- NEWTON'S LAWS OF MOTION
- PRINCIPLE OF CONSERVATION OF MOMENTUM
- WORK, ENERGY AND POWER
- ELASTIC AND INELASTIC COLLISIONS
- MACHINES
- FRICTION AND TERMINAL VELOCITY
- HOMOGENEITY OF EQUATIONS

Dynamics is concerned with the causes or laws of motion.

Newton's laws can be expressed as:

- A body continues in a state of rest or of uniform motion in a straight line unless acted upon by a non-zero resultant force
- The rate of change of momentum of a body is proportional to, and occurs in the direction of, the resultant force
- If a body A exerts a force on a body B, then body B exerts an equal but opposite force on body A

**Observations**

**Law 1**

This expresses the idea of ‘inertia’, which is the ‘built in’
opposition a body has to changes in its state of motion – its reluctant to start
moving or to change its speed or direction if it is moving. The *mass* of a
body is a measure of its inertia – for example, it is much easier to push a
bicycle than a car, because the car has much more mass, and so has much more
inertia.

Momentum is a vector, with the same direction as the
velocity.

The unit of momentum is kg
ms^{-1}.

Momentum is sometimes represented by the symbol ‘p’, so p = mv

**b) The newton (N)**

Suppose that a resultant force F acts on a body of mass m for a time t, and changes its velocity from u to v.

The SI unit of force is the newton (N), defined as:

*One newton is that force which gives a mass of 1kg an acceleration of 1m*s^{-2}

The units used above are the standard SI units for force, mass and acceleration – and these should be used in calculations.

From the last equation we see that: 1 N = 1kg
ms^{-2}

In the statement of Newton’s second law, we say that a force is
*proportional* to the rate of change of momentum, but by suitably defining
the unit of force (as above), the 'constant of proportionality', k, becomes
equal to one, so we now have that force *equals* the rate of change of
momentum.

**c) Impulse**

Suppose that a force F acts on an object for a time t. The product (F*t) is called the impulse of the force on the object.

From this equation, impulse has the unit ‘newton seconds’ (Ns) or,
equivalently, the unit of momentum, kgms^{-1}.

**F-t graphs and impulse**

*The weight of a body equals the force of gravity acting on it*

By experiment, g = 9.8 ms

Now, *mass is a scalar quantity*, just so many kg. Its
value for a particular object does not depend on its location. However, weight
equals the force of gravity on an object so, like all forces, *weight is a
vector quantity*, since it has a direction (downwards). The value of the
weight for a particular mass depends on the value of g. On the earth g =
10 ms^{-2}, but it is
different for different planets.

**Gravitational field strength**

Since W = m g, then g = W/m. In this form g has the units of
N/kg (or Nkg^{-1})

So, g equals the force
acting on a mass of 1 kg, and in this sense we call it ‘gravitational field
strength’

So, g is:

- 'acceleration due to gravity' or 'acceleration of freefall'
(value 10ms
^{-2}) - 'gravitational field strength' (value 10Nkg
^{-1})

So, as far calculations are concerned, numerically g = 10, whichever of the above names or units are used.

**Example**

On the earth the gravitational field strength is 10N/kg, but on
the moon it is 1.6 N/kg.

What is the weight of a 100kg
person (a) on the earth, (b) on the moon?

a) W = m g = 100 * 10 = 1000N

b) W= m
g = 100 * 1.6 = 160N

**Example**

A force of 200N pulls a sledge of mass 50kg and overcomes a constant frictional force of 40N.

- What is the net (i.e. resultant) force on the sledge?
- What is the acceleration of the sledge?
- What is the speed of the sledge after 5 seconds, assuming that it starts from rest?
- How far did it travel in the 5 seconds?

Resultant force = 200 – 40 = 160 N to the right.

**Example**

A car of mass 1000kg travelling at 36km/h is brought to rest over a distance of 20m.

- convert 36km/h to m/s
- calculate the average deceleration
- calculate the average braking force

Notice that the direction of travel has implicitly (i.e. without actually saying so) been taken as positive, because u and s have been made positive. The acceleration comes out negative, which is because the car is decelerating - the acceleration acts in the opposite direction to the motion, and so has the opposite sign to the original velocity.

Note: The equations of motion apply to an object moving with a
*uniform* acceleration (or deceleration). In a situation such as the above,
the acceleration may not be uniform, and the value of acceleration produced is
an *average* value.

(c) F = ma = 1000 * (- 2.5) = - 2500 N

Note: A force that produces deceleration, i.e. slows something down, is called a braking or retarding force.

**Example**

A bullet of mass 20g travelling at 16m/s penetrates a block of wood and is comes to rest in 0.05 seconds. Find:

- the deceleration
- the average retarding force
- the depth of penetration

**PRINCIPLE OF CONSERVATION OF MOMENTUM **(return
to start of page)

Suppose that a body A of mass m_{1} and velocity
u_{1} collides with a body B of mass m_{2} and velocity
u_{2}:

Suppose that after the collision, A has a velocity v

During the collision each body experiences a force.

This is an example of the *principle of conservation of
momentum*:

*When two or more bodies act upon one another, their total momentum remains constant, provided that no external forces act*

**Example**

A trolley of mass 2kg moving at 3ms^{-1} collides head on with a trolley of
mass 1kg moving the opposite direction at 4ms^{-1}.
On impact the objects coalesce (i.e. stick together) and so move with a common
velocity v. Calculate v.

Before the collision:

The diagram indicates that velocity to the right is taken as being positive.

After the collision:

The velocity is positive, so the final velocity is 2/3 ms

**Momentum and explosions**

The above principle applies to explosions. Suppose that a gun fires a bullet:

What is the recoil velocity of the gun, V?Mass of bullet = 20g = 20/1000 = 0.02 kg. Letting velocity to the right be positive:

The minus sign indicates that the recoil velocity is to the left (as
we'd expect), and its magnitude is 1.0ms^{-1}.

Note: Rockets exploit the conservation of momentum - the ejected burnt fuel acquires momentum in one direction and the rocket acquires momentum in the opposite direction - similar to the balloon represented bellow:

**Work**

Suppose that an object is pulled along the ground by applying a force as shown:

From the definition of work, its unit is: N m

Notice that the definition of work means that if s = 0, then W
= 0 no matter how big the force. Thus, if you try to push a bus, and fail to
move it, then you have done no work *on* the bus, no matter how much you
sweat (what you have done is convert stored energy into heat).

**Example**

What work is done when:

- a person pushes a trolley with a force of 500N over a distance of 20m?
- a rocket engine applies a thrust of 300 000N over a distance of 10km

b) Use standard SI units: Thus, 10km = 10 000m. W = F s = 300 000 * 10 000 = 3 000 000 000 J

Note: We can write numbers, especially very large or very small
numbers, in 'standard form' or 'scientific notation'.

This means a number between 1 and 10 multiplied by a power of ten. E.g.
3 000 000 000 J = 3.0 x 10^{9} J

**Principle of conservation of energy**

We say that something has energy if it can (potentially) do
work.

Examples of energy include light, sound, heat,
electrical, motion, coal, oil, gas, tidal, solar, geothermal, nuclear.

Energy in any form is measured in joules (J). So energy and
work are both in joules.

The above principle can be
stated as:

*Energy can exist in a number of different forms and can be converted from one form to another, but it cannot be created or destroyed*

*Work and energy are both scalar quantities *- they do not have an associated direction - each is just a number of joules.

**Potential energy (‘PE’)**

*potential energy is the energy something has due to its position or state*

An object above the ground has PE due to its position. We can determine an equation for gravitational PE:

PE of the mass at height h = work done on it to lift it = force * distance = (m g) * h, so:

What is the PE of a 100g mass raised to a height of 1m (use
g=10 ms^{-2})

We need to use standard SI units: 100g = 100/1000 kg = 0.1 kg. So, PE = mgh = 0.1 * 10 * 1 = 1 J

A small apple weighs about 100grams, and if you lift it by 1m you’ve done 1J of work on it, which has become 1J of potential energy. This gives an idea of what one joule ‘feels’ like.

**Kinetic energy (‘KE’)**

*kinetic energy is energy something has due to its motion*

Assuming that all the work done by the force becomes KE of the mass:

If an object is dropped, its PE changes to KE as it falls. At every point PE + KE equals the original PE. At impact, all the original PE has changed entirely to KE (neglecting air resistance).

If you drop a 100 gram apple from 1m into your hand, you feel the impact of 1J of KE (since it started with 1J of PE).

**Example**

The model car is released and accelerates down the runway, ‘loops-the-loop’, and emerges on the flat at E.

- At A it only has PE
- At B it has PE + KE
- At C it only has KE
- At D it has KE + PE
- At E it only has KE
- Assuming no loss of energy, the KE at E equals the PE at A

Mass, m = 20grams = 20/1000 = 0.02 kg.

**Power**

*power is the rate of doing work or the rate of changing or using energy*

Energy and work are both in joules. So the unit of power is joules per second (Js

*Power is a scalar quantity* - just a
number of joules per second, i.e. watts.

E.g.: A 100W light bulb = a 100 J/s light bulb. It converts electricity to light and heat at a rate of 100 joules per second.

*Other examples of energy changes*:

- A microphone changes sound to electrical energy
- A battery changes chemical energy to electrical energy
- An electric motor changes electrical energy to kinetic energy
- A car engine changes chemical energy to kinetic energy, heat and sound

A 60 kg girl runs up 15 steps, each 15cm high, in 4 seconds.

- What vertical height has she moved through?
- What PE has she gained?
- What power does she use?

(a) Vertical height, h = 15 * 15 cm = 225 cm = 225/100 m = 2.25 m

(b) PE = m g h = 60 * 10 * 2.25 = 1350 J

(c) Power, P = energy/ time = 1350J/4s = 337.5 J/s = 337.5 W

(the actual power used will be greater than this, since some of her body's energy has been changed to heat, as well as PE)

**Power and velocity**

This can
be applied to the previous example:

The force required to move the girl along the slope, F = W sinA = (60*10) * (2.25/L).

So, power = Fv = (600*2.25/L) * (L/4) = 337.5 watts (as before).

**ELASTIC
AND INELASTIC COLLISIONS **(return
to start of page)

We consider two objects on the same straight line, which collide and remain on the original straight line:

According to the principle of conservation of momentum, the total momentum before and after impact will be the same.However, the total kinetic energy (KE) before and after impact
*may or may not *be the same. This is because some KE may be converted to
other forms of energy, such as sound and heat.

*An elastic collision is one in which the total KE before and after the collision are the same*- i.e. KE is conserved in an elastic collision

If the collision in the above diagram is elastic:

**Three equations for two, isolated, colliding
objects**

**Example**

A ball of mass 1kg moving at 5.0 m/s collides head-on with a stationary ball of mass 2kg. If the collision is elastic, what are the velocities of the balls immediately after the collision.

Let’s take velocities to the right to be positive.

If we add the left-hand side of (1) to the left-hand side of (2), and the right hand side of (1) to the right hand side of (2), we get:

Thus, after the collision:

- the 2 kg ball moves at 3.3 ms
^{-1}to the right - the 1 kg ball moves at 1.7 ms
^{-1 }to the left

A machine enables one force (the **effort**) to overcome
another force (the **load**). Some machines are very simple devices.

**Crowbar**

A crowbar is a **lever **– *a device that can turn around
a pivot (or fulcrum or hinge)*

Taking moments about the fulcrum:

Thus, an effort of 200N moves a load of 2000N. The crowbar acts as*
force amplifier*, since the load is bigger than the effort (in the above case
the effort is amplified by a factor of 10).

Notice that from the geometry, if the load moves 1cm (0.01m), the effort moves 10cm (0.1m).

Thus, the above machine enables us to make a more effective use of an
applied force, but the principle of conservation of energy is not broken.

**Forearm**

This is also a lever. The hinge is the elbow joint. The biceps muscle is joined quite close to the elbow joint.

- the effort in this case has to be bigger than the load
- the load moves further than the effort

It is easier to pull or push a trolley up an inclined plane rather than lift it vertically through the same height. But notice that the effort has to be applied over a longer distance than if the trolley were lifted straight upwards. This is made practical use of when roads are built on steep hills.

The road is much less steep than directly up the hill – the distance travelled is greater, but the effort needed is less.

**Efficiency**

The same unit of energy or work must be used on the top and bottom of the right hand side of the definition, and so the units will cancel. Thus,

Efficiency is a measure of how effective something is at doing what it is designed to do.

Filament bulbs are designed to produce light, but are quite inefficient since a lot of the electrical energy they use gets turned into heat. Fluorescent lights are much more efficient, since they produce more light and less heat for a given amount of electrical energy.

Cars are very inefficient. Only about 20% of the fuel they use gets converted into useful motion. The rest is wasted as mainly heat and sound.

**Example**

As indicated, a 2N effort pulls a trolley of weight 10N, 20m along an inclined plane, raising it through a vertical height of 2m.

**FRICTION
AND TERMINAL VELOCITY **(return
to start of page)

Frictional forces oppose motion. Friction can exist between
solids in contact, and can result in wear and in energy wasted as heat. Friction
also exists in fluids (liquids and gases). ‘*Viscosity*’ is a measure of
fluid friction. It determines how readily a liquid will flow, and how readily an
object will fall through a liquid or gas.

Suppose that an object is released from rest in a fluid. When
released, it initially accelerates downwards due to its weight (=force of
gravity on the object). It also experiences two forces which act upwards.
Archimedes’ principle says that when an object is immersed in a fluid it
experiences an upthrust equal to the weight of displaced fluid - this force
stays constant. The second force is the friction of the fluid, and this
*increases as the object moves faster*. Thus, if the object falls for long
enough, a speed will be reached at which:

downward force of gravity = opposing frictional force of the fluid + upthrust due to displaced fluid

When this occurs, there is *no net force on the object*,
so it no longer accelerates, but moves with a constant velocity, called its
terminal velocity, v_{T}.

A parachutist in freefall reaches terminal velocity when the air drag, which increases with speed, equals the parachutist's weight:

When the parachute is released, the parachutist acquires a new, but much lower, terminal velocity.

Consider a small ball bearing released in glycerol. The graph represents the variation of velocity with distance fallen:

A small ball bearing reaches its terminal velocity quite quickly in a viscous
liquid like glycerol. Assuming that the ball bearing reaches it terminal
velocity before reaching the first tape, it can then be timed over the distance
h and the value of v_{T} determined (= h/time).

**HOMOGENEITY OF
EQUATIONS **(return
to start of page)

Something is 'homogeneous' if it is composed of similar or identical parts. For example, tomato soup is homogeneous, being the same throughout its volume.

It can be useful to check that an equation is homogeneous in terms of units:

- the units on both sides of an equation must be the same, and
- each term in the equation must have the same units

In SI units, s is in metres (m) - so each term on the right must 'boil down' to the same unit:

- unit of ut = ms
^{-1 }* s = m s^{0 }= m (anything to the power zero = 1, so s^{0 }= 1) - unit of 1/2 a t
^{2 }= unit of a t^{2}= ms^{-2 }* s^{2 }= m s^{0 }= m

Any quantity can have its unit reduced to its most basic form. For
example:

- Force = ma, so the unit of force, newtons = unit of ma = kg ms
^{-2} - Work = force * distance, so the unit of work, joules = unit of Fs = kg
ms
^{-2}m = kg m^{2}s^{-2}

Using units to check an equation for homogeneity *can* reveal if
there is an error in the equation, but it is not *bound* to do so. For
example:

- an equation may contain two or more errors that compensate for each other
- a factor with no unit may be missing or incorrect - for example, the 1/2 in the above equation of motion has no unit, so if it were missed out, the term that it occurs in would still have the same unit

A Level Physics - Copyright © A
C Haynes 1999 & 2004