As Physics Unit 1 Basic Notes

Units All units in science are derived from seven base units. These are: kilogram (kg), metre (m), second (s), ampere (A), mole (mol), kelvin (K) and candela (cd). eg: Speed is calculated as distance (measured in metres) divided by time (measured in seconds). Thus speed is measured in metres per second (ms-1)

Prefixes Prefixes are used with units to make the numbers more manageable. For example: giga (G) means x 109 e.g. 1 GHz = 109 Hz mega (M) means x 106 kilo (k) means x 103 milli (m) means x 10-3 e.g. 1 mA = 10-3 A micro () means x 10-6 nano (n) means x 10-9

Vectors and Scalars A scalar is a quantity that has magnitude (i.e. size) only - e.g. energy, distance, speed, mass, density, etc. A vector is a quantity that has both magnitude and direction - e.g. force, displacement, velocity, etc.

Displacement, Velocity and Acceleration Displacement s (in m) is distance in a particular direction - i.e. it is a vector. Velocity v (in ms-1) is defined as rate of change of displacement. i.e. velocity = (change in displacement)/(time taken for change)

i.e. v = s/t

REMEMBER!

Thus, it follows that v is the gradient of a graph of s against t - as shown below:

Acceleration a (in ms-2) is defined as rate of change of velocity. i.e. acceleration = (change in velocity)/(time taken for change)

i.e. a = v/t

REMEMBER!

Thus, it follows that a is the gradient of a graph of v against t - as shown below:

N.B. The area under a v-t curve is the distance travelled:

Uniform Acceleration Uniform (i.e. constant) acceleration is an important special case. For example, the acceleration due to gravity near the earth's surface is uniform, and is equal to 9.81 ms-2. For constant acceleration, the velocity-time has a constant gradient - i.e. it is a straight line, as shown below:

Suppose also that s is the displacement after time t. Since v increases with time, the gradient of an s - t graph increases with time, as shown below:

The following equations can be used:

v = u + at s = ut + 1/2at2 v2 = u2 + 2as s = 1/2(u + v)t

Newton's Second Law Provided mass is constant, Newton's Second Law of Motion is effectively expressed by the equation:

F = ma

REMEMBER!

where a is the acceleration (in ms-2) of a body of mass m (in kg) which has a force F (in N) applied to it.

Gravitational Field Strength Gravitational field strength g (in Nkg-1) is defined by the equation:

g = F/m

REMEMBER!

where F is the force (in N) on a body of mass m (in kg) in a gravitational field. g = 9.81 Nkg-1 near the earth's surface. F is also known as the weight W of the body. Therefore g = W/m

i.e: W = mg

Vector Addition It is a simple matter to add scalars - e.g. 1 kg + 5 kg = 6 kg. However, with vectors, you also have to take the direction into account. To add them, the vectors are represented as arrows drawn to scale and placed in order, end to end, as shown below:

In the simple case shown above, the 2 vectors are at right angles - so Pythagoras' Theorem can be used: i.e. sum2 = 32 + 42 = 25, therefore sum = 5 N. The sum of 2 (or more) vectors is often called the resultant. The resultant has the same effect as the 2 vectors put together. The angle that the 5 N vector makes with the 3 N vector is the angle whose tan is 4/3 (i.e. 53 degrees). If the 2 vectors are not at right angles it is possible to add them by doing a scale drawing, as shown below:

If a body is in equilibrium the sum of the forces on it must = 0. i.e. The force vectors must form a closed triangle, as shown in the example below:

Resolving Vectors Resolving a vector means finding 2 vectors at right angles that have the same effect as the original vector. i.e. They must add together to give the original vector, as shown below:

Fx = Fcos Fy = Fsin

The 2 vectors at right angles are called the components of the original vector. The reason for doing this is that it is often easier to solve a problem using 2 vectors at right angles (e.g. vertical and horizontal) rather than one vector which is neither vertical nor horizontal. The example below illustrates the use of components. (The tension T has been replaced by its components.)

Since the body is in equilibrium: Tsin = F Tcos = W The above equations can be used to solve the problem.

Stiffness and Hooke's Law When a force is applied to a string, rope, wire etc it extends. The stiffness k (in Nm-1) of the string, rope, wire etc is the force required to produce unit extension.

i.e: k = F/x where F is the applied force (in N) and x is the extension (in m). If a graph of F against x is a straight line, as shown below .....

..... the stiffness k is the gradient - and is a constant. The sample of material is then said to obey Hooke's Law.

Work

When a force F (in N) moves a body through a distance s (in m) the work done W (in J) on the body is given by the following equation:

W = Fs Sometimes the force F is not in the same direction as the displacement s, as shown in the example below:

In this case the work done is given by the formula: where  is the angle between F and s.

W = Fscos

Energy Energy (in J) is the capacity to do work. i.e. If a body has energy it can move something with a force. There are different forms of energy - with different equations to calculate the energy in each case: Gravitational potential energy (g.p.e.) is the energy a body has because it is in a gravitational field. When one mass is moved further away from another mass its g.p.e. increases. In a uniform gravitational field (such as that near the earth's surface) the increase in gravitational energy Egrav (in J) when a mass m (in kg) is raised through a height h (in m) is given by the formula:

Egrav = mgh

REMEMBER!

where g is the gravitational field strength (9.81 Nkg-1 near the earth's surface). Kinetic energy Ek (in J) is the energy a body has because it is in motion. Ek is given by the formula:

Ek = 1/2mv2

REMEMBER!

where m is the mass of the body (in kg) and v is its velocity (in ms-1).

Conservation of Energy "Energy can change form (be transferred), but cannot be created or destroyed." For example, when a body falls under gravity, as shown below .....

..... it loses gravitational potential energy, but gains the same amount of kinetic energy (assuming no energy is converted to heat through air resistance). i.e. Loss of g.p.e. = gain of k.e. i.e. mgh = 1/2mv2

Power Power P (in watts W) is defined as the rate at which work is done.

i.e. P = W/t where W is the work done (or energy transferred) (in J) and t is the time taken (in s). W = Fs (see above) Therefore P also = Fs/t = Fv (force x velocity).

Projectiles The key to solving projectile problems is: (1) to treat the vertical and horizontal components of the motion separately and (2) to remember that the vertical motion is uniformly accelerated, whereas there is no horizontal acceleration - i.e. the horizontal component of the velocity is constant. The simplest type of situation is shown below:

Suppose we want to calculate the range - and V = 20 ms-1, h = 100 m and g = 9.81 ms-2. Vertical motion: u = 0 (initial vertical velocity = 0 since the projectile is initially travelling horizontally) a = g = 9.81 ms-2 s = 100 m t = time of flight = ? (It is usually a good idea to calculate the time of flight.) Use: s = ut + 1/2at2 Therefore: 100 = 0 + 1/2 x 9.81 x t2 Therefore: t2 = 100/(1/2 x 9.81) = 20.39 Therefore: t = 4.52 s N.B. This is the same as the time the projectile would have taken to drop vertically through a height of 100 m. Horizontal motion: Range = horizontal speed x time of flight (There is no horizontal acceleration because there is no horizontal force - if you neglect air resistance.) Therefore: range = 20 x 4.52 = 90.4 m

The situation shown below can be dealt with using a similar method, but you have to start by working out the horizontal and vertical components of the projectile's initial velocity - i.e. Vcos and Vsin.

Suppose, again, that we want to calculate the range - and that V = 20 ms-1 and  = 50 degrees. Vertical motion: u = +20sin50 = +15.32 ms-1 a = -g = -9.81 ms-1 s = 0 (the final vertical displacement = 0) t = time of flight = ? N.B. Since the initial velocity is upwards and the acceleration due to gravity is downwards, a choice about signs had to be made. "Upwards is +ve and downwards is -ve" was chosen - but the opposite choice would have worked just as well. s = ut + 1/2at2 Therefore: 0 = 15.32t - 1/2 x 9.81 x t2 = 15.32t - 4.905t2 Therefore: t(15.32 - 4.905t) = 0 Therefore: either t = 0 (neglect this solution!) or 15.32 - 4.905t = 0 Therefore: t = 15.32/4.905 = 3.12 s Horizontal motion: Range = 20cos50 x time of flight = 12.86 x 3.12 = 40.1 m

Viscosity Fluids with high viscosity (e.g. honey, treacle, etc) flow slowly through pipes, out of containers, etc. When a body moves through a very viscous fluid there is a large viscous drag force. The reverse is true for fluids with low viscosity (e.g. water, etc). The coefficient of viscosity  of a fluid tells us how viscous it is.  can be related to the viscous drag F on a sphere of radius r travelling at constant velocity v through the fluid, as shown below:

The formula that relates the variables is called Stokes' Law:

F = 6rv The units of  are Nsm-2. e.g.

viscosity of water ~ 1.0 x 10-3 Nsm-2

viscosity of air ~ 1.8 x 10-5 Nsm-2

The viscosities of most fluids are very sensitive to temperature:  goes down as temperature goes up

Terminal Velocity Consider a ball-bearing falling through a liquid. It has 3 forces acting on it, as shown below:

The weight W is a constant force. It is given by: W = mg = density of steel x volume of sphere x g = 4/3r3steelg

where steel = density of steel

The upthrust U is also a constant force. According to Archimedes' Principle: U = weight of fluid displaced = density of fluid x volume of sphere x g = 4/3r3fluidg density of fluid

where fluid =

The viscous drag F, however, increases as the velocity of the sphere increases. According to Stokes' Law: F = 6rv Assume the ball starts from rest at the surface of the fluid - so initially the viscous drag F = 0. Thus, provided W > U, the ball will accelerate downwards to begin with - since there is a net downwards force. However, as the ball accelerates, v increases - so the viscous drag F increases ..... until eventually ..... U + F becomes equal to W At this point there is no net force on the ball - so it no longer accelerates. It has reached terminal velocity. Substituting all the above expressions into ...... U+F=W

(which is true at terminal velocity)

we get ..... 4

/3r3fluidg + 6rv = 4/3r3steelg

An experiment can be done to measure the terminal velocity v, and the above equation can then be re-arranged to find .

Streamlined and Turbulent Flow In streamlined or laminar flow the fluid does not make any abrupt changes in direction or speed. It may be thought of as layered flow. In turbulent flow there is chaos, mixing, eddies, etc. Both cases are illustrated below:

Turbulence often sets in at higher flow rates or if the fluid has to flow around a shape that is not "aerodynamic" (e.g. a block etc.). It is normally undesirable because it heats the fluid and is therefore inefficient.

Properties of Materials Materials deform (i.e. extend or compress) when forces are applied to them. In elastic deformation the material returns to its original dimensions when the the force is removed. In plastic deformation the material remains stretched or compressed when the force is removed. Most materials are elastic for small loads (forces) - but behave plastically if the force is increased beyond a certain point. This is illustrated by the graph below, which is for a typical material (e.g. copper wire):

If the load is removed in the elastic region (0 - A), the material returns to 0 (zero extension/compression). If the load is removed in the plastic region (between A and B), the material returns to C. i.e. It is permanently deformed. Ductile materials can be drawn out into long thin shapes (or wires) - e.g. chewing gum. Malleable materials can be deformed (hammered) into a flat shape - e.g. fudge. Ductility and malleability are both examples of plastic behaviour. Brittle materials crack and shatter easily. They are usually stiff, and break before becoming plastic - e.g. glass. Tough materials are not brittle. They can withstand dynamic loads (shocks). They also deform plastically (and absorb a lot of energy) before breaking, Tough materials in the engineering sense are usually stiff as well - e.g. Kevlar. Stiff materials (e.g. steel) need a large force to produce a small deformation. i.e. Their stiffness k is large (see HFS notes). Hard materials are resistant to scratching and indentation - e.g. diamond.

Stress, Strain and Young Modulus Consider the sample of material shown below, which is being extended by a force F .....

The tensile stress  in the material is defined as follows .....

 = F/A

units of  = Nm-2 or Pa

The tensile strain  in the material is defined as follows .....

 = x/l

Strain is a ratio of two lengths, and  has no units.

(Compressive stress and compressive strain are defined by the same equations as those above. The only difference is that the force arrows are pushing inwards - and x is a compression rather than an extension.) N. B. The advantage of using "stress" and "strain" rather than "force" and "extension" is that a given stress on a given material always produces the same strain in that material. However, the extension produced by a given force is not always the same - it depends on the dimensions of the sample of material. The strength or breaking stress of a material is the stress at which it breaks. It is a constant for a given material.

A graph of stress against strain for a typical material is shown below .....

Note that it is similar in shape to the force-extension graph in "EAT". The limit of proportionality is the end of the straight line section. Beyond this point, stress is no longer proportional to strain. The elastic limit or yield point is where the material makes the transition from elastic to plastic behaviour. Beyond the elastic limit the material no longer returns to its original dimensions when the force (or stress) is removed. The Young modulus E of the material is defined as follows .....

E = stress/strain =  / 

units of E = Nm-2 or Pa

Note that the Young modulus is a constant for a given material. e.g. The Young modulus of steel = 2 x 1011 Nm-2. It is a measure of how stiff the material is. (Recall that: Stiffness k = F/x. k is not a constant for a given material. It depends on the dimensions of the sample.) Since the Young modulus E = stress/strain, it follows that E is also the gradient of the straight part of a stressstrain graph. (It is only valid up to the limit of proportionality.)

Elastic Strain Energy The elastic strain energy Eel stored in a sample of material when it is stretched (or compressed) is the area under the force-extension (or force-compression) graph, as shown below .....

If the material obeys Hooke's Law, the graph is a straight line, as shown below .....

The area then equals the area of a triangle, in which case .....

Eel = Fx / 2 The difference at a molecular level between an unstretched and a stretched polymer (such as rubber, plastics, etc) is shown below .....