NEWTON'S LAWS OF MOTION (return to start of page)

Dynamics is concerned with the causes or laws of motion.

Newton's laws can be expressed as:

  1. 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
  2. The rate of change of momentum of a body is proportional to, and occurs in the direction of, the resultant force
  3. If a body A exerts a force on a body B, then body B exerts an equal but opposite force on body A
Law 3 is sometimes expressed as ‘action and reaction are equal and opposite’ - but note that the ‘action’ and ‘reaction’ forces act on different bodies.


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:

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

d) Weight In the second law, F = m a, we can replace ‘F’ by ‘W’, for weight, and ‘a’ by ‘g’ for acceleration due to gravity, to get:

By experiment, g = 9.8 ms-2, but in calculations we often use g = 10 ms-2, so:

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:

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


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


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

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

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


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

  1. convert 36km/h to m/s
  2. calculate the average deceleration
  3. 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.


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

  1. the deceleration
  2. the average retarding force
  3. the depth of penetration


Suppose that a body A of mass m1 and velocity u1 collides with a body B of mass m2 and velocity u2:

Suppose that after the collision, A has a velocity v1 and B has a velocity v2:
During the collision each body experiences a force.

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

Note: The above steps can be reversed, i.e. we could start with the equation for conservation of momentum and infer Newton’s third law.


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-1to the right.

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, ENERGY AND POWER (return to start of page)


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).


What work is done when:

  1. a person pushes a trolley with a force of 500N over a distance of 20m?
  2. a rocket engine applies a thrust of 300 000N over a distance of 10km
a) W = F s = 500 * 20 = 10 000J
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 109 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:

We can view work as a means of energy transfer. For example, the work done by the force of gravity on a falling object (neglecting air resistance) becomes kinetic energy of the object.

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

Potential energy (‘PE’)

For example, a wound clock spring has PE due to its state of tension.
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’)

The mass m is initially at rest. Force F is applied over the distance s, and the mass acquires a velocity v.
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).


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

In the following calculations, we assume that no energy is lost, and use g = 10 ms-2.

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


Energy and work are both in joules. So the unit of power is joules per second (Js-1). We call this the watt (W).

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

  1. What vertical height has she moved through?
  2. What PE has she gained?
  3. What power does she use?
(assume g = 10 m/s2)
(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:

Let L = length of the slope, which she covers in 4 seconds. So her average speed, v = L/4.
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.

Collisions between snooker balls are almost elastic, but not perfectly. For example, the ‘click’ heard on impact indicates some KE has been changed to sound.

If the collision in the above diagram is elastic:

Three equations for two, isolated, colliding objects



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:

MACHINES (return to start of page)

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


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

Suppose that the load is 2000N, and that the effort is increased until the block just begins to move – the bar is then just in equilibrium, so the moments of the forces (the effort and load) are balanced.

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.


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

Inclined Plane
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.


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 has no units, i.e. it is just a number. Since the energy/work output cannot exceed the energy/work input, efficiency is always less than or equal to 1 and the percentage efficiency less than or equal to 100%.

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.


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, vT.

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 vT 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:

For example, consider one of the equations of motion:  s = ut + 1/2 a t2

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

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

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:

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