/22.2 = 6.31x103kg.4. u = 90.0/3.6 = 25.0ms-1. v = 50.0/3.6 = 13.9ms-1

Δρ = mv - mu = 6.31x103x(13.9-25.0) = -700(negative, because lost momentum) = -7.00x102kgms-1.5.motorcycle: ρ = mv = 180x35.0 = 6.30x103kgms-1 north.car: ρ = mv = 630x10.0 = 6.30x103kgms-1 south.Comparison: both vehicles have the same magnitude ofmomentum, but in opposite directions.(Remember, momentum is a vector)Conservation of Momentum6. ρi = ρf

mA.uA + mB.uB = mA.vA + mB.vBSince the cars lock together, their final velocity is the same.

600x27.0 + 1,500x0 = (600+1,500)xV2,100V = 16,200

v = 7.71ms-1.7. ρi = ρf

mA.uA + mB.uB = mA.vA + mB.vB700x24.5 + 700x8.50 = 0 + 700x VB

VB = (17,150+5,950)/700= 33.0ms-1.

8. ρi = ρfmA.uA + mB.uB = mA.vA + mB.vB

(let north be +ve, south -ve)mAx15.0 - 900x35.0 = mAx6.25 + 900x6.25

8.75xmA = 31,500 + 5,625mA = 37,125/8.75 = 4.24x103kg.

9.a) If they had equal magnitudes of momentum, but oppositedirections, then the sum of their momentum = zero.b) To have equal magnitudes of momentum, the product MxVmust be the same for each (ignoring direction).If one has twice the mass, the other must have twice thevelocity.





34

Worksheet 4 Part B (cont.)Impulse & Momentum10.a) I = F.t = 20x4.0 = 80Ns.b) I = F.t = 150x60 = 9x103Ns.c) I = F.t = 900x22.5 = 20,250 = 2.03x104Ns.11.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.12.a) Δρ = 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.13.Δρ = m(v - u) = 850(0 - 13.9) = -11,815 = -1.18x104kgms-1.(negative because it lost momentum)Change in momentum = Impulse = F.t

t=I/F = -1.18x104/-3,900 (negative force, opposing motion)= 3.03s.

14.a) Δρ = m(v - u) = 750x(8.50 - 0) = 6,375 = 6.38x103kgms-1.b) Δρ = Impulse = F.t, so F = I/t = 6.38x103/0.350

= 1.82x104N.c) Momentum is conserved, so momentum gained by one equalsmomentum lost by by the other.So momentum lost by the other vehicle = 6.38x103kgms-1.d) F = -1.82x104N (by Newton’s 3rd Law)e) Δρ = 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 5a) do not actb) continue moving in a straight line, with constant velocityc) 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 af) chicanes

Practice QuestionsPart A Multiple Choice1. C 5. A 9. A 13. A 17. D2. B 6. D 10. B 14. C 18. D3. C 7. B 11. D 15. C 19. B4. D 8. C 12. B 16. A 20. A

Part B Longer Response QuestionsIn some cases there may be more than one correct answer. Thefollowing “model” answers are correct but not necessarily perfect.21.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 φ = 100/150

φ = 34o

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

c) v = S/t =180/3.00 = 60km/hr, bearing 326o.22.

23.a) Forces in equilibrium means the vector diagram must “close” sothere is no resultant.

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

24.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.



Remember that for full marksin calculations, you need to show

FORMULA, NUMERICAL SUBSTITUTION,APPROPRIATE PRECISION and UNITS

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Answers to Practice Questions (cont.)25.a)

b) Graph shows a directrelationship between force andacceleration.

c) gradient = force/acceleration= 3.0/2.5= 1.2

Trolley is approx. 1.2kg.

26.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 AND overcome thefriction. T = ma + 200 = 400x1.50 + 200 = 800N.

27.a) on diagram

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

28.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.

29.a) Work = change in kinetic energy

F.S = 0.5m(v2 - u2) = 0.5x600x(8.502 - 25.02) = -165,825 JF = -165,825/50.0 = -3.32x103N.(negative force, because opposing the motion)

b) It means that energy cannot be created or destroyed... it neverdisappears or ceases to exist. It simply gets transformed fromone type of energy to another.c) The car lost kinetic energy, but this energy didn’t disappear...it was transformed, mainly into heat, by the brakes.

30. ρi = ρfmA.uA + mB.uB = mA.vA + mB.vB

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

31.a) Δρ = m(v - u) = 600x(3.64 - 15) = -6.82x103kgms-1.(negative, because the change in momentum was southward, or aloss of northward momentum)b) Δρ = Impulse = F.t = -6.82x103

∴ F = -6.82x103/0.200 = -3.41x104N.

32.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 beentransformed into the sound and heat of the collision, but mosthas been used to distort and damage the vehicles.

33.1st Law: A moving object will continue to move in a straight lineat a constant velocity unless acted upon by a net force. If at rest itwill remain at rest unless a force acts on it.In a collision in which a vehicle stops suddenly, an unrestrainedpassenger will continue to move according to 1st Law, and may gothrough the windscreen.

34.a) The car body is designed so that it collapses, one section afteranother, and crumples in like a concertina.b) This extends the time over which the car loses its momentum.Since change of momentum = Impulse = Force x time, then forany given amount of momentum, increasing the time involvedreduces the force acting on the people in the vehicle, anddecreases the risk of injury or death.

35.a) Kinetic energy depends upon both mass and velocity, butvelocity has the biggest contribution, since Ek = 0.5mv2. Thismeans doubling the velocity increases the energy by a factor of 4.Since velocity is so important, it means that reducing speeds cangreatly reduce the energy involved in vehicle accidents, and reducethe incidence of death and injury.b) Speed humps

ChicanesLow speed zones in residential areas, and around schools.



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topicsome students find

newtons 3rd law

newtons 1st law

newtons 2nd law

original kinetic energy