Moving About by Chris Rusnak

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St Johns Park High School SL#802444

Preliminary Physics Topic 3

MOVING ABOUT What 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 ACCELERATION 3. WORK and KINETIC ENERGY 4. MOMENTUM and IMPULSE 5. SAFETY DEVICES in VEHICLES

...all in the context of moving vehicles.

but first, let’s revise... WHAT IS SPEED?

WHAT IS ENERGY?

“Speed” refers to how fast you are going. You will already know that mathematically:

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.

SPEED = distance travelled time taken In this topic, you will extend your understanding of speed to include VELOCITY, which is just a special case of speed.

In this topic the most important energy form you will study is the one associated with moving vehicles...

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

KINETIC ENERGY

Some forces, like gravity and electric/magnetic fields, can exert forces without actually touching things. In this topic you will deal mainly with CONTACT FORCES, which push or pull objects by direct contact.

WHAT MAKES A CAR GO? Overview of Topic:

ENGINE provides ENERGY (from chemical energy in petrol)

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

FORCE causes

In the context of moving vehicles, the most important force is FRICTION. Friction allows a car’s tyres to grip the road to get moving, and for the brakes to stop it again. Without friction the car couldn’t get going, and couldn’t stop if it did! Preliminary Physics Topic 3 copyright © 2005-2007 keep it simple science

ACCELERATION

VELOCITY changes 1

FORCE acts over a distance... “WORK” done

KINETIC ENERGY changes

MOMENTUM changes

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

Displacement-T Time

Net Force

Vector Diagrams

Instantaneous Velocity Motion Graphs

Average Speed or Velocity

Friction a=v-u t

Concept of Force

Measuring Motion

v=S t

F = ma Acceleration

Vectors & Scalars

Newton’s 2nd Law

Force & Acceleration

Speed & Velocity

MOVING ABOUT

Centripital Force

F = m v2 r

Work & Kinetic Energy

Energy Transformations

Safety Devices in Vehicles

Momentum & Impulse

Conservation of Momentum in Collisions

Physics of Safety devices

Mass & Weight

Inertia & Newton’s 1st Law •safety belts •crumple zones •air bags

Equivalence of Work & Energy

Momentum

Impulse of a Force

Law of Conservation of Energy W = F.s

p = mv

Ek = 1 mv2 2

Newton’s 3rd Law I = F.t 2

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1. SPEED & VELOCITY Average Speed for a Journey If you travelled in a car a distance of 300 km in exactly 4 hours, then your “average speed” was: average speed

distance travelled time taken

300 4

75 km/hr (km.hr-1)

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.

Distance-Time Graphs

Speed-Time Graphs

Perhaps your journey was as shown by this graph.

The same journey could also be represented by a different graph, showing the SPEED at different times:

Start at the bottom-left of the graph and consider each section A, B, C and D. Graph section D Distance-Time Graph Travelled 150km in 1.5 hr: gradient = distance time Speed = 100 km/hr 300

Study this graph carefully and compared it with the other...

2 3 TIME (hours)

100 4

Stopped. Speed scale reads zero. 2 TIME (hr)

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

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

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

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

SPEED-TIME GRAPHS show the SPEED of a moving object at each TIME.

DISTANCE-TIME GRAPHS show the DISTANCE (from the starting point) at each TIME. The GRADIENT at any point equals INSTANTANEOUS SPEED.

The speed at any time can be read from the vertical scale of the graph. A horizontal section means that the object was moving at constant speed.

B 0

So although the average speed for the entire journey was 75km/hr, in fact you never actually moved at that speed.

“Flat” parts DO NOT mean stopped, but mean constant speed

gradient = zero i.e. stopped

SPEED (km/hr) 40 60 80

DISTANCE TRAVELLED (km) 100 150 200 250

These graphs represent the same journey

C B

Graph section A Travelled 100 km in 1.0 hour: Speed =100 km/hr

Graph section B Zero distance moved in 0.5 hr: Speed= zero.

= speed

Graph section C Travelled 50 km in 1.0 hr: Speed=50 km/hr

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

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Scalars & Vectors A Scalar quantity is something that has a size (magnitude) but no particular direction. A Vector quantity has both size (magnitude) AND DIRECTION.

BUT, consider the “NET journey”: at the end of the journey 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.

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

Notice that both displacement and velocity have a direction (“east”) specified.... they are VECTORS!

Now you must learn the vector equivalents: “Displacement” = distance in a given direction, and “Velocity” = speed in a given direction. Consider this journey: drove 60 km EAST in 1 hour START

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

then

drove 30 km WEST in 0.5 hour.

Note: The symbol “S” is used for Displacement

Average = Displacement Velocity time

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

Vav = S t

TRY THE WORKSHEET at the end of this section

MORE GRAPHS... Displacement - Time

...and the corresponding Velocity

Refer back to the Distance -Time graph on previous page.

south -5 50

ive sit tp o

Back at starting point. (Displacement = 0 ) 1

2 TIME (hours)

In vector terms; displacement north is positive (+) displacement south is negative ( - ) In section D: displacement = -150 km (south) velocity = displacement time = -150 /1.5 = -100 km/hr

(i.e. 100km/hr southward)

-1 100

ati

ad ien

neg

Displacement NORTH (km) 0 50 100

nt die Gra

Velocity (km/hr)

150

Downsloping line means travelling SOUTH

The Displacement - Time Graph would be:

Positive values mean north-bound velocity

north 50

100

What if the 300km journey had been 150 km north (graph sections A, B & C) then 150 km south (section D)?

- Time Graph:

TIME (hrs)

B 1 Zero velocity: means stopped

Negative value: south-bound velocity D

The velocity values for each part of this graph are equal to the gradients of the corresponding parts of the Displacement - Time Graph. 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 information the “average” gives you. The instant-by-instant Physics of the journey is in the graph details. 4

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Prac Work: Measuring Motion You will probably experience one or more of these commonly used ways to measure motion in the laboratory. • “Ticker-timer”: a moving object drags a paper strip on which dots get printed (usually every 0.02 second) as it goes. The gap between dots is a record of displacement and time. This allows you to calculate the velocity over every 0.02 s. It’s still an average, but over such small time intervals it approximates the instantaneous velocity. Every time the hammer hits the moving strip of paper it leaves a dot. The string of dots can be analysed to study the motion of the trolley.

• Tape Measure & Stopwatch. The simplest method of all: measure the distance or displacement involved, and the time taken. Then use speed (velocity) = distance (displacement) time However, this can only give you the AVERAGE speed or velocity. In Physics we often need to consider INSTANTANEOUS velocity.

Moving lab. trolley drags a strip of paper behind it

• Electronic or Computer Timing devices use either “Light Gates” or a “SONAR” device to record displacements and times for you. Once again, any velocities calculated are averages, but the time intervals are so short (e.g. as small as 0.001 s) that the velocity calculated is essentially instantaneous.

“Ticker-ttimer” device has a small hammer which vibrates up and down every 0.02 sec.

Worksheet 1 Part A Fill in the blanks. Check answers at the back The average speed of a moving object is equal to the a)................................... travelled, divided by b)....................... taken. On a Distance-Time graph, the c)................................ of the graph is equal to speed. A horizontal graph means d)................................. .................................................

The vector equivalent of distance is called j)......................................, and refers to distance in a particular k).......................................... For example, if displacement was being measured in the north direction, then a distance southward would be considered as l)............................... displacement. On a displacement-time graph, movement south would result in the graph sloping m)................................... to the right and having a negative n)............................................

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

The vector equivalent of speed is o).............................................. The average velocity is equal to p)........................................ divided by q)...................................... Instantaneous velocity refers to r)..................................................................

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

Laboratory methods for measuring motion include: • using a tape measure and stopwatch. This allows calculation of s)................................................ only. • “Ticker-timers” record both t)............................... and .......................... on a paper tape. Average velocity can be calculated for short time intervals which are approximately equal to u)..................................................... velocity. • Electronic or Computer-based devices often use v)........................ or ..................................................... to gather displacement, time and velocity data at very short time intervals.

COMPLETED WORKSHEETS BECOME SECTION SUMMARIES

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Worksheet 1 Part 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.

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

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.

b) The driver has a “micro-sleep” for 5.00 s. How far will the car travel in this time? c) At this velocity, how long does it take (in seconds) to travel 1.00km (1,000m)? 4. The Velocity - Time Graph shows a journey in an east-west line, by motorcycle. 60 40

(km/hr) East

Time (hr)

B 1.0

2.0

-2 20 -4 40

VELOCITY

200

400

-6 60

West

600

Displacement north (km)

800

2. An aircraft took off from town P and flew due north to town Q where it stopped to re-fuel. It then flew due south to town R. The trip is summarized by the following graph.

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.

Time (hr)

P -4 400 -2 200

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

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 from t=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 (including direction) at t=5 hr? h) What was the: i) total distance ) ii) average speed ) over the whole iii) total displacement ) 6 hours? iv) average velocity ) i) Construct a Velocity- Time Graph for this flight.

c) Find

UNITS OF MEASUREMENT So far all examples have used the familiar km/hr for speed or velocity. The correct S.I. units are metres per second (ms-1). You need to be able to work in both, and convert from one to the other..... here’s how:

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

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 car travelling 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?

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.

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 3.6 -1 to convert ms km/hr multiply by 3.6 Preliminary Physics Topic 3 copyright © 2005-2007 keep it simple science

i) total distance covered ii) average speed iii) total displacement iv) average velocity for the 2 hour trip.

FULLY WORKED SOLUTIONS IN THE ANSWERS SECTION 6

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2. FORCE & ACCELERATION Graphs of Accelerating Vehicles

Change of Velocity = Acceleration

You may have done laboratory work to study the motion of an accelerating trolley. If you used a “Ticker-timer”, the paper tape records would look something like this:

Vehicles 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 to a change in a quantity

Tape of trolley accelerating... dots get further apart

v = final velocity u = initial velocity t = time involved

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

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

The graphs that result from acceleration are as follows:

ms-2:

Explanation: Imagine a car that accelerates at 1 Start 1 sec. later 1 sec.later 1sec.later v=2 ms-1 v=3ms-1 v =0 v = 1 ms-1

Remember, Gradient equals Velocity

Displacement

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.

+ -

THIS CAR IS SLOWING DOWN... DECELERATING

ACCELERATION VECTOR

Velocity decreasing

Velocity

Velocity = 0 ∴ Stopped!

Time

v = u + at = 25.0 + (-1.50) x 12.0 = 25.0 - 18.0 = 7.00 ms-1.

Gradient positive

Gradient negative

On a Velocity-T Time Graph Gradient = Acceleration

A common error is to think that this means the object is moving backwards. Wrong! It is moving forward, but slowing down.

g in at ler ce De

Velocity increasing

Example Problem 2 A car moving at 25.0 ms-1 applied its brakes producing an acceleration of -1.50 ms-2 (i.e. deceleration) lasting for 12.0 s. What was its final velocity? so

Gradients increasing (curve gets steeper)

Constant Velocity

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

a = v - u, t

Gradient constant (straight line)

VELOCITY-T TIME GRAPH

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

Solution:

ing rat ele c De

Time

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

Solution:

Gradients decreasing (curve flattens out)

Ac ce ler at in g

VELOCITY VECTOR

DISPLACEMENT-T TIME GRAPH

Co Ve nsta loc nt ity

a=v-u t

Δ (Greek letter “delta”) refers

Ac ce le ra tin g

a = Δv Δt

Tape of trolley moving at constant velocity (for comparison)

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Force Causes Acceleration

PRESSING ON THE ACCELERATOR...

A simple definition of “Force” is that it is when something pushes or pulls on something else. However, in the context of moving vehicles,

Vertical forces are balanced, and cancel

Weight

Thrust Force Increases

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)

Friction & Air Resistance

Horizontal forces are UNBALANCED

This car will SPEED UP

Reaction Force

To actually result in a change of velocity, the force must be and

External

TURNING THE STEERING WHEEL...

Unbalanced (Net) Force

Vertical forces are balanced, and cancel

For example, if you were inside a moving car and kicked the dashboard, this force would have NO EFFECT on the car’s motion... This is an “Internal Force” and cannot cause acceleration.

Weight

Forward & Back Forces are balanced and cancel

Thrust Force from Engine Sideways Forces become UNBALANCED (These would be equal if wheel not turned)

WEIGHT FORCE Car pushes on Earth Friction and Air Resistance

Thrust from Engine

REACTION FORCE Earth pushes back

Friction & Air Resistance

Reaction Force

This car will turn a corner at constant speed (but new velocity since new direction)

GOING UP A HILL (without pushing harder on accelerator)

BALANCED & UNBALANCED FORCES

Weight (still vertical)

The car above has a number of forces acting on it, but they are BALANCED... those acting in the same line are equal and opposite, and cancel each other out. This car will not alter its velocity or direction; it will not accelerate. It is either travelling at a constant velocity, or it is stationary. EXAMPLES OF BALANCED UNBALANCED FORCE FORCES SITUATION

Friction still the same

Engine Thrust still the same

Part of the Weight Force acts to cancel some of the thrust

Reaction Force is not vertical, and no longer cancels the weight completely... UNBALANCED

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

PASSING OVER AN ICY PATCH ON THE ROAD Opposite Forces are BALANCED and cancel

Weight

PRESSING ON THE BRAKES... Vertical forces are balanced, and cancel

Virtually no Thrust Force because tyres can’t grip on ice This car will continue in a straight line, at a constant velocity... whether the driver wants to or not...

Virtually

Thrust Force decreases as

no Friction on Ice

Friction Increases as Brakes are applied

accelerator is released

Reaction Force cancels Weight

Horizontal forces are UNBALANCED

Car is out of control; Can’t stop... Can’t turn...

Weight

Reaction Force

This car will SLOW DOWN (Decelerate)

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Newton’s 2nd Law of Motion

Mass & Weight

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

“Mass” is a measure of the amount of matter in an object. In terms of force and acceleration, mass is the stuff that tries to prevent acceleration... the more mass there is, the less acceleration an applied force will produce. The mass of an object is the same wherever it might be.

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

“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.).

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 m

F = m a, so Weight = mg For example, consider an astronaut with a mass of 80kg Mass (kg) Weight (N) Astronaut on Earth 80 800 Astronaut in orbit 80 zero Astronaut on Moon 80 133

F=ma

UNCHANGED

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

Verifying 2nd Law You may have done laboratory work similar to this: Extra masses on trolley

Power Pack

The acceleration of the trolley is determined by analysing the displacement & time data from the ticker tape record. This is repeated several times, tranferring some of the “extra” masses from trolley to the hanging weight each time.

Force v Acceleration Graph

Final Results and Conclusions • Within experimental error, the graph shows a straight line. This proves there is a direct relationship between the force applied, and the acceleration produced. • The gradient of the graph will be found to be equal to the mass of the total system (i.e. trolley + masses) in kilograms: Force = Mass Acceleration

Force (weight on string) (N)

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

FI T”

Weight = mg Weight on string causes trolley to accelerate

“B ES T

Lab.Trolley

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

O F

Paper tape

LI N E

Tickertimer device

Gradient =

CHANGES

Study this information to get the idea. The confusion about mass and weight has been caused by the unfortunate choice by society to talk about the “weight” of things, but then measure it in kg... it should be in N!!

Units:

Find Gradient of line

F =m a and therefore,

Acceleration

F=ma

(ms-22)

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Vector Analysis

Vectors in Equilibrium

Force is a vector quantity, the same as velocity and acceleration. To fully describe a force, you must state the direction of the force.

It is often the case that 2 or more vectors might all cancel each other out so the “resultant” is zero. In fact this is always the case when a vehicle is moving in a straight line with a constant velocity.

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

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

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

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

Force B 30N west

Force A 20N east

“Lift” Force (on wings)

Air Resistance “Drag”

The sum of these 2 vectors is a single force: “Resultant” 10N west

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

The vector diagram for this plane must be:

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)

Thrust

Lift

However, if the forces are acting in totally different directions, the problem is more complicated. Force B 30N south

Example: Force A 20N east A = 20

nt lta su Re

B = 30

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

First, sketch these vectors “head-totail”.

You may have done laboratory work to measure some vectors and their sum. A common experiment is shown in the photo: Three Force Vectors in Equilibrium

Next, connect the beginning to the end, to from a rightangled triangle.

202

+ = + = 400 + 900 = 1300 ∴R = Sq.Root(1300) ≅ 36N (approximately)

B C

302

F=mg

C A

The angles between strings A, B & C need to be measured.

The 3 vectors can then be analysed

The vectors can be analysed either by accurate scale drawing, or by mathematics (e.g. Sine Rule in triangle).

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

Tension Forces in strings A & B measured by Spring Balances

Use Pythagorus’s Theorem to find the size of the “Resultant” force: A2

Weight

Drag

The 3rd side is the “Resultant” vector.

R2 =

Thrust from engine

Weight Force

It will be found (within experimental error) that these vectos add to zero. They are in equilibrium. www.keepitsimplescience.com.au

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More Examples of Vectors So far, all the examples given of vectors have been forces. Vector analysis could involve any vector quantity, of course... displacement, velocity, acceleration or force. Displacement Vectors

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

Velocity Vectors

Where is it in relation to its starting point?

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? nt ulta Reslocity Ve

200km

5.0 100km

Di Res sp ul lac tan em t en t

R2 = 2002 + 1002 = 50,000 ∴ R = Sq.Root (50,000) = 224 km

1.8

R2 = 5.02 + 1.82 = 28.24 ∴ R = Sq.Root (28.24) = 5.3ms-11

Tan φ = 100/200 = 0.500 ∴ φ ≅ 27o

Tan φ = 1.8/5.0 = 0.36 ∴ φ ≅ 20o (this angle is 70o clockwise from north, ∴ bearing = 70o)

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

27o

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

A Special Force: Friction Often 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: acceleration

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?

Thrust Force

Solution: The net, unbalanced force causes acceleration.

This net force is the vector sum of all forces acting: Net Force = Engine thrust + Friction 1,250 = 1,700 + FF FF = 1,250 - 1,700 ∴ Friction = -450N (the negative value simply means that friction is in the opposite direction to the car’s motion)

Friction Force

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Two Masses Hanging on Strings

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 that it pulls in both directions at once.

Tension force in top string must hold up both weights, so T1 = (5+2) = 7N Tension force T1 simultaneously pulls down on the top support (assumed immovable) and pulls the round object upwards.

Consider these 2 examples:

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

Tension forces acts in both directions in each string

Tension force T2 simultaneously pulls down on the round object and pulls the rectangular object upwards.

T1 Object 1 Weight Force

500g (0.5kg) T2

Nothing is moving, so all forces must be in equilibrium. i.e. Net Force = zero, but can we prove it?

mg = 0.5 x 10 = 5N

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

Round Object Force T1 is pulling it upwards, while its weight and T2 pull it downwards.

weight = mg

Acceleration due to gravity has been taken as 10ms-2 for simplicity (KISS Principle)

T2 200g (0.2kg)

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

Object 2 Weight Force mg = 0.2 x 10 = 2N

ΣF = T2 + mg = (+2) + (-2 2) = zero

It all works! If you undertstand the tension forces acting, you can explain that this system is not moving because the net forces add up to zero.

weight = mg

Example 2

Rectangular Object Force T2 pulls it up, while its weight pulls downwards.

Tension Under Acceleration

Engine 20,000kg

Carriage 5,000kg

This train engine produces 35,000N net thrust force. Problem a) What is the acceleration? b) What tension force acts in the coupling between engine and carriage? c) What is the net force acting on the engine alone? Coupling

Solution: The net force must accelerate the entire mass of 25,000kg. a) F= m a ∴ a= F/m = 35,000/25,000 = 1.4 ms-22

b) Tension in coupling must cause the carriage to accelerate. F=ma = 5,000 x 1.4 = 7,000N

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

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

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Turning Corners - Circular Motion

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

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

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

To keep things as simple as possible (K.I.S.S.) let’s assume that all the corners being turned are circular. The velocity vectors at any instant are tangents to the circle. V V V

F Path of a vehicle turning a circular corner

Wheel turned

Centripital Force

The force causing the turning is always toward the centre of the circle. This is called “Centripital” force

Instantaneous Velocity vector

Even though the speed may be constant, the vehicle is constantly accelerating because its direction is constantly changing.

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

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

If the centripital force required for a particular corner exceeds the friction “grip” of the tyres, then the vehicle will not make it, and may “spin out” and crash.

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

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 velocity vector is constantly changing, but at any instant it is a tangent to the circle, and therefore, at right angles to the acceleration and force vectors. Centripital Acceleration Centripital Force

ac = v2 R

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

Fc = m v2 R

Fc = m v2 = 900 x 302 / 50 R = 16,200N

Solution

Where R = radius of the circle (in metres) V = instantaneous velocity (or “orbital speed”)(ms-1) m = mass of vehicle (in kg) Example Problem 2 The maximum frictional force possible from each tyre of this 750kg car is 5,000N.

Solution: Max. Force possible from 4 tyres = 4x5,000 =20,000N Centripital Force cannot exceed this value. Fc = m v2 / R,

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

Path car will take

so v2 = FcR /m = 20,000 x 40 /750 v2 = 1067 ∴ v = Sq.Root(1067) ≅ 33ms-1 (This is almost 120 km/hr)

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Worksheet 2

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COMPLETED WORKSHEETS BECOME SECTION SUMMARIES Part B Practice Problems

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

Acceleration Problems 1. Starting from rest (i.e. u=0) a car reaches 22.5 ms-1 in 8.20 s. What is the rate of acceleration? 2. A truck decelerated at -2.60 ms-2. It came to a stop (v = 0) in 4.80 s. How fast was it going when the brakes were applied? 3. A car was travelling at 12.0ms-1. How long would it take for it to reach 22.5ms-1, if it accelerated at 1.75ms-2?

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

4. A spacecraft was travelling in space at 850ms-1 when its “retro rockets” began to fire, producing a constant deceleration of 50.0ms-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.

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

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

Velocity (ms-11) 40 60

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 arithmetic way if they act x)................................................................................... If vectors are in different directions, they must be added using a vector diagram (in which vectors are joined y)...................... to ...................................). This diagram can then be analysed mathematically using z)....................................... and/or trigonometry to find the “aa)........................................” vector. The complete answer must contain both the magnitude and ab)............................................ of the resultant.

Time (s) a) Find the rate of acceleration of the racer.

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

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?

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

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.

When a vehicle turns a corner it is accelerating, because the aj)............................................ keeps changing. The force causing this is called ak)............................................ force, and is directed at the al)..................................... of a circle. The instantaneous velocity vector is a am)...................................... to the circle. Preliminary Physics Topic 3 copyright © 2005-2007 keep it simple science

Worksheet continued next page...

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Worksheet 2 Part B

Vector Analysis 15. Find the resultant force, if a 25N force pushes eastward, and a 40N force pushes northward. (Remember to find magnitude AND direction)

Practice Problems (continued)

Newton’s 2nd Law 6. What force is required to cause a 600kg car to accelerate at 2.65ms-2?

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

7. A 120kg motorcycle and its 60kg rider are accelerating at 4.50ms-2. What net force must be acting? 8. A 500N force acts on a truck with mass 3,500kg. What acceleration is produced?

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 is the plane’s true velocity?

9. What is the mass of a vehicle which accelerates at 3.20ms-2 when a force of 1.25x103N acts on it?

18. A ship sailed 300km due east, then 200km due south, then 150km west. Where is it in relation to the starting point?

10. A truck with mass 8.00x103kg is travelling at 22.5ms-1 when the 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?

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

11. A 60kg cyclist exerts a net force of 100N pedalling his 15kg bike 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 in the 10s? c) What is the final velocity in km/hr?

Friction, Tension and Turning Corners 20. The engine of an 850kg car is producing a “thrust” force of 2.25x103N. 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 to produce 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 a frictional force of magnitude 450N. (ie total 900N) What acceleration is achieved when the engine produces the force calculated in (a)? d) What tension force acts in the tow-bar? (Hint: Tension must overcome the friction on the van AND cause acceleration... careful!)

Mass and Weight 12. A space capsule, ready for launch has a mass of 25,000kg. Of this, 80% is fuel. By the time it reaches Earth orbit it has burned three-quarters of the fuel. Later, it proceeds to the Moon and lands, 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?

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

13. In a laboratory experiment, a 500g trolley is attached by a string to 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?

23. a) The maximum “grip” force of each tyre on a 1,000kg car is 4,500N. What is the tightest turn (in terms of radius of curve) the car can negotiate at 90 km/hr? (Hint: velocity units?) b) The same car comes to a curve with double this radius, (ie a much gentler curve) but it is travelling at double the speed. Can it make it?

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

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

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3. WORK & KINETIC ENERGY The Concept of “Work”

Energy of a Moving Vehicle

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

You will be already aware that any moving object possesses “Kinetic Energy”. The “bigger” the object, and the faster it moves, the more energy it has.

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

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

Ek = 1 mv2 2

Kinetic Energy

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

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

From this equation you would expect that the units of work 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...

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

Work & Energy are Equivalent WORK = ENERGY This means, for example, if a vehicle’s engine exerts a FORCE, we can now calculate the effects of the force in various ways:

Effect of Mass & Velocity on Kinetic Energy Some simple example calculations can make an important point:

Initial velocity u=0

Mass 1,000kg

F = 1,000N

Velocity 10ms-11

Force from Engine acts this way

Calculation 1 How much Ek does this vehicle have? Ek = 0.5mv2 Calculation 2 What if you double the mass? (same velocity) Ek = 0.5mv2 = 0.5 x 2,000x 102 = 100,000 J (or 100 kJ) So, 2X the mass gives 2X the Kinetic Energy.

Force causes acceleration

= 0.5 x 1,000 x 102 = 50,000 J (or 50 kJ)

m= 500kg Force is applied over a distance of 100m. It takes 10 sec to move this 100m distance.

F = ma 1,000 = 500 x a ∴ a = 2.0 ms-22 The acceleration goes on for 10s

Calculation 3 What if you double the velocity? (same mass)

v = u + at = 0 + 2 x 10 v = 20 ms-11

Ek = 0.5mv2 = 0.5 x 1,000 x 202 = 200,000 J (or 200 kJ)

Notice how 2 totally different calculations give the same result... ..don’t you just love it when things work?!

So, 2X the velocity gives 4X the Kinetic Energy !!!

W = F.S

Work

Force acting over a distance does work which increases the car’s Kinetic Energy (and velocity) Work, W= F.S = 1,000 x100 = 100,000 J Work = Gain of Ek Done Ek = 0.5 m v2 100,000 =0.5x500x v2 v2 = 400 ∴ v = 20 ms-11

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Work is Done to Slow Down, Too

Energy Transformations

In the previous example, the force applied by the car’s engine, via the gearbox, axles and wheels, was used to increase the car’s Kinetic Energy and velocity.

Energy can be changed from one form into another, and does so frequently. Electricity

Sound

What about when the car slows down? Initial Velocity u = 30 ms-11

BRAKES APPLIED

mass=500kg

Electricity

Final Velocity v = 10 ms-11

We find electricity very useful because it can be easily transformed into many other types of energy.

Displacement = 100m during braking

WHAT FORCE ACTED IN THE BRAKES? Change in Kinetic Energy

Light

= Final Ek - Initial Ek

ΔEk = 0.5mv2 - 0.5mu2 = 0.5x500x102-0 0.5x500x302 = 25,000 - 225,000 ΔEk = -2 200,000 J

Chemical Potential Energy

Heat Light

The Law of Conservation of Energy This is a very grand-sounding title for a very simple concept:

This energy change must equal the WORK DONE by the brakes to slow the car down.

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

W = F.S -2 200,000 = F x 100 ∴ F = -2 2,000N The brakes applied a force of -2,000N Work

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

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

Energy Transformation When Accelerating When the car engine does “Work” to accelerate the car, the energy transformation is: CHEMICAL POTENTIAL ENERGY (in petrol)

You could also calculate the acceleration: F=ma -2 2000 = 500 x a ∴ a = -4 4.0 ms-22

KINETIC ENERGY

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

The negative shows that this is a deceleration.

Energy Transformation When Braking When the brakes do “Work” to slow the car down, the main energy transformation is:

TRY THE WORKSHEET at the end of this section

KINETIC ENERGY

HEAT ENERGY

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Energy Transformations in a Collision When a “bouncy” ball collides with a wall it will bounce off again. A lot of its original Kinetic Energy is “conserved”, meaning that after the collision, it is still in the form of Kinetic Energy. A collision in which 100% of the Ek is conserved is said to be an “Elastic Collision”. True elastic collisions occur only at the atomic level, such as the particles in a gas bouncing off each other. Even a really “bouncy” ball will lose some of its Ek with each bounce, and so is not truly “elastic”. The energy itself is not lost, but transformed into other energy types, such as heat.

The Law of Conservation of Energy demands that the Kinetic Energy of a moving vehicle cannot just disappear when the vehicle collides with something and stops suddenly.

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

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

There is still some Kinetic Energy here, but it be converted into “damage” when he lands!

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

Sound & Heat produced

And remember... double the speed means 4 times as much energy to be converted into death and destruction! As they say, “Speed Kills”.

Vehicles distorted by the energy absorbed

Example Calculation: Energy, Work & Force in a Collision Solution Velocity =140 km/hr = 140/3.6 v = 38.9 ms-1 Kinetic Energy Ek = 0.5m v2 = 0.5 x 820 x 38.92 = 6.20 x 105J This energy is equivalent to the work done on the car, causing damage.

Calculate the force which acted on this car The driver of this 820kg car lost control at 140km/hr and hit a solid rock embankment. The vehicle’s structure was badly distorted. It is estimated that the “work done” on the car was due to a force which acted over a distance of only about 2.50m, in a fraction Photo: Dan Mitchell of a second. Preliminary Physics Topic 3 copyright © 2005-2007 keep it simple science

Work W = F.S 6.20x105 = F x 2.50 ∴ F = 6.20x105/2.50 = 2.48x105N i.e. a force of 248,000N This is equivalent to being underneath a 25 Tonne weight !!

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Worksheet 3 Part A Fill in the blanks. Check answers at the back.

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

Any moving object possesses a).............................. energy. The 2 factors which determine how much energy a moving object has, are its b)............................ and its c)...................................... Their effects are not equal however; if the mass is doubled, then the Ek is d)........................................, but if velocity is doubled then the Ek is e)........................................... Energy is a f)......................................... (vector/scalar) and the unit is the g)................................

4. A 5,500kg truck was travelling at 20.0ms-1, but then slowed down, losing 5.00x105J of kinetic energy as it did so. What was its new velocity? 5. How much work is done in each case? a) A 50N force acts on an object over a distance of 4.5m. b) A 4.0kg mass accelerates at 1.5ms-2, over a displacement of 3.2m. c) Over a 50m distance, a 30N force acts on a 6.0kg mass.

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

6. The engine of a 900kg car provides a force of 1,200N. If this force acts to accelerate the car from rest (u=zero) over a 75.0m displacement, a) how much work is done on the car? b) How much kinetic energy does it gain? c) What is the car’s final velocity? d) Find the acceleration of the car, using F=ma. e) How long did it accelerate for?

The Law of k)..................................... of Energy states that “Energy cannot be l)................................ nor.............................., but can be m).................................................................................. The important energy transformation in an accelerating vehicle is n)............................ ......................................... energy (in the petrol) is converted into o).................................................. energy. When braking, the p)........................................ energy of the vehicle is mostly converted into q)...................................... energy in the brakes. In a collision, most of the Ek possessed by the moving vehicle is used to “do work” and cause r)............................................ to the vehicle and its occupants.

7. A fully laden truck with mass 10,000kg is travelling at 25.0ms-1 when the engine is switched off and it is allowed to “coast” on a level road. Over a distance of 250m it gradually 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, and hence 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 attempt on the World Stupidity Record. When fired, the rocket provided 4,000N of thrust for just 5.20s.

COMPLETED WORKSHEETS BECOME SECTION SUMMARIES

Part B Practice Problems: Kinetic Energy & Work

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, calculate the distance covered during the acceleration.

1. Calculate the Ek possessed by a) 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 factor of 3. By what factor did the Ek increase?

Just after attaining maximum velocity, the bicycle and rider struck 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 bike and rider over this distance.

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4. MOMENTUM & IMPULSE Conservation of Momentum in a Collision

Momentum

Kinetic Energy can only be “conserved” in an elastic collision, which only happens at the atomic scale. In “reallife” vehicle collisions most of the kinetic energy is transformed into heat and distortion to the vehicles.

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

= mass x velocity

Unlike kinetic energy, momentum is always conserved.

=mv

When 2 vehicles collide:

The symbol used for momentum ( Greek letter “rho”.

ρ ) is the

Total Momentum = Total Momentum before Collision after Collision

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

Car “A”

Car “B”

Mass = mA Initial Velocity = uA

Example Compare the Momentum of these Two Vehicles

Total Momentum before collision

Bicycle

= 100 x 1.50 = 150 kgms-1 east

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

Velocity 1.50ms-11 east

Final Velocity = vA

Car 600kg

ρi = mA.uA + mB.uB

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

MASS 100kg

=mv

Mass = mB Initial Velocity = uB

Final Velocity = vB

=mv

= 600 x 25.0 = 15,000 kgms-1 south

Total Momentum after collision

ρf = mA.vA + mB.vB

(Again, the same (+ve) and ( -ve) signs as before need to be assigned for opposite directions) 25.0ms-11, south

Conservation of Momentum

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

Total Momentum = Total Momentum before Collision after Collision

ρi

car 15,000kgms-1 south

ρf

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

The momentum vectors are also in totally different directions. bike 150kgms-1 east

STUDY THE EXAMPLES next page. TRY THE WORKSHEET at the end of this section 20

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Example 3 Collision with an Immovable Object e.g. Rock Cliff Example 1 Collision with a Stationary Vehicle Car “A”

Mass = mA = 500kg Initial Velocity = uA = 20.0ms-1

Car “A” Car “B” Mass = mA = 500kg Initial Velocity = uA = 20.0ms-1

Mass = mB = 750kg Initial Velocity = uB = 0

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

Final Velocity = vB Final Velocity = vA = 0

Car A stops, Car B moves. What is Car B’s velocity?

Momentum must be conserved, so the intitial momentum (10,000kgms-1) still exists. It has been absorbed by the Earth, so the Earth’s rotation has been changed. However, the immense mass of the Earth means that its velocity has been altered by such a tiny amount that it is not noticeable.

ρi = ρf mA.uA + mB.uB = mA.vA + mB.vB 500x20.0 + 750x0 = 500x0 + 750x vB 10,000 + 0 = 0 + 750 vB ∴ vB = 10,000/750 = 13.3ms-1 Car B moves forward at 13.3 ms-1 In every example, the Momentum is conserved. If you calculate the Total Kinetic Energy before and after each collision, you will see that it is NOT conserved in any of the cases. The “missing” energy is used to damage and destroy the vehicles.

Examples of Conservation of Momentum in Collisions

(+ve) and ( -v ve) signs must be assigned

Example 2 Head-on Collision. Vehicles “lock” together. Car “A” east-bound (+ve)

Mass = mA = 500kg Initial Velocity = uA = 20.0ms-1 Cars lock together

Conservation of Momentum often goes against common sense. After a vehicle collision, things usually stop moving almost immediately. This is because of friction acting on damaged vehicles with broken axles dragging on the ground, etc. In the instant after the collision however, the momentum HAS been conserved.

Example 4 Collision with a Vehicle Moving in the Same Direction

Car “A”

Car “B”

Car “B” west-bound ( -ve) Mass = mA = 500kg Initial Velocity = uA = 20.0ms-1

Mass = mB = 750kg Initial Velocity = uB = (-)25.0ms-1

Final Velocity vB = 15.0ms-1

What is Final Velocity?

Car B is jolted forward at new velocity =15.0ms -11 What is Car A’s final velocity?

ρi = ρf mA.uA + mB.uB = mA.vA + mB.vB Since the cars lock together, their final velocity is the same 500x20.0 + 750x (-25.0) = (500 + 750) x v 10,000 - 18,750 = 1250 v ∴ v = -8,750/1250 = -7.00ms-1

ρi = ρf mA.uA + mB.uB = mA.vA + mB.vB 500x20.0 + 750x10.0 = 500.vA + 750x15.0 10,000 + 7,500 = 500vA + 11,250 17,500 - 11,250 = 500vA ∴ vA = 6,250/500 = 12.5ms-1

Both cars move at 7.00ms-1 west

Mass = mB = 750kg Initial Velocity = uB = 10.0ms-1

Car A continues to moves forward, but at slower velocity of 12.5 ms-1 21

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Newton’s 3rd Law of Motion

Impulse of a Force

It was Sir Isaac Newton who figured out WHY momentum must be conserved in a collision. It is because, when one object collides with another, it exerts a force on the other object, and that one pushes back!

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

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

I = F.t

Impulse = Force x Time

If Force is in newtons (N), and time is in seconds (s) Then the units for Impulse will be “newton-s seconds” (N.s)

Newton’s 3rd Law explains quite a few things... Reaction force pushes rocket forward

Action Force pushes on exhaust gasses, accelerating them backwards

So what? Well, study the maths...

Action force pushes bullet

Reaction force kicks back

Start with Newton’s 2nd Law,

F = ma

Now

F = m(v - u) t

a= v-u t

Multiply both sides by “t”

Weight force pushes on Earth,

F.t = m(v - u) F.t = mv - mu

Impulse Change in Momentum

Reaction force pushes back

Impulse = Change in Momentum F.t = mv - mu

... including Conservation of Momentum.

This means that the unit of Impulse (N.s) must be the same as the unit of Momentum (kg.ms-11)

In a collision between moving Car A and stationary Car B

These units are inter-c changeable Action Force A pushes on B

Reaction Force B pushes back on A with equal force

This turns out to be a very useful relationship.

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

Example A car driver applied the brakes for 6.00s and slowed his 800kg vehicle from 25.0ms-1 down to 10.0ms-1.

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

What was the average force applied by the brakes? Solution

and momentum = momentum lost by A gained by B Since the momentum lost by one is equal to that gained by the other, it follows that the total amount of momentum has not changed, and therefore that

The braking force was -2 2,000N Note that the answer is negative, indicating that the force is acting against the motion, causing deceleration.

Impulse = Change in Momentum F.t = mv - mu = m(v - u) F x 6.00 = 800 (10.0 - 25.0) ∴ F = -1 12,000 / 6.00 = -2 2,000N

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Worksheet 4 Part A

Conservation of Momentum 6. A 600kg car is travelling at 27.0ms-1, when it collides with a stationary 1,500kg utility. The vehicles lock together on impact. Find the velocity of the wreckage immediately after impact.

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

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

7. Two identical 700kg cars are travelling in the same direction, but at different speeds. One is moving with a velocity of 24.5ms-1 and fails to notice the other in front doing just 8.50ms-1. The “rearend” collision stops the back car instantly. Find the velocity of the front car immediately after the collision.

In any collision, momentum is e)........................................................ This means that the total momentum before the collision is equal to 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 been h).................................. However, this is because of i).......................... acting on the wreckage. In fact, in the instant following the collision, momentum has been j).........................................................

8. A truck is heading north at 15.0ms-1 when it has a head-on collision 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.

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

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 true about their velocities?

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

Impulse & Momentum 10. 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 100N force be applied? c) How much force is needed to achieve 1240Ns of impulse in a time of 32.5s?

COMPLETED WORKSHEETS BECOME SECTION SUMMARIES 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.

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?

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

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

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

14. In a rear-end collision, the stationary car is jolted forward with a new velocity of 8.50ms-1 in the instant after collision. The car’s mass 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 at 10.5ms-1 before the collision. What was its velocity immediately after collision?

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.

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5. SAFETY DEVICES in VEHICLES Inertia

Newton’s 1st Law of Motion

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

Finally, we get to 1st Law! The 2nd and 3rd Laws are all about the things that happen when forces act, but 1st Law is about what happens when forces DON’T act.

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

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.

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 2nd Law that it is mass that “resists” accelerations... the bigger the mass, the less acceleration occurs. Now we can say that this is because of inertia.

Essentially 1st Law means that, if no net force occurs, then motion cannot change... no acceleration, no change in momentum is possible.

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

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

Sudden Acceleration Forward We feel pressed-back in the seat

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

acceleration

The explanation is, of course, FRICTION and air resistance. In all everyday situations there is always some friction acting against the motion. Engine off... car coasting

of car

Weight force W = mg

Sudden Deceleration We feel thrown forward

retards motion Reaction force equals weight

FORCES BALANCED NO NET FORCE VELOCITY CONSTANT

of car

Inertia!

The ultimate in inertia effects occur in collisions. The unfortunate rider has NOT been “thrown forward”. His inertia has simply kept him in motion after his bike stopped.

Weight force W = mg Friction retards motion Reaction force equals weight

Loose objects seem to fly forwards

In fact, our bodies, and the loose objects, are simply trying to remain in motion, while the car decelerates around us.

deceleration

We are used to the fact that to maintain a constant speed forward, the engine must supply a force. Engine pushing car with force equal to Friction

In fact, our bodies, and the loose objects, are simply trying to stay where they were, while the car accelerates forward. Inertia!

Friction FORCES UNBALANCED NET FORCE CAUSES DECELERATION

Loose objects seem to fly backwards

Bike stops suddenly

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The Physics of Safety Devices

Energy & Momentum in Collisions

In a vehicle collision, most of the injuries to people are caused by inertia.

In 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

Typically, when a car hits something there is a rapid deceleration. The car comes to a sudden halt, but the inertia of the driver and passengers causes them to keep moving forward, with tragic results:

and • Momentum changes as the vehicle changes speed, and this is equal to IMPULSE = Force x time

• 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

The effect of most safety devices is to maximise the time and distance over which these changes occur, because this will minimise the force. Example Calculation 1 In a collision at 50km/hr (approx 14ms-1), a 75kg passenger is brought to rest (on the dashboard) in a time of 0.25s What force acts on the person’s body?

CRUMPLE ZONE in Car Body In collision, the car structure collapses, one section after the other

Solution

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

This distortion absorbs the kinetic energy and increases the time to come to rest

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

SEAT BELTS Seat Belts restrain people, and prevent their inertia from throwing them into the dash, or through the windscreen. The belt has a little “give”, and stretches to increase the time of momentum change... less force acts!

Impulse = Change in Momentum F.t = m(v - u) F x 0.25 = 75 (0 - 14) ∴ F = - 4,200 N Lethal Force!

Solution Safety Devices Increase the Time & Distance of Collision.

Impulse = Change in Momentum F.t = m(v - u) F x 1.25 = 75 (0 - 14) ∴ F = - 840 N Survivable! CONCLUSION

This Decreases the Forces Acting on People

Other Strategies... Reducing Speed Crumple Zones, Seatbelts and Air Bags all help to reduce the effects of a collision. AIR BAGS Air bags are “triggered” by inertia, and set off a chemical explosion that releases a gas to inflate the bag.

Another strategy is to reduce vehicle speed, so that vehicles generally have less Kinetic Energy and less Momentum to lose in a collision. It also gives drivers more time to react to danger and perhaps avoid the collision.

This cushions the person (especially their head) and slows down their change of momentum... less force acts!

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

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Worksheet 5 Fill in the blanks. Check your answers at the back.

In a collision, most injuries are due to t).................................... When a car stops abruptly in a collision, the pasengers’ inertia keeps them moving into the dash, or through the u)......................... ...............................

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

Most safety devices such as v)........................................, ........................................ and .................................................... work by increasing the w)................................... over which the person stops moving. This helps by reducing the x)............................................ acting on their body. Since “Impulse” equals change in y)..............................................., then for any given amount of momentum, the larger the z)................................... involved, the aa)...................................... the force acting.

Observation of everyday events seems to contradict 1st Law. For example, we observe that vehicles need to be powered to maintain f)................................................., and that they slow down and stop when no forces seem to be acting. This is because we don’t see g)............................................... acting. To maintain a constant speed, a car’s engine must supply force equal to h)....................................... Then, and only then, are the forces i)............................................. and there is there NO net force: 1st Law is obeyed.

Another strategy to minimise the effects of vehicle accidents is to reduce driving speeds, because less speed means less ab)..................................... energy and ac).............................................. to be lost in a collision. Strategies to slow traffic down include lower speed limits in ad)............................. ................................. and the installation of ae)................................................. and ............................................................

j)..................................... is the tendency of any object to resist any k)......................... .................................................... Inertia is a property associated with l)................................, the “stuff ” that resists m)................................................ when a force acts. When a car accelerates forward, it feels as if you are being n).............................................. In reality, your o).................................... is trying to keep you stationary, while the car p).................................. around you. In a sudden stop you feel as if you are q).................... ..............................................., but really your inertia is trying to r).................................................................................... while the car s)............................................... around you.

COMPLETED WORKSHEETS BECOME SECTION SUMMARIES

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CONCEPT DIAGRAM (“Mind Map”) OF TOPIC Some students find that memorizing the OUTLINE of a topic helps them learn and remember the concepts and important facts. Practise on this blank version.

MOVING ABOUT

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Practice Questions

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

These are not intended to be "HSC style" questions, but to challenge your basic knowledge and understanding of the topic, and remind you of what you NEED to know at the K.I.S.S. principle level.

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

When you have confidently mastered this level, it is strongly recommended you work on questions from past exam papers.

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

DISTANCE TRAVELLED

Part A Multiple Choice 1. Which part of this graph (A, B, C or D) indicates an object moving, but with a lower velocity than elsewhere?

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

8. A laboratory trolley is found to have 5 different forces acting on it. Four of them are known: • 0.75N weight force, vertically down • 0.75N reaction force, vertically up • 3.2N east • 2.5N east The trolley is motionless. The 5th force must be: A. 7.2N in all directions B. 0.7N west C. 5.7N west D. 0.7N east

C B

TIME

The average speed over the first 3 seconds (in ms-1) is: A. 0.75 B. 1.3 C. 2.0 D. 1.0 This Velocity-Time graph refers to Q3, 4 and 5. Velocity North

It shows the motion of an object travelling north.

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

C B

3. In section D of this graph, the 0 1 2 Time (sec) object’s motion is best described as: A. moving southward at constant velocity. B. moving southward, and decelerating. C. moving northward, and decelerating. D. moving northward at constant velocity.

10. In an experiment, a 700gram trolley was found to accelerate at 1.70ms-2. What net force must have acted on it? A. 1190N B. 1.19N C. 412N D. 2.4N 3

11.

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

4. In the first 3 seconds of this motion, the time when the object was stationary was: A. graph section A B. graph section B C. time = 2.5s D. time = zero

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

5. An instant of time when the acceleration is zero is: A. t = 1.25s B. t = 2.0s C. t = 4.0s D. never

13. A vehicle has mass “M” and velocity “V”. Another vehicle has mass “2M” and velocity “2V”. The ratio between their kinetic energies would be: A. 8:1 B. 4:1 C. 2:1 D. 1:1 28

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Longer Response Questions

14. The work done on a vehicle is equivalent to the A. acceleration of the vehicle B. change in momentum of the vehicle C. change in kinetic energy of the vehicle D. force multiplied by time for which it acts

Mark values shown are suggestions only, and are to give you an idea of how detailed an answer is appropriate. Remember that for full marks in calculations, you need to show FORMULA, NUMERICAL SUBSTITUTION, APPROPRIATE PRECISION and UNITS

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.

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

16. Just before a “head-on” collision, the momentum vectors of 2 cars could be represented as follows: car P car Q 5,000kgms-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 after the collision? C. 10,000kgms-1

B. 10,000kgms-1

D. 15,000kgms-1

Displacement south (-ve) north (+ve)

A. 20,000kgms-1

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

17. The”Conservation of Momentum” in a collision is a consequence of: A. Law of Conservation of Energy B. Newton’s 1st Law of Motion C. Newton’s 2nd Law of Motion D. Newton’s 3rd Law of Motion

Time A B

Sketch the corresponding Velocity-Time graph for the same journey. There is no need to show any numerical values on the axes, 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

18. Most safety devices in modern cars are designed to reduce the effects of a collision by: A. reducing the time duration of the collision. B. increasing the change of momentum involved. C. decreasing the distance over which the forces act. D. increasing the time duration of the collision.

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)

19. As the car accelerated when the traffic lights changed, a book on the dashboard “jumped back” into Sally’s lap. She immediately thought of several possible explanations for the motion of the book. 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.

24. (4 marks) a) Calculate the net force acting on a 2.50kg trolley that accelerates from rest to 3.50ms-1 in 5.00s. b) The trolley is being pulled by a string. The tension in the string is found to be 2.20N. What force of friction is acting?

20. Which of the following shows a correct relationship? A. Change in Momentum = Impulse B. Change in Kinetic Energy = Impulse C. Change in Momentum = Work done D. Change in Kinetic Energy = Change in Momentum

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29. (5 marks) A 600kg car braked from a velocity of 25.0ms-1 to 8.50ms-1 over a 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 this car’s Kinetic Energy as it slowed down.

25. (7 marks) In a laboratory experiment, a trolley of fixed mass was accelerated by different forces. The acceleration was measured in each case. Results: Force Applied (N) Acceleration (ms-2) 1.5 1.2 2.5 1.9 3.0 2.3 4.5 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.

30. (4 marks) A 600kg car, heading north at 15.0ms-1 collided head-on with a 500kg car heading south at 10.0ms-1. The vehicles locked together in the collision. Find the velocity (including direction) of the wreckage immediately after the collision.

26. (6 marks) A broken-down car is being towed as shown in the diagram. 750kg

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 the average force that acted on the north-bound car.

400kg

a = 1.50ms-22

Friction = -2 200N

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

Both cars are accelerating at 1.50ms-2. Someone accidentally left the hand brake on in the car being towed, causing a friction force of 200N to act as shown. Other friction 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 the driver’s left, at constant speed. a) Mark clearly on the diagram (and label) vectors to represent i) instantaneous velocity ii) acceleration iii) any net, unbalanced force

33. ( 3 marks) State Newton’s 1st Law of Motion, and use it to explain why an unrestrained passenger may go through a car windscreen during a collision. 34. (3 marks) One of the important safety features of modern motor vehicles is the “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 the car during a collision.

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 the road.

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

The maximum “grip” possible from each tyre is 2,500N. c) Explain what will happen, and why, if the curve becomes tighter... e.g. radius decreases to 23.0m. 28. (6 marks) An alien creature has a weight of 5.50x103N on his/her/its home planet where g=15.3ms-2. a) What is this creature’s mass? b) What will he/she/it weigh on Earth, where g=9.81ms-2? c) The creature’s personal propulsion device can exert a net force of 2.50x104N. What acceleration can the alien achieve while wearing the device? (Assume no friction, and that the device itself has neglible mass)

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Answer Section

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.

Worksheet 1 Part A a) distance b) time c) gradient d) stationary, not moving e) horizontal line f) zero on the speed scale g) scalar h) vector i) magnitude and direction j) displacement k) direction l) negative m) down n) gradient o) velocity p) displacement q) time r) velocity at a particular instant of time s) average velocity t) displacement and time u) instantaneous v) sonar or “light gates”

Displacement (km) East

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 zero C = 60x0.2 = 12km D = -50x0.8 = -40km b) graph c) i) 16+12+40 = 68km ii) Sp = 68/2 = 34km/hr iii) S = 16+12-40 = -12km iv) V = S/t = -12/2 Time (hr) = -6km/hr 1.0 2.0 (i.e. 6km/hr west)

200

West

-10

Part B 1. a) 200+100 = 300km b) +200 + (-100) = 100km north c) 5hr d) Speed = dist/time = 300/5 = 60km/hr e) V = S/t = 100/5 = 20km/hr f) graph

Time (hr)

6. 1st leg: S = V.t 2nd leg: 3rd leg: 4th leg:

100

Velocity (km/hr) South North

= 460x2.50 = 1,150km west = 105x(50x60) =315,000m =315km east = 325x3.25 ≅ 1,056km west = 125x(5.50x60x60) = 2,475,000m = 2,475km east Let east be (+ve), west be ( -ve) Final displacement = -1,150 + 315 -1,056 + 2,475 = +575 km (east) of starting point.

Time (hr) 1

-100

g) from graph: i) gradient = 200/3 ≅ 67 km/hr ii) gradient = zero iii) gradient = -100/1 = -100km/hr (i.e. 100km/hr south) h) graph

100

Displacement North (km)

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

2. a) 600km b) 1.5hr c) V = S/t = 600/1.5 = 400km/hr north d) gradient = -900/3 = -300 e) Flight from Q to R f) R is 300km south of P g) Position = over town P. Velocity = 300km/hr south h) i) distance = 1,500km ii) Speed = 1,500/6 = 250km/hr iii) Final displacement = 300 km south iv) V = S/t = 300/6 = 50km/hr south i) graph

Worksheet 2

100 0

Time (hr) 2

-100

-300

South

Velocity (km/hr)

200

300

North

400

Part A a) velocity b) slowing down c) direction d) vector e) negative f) ms-2 g) curve h) sloping, straight line i) horizontal j) gradient k) negative l) force m) external n) unbalanced o) 2nd p) net force applied q) inversely r) newton s) kg t) ms-2 u) matter v) force w) gravity x) in the same line y) head to tail z) Pythagorus’s Theorem aa) Resultant ab) direction ac) equilibrium ad) net force ae) in a straight line at constant velocity af) acceleration ag) opposes ah) Tension ai) both aj) direction ak) centripital al) centre am) tangent

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16. R2 = 302 + 102 R = sq.root(1,000) = 32

Worksheet 2 (cont) Part B Practice Problems

1. a = (v - u)/t = (22.5-0)/8.20 = 2.74ms-2 2. u = v - at = 0-( -2.60x4.80) = 12.5ms-1 3. a = (v - u)/t \ t = (v - u)/a = (22.5 - 12.0)/1.75 = 6.00s 4. v = u + at = 850 + (-50.0)x20.0 = -150ms-1 The final negative velocity means it is 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/hr c) 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: curves down to horizontal Constant Velocity: straight line

R φ 10 20

Tan φ = 20/200 R2 = 2002 + 202 R = sq.root(40,400) φ = 6o = 201

17.

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.

200 φ

18. R2 = 1502 + 2002 R = sq.root(62,500) = 250 Tan φ = 200/150 φ = 53o

300 φ

200

R 150

R = 250km, 53o S of E (bearing 143o). 3.85

19. Not accelerating means there is NO net force, The 3 forces must be in equilibrium F F2 = 5.252 + 3.852 Tan φ = 5.25/3.85 o F = sq.root(42.385) φ = 54 = 6.51 3rd Force = 6.51N, 54o S of E (bearing 144o).

Note: Although slowing down, the vehicle continues to move away from the start, so the Displacement-Time graph never shows a negative gradient. Newtons 2nd Law 6. 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 & Weight 12. 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 capsule remain. Mass in orbit = 10,000kg. ii) In orbit (free fall) weight = zero. c) i) No fuel left, so mass = 5,000kg. ii) W=mg = 5,000x1.7 =8,500N = 8.5x103N. 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 Analysis 15. R2 = 402 + 252 Tan φ = 25/40 ∴ R = sq.root(2,225) φ = 32o φ R = 47N Resultant = 47N at 32o bearing Preliminary Physics Topic 3 copyright © 2005-2007 keep it simple science

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

Acceleration: curves up from horizontal

Tan φ = 30/10 φ = 72o

5.25

Friction, Tension & Turning Corners 20. Net Force: F= ma = 850x2.15 = 1.83x103N. Net Force = “Thrust” + Friction 1.83x103 = 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) = 4.35x103N F=ma, so a=F/m = 4.35x103/1,500 = 2.90ms-2. d) Tension must overcome 450N of friction and accelerate the van 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 = 35,714N Since the maximum grip of the tyres is only 18,000N, the tyres cannot provide the force needed to to turn this corner... car will “spin out”.

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keep it simple science 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.

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 4

Worksheet 3

Part A a) mass b) velocity c) vector d) kg.ms-1 e) conserved f) total momentum after collision g) stops moving h) lost i) friction j) conserved k) For every “action” force there is an equal, opposite “reaction” force. l) recoil (kick) m) momentum n) Impluse o) time p) newton-seconds (N.s) Part B Practice Problems 1. 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 of momentum, but in opposite directions. (Remember, momentum is a vector) Conservation of Momentum 6. ρi = ρf mA.uA + mB.uB = mA.vA + mB.vB

Part A a) kinetic b) mass c) velocity d) doubled e) quadrupled (4X) f) scalar g) joule ( J) h) force i) distance j) kinetic energy k) Conservation l) created nor destroyed m) transformed (into other forms of energy) n) (chemical) potential o) kinetic p) kinetic q) heat r) distortion/damage Part B Practice Problems 1. 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)/2,750 v = 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.m c) 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) Friction d) 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.

Since the cars lock together, their final velocity is the same.

600x27.0 + 1,500x0 = (600+1,500)xV 2,100V = 16,200 v = 7.71ms-1. 7. ρi = ρf mA.uA + mB.uB = mA.vA + mB.vB 700x24.5 + 700x8.50 = 0 + 700x VB VB = (17,150+5,950)/700 = 33.0ms-1. 8. ρi = ρf mA.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 + 5,625 mA = 37,125/8.75 = 4.24x103kg. 9. a) If they had equal magnitudes of momentum, but opposite directions, then the sum of their momentum = zero. b) To have equal magnitudes of momentum, the product MxV must be the same for each (ignoring direction). If one has twice the mass, the other must have twice the velocity.

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keep it simple science Worksheet 4 Part B (cont.) Impulse & Momentum 10. 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 equals momentum 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)

Remember that for full marks in calculations, you need to show FORMULA, NUMERICAL SUBSTITUTION, APPROPRIATE PRECISION and UNITS

(-ve)

a) do not act b) continue moving in a straight line, with constant velocity c) acted upon by net force d) remain at rest e) acted upon by net force f) constant speed g) friction/retarding forces h) friction i) balanced/in equilibrium j) Inertia k) change of motion l) mass m) acceleration n) pushed backwards o) inertia p) accelerates q) flung forward r) keep you moving forward s) decelerates t) inertia u) windscreen v) seatbelts, airbags & crumple zones w) time (and distance) x) Force y) momentum z) time aa) smaller ab) kinetic energy ac) momentum ad) residential areas ae) speed humps af) chicanes

13. A 14. C 15. C 16. A

B. stopped zero velocity A. constant negative velocity

Weight

Lift Drag

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

Lift increased

Weight same Drag same

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

17. D 18. D 19. B 20. A

time

Thrust

Practice Questions Multiple Choice 5. A 9. A 6. D 10. B 7. B 11. D 8. C 12. B

D. constant positive velocity

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

Worksheet 5

Part A 1. C 2. B 3. C 4. D

C. accelerating

Velocity 0

(+ve)

Part B Longer Response Questions In some cases there may be more than one correct answer. The following “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 100 R2 = 1002 + 1502 ∴R = sq.root(32,500) = 180km 150 Tan φ = 100/150 R φ = 34o φ Displacement = 180km, 34o W of N (bearing 326o) c) v = S/t =180/3.00 = 60km/hr, bearing 326o. 22.

Net Force = Tension + Friction 1.75 = 2.20 + F Friction = -0.45N.

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keep it simple science Answers to Practice Questions (cont.) 25. a)

b) Graph shows a direct relationship between force and acceleration.

Force (N)

3.0

2.5

31. a) Δρ = m(v - u) = 600x(3.64 - 15) = -6.82x103kgms-1. (negative, because the change in momentum was southward, or a loss of northward momentum) b) Δρ = Impulse = F.t = -6.82x103 ∴ F = -6.82x103/0.200 = -3.41x104N.

Acceleration (ms-2)

32. a) Ek = 0.5mv2 northbound car southbound car Ek = 0.5x600x15.02 Ek =0.5x500x10.02 = 67,500 J = 25,000 J Total Ek = 92,500 = 9.25x104J. b) After collision, velocity = 3.64ms-1 Ek = 0.5x(600+500)x3.642 = 7.29x103 J. c) Over 90% of the original kinetic energy is gone. Some has been transformed into the sound and heat of the collision, but most has been used to distort and damage the vehicles.

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: Thrust Net force Friction Net F = Thrust + Friction 1.73x103 = Thrust + (-200) Thrust = 1.73x103 + 200 = 1.93x103N. c) Tension must accelerate the towed car AND overcome the friction. T = ma + 200 = 400x1.50 + 200 = 800N.

33. 1st Law: A moving object will continue to move in a straight line at a constant velocity unless acted upon by a net force. If at rest it will remain at rest unless a force acts on it. In a collision in which a vehicle stops suddenly, an unrestrained passenger will continue to move according to 1st Law, and may go through the windscreen.

27. a) on diagram b) F = mv2/R = 500x22.02/25.0 = 9.68x103N.

34. a) The car body is designed so that it collapses, one section after another, 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 for any given amount of momentum, increasing the time involved reduces the force acting on the people in the vehicle, and decreases the risk of injury or death.

ii) a

i) V

iii) F (toward centre of circle)

tangent to circle

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.

35. a) Kinetic energy depends upon both mass and velocity, but velocity has the biggest contribution, since Ek = 0.5mv2. This means doubling the velocity increases the energy by a factor of 4. Since velocity is so important, it means that reducing speeds can greatly reduce the energy involved in vehicle accidents, and reduce the incidence of death and injury. b) Speed humps Chicanes Low speed zones in residential areas, and around schools.

29. a) Work = change in kinetic energy F.S = 0.5m(v2 - u2) = 0.5x600x(8.502 - 25.02) = -165,825 J F = -165,825/50.0 = -3.32x103N. (negative force, because opposing the motion) b) It means that energy cannot be created or destroyed... it never disappears or ceases to exist. It simply gets transformed from one 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. ρi = ρf mA.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)xV 9,000 - 5,000 = 1,100V ∴ V = 4,000/1,100 = 3.64ms-1 north (since answer is +ve)

30.

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