vce physics unit 3 topic 1 motion in 1 & 2 dimensions

VCE PHYSICS

Unit 3

Topic 1

Motion in 1 & 2 Dimensions

Unit OutlineUnit OutlineUnit OutlineTo achieve this outcome the student should demonstrate the knowledge and skills to:• apply Newton’s laws of motion to situations involving two or more forces acting along a straight line and in

two dimensions;• analyse the uniform circular motion of an object moving in a horizontal plane (FNET = mv2/R) such as a

vehicle moving around a circular road; a vehicle moving around a banked track; an object on the end of a string.

• Apply Newton’s 2nd Law to circular motion in a vertical plane; consider forces at the highest and lowest positions only;

• investigate and analyse the motion of projectiles near the Earth’s surface including a qualitative description of the effects of air resistance;apply laws of energy and momentum conservation in isolated systems;

• analyse impulse (momentum transfer) in an isolated system, for collisions between objects moving along a straight line (FΔt = mΔt);

• apply the concept of work done by a constant forcework done = constant force x distance moved in the direction of the forcework done = area under force distance graph

• analyse relative velocity of objects along a straight line and in two dimensions;• analyse transformations of energy between: kinetic energy; strain potential energy; gravitational potential

energy; and energy dissipated to the environment considered as a combination of heat, sound and deformation of material;

kinetic energy i.e. ½ mv2; elastic and inelastic collisions in terms of conservation of kinetic energy

strain potential energy i.e. area under force-distance graph including ideal springs obeying Hooke’s Law ½ kx2

gravitational potential energy i.e. mgΔh or from area under force distance graph and area under field distance graph multiplied by mass

• apply gravitational field and gravitational force concepts g = GM/r2 and F = GM1M2/r2

• apply the concepts of weight (W = mg), apparent weight (reaction force, N) , weightlessness (W = 0) and apparent weightlessness (N = 0)

• model satellite motion (artificial, moon, planet) as uniform circular orbital motion (a = v2/r = 4π2r/T2)• identify and apply safe and responsible practices when working with moving objects and equipment in

investigations of motion.

Chapter 1

Topics covered:

• The S.I. System.

• Position.

• Scalars &Vectors.

• Vector Addition & Components.

1.0 The S. I. System

S.I. System:7 Fundamental

Units

Length:Unit

Metre (m)

Mass:Unit

Kilogram(kg)

Time:Unit

Second (s)

Electric Current:

UnitAmpere (A)

Temperature:Unit

Kelvin (K)

LuminousIntensity:

UnitCandela (cd)

Amount ofSubstance:

UnitMole (M)

The system of units used in Physics is the “Systeme Internationale d’Units” or more simply the S. I. System. The system has two important characteristics;Different units for the same physical quantity are related by factors of 10.(eg. mm; cm; km)

Velocity:Unitms-1

Velocity - Derived Unit

Force:Unit

N = kg ms-2

Force - Derived Unit

The system is based on 7 Fundamental Units, each of which is strictly defined.

All other units, so called DERIVED UNITS, are simply combinations of 2 or more of the Fundamental Units.

1.1 Position

Thus a number called -15 is 15 units to the left of 0 on the number line.

0-5-10-15-20-25-30 5 10 15 20 25 30

The Number Line

- 15 Units + 30 Units

A number called +30 is 30 units to the right of 0.

To specify the POSITION of an object, a point of ORIGIN needs to be defined. It is from this point all measurements can be taken.

For example on the number line below the point labelled 0 is the origin and all measurements are related to that point.

1.2 Scalars & Vectors• Before proceeding, it is

important to define two general classes of quantities.

• 1. SCALAR QUANTITIES:• These are COMPLETELY

specified by:• A MAGNITUDE (ie a NUMBER)• and A UNIT• Examples of Scalar Quantities

would be: Temperature (17oC), Age (16 years), Mass (2.5 kg), Distance (150 m).

• 2. VECTOR QUANTITIES:

• These are COMPLETELY specified by:• A MAGNITUDE (ie. A NUMBER)• and A UNIT• and A DIRECTION

• Examples of Vector Quantities would be: Displacement (2.7 km, West), Force (15 N, Downward), Acceleration (1.5 ms-2, S.E.)

VECTORS ARE GENERALLY REPRESENTED BY ARROWS:The length of the arrow represents the magnitude of the vector.

The orientation of the arrow represents the direction of the vector.

N 15 A vector of magnitude 15 unitsdirected East

1.3 Vector Addition

N

450

SINGLE VECTORDIAGRAM

5

A Vector of:Magnitude; 5 units Direction; NE or N45E or 45T

VECTOR ADDITION

Two Forces act at the Centre of Mass of a body.

Centre of Mass

The first of 4N East and the second of 3N South

5

4

3

4

35

Magnitude = 42 + 32

= 25

Direction:Sin = 3/5 = Sin-1 3/5 = 36.90

THE RESULTANT FORCE HAS A MAGNITUDE OF 5 N DIRECTED AT E 36.90 S

Which way will the body move ?

In a direction, and with a force, that is the sum of the 2 vectors

1.4 Vector SubtractionAn object moving East at 8.0 ms-1

8.0 ms-1 E

8.0 ms-1 S

changes its velocity to 8.0 ms-1 South

What is the object’s change in velocity ?

The velocity change (v) is given by vf - vi

(- vI ) is a negative vector. It can be converted to a positive one by reversing its direction.Then, by performing a vector addition, the velocity change v can be obtained.

vf

vi

8.0 ms-1 W

- vi

8.0 ms-1 Svf

v

Magnitude v = 82 + 82

= 128

= 11.3 ms-1

Direction:Tan = 8/8 = 1.0 = Tan-1 1.0 = 450

THE CHANGE IN VELOCITY = 11.3 ms-1 AT S 450W

1.5 Vector Components A “Jump Jet” is launched from a 150 ramp at a velocity of 40 ms-1

What are the vertical

V VERTICAL

and horizontal components of its velocity ?

150

40 ms-1

V VERTICAL and VHORIZONTAL are the COMPONENTS of the plane’s velocity.

Vertical Component:V VERTICAL = 40 Sin 150

= 10.4 ms-1

Horizontal Component:VHORIZONTAL = 40 Cos 150

= 38.6 ms-1

VHORIZONTAL

Motion - Revision Questions

Question type:

Adam is testing a trampoline. The diagrams show Adam at successive stages of his downward motion.

Figure C shows Adam at a time when he is travelling DOWNWARDS and SLOWING DOWN.

Vectors

A: Acc is UPWARD. In order to meet the requirements set - travelling downward BUT slowing down, he must be decelerating ie. Accelerating in a direction opposite to his velocity. Thus acc is upward.

Q1: What is the direction of Adam’s acceleration at the time shown in Figure C ? Explain your answer.

A B C D

Chapter 2

Topics covered:

• Distance versus Displacement.

• Speed versus Velocity.

• Acceleration.

• Graphical Representations.

2.0 Distance vs Displacement• Distance is a Scalar Quantity

having a magnitude and a unit. The S.I. unit for Distance is the metre (m)

• Distance is best thought of as: “How far you have travelled in your journey”.

• Displacement is a Vector Quantity having a magnitude, a unit and a direction. The S.I. unit for Displacement is the metre (m), plus a direction

• Displacement is best thought of as: “How far from your starting point you are at the end of your journey”.

• Distance and Displacement may or may not be numerically equal, depending on the nature of the journey.

JOURNEY No 1.Distance = Displacement

Start Finish

100 m

+ve direction

At the end of the run:Distance = 100 m.Displacement = +100 m

JOURNEY No 2.Distance Displacement

At the end of the one lap run:Distance = 400 m.Displacement = 0 m

Start/Finish

400 m track

2.1 Speed vs Velocity• Speed is defined as the Time Rate

of Change of Distance.• Speed is a Scalar Quantity.• Mathematically:

Speed = Distance/Time• The S.I. unit for Speed is

metres/sec (ms-1)

• Velocity is defined as the Time Rate of Change of Displacement.

• Velocity is a Vector Quantity.• Mathematically:

Velocity = Displacement/Time• The S.I. unit for Velocity is

metres/sec (ms-1), plus a direction

INSTANTANEOUS vs AVERAGE VELOCITY

• The term velocity can be misleading unless a specific label is attached.

• The label indicates whether the velocity is an Average value calculated over a long period of time OR an Instantaneous value calculated at any instant of time.

• A simple example illustrates:• A journey of 40 km across the suburbs

takes 1 hour; • VAV = 40/1 = 40 kmh-1 • BUT VINST could be anything from 0

kmh-1 (stopped at traffic lights) to VINST = 100 kmh-1 (travelling along the freeway).

IN ALL CALCULATIONS AND EQUATIONS USED IN THE COURSE, ASSUME INSTANTANEOUS VALUES ARE REQUIRED UNLESS OTHERWISE STATED.

2.2 Some Common Speeds

Event Speed (ms-1) Speed (kmh-1)

1. Grass Growing 5.0 x 10-8 1.8 x 10-7

2. Walking Pace 1 - 2 4 - 8

3. Marathon Runner

5 18

4. 100 m Sprinter

10 36

5. Suburban Speed Limit

16.7 60

6. Freeway Speed Limit

30.6 110

7. Boeing 737 Cruising

246 886

8. Speed of Light 2.99 x 108 1.1 x 109

2.3 Acceleration• Acceleration is defined as the Time

Rate of Change of Velocity.

• Acceleration is a Vector Quantity.

• Mathematically: Acceleration = Velocity/Time

• The S.I. unit for acceleration is metres/sec/sec (ms-2)

• Since acceleration is a vector quantity, a body travelling with a constant speed but in a constantly changing direction must be accelerating.

• So a cyclist travelling around a corner at constant speed is, in fact, accelerating ! (More of this later).

ACCELERATING VEHICLE

v

v

a

a

DECELERATING VEHICLE

The velocity and accelerationare in the same direction

The velocity and accelerationare in opposite directions.

2.4 Graphical Representations• Much of the information delivered in this Physics course is presented graphically.• Generally, graphs “tell a story” and you need to develop the ability to “read” the story the

graph is telling.• There are two basic families of graphs you should be familiar with:(a) Sketch Graphs, paint a broad brush, general picture of the relationship between the

quantities graphed. (b) Numerical Graphs from which exact relationships may be deduced and/or exact values

may be calculated.

Distance

Time

The Story:As time passes, the distance of the object from its starting point does not change. This is the graph a stationary object.

Distance

TimeThe Story:As time passes, the distance of the object from its starting point is increasing in a uniform manner (the slope is constant). This is the graph an object moving at constant speed.

Velocity

Time

The Story:As time passes, the velocity of the object is increasing in a uniform manner (the slope is constant). This is the graph a constantly accelerating object

Displacement

Time

The Story:As time passes, the displacement of the object is increasing more quickly (the slope is increasing at a constant rate).

This is the graph a constantly accelerating object

SKETCH GRAPHS


Question type:In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant deceleration.The graphs A to F below should be used to answer the questions below. The horizontal axis represents time and the vertical axis could be velocity or distance.

Q2: Which of the graphs, A to F, represents the velocity time graph for the entire journey ?

A: Graph B

Q3: Which of the graphs, A to F, best represents the distance time graph of the car for the entire journey ?

A B C D E F

A: Graph E

Sketch Graphs

2.5 Exact Graphical Relationships

• You are required to be familiar with graphs of:

• Distance or Displacement Versus Time;

• Speed or Velocity Versus Time and

• Acceleration Versus Time

• These graph types and the exact information obtainable from them can be summarised in the table given below. Put this table on your cheat sheet.

Distance orDisplacement versus Time

Graph Type Read directly from Graph

Slope of Graphgives:

Area under Graph gives:

Distance or

Displacement

Speed or

Velocity

No UsefulInformation

No UsefulInformation

Speed or Velocityversus Time

Speed or

VelocityAcceleration

Distance or

Displacement

Acceleration versus Time

Acceleration Velocity

2.6 The Equations of Motion• These are a series of equations

linking velocity, acceleration, displacement and time.

• THE EQUATIONS CAN ONLY BE USED IN SITUATIONS WHERE THE ACCELERATION IS CONSTANT.

• The 3 most important of these equations are:

• 1. v = u + at

• 2. v2 = u2 +2ax

• 3. x = ut + ½at2

• where u = Initial Velocity (ms-1) v = Final Velocity (ms-1) a = Acceleration (ms-2) x = Displacement (m) t = Time (s)

• HINTS FOR EQUATIONS’ USEu =

v =

a =

x =

t =

Questions are often asked which require a 2 step process to get to the answer, ie. A value for acceleration may be needed before the final velocity can be found.

Sometimes it is necessary to choose a positive direction ie. up or down for vertical motion questions or left or right for horizontal motion questions.

When using these equations always list out the information supplied in the question and what you are required to calculate; then choose the appropriate equation to use.

+ve

Make a list like this.


Question type:

In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant deceleration.

Q4: Calculate the acceleration of the car for the first 400 m.

A: Firstly, list information:u = 0v = ?a = ?x = 400 mt = 19 s

Choose the appropriate equation: x = ut + ½at2

400 = 0 + ½a(19)2

a = 2.22 ms-2

Equations of Motion


Question type:In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant deceleration.

Q5: Calculate the average speed for the entire journey, covering both the accelerating and braking sections.

A: Average Speed = Total Distance Total Time

For the accelerated part of journey:Distance = 400 m Time = 19 secFor the braking part of journey:Distance = needs to be calculated Time = 5.1 sec

To get braking distance, use Eqns of Motionu = ?v = 0a = ?x = ?t = 5.1s

List does not contain enough info the calculate s

Need to know u, the initial speed for the braking section which equals the final speed for the accelerating section.

For accelerating section:u = 0v = ?a = 2.2 ms-2

x = 400 mt = 19 s

v = u + at = 0 + (2.2)(19) = 41.8 ms-1

Braking list becomesu = 41.8 ms-1 v = 0a = ?x = ?t = 5.1s

Still not enough info

Need to calc accv = u + at0 = 41.8 + a(5.1)a = - 8.2 ms-2

Now can calc sx = ut + ½at2

= (41.8)(5.1) + ½(-8.2)(5.1)2

= 213.2 - 106.6 = 106.6 m

Total Distance = 400 + 106.6 = 506.6 m Total Time = 19 + 5.1 = 24.1 sSo, Average Speed = 506.6/24.1 = 21 ms-1

Average Speed

Chapter 3

Topics covered:

• Newton’s Laws.

• Force in Two Dimensions.

• Momentum and Impulse.

• Conservation of Momentum.

3.0 Newtonian Motion• Sir Isaac Newton (1642 - 1727) was

unique for a number of reasons, but mostly because he developed a set of laws describing the motion of objects in the universe.

• Prior to Newton, scientists believed that a set of laws existed which explained motion on Earth and these laws had to be modified to describe motions in all other parts of the universe.

• Newton was the first scientist to realise that all motion anywhere in the universe could be described by a single set of laws which then had to be modified for use in the friction riddled confines of the Earth. Isaac Newton

Aged 26

3.1 Newton’s Laws

LAW 3. For every action there is an equal

and opposite reaction

INERTIA – That property possessed by all bodies with mass whereby they tend to resist changes to their motion. It is associated with an object’s mass – more mass, more inertia.Inertia is NOT a force.

Newton developed 3 laws covering motion in the universe, they are:

LAW 1. THE LAW OF INERTIA. A body will remain at rest, or in a state of uniform motion, unless acted upon by a net external force.

LAW 2. The acceleration of a body is directly proportional to the net force applied and inversely proportional to its mass. (a = F/m)

LARGE MASS – LARGE INERTIA

small mass – small inertia


Question type:A seaplane of mass 2200 kg takes off from a smooth lake as shown. It starts from rest,

and is driven by a CONSTANT force generated by the propeller. After travelling a distance of 500 m, the seaplane is travelling at a constant speed, and then it lifts off after travelling a further 100 m.

The total force opposing the motion of the seaplane is not constant. The graph shows the TOTAL FORCE OPPOSING THE MOTION of the seaplane as a function of the distance travelled.

Q6: What is the magnitude of the net force acting on the seaplane after it has travelled a distance of 500 m from the start ?

A: At d = 500 m the plane is travelling at CONSTANT VELOCITY,

So ΣF = 0

Newton’s Laws

Total OpposingForce (N)

Distance (m)0

2000

4000

6000

8000

10000

100 200 300 400 500 600


Question type:

A: At d = 500 m the seaplane is subject to 0 net force (see previous question).

Thus, Driving Force = Opposing Force = 10,000 N (read from graph).

At d = 200 m the total opposing force = 2000 N (read from graph)

So ΣF = 10,000 - 2000 = 8000 N

Now, we know that ΣF = ma

So, a = ΣF/m = 8000/2200 = 3.64 ms-2

Newton’s 2nd Law

Q7: What is the magnitude of the seaplane’s acceleration at the 200 m mark ?


Distance (m)0

2000

4000

6000

8000

10000

100 200 300 400 500 600


Question type:Q8: Estimate the work done by the seaplane against the opposing forces in travelling for a distance of 500 m.

Total number of squares (up to d = 500 m) = 9 whole squares (x)+ 6 part squares (p) = 12 whole squares.

So Work done = (12) x (2 x 105) = 2.4 x 106 J


Distance (m)0

2000

4000

6000

8000

10000

100 200 300 400 500 600

x x

x x

x x

x xx

p

p

p

p

p

p

A: Work = Force x Distance = Area under F vs d graph.

Area needs to be calculated by “counting squares”.Each square has area = 2000 x 100 = 2 x 105 J

Work


Question type:In Figure 2, a car of mass 1000 kg is being towed on a level road by a van of mass 2000 kg. There is a constant retarding force, due to air resistance and friction, of 500 N on the van, and 300 N on the car. The vehicles are travelling at a constant speed.

Q9: What is the magnitude of the force driving the van?

Q10: What is the value of the tension, T, in the towbar?

Figure 2

As the vehicles are travelling at constant speed, a = 0 and thus ΣF = 0 Thus driving force = total retarding forceThus driving Force = 800 N

Looking at the towed car alone the forces acting are tension in the towbar and the retarding force.Since the car is travelling at constant velocity, ΣF = 0, so tension = retarding force = 300 N

Newtons 2nd Law


Question type:The figure shows a cyclist with the bicycle wheels in CONTACT with the road surface. The cyclist is about to start accelerating forward.

Q11: Explain, with the aid of a clear force diagram, how the rotation of the wheels result in the cyclist accelerating forwards.

A: The wheels rotate in the direction shown.

FTR

FRT

The force labelled FTR is the force the tyre exerts on the road. This force is directed in the opposite direction to the acceleration and thus cannot be the force producing that acceleration.

a

The force labelled FRT is the Newton 3 reaction force arising from the action of FTR. It is this force (directed in the same direction as the acceleration) that actually produces the acceleration of the bike and rider.

Newton’s 3rd Law

3.2 Newton’s Laws Restrictions & Consequences

The action and reaction law requires there to be TWO bodies interacting, the ACTION force acting on one body and the REACTION acting on the other.Deciding when an Action / Reaction situation exists can be done by answering the question: Does the second (reaction) force disappear immediately the first (action) force disappears ? If the answer is yes, you have an action reaction pair.

CONSEQUENCES

As far as Newton’s 1st law is concerned “rest” and “uniform motion” are the same state. You cannot perform any test which can show whether you are stationary or moving at constant velocity.

RESTRICTIONSThe laws only apply at speedsmuch, much less than thespeed of light.The laws apply equally in ALLinertial frames of reference.

FRAME OF REFERENCE ?A frame of reference is best described as “your point of observation.” An inertial frame of reference is one that is either stationary or moving with constant velocity. A non inertial frame of reference is accelerating.Humans in non inertial frames tend to invent “fictitious forces” to explain their experiences


Question type:

A train is travelling at a constant velocity on a level track. Lee is standing in the train, facing the front, and throws a ball vertically up in the air, and observes its motion.

Q12: Describe the motion of the ball as seen by Lee.

Sam, who is standing at a level crossing, sees Lee throw the ball into the air.

Q13: Describe and explain the motion of the ball as seen by Sam.

Lee sees the ball move straight up and down.

From Sam’s point of view, the ball follows a parabolic path made up of the vertical motion imparted by Lee and the horizontal motion due to the train.

Relative Motion

3.3 Force in Two Dimensions • A Force is either a Push or a Pull.• A Force is either a CONTACT type

force or a FIELD type force.• Force is a Vector quantity having both

a magnitude, a unit and a direction.• Force is NOT one of the 7 fundamental

units of the S.I. System and thus it is a Derived quantity.

• The unit for Force is kgms-2. This was assigned the name the NEWTON (N), in honour of Sir Isaac.

• Forces can act in any direction and the TOTAL, NET or RESULTANT force is the vector sum of all forces acting on a body.

• The body will then ACCELERATE in the direction of the RESULTANT FORCE, according to Newton 2.

F1

F2

F3

F4

AN OBJECT UNDER THEACTION OF 4 FORCES

F1

F2

F3 F4

In which direction willthe object accelerate ?

Perform a vector additionof the forces. The

Resultant Force (FRES ) will give the direction of the acc.

FRES

FRES

The object will be subject to FRES and accelerate in

that direction


Question type: A cyclist is towing a small trailer along a level bike track (Figure 1). The cyclist and bike have a mass of90 kg, and the trailer has a mass of 40 kg. There are opposing constant forces of 190 N on the rider and bike, and 70 N on the trailer. These opposing forces do not depend on the speed of the bike.

The bike and trailer are initially travelling at a constant speed of 6.0 m s-1.

Q14. What driving force is being exerted on the road by the rear tyre of the bicycle?

Force in 1 & 2 Dimensions

A: Constant speed implies ΣF = 0; So driving force = total retarding forceTotal retarding force = 190 + 70 = 260 N = driving force

3.4 Momentum & Impulse• Newton called Momentum the

“quality of motion” and it is a measure of a body’s translational motion - its tendency to continue moving in a particular direction.

• Roughly speaking, a body’s momentum indicates which way the body is heading and just how difficult it was to get the body moving with its current velocity.

• Momentum is a Vector Quantity.• Mathematically:

Momentum (p) = m.v where, m = mass (kg), v = velocity (ms-1) p = momentum (kgms-

1)

• Impulse is the “transfer mechanism” for momentum.

• In order to change the momentum of a body you need to apply a force for a certain length of time to produce the change.

• Impulse is a Vector Quantity.• Mathematically:

Impulse (I) = F.t whereF = Force (N)

t = time (s) I = Impulse (Ns)

• From Newton 2 (F = ma) and the definition of acceleration, (a = v/t), we get:

F = mv/t Ft = mv

• Thus: Impulse = Momentum


Question type:A small truck of mass 3.0 tonne collides with a stationary car of mass 1.0 tonne. They remain locked together as they move off. The speed immediately after the collision was known to be 7.0 ms-1 from the jammed reading on the car speedometer. Robin, one of the police investigating the crash, uses conservation of momentum to estimate the speed of the truck before the collision.

Q15 : What value did Robin obtain?

A: PBEFORE = PAFTER

PBEFORE = (3000)(x)

PAFTER = (3000 + 1000)(7.0)

So, 3000x = 28,000 x = 9.3 ms-1

The calculated value is questioned by the other investigator, Chris, who believes that conservation of momentum only applies in elastic collisions.Q16: Explain why Chris’s comment is wrong. A: Momentum is conserved in ALL

types of collisions whether they be elastic or inelastic. KE is not conserved in this type (inelastic) collision.

Momentum


Question type:A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides with a truck that is stationary at a set of traffic lights. After the collision they are stuck together and move off with a speed of 2.0 ms–1

Q17 : How much momentum did the car transfer to the truck?

Q18 : What is the mass of the truck?

A: Mom is ALWAYS conserved. Mom of car before = (1000)(5) = 5000 kgms-1

Mom of car after = (1000)(2) = 2000 kgms-1

Since mom is conserved Mom loss by car = Mom gain by truck So, Mom transferred to truck = 3000 kgms-1

A: p = mv so, m = p/v = 3000/2 = 1500 kg

Momentum


Question type:

Q19: If the collision took place over a period of 0.3 s, what was the average force exerted by the car on the truck?

For the truck Impulse = Change in momentum Ft = mv F(0.3) = 3000 F = 10,000 N

Impulse

A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides with a truck that is stationary at a set of traffic lights. After the collision they are stuck together and move off with a speed of 2.0 ms–1


Question type:A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and move off as one.

X Y6.0 ms-1

10 tonnes 5 tonnes

stationary

Before Collision

X Y

v ms-1

After Collision

Q20: Calculate the final speed of the joined railway trucks after collision.

A: In ALL collisions Momentum is conserved.So Mom before collision = Mom after collisionMom before = (10 x 103)(6.0) + (5 x 103)(0)Mom after = (15x 103)(v) So v = 60,000/15,000

= 4.0 ms-1

Momentum


Question type:

X Y6.0 ms-1

10 tonnes 5 tonnes

stationary

Before Collision

X Y

4.0 ms-1

After Collision

Q21: Calculate the magnitude of the total impulse that truck Y exerts on truck X

A: Impulse = Change in MomentumTruck X’s change in momentum = Final momentum – Initial Momentum

= (10 x 103)(4.0) – (10 x 103)(6.0) = -2.0 x 104 kgms-1

The mechanism for this change in momentum is the impulse supplied by Truck YSo, I = 2.0 x 104 Ns

A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and move off as one at a speed of 4.0 ms-1.

Impulse

3.5 Elastic & Inelastic Collisions

(a) ELASTIC COLLISIONS, where BOTH Momentum AND Kinetic Energy are conserved. (Very few collisions are of this type). If anywhere, these will most likely occur on the atomic or subatomic level.

INELASTIC COLLISION

(b) INELASTIC COLLISIONS, where Momentum is conserved BUT Kinetic Energy is NOT conserved. (Most collisions are of this type).The “lost” Kinetic Energy has been converted to other forms of energy eg, heat, sound, light.

All collisions (eg, cars with trees, cyclists with the footpath, neutrons with uranium atoms, bowling balls with pins etc.) fall into one of 2 categories:

Neutronn

ELASTIC COLLISION

U


Question type:

Q22: Use a calculation to show whether the collision was elastic or inelastic.

A: Total KE before collision = ½ mv2 = ½ (3000)(9.3)2

= 129735 J

Total KE after collision = ½ mv2 = ½ (4000)(7.0)2

= 98,000 J

KE is NOT conserved. So collision is INELASTIC

Elastic/Inelastic Collisions

A small truck of mass 3.0 tonne collides with a stationary car of mass 1.0 tonne. They remain locked together as they move off. The speed immediately after the collision was known to be 7.0 ms-1 from the jammed reading on the car speedometer. Robin, one of the police investigating the crash, uses conservation of momentum to estimate the speed of the truck before the collision at 9.3 ms-1


Question type:

X Y6.0 ms-1

10 tonnes 5 tonnes

stationary

Before Collision

X Y

4.0 ms-1

After Collision

Q23: Explain why this collision is an example of an inelastic collision. Calculate specific numerical values to justify your answer.A: In Inelastic collisions Momentum is conserved BUT Kinetic Energy is not.For this collisionpBEFORE = (10 x 103)(6.0) + (5 x 103)(0.0) = 6 x 104 kgms-1

pAFTER = (15 x 103)(4.0) = 6 x 104 kgms-1

KEBEFORE = ½mv2 = ½(10 x 103)(6.0)2 = 1.8 x 105 JKEAFTER = ½mv2 = ½(15 x 103)(4.0)2 = 1.2 x 105 JThus Mom IS conserved but KE is NOT conserved, therefore this is an inelastic collision

A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and move off as one at 4.0 ms-1.

Inelastic Collisions

3.5 Conservation of Momentum• The Law of Conservation of

Momentum states that in an isolated system, (one subject to no outside influences), the total momentum is conserved.

• So in a collision (say a car hitting a tree and coming to a stop), the total momentum before the collision = total momentum after the collision.

Velocity = v

Mass of car pluspassengers = m

Momentump = mv

Mass of car pluspassengers = m

v = 0 Momentump = 0

THE CONSERVATION LAW DOES NOT ALLOW MOMENTUM TO DISAPPEAR.THE APPARENTLY “LOST” MOMENTUM HAS, IF FACT, BEEN TRANSFERRED THROUGH THE TREE TO THE EARTH.Since the Earth has a huge mass (6.0 x 1024 kg) the change in its velocity is negligible.

3.6 The Physics of Crumple Zones & Air Bags

A car crashes into a concrete barrier.The change in momentum suffered by the car (and passengers) is a fixed quantity.So, Impulse, (the product of F and t), is also fixed. However, individual values of F and t can vary as long as their product is always the same.So if t is made longer, consequently F must be smaller.Crumple Zones increase the time (t) of the collision.So, F is reduced and the passengers are less likely to be injured.The same logic can also be applied to Air Bags

The air bag increases the time it takes for the person to stop.So the force they must absorb is lessened. So they are less likely to be seriously injured.

A further benefit is this lesser force is distributed over a larger area


Question type:

In a car the driver’s head is moving horizontally at 8.0 ms-1 and collides with an air bag as shown. The time taken for the driver’s head to come to a complete stop is 1.6 x 10-1 s. This collision may be modelled as a simple horizontal collision between the head of mass 7.0 kg and the air bag.

Q24: Calculate the magnitude of the average contact force that the air bag exerts on the driver’s head during this collision.

A: Impulse = Change in Momentum FΔt = Δ(mv) So F = Δ(mv)/Δt = (7.0)(8.0)/(1.6 x 10-1) = 350 N

Momentum/Impulse


Question type:

Q25: Explain why the driver is less likely to suffer a head injury in a collision with the air bag than if his head collided with the car dashboard, or other hard surface.A: The change in momentum suffered by the driver’s head is a FIXED quantity no

matter how his head is brought to rest.• Therefore the product of F and t (ie Impulse) is also a fixed quantity.• However the individual values of F and t may be varied as long as their

product always remains the same. • The air bag increases the time over which the collision occurs, therefore

reducing the size of the force the head must absorb so reducing the risk of injury.

• The air bag also spreads the force over a larger area, reducing injury risk.• Without the air bag the driver’s head may hit a hard surface decreasing the

time to stop his head and necessarily increasing the force experienced and thus the likelihood of injury.

• In addition the force will be applied over a much smaller area increasing the likelihood of severe injury

In a car the driver’s head is moving horizontally at 8.0 ms-1 and collides with an air bag as shown. The time taken for the driver’s head to come to a complete stop is 1.6 x 10-1 s. This collision may be modelled as a simple horizontal collision between the head of mass 7.0 kg and the air bag.

Air Bags/Crumple Zones

Chapter 4

Topics covered:

• Centre of Mass.

• Weight.

• Reaction Force.

• Bouncing Balls.

• Friction.

• Various Force Applications

4.0 Centre of MassIn dealing with large objects it is useful to think of all the object’s mass being concentrated at one point, called the Centre of Mass of the object.The C of M of the object is the point around which it will spin if a torque or turning force is applied to the object.

Centre of Mass

For oddly shaped objects eg. a boomerang, the C of M may fall outside the perimeter of the object.

C of M

For regularly shaped objects eg. squares or rectangles, cubes or spheres the Centre of Mass of the object is in the geometric centre of the object

4.1 Centre of Mass - SystemsFor a system of 2 or more bodies, the

position of the C of M may be determined

from the formula:

50 kg beam

5.0 m

A

Centre of Massof each mass

1.0 m 1.0 m

30 kg mass

WHAT IS THE CENTRE OF MASS OF THIS SYSTEM?

CHOSEN REFERENCE POINT IS A

X C of M = [(50 x 2.5) + (30 x 3.5)]

(50 + 30)

2.5 m

3.5 m

= 2.875 m from ACentre of Mass

of System

2.875 m

XCofM = (m1x1 + m2x2 + m3x3 + …)

(m1 + m2 + m3 +…)

XC of M is the position of the Centre of Mass of the System as measured from a CHOSEN REFERENCE POINT.m1,m2,m3, etc are the masses of the individual components of the Systemx1, x2, x3 etc are the distances measured from the CHOSEN REFERENCE POINT to the centres of mass of the system’s components.

4.2 Weight• The effect of a Gravitational Field on a

Mass is called its WEIGHT.• Weight is a FORCE and therefore a

Vector quantity.• Mathematically:

W = mg where W = Weight (N) m = mass (kg) g = Gravitational Field Strength (Nkg-1)

• Weight acts through the Centre of Mass of the body and is directed along the line joining the centres of the the two bodies between which the Gravitational Field is generated.

• On Earth, the Gravitational Field of Strength 9.8 Nkg-1 gives any mass under its influence alone an acceleration of 9.8 ms-2

Centre of Mass

Object of Mass (m)

Weight = mg

Weight Force acts through the Centre of Mass and is directed toward the Centre

of the Earth

4.3 Reaction Force• Any stationary object which is

under the influence of the Earth’s Gravitational Field must be subject to a force equal in size, but opposite in direction to, its weight.

• This equal but opposite force is called the NORMAL REACTION FORCE.

• The Normal Reaction Force only arises from the action of the weight force and DOES NOT EXIST AS AN ISOLATED FORCE IN ITS OWN RIGHT.

Table

R - Normal Reaction Force

W - Weight

ObjectMass = m

R acts upwards from the boundary between the table and the mass through the Centre of Mass

W acts downwards from the Centre of Mass toward the centre of the Earth

R and W are NOT an ACTION - REACTION PAIRWHY ?

4.4 Bouncing Balls.When a ball, falling under the action of gravity, arrives at a hard surface, it is the Normal Reaction Force (R), which provides;

(a) The Force required to decelerate the ball to a stop and then,

(b) The Force needed to accelerate the ball away from the surface.

v = 0

a = g

W

The ball is dropped:Velocity = 0Acceleration = g

v

a = g

W

v

a = g

W

W

v

a = g

W

R

av

Ball strikes surface, R arises and begins todecelerate ball.R > W (acc now up)

W

v = 0

R

aa = g

W

v

a = g

W

v

a = g

W

v

a = g

W

v = 0

4.5 Friction• Friction is the most unusual of all

forces as it cannot start an object moving.

• Friction can, however, slow or stop an object once it is moving or prevent it from starting to move.

• Slowing a moving object is the result of DYNAMIC FRICTION, while stopping an object from starting to move is the result of STATIC FRICTION.

• Generally speaking, for the same pair of surfaces, Static Friction Dynamic Friction.

• The size of the frictional force depends upon:

(a) The Roughness of the surfaces measured by the Coefficient of Friction ()

(b) The Separation of the surfaces.• Friction does NOT depend on the

AREA in contact.

W

R

F (Pulling Force)

Fr

(Static Friction)

OBJECT HELD STATIONARY BY STATIC FRICTION

The size of the Frictional Force Fr is calculated from:

Fr = R

4.6 Friction Applications• Friction is NOT always a

hindrance to living on Earth, often it is vital for movement over the Earth’s surface.

• Consider the following:

• A car is accelerating along a Rough (meaning having Friction) Road Surface.

FTR

Force of TYRE on

ROAD.ACTIONFORCE

Direction of rotation of Driven Wheel

FRT

Force of ROAD on

TYRE.REACTION

FORCE

Acceleration

In order for the Car to accelerate in the direction shown a force must exist in that direction.

Thus it is the Force of the Road on theTyre (FRT) which gives rise to the acceleration

The frictional force between the tyre and the road (FTR) is directed backwards. This force CANNOT provide the forward propulsion.

IT IS THE REACTION FORCE WHICH PRODUCES THE ACCELERATION

4.7 Various Force Situations: Stationary and Falling Bodies

MASS - STATIONARY ON A TABLE

R Normal Reaction ForceActs from the point ofcontact b/w Mass &Table

Centre of Mass

W = mg Weight acts directly downfrom the Centre of Mass

+ ve direction

a = 0

From Newton 2 F = ma

F = R - W = mabut a = 0

So, R - W = 0and R = W

MASS - FALLING WITH AIR RESISTANCE

W = mg

FR

F = ma F = W - FR

So, W - FR = maand FR = ma - mg

= m(a - g)

+ve

a

4.8 Various Force Situations: Lifts

a

+veR

W = mg

LIFT - ACCELERATING UPWARDS

For the Mass (m)F = ma

F = R – Wso, R - W = ma

and R = W + ma = mg + maThus R = m(g + a)

Mass (m) on Floor of Lift

LIFT - ACCELERATING DOWNWARDS

R

W = mg

Mass (m) on Floor of Lift

For the Mass (m)F = ma

F = W - Rso ,W - R = ma

and R = W - ma = mg - ma

Thus R = m(g - a)

a

+ve

R is a measure of the “apparent weight” of the mass. When accelerating upward the mass is “heavier” than normal, by an amount (ma), and when accelerating downward it is “lighter” than normal, by (ma).

4.9 Various Force Situations: Inclined Planes

R

W = mg

a

mg Sin

mg Cos

Body of mass ( m) on an inclined planeWeight and Normal Reaction act on massWeight broken up into 2 componentsForce triangle and inclined plane triangle

are similar, ’s are equal.

Perpendicular Component (mg Cos ) = R

Parallel (to the plane) component = mgSin Parallel Component can be transferred to act through the C of M. IT IS THIS COMPONENT THAT ACCELERATES THE MASS DOWN THE PLANE

W

Ra

W

aR

W

aR

W

Ra

Notice the Component of the Weight parallel to the plane (mg Sin) increases as the plane gets steeper thus increasing the acceleration of the mass down the plane.

4.10 Various Force Situations:Connected Masses

m1

T

Ta+ve

F = ma

For m1: T = m1a

For m2 : W - T = m2a

Frictionless Table

Frictionless Pulley

m2

W = m2 g

m2

m1

T

T

W = m2g

W = m1g

a +ve

Frictionless Pulley

Masses connected byLight Inextensible String. (Tension (T)

the same everywhere)

Light inextensible String. (Tension (T)

the same everywhere)

For m1: T - m1g = m1a

For m2: m2g - T = m2a

Solve simultaneous equations to get acceleration

Chapter 5

Topics covered:

• Work.

• Work & Energy.

• Power.

• Energy Types.

• Conservation of Energy.

5.0 Work• The term WORK has a very strict

definition in the physics world.

• If a FORCE moves an object through a DISTANCE, WORK has been done on that object.

• Mathematically: W = F.d

where W = Work (Joules) F = Force (N) d = Distance (m)

• This formula can only be used if the Force remains constant through the course of doing the work.

• Work is a SCALAR quantity.

• No work has been done if the force has not caused the object to move.

• No work has been done if the object begins and ends its movement in the same place. ie travels in a circle.

Area = Work done by the Force

Work = ½F x d

Force

distanced

F

If the Force varies in the course of doing the Work (as in stretching a spring), the work done can only be calculated from the area under the Force versus distance graph.

WORK DONE BY A VARIABLE FORCE

5.1 Work & EnergyIn the Physics world, Work and Energy are intimately related.Energy is very difficult to define. It is easy to say what energy can do but not so easy to say what it is.Thus energy is defined in terms of work.

WORK DONE = ENERGY TRANSFERRED

An object is said to possess energy if it has the ability or capacity to do work.Work is the “Transfer Mechanism” for Energy meaning that if some work has been done on an object the amount of energy it possesses has been changed.


Question type:

Q26 : How much energy is stored in the spring when it is compressed?

A: Compressing the spring requires work to be done on it. (this work is stored as elastic potential energy in the spring)Work done = area under graph up to a compression of 0.2 m = ½ (0.2)(1000)

= 100 J

A model rocket of mass 0.20 kg is launched by means of a spring, as shown in Figure 1. The spring is initially compressed by 20 cm, and the rocket leaves the spring as it reaches its natural length. The force-compression characteristic of the spring is shown in Figure 2.

Work and Energy


Question type:

Q27: What is the speed of the rocket as it leaves the spring?

A: All the elastic potential energy stored at max compression (100 J) will be converted to Kinetic Energy at release. So

100 = ½ mv2

v = √1000 = 31.6 ms-1

Energy Conversion


Question type:

Q28: What is the maximum height, above the spring, reached by the rocket? You should ignore air resistance on the way up since the rocket is very narrow.

u = 31.6 ms-1

v = 0a = -10 ms-2

x = ?t = ?

+ve

v2 = u2 +2ax0 = (31.6)2 + 2(-10)xx = 1000/20 = 50 m

Equations of Motion


Question type:

Retarding Force (R) (Newtons)

time (s)

When the rocket reaches its maximum height, the parachute opens and the system begins to fall. In the following questions you should still ignore the effects of air resistance on the rocket, but of course it is critical to the force on the parachute. This retarding force due to the parachute is shown as R in Figure 3, and its variation as a function of time after the parachute opened is shown in Figure 4.

Q29: What is the acceleration of the rocket at a time 5 s after the parachute opens?

R

W

+veA: At t = 5.0 s, R = 1.8 NWeight of Rocket W = mg = (0.2)(10) = 2.0 N ΣF = ma

a = ΣF/m = (2.0 – 1.8)/ 0.2 = 1.0ms-2

Newtons 2nd Law


Question type:In a storeroom a small box of mass 30.0 kg is loaded onto a slide from the second floor, and slides from rest to the ground floor below, as shown in Figure 4. The slide has a linear length of 6.0 m, and is designed to provide a constant friction force of 50 N on the box. The box reaches the end of the slide with a speed of 8.0 m s–1

Q30: What is the height, h, between the floors?

Figure 4

At the second floor the box has Potential Energy = mghOn reaching the ground floor this has been converted to Kinetic Energy (½mv2) plus the work done against friction in moving down the slope

thus, mgh = ½ mv2 + Fd (30.0)(10)h = ½ (30)(8.0)2 + 50(6.0)

h = 4.2 m

Work and Energy

5.2 Energy TypesKINETIC ENERGY

• The Energy of Motion. It is the energy possessed by moving objects

• Mathematically: K.E. = ½mv2 where, K.E. = Kinetic Energy (J ) m = mass (kg) v = velocity (ms-1)

GRAVITATIONAL POTENTIAL ENERGY• The Energy of Position. The energy

possessed by an object due to its position.

• Mathematically: P.E. = mgh where, P.E. = Grav. P. E. (J ) m = mass (kg) g = gravitational field

strength (N kg-1) h = height above zero point (m)

• The zero point for measuring height is usually, but not always, the surface of the Earth.

ELASTIC POTENTIAL ENERGY•The energy stored in elastic materials (eg. Springs & rubber bands)•Mathematically:

Es = ½ kx2 where, Es = Elastic P. E. (J )

k = Spring Constant (N kg-1) x = Extension or

Compression (m)

Energy is not a directional quantity so all these forms of energy are Scalar Quantities.


Question type:

The box then slides along the frictionless floor, and is momentarily stopped by a spring of stiffness 30 000 N m–1

Q31: How far has the spring compressed when the box has come to rest?

KE of box is converted to Elastic Potential Energy in the spring

Thus ½ mv2 = ½ kx2

½ (30.0)(8.0)2 = ½ (30,000)(x)2

x = 0.25 m

Figure 4

Elastic Potential Energy

5.3 Power• Power is defined as the Time rate of

doing Work.• Mathematically:

P = W/t where,

• P = Power (W - Watts) W = Work (J ) t = time (s)

• Power is a Scalar Quantity• SINCE WORK DONE

= ENERGY TRANSFERRED• Power can also be defined as the Time

rate of Energy Transfer.• Mathematically:

P = E/t where, P = Power (W - Watts)

E = Energy (J ) t = time (s)

• It is useful to note that since W = F.d and v = d/t the original power formula

P = W/t = (F.d)/t =F.v • This allows the calculation of the

power of a body moving at constant velocity

5.4 Conservation of Energy

• The law of conservation of energy states:ENERGY CAN BE NEITHER CREATED NOR DESTROYED BUT ONLY TRANSFORMED FROM ONE FORM TO ANOTHER.

• Energy conversion processes convert useful, ORDERED energy (energy capable of doing useful work) into useless, DISORDERED energy. This disordered energy is sometimes called Thermal Energy, or Low Grade Unrecoverable Heat.

The percentage efficiency of energy transfers is calculated from: %Eff = EOUT/EIN x 100/1No energy transfer process can be 100% efficient in our friction ridden world.If 100% efficiency could be attained perpetual motion machines would then be possible.

5.5 “Seeing” Energy (1)

As KE leaves the object it slows down: as in a car slowing down when you lift off the accelerator pedal.

With a little practice you can “watch” energy flow through a system just as an accountant can “watch” money flow through an economy.

The most obvious form of energy is Kinetic Energy, the energy of motion. It is easy to see when kinetic energy is transferred to or from an object. A water polo ball speeds up when

you do work on it during the action of throwing, you are transferring energy from your body into the ball, where the energy shows up in the motion of the ball.

5.6 Seeing Energy (2)Potential Energy is more difficult to “see”.It can take many different forms, as shown in the table below. In each case nothing is moving; but because the objects still have a great potential to do work, they possess Potential Energy.

At certain points all a body's energy may be potential whilst at another all its energy may be kinetic. Recognizing these points will enable the solutions to most problems in this area.

Form of Potential Energy Example

1. Gravitational Potential Energy A person standing at the top of a building

2. Elastic Potential Energy A wound clock spring

3. Electrostatic Potential Energy A cloud in a thunderstorm

4. Chemical Potential Energy A firecracker

5. Nuclear Potential Energy Uranium

Chapter 6Topics covered:• The Experience of Acceleration.• Circular Motion. • Centripetal Force.• Vertical Circles.• Projectile Motion.• Projection Angles• Projectile Graphs• Real Life Projectiles

6.0 The Experience of Acceleration

Nothing is more central to the understanding of the laws of motion than understanding the relationship between force and acceleration. Up to now we have looked at forces and noticed they can produce accelerations (Newtons 2nd Law). Now we will reverse the process – looking at accelerations and noticing that they require a force. For you to accelerate, something must push or pull on you.

The “backward” force you experience when a car accelerates is caused by your body’s “inertia”, its tendency not to want to accelerate.

Just where and how that force is exerted on you determines how you “feel” when you accelerate.

The car and your seat are accelerating, and since the seat back acts to keep you from falling “through” its surface, it provides a forward support force which causes you to accelerate forward. But the seat can’t exert a force uniformly throughout your body. Instead, it pushes only on your back and your back then pushes on your bones, internal organs and tissues to make them accelerate forward.The experience of the accelerating car seat is very similar to the experience of “gravity” when you stand still on the Earth’s surface. In this situation you feel “heavy” ie. you experience your weight

6.1 Fictitious Forces

When the car seat is causing you to accelerate forward, you also feel “heavy”; your body senses all the internal forces needed to accelerate its pieces forward, and you interpret these sensations as “weight”. This time you experience the weight directed toward the back of the car.

The gravity-like “force” that you experience as you accelerate is truly indistinguishable from the force of gravity. No laboratory instrument can determine directly whether you are experiencing gravity or simply accelerating.

However despite the convincing sensations, the backward, heavy feeling in your gut as you accelerate is not due to a real force. This experience of acceleration is explained by the supposed action of a Fictitious Force.

This Fictitious Force always points in the direction opposite to the acceleration that causes it and its strength is proportional to the acceleration.

aFFict

6.2 Apparent Weight

Apparent Weight

Weight

aFictitious Force

With the car stationary the only force you are subject to is your Weight

a

WeightApparent Weight

Fictitious Force

a

WeightApparent Weight

Fictitious Force

The larger the Acceleration, the larger the Fictitious force, the more “backward” your Apparent Weight becomes.

The vector sum of the Weight and Fictitious force produces the Apparent Weight youexperience whilst accelerating.

As the car accelerates you are “pushed back” in your seat by the fictitious force

6.3 Circular Motion• Newton’s 1st law says, in part, an

object experiencing no net force will travel in a straight line at constant speed.

• Thus, in order to make an object travel in a circle ie. constantly change the direction of its velocity, a force must be constantly applied to it. This means the object is constantly accelerating.

• The number of times an object spins around per second is called its Frequency (f).

• An object travelling around a circle completes one full spin in a time interval called a Period (T). Period and frequency (f) are the inverse of one another. Thus:

T = 1/f• T =Period (sec)• f = Frequency (Hz or sec-1)

V

R

CIRCLE OF RADIUS = R

Circumference of circle = 2 R

Time for one revolution = Period = T

Speed = Distance , So v = 2 R = 2 Rf Time T

6.4 Centripetal Acceleration

The size of that acceleration (called the CENTRIPETAL (or centre seeking) ACCELERATION) is given by:

aC = v2/R where aC =centripetal acc (ms-2) v = linear velocity (ms-1) R = radius of circle (m)Substituting for v = 2R/T gives aC = 4 2 R/T2

CIRCLE VIEWED FROM ABOVE

vi

vf

aC = vf - vi

- vi

aC

THE CENTRIPETAL ACCELERATION, aC ,IS ALWAYS DIRECTED TOWARD THE CENTRE OF THE CIRCLE.

Since the direction of the velocity at any point is the tangent to the circle at that point, the direction of the velocity is constantly changing, thus an object travelling in a circular path is subject to a constant acceleration.

6.5 Centripetal Force

aCFC

The fact that an acceleration exists, directed toward the centre of the circlerequires there to be a force acting in the same direction.This force is called the CENTRIPETAL FORCE (FC) and is defined in terms of Newton’s 2nd law.

Mathematically:FC = maC = mv2/R = m42R/T2

Centripetal Force is NOT a REAL FORCE in its own right, but is supplied by other real, measurable forces.In the case opposite, the girl requires a Centripetal Force to travel her circular path

She, of course, will use the idea of a Fictitious Force, centrifugal force, to explain the “outward” pull she feels while on the constantly accelerating ride.

Linear Velocity

Direction of Rotation

Centre of Mass

FC

Fictitious Force

Girl Riding a Playground Ride T

The Tension (T) of her muscular grip on the pivot pole of the ride provides the Centripetal Force.


Question type:Mark Webber and his Formula 1 racing car are taking a corner at the Australian Grand Prix. A camera views the racing car head on at point X on the bend where it is travelling at constant speed. At this point the radius of curvature is 36.0 m. The total mass of the car and driver is 800 kg.

FCQ32: On the diagram showing the camera’s view of the racing car, draw an arrow to represent the direction of the NET force acting on the racing car at this instant.

The magnitude of the horizontal force on the car is 6400 N.

Q33: Calculate the speed of the car.

A: FC = mv2/R So v =√FCR/m

= √(6400)(36.0)/(800) = 17 ms-1

Centripetal Force

36.0 m

X

Camera

Camera's head on viewof racing car at point X


Question type:

A: (a) Newton 1 states all objects will continue in a straight line unless acted upon by a net force.

In order for the car to travel in a circular path a constant force directed at right angles to the direction of motion (toward the centre of the circle) must exist.

(b) The horizontal force arises from the frictional force exerted BY THE ROAD ON THE TYRES and is directed toward the centre of the circle the car is travelling in.

Centripetal Force

Q34: Referring to the racing car from the previous slide, explain:(a) Why the car needs a horizontal force to turn the corner.(b) Where this force comes from.


Question type:The safe speed for a train taking a curve on level ground is determined by the force that the rails can take before they move sideways relative to the ground. From time to time trains derail because they take curves at speeds greater than that recommended for safe travel.Figure 5 shows a train at position P taking a curve on horizontal ground, at a constant speed, in the direction shown by the arrow.

Q35: At point P shown on the figure, draw an arrow that shows the direction of the force exerted by the rails on the wheels of the train.

Fc

Centripetal Force


Question type:The radius of curvature of a track that is safe at 60 km/h is approximately 200 m.

Q36: What is the radius of curvature of a track that would be safe at a speed of 120 km/h, assuming that the track is constructed to the same strength as for a 60 km/h curve?

This can be solved as a ratio question

m(60)2 = m(120)2

200 x

Fc = mv2

R

A: Centripetal Force

x = 800 m

A: B

Centripetal Force

Q37: At point Q the driver applies the brakes to slow down the train on the curve.Which of the arrows (A to D) indicates the direction of the net force exerted on the wheels by the rails?

6.6 Centrifugal Force

Fictitious forces, such as Centrifugal Force, do NOT contribute to the net force experienced by an object. Thus if the girl lets go she will fly off the ride in the direction of the linear velocity, NOT in the direction of the fictitious force.

Despite its fictitious nature, centrifugal force creates a compelling sensation of gravity like force. The girl on the ride feels as though gravity is pulling her outward as well as down and must hold the handle tight in order not to fall off.

A stationary object on Earth experiences a weight force of 1g.

The fictitious centrifugal force experienced by 5 kg of clothes during the spin cycle in a washing machine, travelling in a 0.25 m radius circle at 20 ms-1 is about 163 g’s and they have an apparent weight of 815 kg.Linear Velocity

Anyone who talks about centrifugal force as being a real, measurable force is talking rubbish and does not understand physics, be wary of ALL they say !!!!!

6.7 Banked CornersRace track and road designers often build their tracks or roads with “banking” on the corners. This design feature is used to enhance the safety of the track or road allowing users to round corners at higher speeds (and with a greater margin of safety) than they could if the corner was not banked.

W

R

FRTFRT

FC FC

FC

v

Any vehicle travelling around a corner with velocity v needs a centripetal force (FC) acting toward the centre of the cornerIn the case of the flat, non banked, corner this centripetal force (FC) is supplied by the friction of the road against the tyres (FRT)

With the banked corner the centripetal force has an extracomponent - that being a component of the car’s weight force acting toward the centre of the corner.

W

R

FRT

FRT

Component of W acting toward centre of corner

This gives a larger overall Centripetal Force (FC) larger than in the non banked case.

F LATCORNER BANKED

CORNER

The larger FC allows the car either a greater margin for safety or a faster speed around

a banked corner compared to a flat corner of the same radius

6.8 Vertical CirclesIn theoretical situations, the TOTAL ENERGY REMAINS CONSTANT, but varies between Kinetic and Potential, depending on where in the circle you choose to look.

h = 0

A soccer ball of mass 0.5 kg moves along a friction free track at 20.0 ms-1 toward a vertical circle 10.0 m in diameter.

Its initial K.E. = ½mv2 = ½(0.5)(20)2 = 100 J

5.0 m

PE = mgh =(0.5)(9.8)(5.0) = 24.5 J KE = 100 - 24.5 = 75.5 J

PE = 49 JKE = 51 JPE = 24.5 JKE = 75.5 J

10 m

KE = 100 JPE = 0 J

KE =100JPE = 0

v = 20 ms-1

Objects travelling in vertical circles are subject to the acceleration due to gravity, thus their speed will vary depending where in their motion observations are made.

Analysis of this type of motion is based on energy considerations and the fact that the motion takes place in a uniform gravitational field.

vHORIZ

vVERT

6.9 Projectile Motion

vHORIZ

vVERT

vHORIZ

vVERT

vHORIZ

vVERT

vHORIZ vHORIZ

vVERT

vHORIZ remains constant throughout

vVERT increases under the action of gravity

The only common factor between the the two motions is THE TIME OF FLIGHT (t).

Projectile Motion is the motion which objects, launched at some angle to the Earth’s gravitational field, will undergo.

Projectile motion is a combination of TWO INDEPENDENT MOTIONS.(a) HORIZONTAL MOTION which is CONSTANT VELOCITY motion. (b) VERTICAL MOTION which is CONSTANT ACCELERATION motion.

The horizontal motion, being constant velocity motion, is covered by only one equation: v = d/t.

The vertical motion, being constantly accelerated motion, is covered by the Equations of Motion


Question type:

AB

C

D

EF

GH

AB

C

D

EF

GH

XY

Direction of Motion

Projectile Motion

Q38: Which of the directions (A - H), best shows the VELOCITY of the car at X ?

A car takes off from a ramp and the path of its centre of mass through the air is shown below.First model the motion of the car assuming that air resistance is small enough to neglect.

A: Direction C.Q39: Which of the directions (A - H), best shows the VELOCITY of the car at Y ?

A: Direction D


Question type:Q40: Which of the directions (A - H), best shows the ACCELERATION of the car at X ?

A: Direction E.

Q: Which of the directions (A - H), best shows the ACCELERATION of the car at Y ?

A: Direction E.

Now suppose that AIR RESISTANCE CANNOT BE NEGLECTED.

Q41: Which of the directions (A - H), best shows the ACCELERATION of the car at X ?

A: Direction F.

Projectile Motion

AB

C

D

EF

GH

AB

C

D

EF

GH

XY

Direction of Motion

Figure 1Q42: What is the value of h in Figure 1?

A bushwalker is stranded while walking. Search and rescue officers drop an emergency package from a helicopter to the bushwalker. They release the package when the helicopter is a height (h) above the ground, and directly above the bushwalker. The helicopter is moving with a velocity of 10 ms–1 at an angle of 30° to the horizontal, as shown in Figure 1. The package lands on the ground 3.0 s after its release. Ignore air resistance in your calculations.


Question type:

A: Can approach in a number of ways.

One way is: Find how long it takes to rise and fall back to same vertical height (h)Then subtract from total time to give a time for vertical fall under gravity with an initial velocity of 10 ms-1 directed at 30o below the horizontal

Projectile Motion


Question type:

Figure 1

h

1. Time to get back to hSince this part of the journey is symmetrical time to reach max height = ½ time for this part of journey.

Analyse vertical motion

u = - 10 Sin 30o ms-1

v = 0a = 10 ms-2

s = ?t = ?

v = u + at0 = -10 Sin 300 + 10tt = 5/10 = 0.5 sec.

So, total time to get back to h = 1.0 s

2. Value of h

again analyse vertical motion

+ve

+ve

u = 10 Sin 30o

v = ?a = 10 ms-2

s = ht = 2.0 s

s = ut + ½ at2

h = (5)(2) + ½ (10)(4)h = 30 m

Projectile Motion


Question type:

Q43: Assuming that the helicopter continues to fly with its initial velocity, where is it when the package lands? Which one of the statements below is most correct?

A. It is directly above the package. B. It is directly above a point that is 15 m beyond the package. C. It is directly above a point that is 26 m beyond the package. D. It is directly above a point that is 30 m from the bushwalker.

Figure 1

Both the package and the helicopter have the same (constant) horizontal velocity, so the package will always be directly below the helicopter. So, alternative A is correct

Projectile Motion


Question type:Q44: Which of the graphs below best represents the speed of the package as a function of time?

A: The key to the question is SPEED, this is the sum of both vertical and horizontal. Speed never falls to zero (there is always a horizontal component), so C and D are outA shows the object’s acceleration falling as it nears the ground – not so ! So B must be the correct answer

Sketch Graphs


Question type:Fred is playing tennis on the deck of a moving ship. He serves the ball so that it leaves the racket 3.0 m above the deck and travels perpendicular to the direction of motion of the ship. The ball leaves the racket at an angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 ms-1. You may ignore air resistance in the following questions.

Q45: With what speed, relative to the deck, did the ball leave Fred’s racket? Give your answer to three significant figures.

30 ms-1

x

8o

A: Assuming projectile behaviour:VHORIZONTAL = 30 ms-1 throughout the flight

30 ms-1

thus, Cos 8 = 30/x x = 30/Cos 8 = 30.3 ms-1

x = initial speed

Projectile Motion


Question type:

+ve

u = VVERT = 30.3 Sin 8v = 0a = -10 ms-2

s = ?t = ?

v2 = u2 + 2as0 = (30.3 Sin 8)2 + 2(-10)ss = 0.89 m

So Height above the DECK = 3.0 + 0.89 = 3.89 m

A: Using the equations of motion for the balls flight from racquet to point of max height, where VVERT = 0

30 ms-1

x

8o

30 ms-1

VVERT

Q46: At its highest point, how far was the ball above the deck?

Projectile Motion


Question type:The ship is travelling straight ahead at a

velocity of 10 ms-1

Q47: When the ball is at its highest point at what speed is it moving relative to the ocean?

30 ms-1

10 ms-1

A: V BALL REL OCEAN = VSHIP + VBALL

VBALL+VSHIP=

30

VBALL REL OCEAN

10 θ

VBALL REL OCEAN = √(30)2 + (10)2

= 31.6 ms-1

Q48: at what angle is the ball travelling relative to the direction of the ship’s travel?

Tan θ = 30/10 = 3.0 θ = 71.60

Relative Motion

6.10 Projection Angles

• There are two basic types of projectile motion:

(a) Objects initially projected horizontally from some point above the Earth’s surface. As shown previously .

(b) Objects initially projected at some angle above the horizontal as shown here.

v

vHORIZ

vVERT

vHORIZ = v Cos This does not vary

during the flight

vVERT = v Sin This does vary from +v Sin at t = 0 to zero at t = ½t to - v Sin at t = t

Flight can be broken up into 2 equal parts each taking exactly

½t to complete

Total flight time = tAt this point all velocity is horizontal

v

+ve

6.11 Projectiles – Time, Height & Range

h

Range (R)θ

v

An object is fired from ground level with a velocity v at an angle θ to the horizontalIt reaches a maximum vertical height, hand finally lands at the same horizontal level as it started, having covered a horizontal distance of R

1. Time to reach Maximum Height (h)Upward is +ve.

u = v sin θv = 0a = -gs = ?t = ?

use eqns of motion: v = u + at 0 = v sin θ – gt t = v sin θ g

2. Total Time of FlightTotal time = (time to reach h) x 2 = 2 v sin θ

g

3. Maximum Height Upward is +ve

u = v sin θv = 0a = -gs = ht = v sin θ

g

use eqns of motion v2 = u2 + 2as0 = v2 sin2θ – 2ghh = v2 sin2θ 2g

4. Range of ProjectileHorizontally vH = dH / tvH = v cos θdH = Rt = 2v sinθ g

R = 2v sinθ x v cos θ g = v2 x 2 sin θcos θ g = v2 sin 2θ g

5. Maximum Rangeoccurs when sin 2θ = 1 2θ = 900

or θ = 450

6.12 Real Life Projectiles• The mathematical analysis of projectile

motion ignores the effects of friction, in particular air resistance.

• The real world trajectory differs from the theoretical trajectory in three main ways.

Theoretical

Actual

Loss of Range

Loss of height

Path no longer symmetrical

(3) The object will fall more steeply than it rises, the path of the projectile is no longer symmetrical around its highest point.

(2) The actual height achieved will be less, due to the effect of air resistance on the upward vertical velocity.

(1) The actual range (horizontal distance) covered in the real world is less because of the reduction of the horizontal component of velocity due to air resistance.

Chapter 7

Topics covered:

• Elastic Materials and Hooke’s Law.

• Elastic Potential Energy.

7.0 Elastic Materials & Hooke’s Law

Elastic materials such as springs display elastic behaviour. This means that when they are deformed, stretched or compressed, they change their shape or condition.

When the force causing the deformation is removed, they return to their original shape or condition.

Such materials are able to store energy while deformed and release that energy when allowed to return to their original condition.

The relation between Force and the Extension and/or Compression of an elastic material is called Hooke’s Law:

whereF = Restoring Force (N), k = Spring Constant (Nm-1) ∆x = Extension and/or Compression (m)

MathematicallyF = - k(∆x)

The negative sign indicates that the restoring force F acts in the opposite direction to the extension or compression ∆x

7.1 Potential Energy in Elastic Materials

• Elastic materials store energy when they are deformed and release that energy when they return to their original condition.

• The amount of energy stored can be found from the Elastic Potential Energy Formula: ES = ½kx2 where

ES = Elastic P. E. (J ) k = Spring Constant (Nm-1) x = extension and or compression (m)

• For materials which display “irregular” behaviour, the Potential Energy stored can only be found from the area under the Force vs Extension (or Compression) graph.

Area = ½Fx = ½kx2

= Energy stored up to extension x

Force

Extension

Slope = Spring constant (k)

x

F

“REGULAR” ELASTIC BEHAVIOUR

“IRREGULAR” ELASTIC BEHAVIOUR

Extension

x

Force

F

Area = Energy stored up to xArea is found by “counting squares”

Chapter 8Topics covered:•Universal Gravitation•Gravitational Attraction•Circular Orbits•Maths of Circular Orbits•Kepler’s Laws•Satellites in Space•Energy Transfers

8.0 Law of Universal Gravitation• Gravity is the most well known of

all the Natural Forces.• We live with the effects of gravity

every day and would lead completely different lives if gravity was not present. It is gravity alone which gives us our sense of “up and down”.

• Gravity is a FIELD force, it acts at a distance.

• It is ALWAYS an ATTRACTIVE force.

• Gravity arises from the interaction of 2 masses.

• The size of the gravitational force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between them.

Mathematically: Fg = GM1M2

R2

where Fg = Gravitational Force (N) G = Universal Gravitational

Constant (6.67 x 10-11 Nm2kg-2) M1, M2 = masses (kg) R = Separation of the masses (m)EACH of the bodies experiences the

SAME FORCE even though they may have vastly different masses.


Question type:

Newton was the first person to quantify the gravitational force between two masses M and m, with their centres of mass separated by a distance R asF= GMm R2

where G is the universal gravitational constant, and has a value of 6.67 × 10-11 N m2 kg2.For a mass m on the surface of Earth (mass M) this becomes F = gm, where g = GM R2

Q49: Which one of the expressions (A to D) does not describe the term g?A. g is the gravitational field at the surface of Earth.B. g is the force that a mass m feels at the surface of Earth.C. g is the force experienced by a mass of 1 kg at the surface of Earth.D. g is the acceleration of a free body at the surface of Earth.

A: B

Universal Gravitation


Question type: The radius of the orbit of Earth in its circular motion around the Sun is 1.5 × 1011 m

(Figure 3).

Q50: Indicate on the diagram, with an arrow, the direction of the acceleration of Earth.

Q51: Calculate the mass of the Sun. Take the value of the gravitational constant G = 6.67 × 10–11 N m2 kg–2.

Question assumes you know the period of rotation of the earth around the sunie. 365 days = 365 x 24 x 60 x 60 = 31536000 s (3.15 x 107 s)

FC = Fg

me4π2 R/T2 = Gmems/R2

ms = 4π2R3

GT2

ms = 4(3.14)2(1.5 x 1011)3

6.67 x 10-11(3.15 x 107)2

ms = 2 x 1030 kg



Question type:

A: The centripetal force (FC) required by Nato III to complete its circular orbit is supplied by the force of gravitational attraction (Fg) between Earth and Nato III.

Thus Fg = FC

Thus GMem/R2 = mv2/R

So v = √GMe/R

= √(6.67 x 10-11)(5.98 x 1024) --------------------------------------

(4.22 x 107)

= 3074 ms-1

= 3.1 x 103 ms-1


Q52: What is the speed of Nato III in its orbit ?

EarthNato III

R

Nato III is a communication satellite that has a mass of 310 kg and orbits Earth at a constant speed at a radius R = 4.22 x 107 m from the centre of Earth.


Question type:

Q53: Which ONE of the following statements (A - D) about Nato III is correct ?

A: The net force acting on Nato III is zero and therefore it does not accelerate.

B: The speed is constant and therefore the net force acting on Nato III is zero.

C: The is a net force acting on Nato III and therefore it is accelerating.

D: There is a net force acting on Nato III, but it has zero acceleration.

A: Alt C is correct

Circular Motion

EarthNato III

R

8.1 Gravitational Attraction

Earth

MoonR = 384,000 km = 3.84 x 108 m

Mass of MoonMM= 7.34 x 1022 kgMass of Earth

ME = 5.98 x 1024 kg

= 1.99 x 1020 N

FG = GMEMM

R2

= (6.67 x 10-11)(5.98 x 1024)(7.34 x 1022)

(3.84 x 108)2

What Gravitational Force do the Earth and the Moon experience because of their proximity to one another ?

The Earth experiences an attractive force of 1.99 x 1020 N directed toward the Moonwhile AT THE SAME TIME the Moon experiences the same force directed toward Earth.

Note the distance (R) is measuredfrom the Centre of Mass of each body

FG FG


Question type:

Q54: What is the magnitude of the force exerted by Earth on a water molecule of mass 3.0 × 10-26 kg at the surface of Earth?

A: F = GMm R2

= (6.67 x 10-11)( 5.98 x 1024)(3.0 x 10-26)

(6.37 x 106)2

= 2.95 x 10-25 N

Gravitational Force


Question type:

Nato III is a communication satellite that has a mass of 310 kg and orbits Earth at a constant speed at a radius R = 4.22 x 107 m from the centre of Earth.

Q55: Calculate the magnitude of the Earth’s gravitational field at the orbit radius, R = 4.22 x 107 m, of Nato III. Give your answer to 3 sig figs. You MUST show your working. G = 6.67 x 10-11 Nm2kg-2 Me = 5.98 x 1024 kg.

A: g = GMe/R2

= (6.67 x 10-11)(5.98 x 1024) ------------------------------------------

(4.22 x 107)2

= 0.224 Nkg-1

Gravitational Field Strength

EarthNato III

R

8.2 Circular Orbits under Gravity

Thus the Centripetal Force (FC) needed for circular motion is supplied by the Gravitational attraction (Fg) between the planet and its moon.

This interaction is the GRAVITATIONAL ATTRACTION between the Planet and its Moon.

This Centripetal Force must be supplied by some kind of interaction between the planet and its moon.

If, for example, a moon travels in a circular orbit around its host planet, it must be subject to a Centripetal Force.

Planet

MoonFC

FGFG

On the large scale of planets, moons, stars and other bodies in the universe, their motions are determined by the gravitational attractions between them.

8.3 Mathematics of Circular Orbits

Planet Moon

FC

FGFG

The Centripetal Force (FC) the moon is subject to is given by:

FC = MMv2/R

Remember the velocity of an object travelling in a circle is:

v = 2R/T v2 = 4 2R2/T2

Substituting for v2 we get:

FC = MM 42R/T2

This Centripetal Force is supplied by the Gravitational Force between the the Planet and the Moon:

FG = GMMMp/R2

GMMMP/R2 = MM42R/T2

Rearranging we get:

R3/T2 = GMP/42

The terms G, MP, and are all constants so their ratio is constant.

The ratio R3/T2 is also a constant

8.4 Kepler’s Laws

His other 2 laws establish planets’ speeds and the elliptical nature of planet orbits.

R = 3 GMPT2

42

Notice that the radius of orbit of any satellite, while it does depend on the mass of the planet around which it circulates, DOES NOT depend its own mass. This is true of any satellites whether man made of naturally occurring.

Johannes Kepler (1570 - 1630) discovered the laws governing planetary motion which describe the movement of the planets in our solar system.

In the previous slide we found the ratio R3/T2 was a constant, (its value is roughly 3.0 x 1018 ).

In 1615 Kepler discovered that the ratio of the of the cube of the average sun - planet distance (R3) to the square of its period (T2) was a constant for ALL planets in our solar system, this became known as Kepler’s Third Law.

The 3rd law: R3/T2 = GMP/42 can be rewritten as:


Question type:

A satellite in a circular orbit of radius 3.8 × 108 m around Earth has a period of 2.36 × 106 s.

Q56: Calculate the mass of Earth. You must show your working.

A: R3 = GMe

T2 4π2

Me = (4π2) (3.8 x 108)3

(6.67 x 10-11) (2.36 x 106)2

= 5.83 x 1024 kg

Kepler’s 3rd Law

8.5 Satellites in Space. A satellite moving through space will often use the gravitational field of a planet, like Jupiter or star like our Sun to help propel it through space.

A

B

Path of Satellite around Planet

At point A, the satellite comes under the influence of the gravitational field of the planet. The field does Work ON the satellite, accelerating it toward the planet.

By the time it has reached B, the satellite has increased its Kinetic Energy, and hence its Speed, sufficiently to pass around, rather than crash into, the planet.

The satellite flies past the planet leaving with greater speed, having “stolen” some of the energy stored in the planet’s field.

This is the normal way of sending satellites to the outer planets and even outside our solar system.

Jupiter

8.6 Energy Transfers in Gravitational Fields

The Gravitational Field vs Distance graph for Jupiter showing positions A and B for the satellite is shown.

g(Nkg-1)

Distance R (m)

Area gives: 1. Work by the field on 1 kg of satellite moving from A to B.

2. The increase in Kinetic Energy possessed by 1 kg of satellite in moving from A to B.

3. The loss in Potential Energy of 1 kg of satellite moving from A to B

RJ AB

For the satellite with a mass >1 kg, total energy possessed = area x mass

In exam questions, the area normally needs to be found by the “counting the squares” method.

Since Work Done = Energy Transferred, the area under the graph represents the work done by the field on, and the change in energy possessed by, “1 kg of satellite mass” in moving from A to B.


Question type:

A: W = mg

= (68)(8.7)

= 592 N

When the astronaut wishes to rest he has to lie down and strap himself into bed.

Q58: What is the magnitude of the force that the bed exerts on the astronaut before he begins to fasten the strap ?

A: 0 N

Weight & Weightlessness

The Russian space station MIR (Russian meaning - peace) is in a circular orbit around the Earth at a height where the Gravitational Field Strength is 8.7 Nkg-1

Q57: Calculate the magnitude of the gravitational force exerted by Earth on the astronaut of mass 68 kg on MIR


Question type:

A: Weight is the action of a gravitational field acting on a mass.

For true weightlessness to exist the gravitational field strength needs to be zero.

There is no place in the universe where the gravitational field strength is zero, so nothing is truly “weightless”.

The astronaut actually does have weight, as calculated earlier.

The astronaut’s “apparent weight” is zero because he is not subject to a reaction force inside the spacecraft since both he and the spacecraft are in a state of constant free fall.

Newspaper articles about astronauts in orbit sometimes speak about zero gravity when describing weightlessness.

Q59: Explain why the astronaut in the orbiting MIR is not really weightless.

Weightlessness

Ollie Leitl 2009

vce physics unit 3 topic 1 motion in 1 & 2 dimensions

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