Reference: Nelson Physics VCE Units1 & 2 Chapters 13 – 16 Pages 283 – 410

Prac #1: EXPT 2.1 Laboratory Techniques in Physics

Reference: Nelson Physics VCE Units1 & 2 Appendix 3 Page 484

A scalar is a quantity which has only a size.

eg. distance, temperature, time, mass

A vector is a quantity which has both a size and a direction.

eg. displacement, velocity, acceleration, force

34 m north, 3 m s^-1 west, 6m s^-2 south, 12 N north 10° west

To distinguish the scalar F from the vector F we write vectors in one of two ways. F and are the vector F.

If we have two vectors a and b, to add them we put the vectors head to tail and the result of the addition is the vector drawn from the starting point to the finishing point.

Vector subtraction is easiest to do if you think of it in terms of addition.

Consider a - b which can be expressed as a + (-b).

-b means the same magnitude, but opposite direction.

Example. Find a -b

Just as two vectors can be added to give another. A vector can be split into two or more other vectors. These are called components or resolutes.

Example

Problem set #1: Gardiner & McKittrick Page 23 Set 8 All Questions

Kinematics is the study of the motion of objects. This involves a study of position, displacement, velocity, acceleration and time.

The position of an object tells us where the object is situated. The symbol is x and the unit is metres (m).

The displacement of an object is the change in its position and the direction of that change.

Example

Note The displacement is independent of how you got there, the distance travelled or the path taken.

Example

The mathematical way of writing "change of" is to use the symbol D (delta).

Thus change of position = D x

Since displacement has a magnitude and a direction it is a vector and is written as D x or or . The unit is metres.

Velocity is a quantity which tells us how fast an object is travelling and also the direction of travel. Thus velocity is a vector and is denoted by .

The magnitude part of velocity is called speed an is denoted by s or v.

The velocity of an object is calculated by using the displacement and the time taken.

Thus

This equation tells us the average velocity over the time D t.

The units of velocity are metres per second, written as ms^-1

Acceleration is a quantity which tells us about the change in velocity of an object and is a vector. Acceleration is defined as the change in velocity over time.

Thus

The units of acceleration are ms^-2.

Problem Set#2: Text Page 300 Questions 2 – 6

Consider the position-time graph of a person who walks 200 m to the shop at a uniform rate in 50 seconds.

The gradient of a graph is

In this case the gradient is which is the velocity (speed), which in this case is 4 ms^-1.

Thus the gradient of an x-t graph is the velocity.

Now look at the following v-t graph.

The gradient of a graph is which is which is the

acceleration. In this case it is 5 ms^-2.

Thus the gradient of an v-t graph is the acceleration.

Now consider what happens when you have an a-t graph and want to find out velocity.

We know

\ D v = a D t

but a D t is the area under the graph.

Thus the area under an a-t graph is equal to (D v) the change in velocity (speed).

To be able to find the actual velocity at any time, we need to know the velocity at one point in time. Usually the initial or final.

Now consider a v-t graph

We know that

\ D x = v D t

but v D t is the area under the graph.

Thus the area under an v-t graph is equal to (D x) the change in position.

As before to be able to find the actual position at any time we need to know the position at one point in time. Usually the initial or final.

Problem set #3: Text Page 299 Questions 1, 7 – 29

We know the shape of a v-t graph when the acceleration is constant. It is a straight line as shown.

Definitions

a = acceleration

d = displacement

v = final velocity

u = initial velocity

D t = Change in time = t₂ -t₁

By taking the gradient of the v-t graph, we can obtain an expression for the acceleration.

\ v - u = a D t

or v = u + a D t Formula 1

By finding the area under the graph, we obtain an expression for the displacement (s). This can be done in two ways:

1. Breaking the area into a rectangle and a triangle.

s = Area of rectangle + Area of triangle

s = u D t + ½(v -u)D t

Substituting formula 1

s = u D t + ½(u + a D t - u)D t

= u D t + ½(a D t)D t

So s = u D t + ½a D t² Formula 2

2. Using the formula for a trapezium. Area = ½(b + c)h.

In our case b = u, c = v, h = D t

So s = ½(u + v)D t Formula 3

If we rearrange formula 1 we get D t =

Substituting this into formula 3 we get:

s = ½(u + v)

2as = (u + v)(v - u)

2as = u v - u² + v² - u v

2as = v² - u²

So v² = u² + 2as Formula 4

These four formulae are known as the constant acceleration formulae.

Example.

A bobsled is sliding down a hill with an acceleration of 3 m/s². It starts from rest, if we ignore air resistance, find:

a) the speed after 5 seconds

b) how far it has travelled in 5 seconds

c) the time taken to reach 30 m/s

d) the distance travelled when it reaches 45 m/s

Problem Set #4: Text Page 303 Questions 30 – 47

Dynamics is the study of the causes of motion, in particular the forces acting on a body.

Prac #2: EXPT 2.2 Feeling Newtons

There are three states of motion, they are:

1) object at rest

2) object travelling at constant velocity

3) object undergoing acceleration

In dynamics we will study what causes these states of motion.

An object will remain at rest or in uniform motion in a straight line unless acted upon by a force.

Note 1) this applies in any particular direction.

2) if F = 0, v = constant (may be non zero)

Problem set #5: Dynamics sheet #1

Prac #3: EXPT 2.3 Ticker Tape Vibrator

Prac #4: EXPT 2.4 Force and Acceleration

Prac #5: EXPT 2.5 Mass and Acceleration

When a constant force acts on a body we notice that the body undergoes constant acceleration. If we change the size of the constant force the size of the acceleration changes, such that F a a.

Applying a given force to different objects results in differing accelerations. The property of the object that causes this variation is the inertial mass (m_i) and we have

F = m_i a

which strictly speaking should be written as

S F = m a

Force is a vector and behaves as any vector would.

Units Force has the unit of Newton (N)

mass has the unit of Kilogram (Kg)

acceleration has the unit of ms^-2

1 Newton » weight of an apple

Note 1) Direction of a is direction of S F

2) If a = 0 then S F = 0

3) If a = 0 in any direction,

then S F in that direction = 0

Example 1 A force of 6 N accelerates a mass of 3 Kg.

What acceleration results?

Problem Set #6: Text Page 347 Questions 5 – 16

Mass is a measure of how much material is contained in an object and is measured in Kg.

Weight is a measure of the pull of gravity on an object, it is a force and is measured in N.

Mass and weight are not the same thing, however they are related. An object in a gravitational field of strength g feels a force, called weight w, that depends on a property of an object called gravitational mass (m_g) and w = m_g g. m_g is found to be related to m_i by m_g = m_i.

1. Friction

A force opposing the motion of a body.

eg. table-floor, car-air

2. Weight

The force by which a mass is attracted to the Earth. It is proportional to the bodies gravitational mass (m_g), the constant of proportionality being the gravitational field strength (g). (g µ 9.8 N Kg^-1, Book uses 10 N Kg^-1)

W = m g

3. Reaction

When a body sits on a surface there is a force acting upwards to prevent it falling downwards. This force is called the reaction force.

The reaction force leads us to Newton's third law, which states:

If one object exerts a force on another then there is an equal and opposite force (reaction) on the first object by the second. i.e. for every action there is an equal and opposite reaction.

Example 1

Example 2

Problem set #7: Text Page 347 Questions 17 – 43

Hand Out: Set of worked examples

Prac #6: EXPT 2.6 Acceleration on an Inclined Plane

Prac #7: EXPT 2.7 Acceleration Due to Gravity

We say that an object is in free fall if the only force acting on it is its weight ( not strictly true there is air resistance also). eg. dropping a ball from a building.

For a body travelling in the vertical direction close to the Earth's surface, with no friction.

We have S F = m a

The only force acting is weight

\ S F = W

W = m a

but W = m g

so m g = m a

\ a = g

So for vertical motion the acceleration is constant and is equal to g , the gravitational field strength (9.8 N Kg^-1). Since the acceleration is constant we can use the constant acceleration formulae.

Air resistance is opposite to the direction of travel. It is down if the object is travelling up, and up if the object is travelling down. So a diagram is needed to be drawn, S F and go from there.

However air resistance is more complicated than that because it is not constant, but proportional to the velocity. It increases as the velocity increases.

Lets look at the example of an object falling downwards. As it falls downwards the velocity increases and the air resistance in creases accordingly. After awhile the air resistance will equal the weight force ( in size opposite in direction) and the following will happen.

AR = W

then S F = W - AR = 0

from S F = m a

we get a = 0

and if a = 0, then the velocity must be constant. This is known as the terminal velocity.

Examples Parachute, feather, table tennis ball, car

1. No Friction

When a body is thrown up it comes to rest momentarily at the top of the path. a = g still.

2. If air resistance = 0, then a = g for all masses.
3. Book uses g = 10 N Kg^-1

Example 1

A rock with a mass of 5 Kg is dropped from a height of 45 m. It starts at rest, air resistance is zero and g = 10 N Kg^-1.

a) What is it's acceleration 2.0 sec. after it is released?

b) How fast will it be going after 2.0 sec.?

c) How long will it take to reach the ground?

d) How fast will it be going just before it hits the ground?

e) What would the acceleration be if the air resistance was 3 N?

Example 2

The initial speed of a stone projected vertically upwards is 21.0 m/s.

a) What is the maximum height reached?

b) How long does it take to reach the highest point and what is the total time it is in the air?

c) How long is the stone more than 10 m above the ground?

Problem Set#8: Text Page 379 Questions 1 – 46

Gardiner & McKittrick Page 32 Set 12 Questions 1 - 11

If an object is moving with some speed, then to change its speed or direction a force must be applied, even then the object will be reluctant to change. This reluctance of a body to change its velocity is known as momentum.

The momentum of a body is defined by:

Momentum = mass of body ´ velocity of body

Momentum is a vector and should be treated as any vector would. The units of momentum are Kg ms^-1.

Example:

At what speed must a 60 Kg athlete be running if his or her momentum is to equal that of a 1000 Kg car travelling at a constant speed of 5 Km hr^-1.?

Speed of the car = 5 Km hr^-1 = = 1.39 m s^-1

Momentum of car

p = m v

= 1000 ´ 1.39

= 1.39 ´ 10³ Kg m s^-1

Athlete

p = 1.39 ´ 10³ Kg m/s m = 60 Kg

p = m v

1.39 ´ 10³ = 60 ´ v

v = 1.39 ´ 10³

60

= 23.15 m s^-1

Note: In a closed system (of two or more objects) the momentum of the system remains the same before and after the collision.

To change a bodies momentum a force must be applied. Usually this force will be in the form of a collision, of fairly short duration and its size may vary. This type of force is called an impulse. Impulse is defined as the change in momentum.

i.e. Impulse = D Momentum

I = D p

The average impulsive force is defined as the rate of change of momentum.

i.e.

or

\ I = F_ave D t

Example:

When served, a tennis ball leaves the racquet with a speed of 45 m/s. The impact time between the racquet and the ball is 5 x 10^-3 s. The mass of the ball is 0.06 Kg. Find the average impulsive force which acts on the ball. The speed of the ball the speed of the ball before being hit is zero.

v_f = 45 v_i = 0 m = 0.06 t = 5 x 10^-3

p_f = 0.06 x 45 = 2.7

p_i = 0.06 x 0 = 0

D p = p_f - p_i = 2.7 - 0 = 2.7

=

= 540 N

Problem Set #9: Text Page 3 Questions 44 – 83

Prac #8: EXPT 2.8 Work Done along an Inclined Plane

From the ideas we have about work we can say that work requires some effort. E.g. pushing a wheel barrow. Thus some force is being applied over some distance.

From common sense we get

Work = Force ´ Displacement

or W = F ´ d

The units for work are the Joule (J)

Example:

Find the work done against a frictional force of 7 N. If the fridge is kept at a constant velocity of 2 m s^-1 for a distance of 10 m.

Suppose we have a force-distance graph that looks like this

How do we calculate the work done?

Firstly we replace the curve by a series of steps of constant force.

Then what we have is a series of small rectangles, with an area of f ´ d.

So if we add up all the rectangles.

We get, S F(x) dx or simply S (F x d) which is the area under the graph.

So Work done = area under F-d graph

This follows for all F-d graphs even those where F is not constant.

Examples: Find the work done in the following cases

1.

2.

Work is a scalar but force and displacement are vectors. What happens if the force applied and the displacement are not in the same direction? We then need to take the component of the force in the direction of the displacement.

So Work = distance ´ component of force in direction of displacement

Work = d ´ F_{| |}

Example:

Find the work done by the force if the distance moved is 6 m.

Problem Set#10: Text Page 400 Questions 1 – 17

Gardiner & McKittrick Page 42 Set 18 Questions 1 - 8

When we do work, such as pushing a wheel barrow, we get tired or use up the quantity known as energy. Energy does not disappear, but is either

1) transferred to another object

or 2) transformed into another kind.

Thus we formulate the principle of conservation of energy which states that

Energy is neither created or destroyed

we say that

Work done by an object = loss of energy by that object

and Work done on an object = gain in energy by that object

or W = D E

The units for energy are the same as the units for work, the Joule (J).

Example:

If a man does 200 J of work pushing a wheel barrow, he loses 200 J of energy and the wheel barrow gains 200 J of energy.

We define kinetic energy as the energy a body has when it is in motion. We can derive an expression for kinetic energy.

Consider an object of mass, m, originally at rest being acted upon by a force of F N for a distance of d m. No friction.

We have

Work done = energy gain = final K.E. (in this case)

\ Final K.E. = F ´ d

= ma ´ d (eqⁿ 1)

Evaluating the accⁿ using constant accⁿ formula

v² = u² + 2ad

u = 0 v = v d = d a = a

v² = 0 + 2ad

v² = 2ad

Þ a = v²

2d

Now plug back into eqⁿ 1

Final K.E. = m v² x d

2d

= ½ m v²

so Kinetic Energy = ½mv²

In fact work done = change in kinetic energy

D K.E. = ½m v² - ½m u²

Examples:

1. A body of mass 6 Kg has a speed of 3 m s^-1. What is its K.E.?

2. A body of mass 4 Kg with a speed of 3 m s^-1 accelerates to a speed of 6 m s^-1. What is

a) the change in K.E.

b) the work done on the body

Problem Set#11: Text Page 402 Questions 26 – 36

Gardiner & McKittrick Page 42 Set 18 Questions 9 - 20

The potential energy is the energy stored within a body.

Springs can store energy when they are stretched or compressed. We can store the energy in the spring by applying a force to alter its length, thus we are doing work on the spring.

we have

Energy stored = potential energy of spring = work done on spring

In about 1675 Robert Hooke noticed that the more you stretch a spring from it's natural length, the stronger the force needed.

i.e. F a D x

we write F = k x Hooke's law

where k = spring constant (unit N m^-1)

we get a graph which looks like

Now P.E. of spring = Work done on it

We can work out the work done on a spring in stretching it x metre from the force-distance graph.

Work done = area under graph (can't use w = f x d because force not constant)

= ½ F x

But F = k x

So work done = ½ k x x

= ½ k x²

\ P.E. of spring = ½ k x² (Joule)

Note: For a spring compressed and then released D P.E. (spring) = D K.E. (body). Conservation of energy. E_total = K.E. + P.E.

Example 1. Find the P.E. of the compressed spring

Example 2.

For a spring with k = 5 N m^-1. Find

a) D P.E. when compressed from 0à 20 cm

b) D P.E. when compressed from 20à 40 cm

c) If compressed to 20 cm and a body is placed there and let go. What is the K.E. as it passes zero compression?

Problem set#12: Gardiner & McKittrick Page 47 Set 20 Questions 1 - 8

When an object is raised above the surface of the Earth energy is stored.

To raise a body above the ground we must do work against the weight force.

Let us raise a mass, m, h metre above the ground

P.E. = work done against weight force

= F ´ d

= m g h (Joule)

\ P.E. = m g h (Joule)

Examples

1) A mass of 5 Kg is raised 6 m above the ground. What is it's P.E.?

2) A mass of 3 Kg is 7 m above the ground. If it is released, what is it's K.E. just before it hits the ground? What is it's speed?

Problem Set#13: Text Page 403 Questions 37 – 58

The rate at which work is done on, or by a body is called power.

Power = Work done

Time taken

or

Units of power are joule per second = Watt (W)

Problem Set#14: Text Page 401 Questions 18 – 25

Gardiner & McKittrick Page 42 Set 18 Questions 21 -25