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FHSST Physics/Forces/Newton's Laws of Motion

=Newton's Laws of Motion= Our current laws of motion were discovered by Sir Isaac Newton. It is said that Sir Isaac Newton started to think about the nature of motion and gravitation after being struck on the head by a falling apple.

Newton discovered 3 laws describing motion:

First Law
Newton's first law basically says that a force has to be applied to an object to make it move or to make it stop. The first part of that statement definitely makes sense. The only way I can make something move is to have something give it a push. The second part of that statement might not be quite as easy to just take as fact. We've all witnessed objects slow down when nobody is pushing them. How then can we say that the only way to stop an object's motion is with a force? The answer is that there are forces that we don't always see. Most of the time, the force that we don't see is the force of friction.

Friction is the force that resists motion when two things are sliding past one another. To understand what friction is, think about sandpaper. If you try to rub 2 pieces of sandpaper together, it will be hard to get them to slide. This same phenomenon happens between all objects to some degree. This frictional force is what slows objects down or stops their motion.

Second Law
The law is represented in the following basic form (the system of measurement is chosen such that constant of proportionality is 1):

 * $$\vec F=\frac{d(m\vec v)}{dt}$$

The product of mass and velocity i.e. $$m\vec v$$ is called the momentum. The net force on a particle is, thus, equal to rate change of momentum of the particle with time. Generally mass of the object under consideration is constant and thus can be taken out of the derivative:

 * $$\begin{align}\vec F&=m\frac{d\vec v}{dt}\\\vec F&=m\vec a\end{align}$$

Force is equal to mass times acceleration. This version of Newton's Second Law of Motion assumes that the mass of the body does not change with time, and as such, does not represent a general mathematical form of the Law. Consequently, this equation cannot, for example, be applied to the motion of a rocket, which loses its mass (the lost mass is ejected at the rear of the rocket) with the passage of time.

It makes sense that the direction of the acceleration is in the direction of the resultant force. If you push something away from you it doesn't move toward you unless of course there is another force acting on the object towards you!

Worked Example 16 Newton's Second Law
Question: A block of mass 10 kg is accelerating at 2 m·s−2. What is the magnitude of the net force acting on the block?

 * Answer

Step 1:

We are given

 * the block's mass
 * the block's acceleration

all in the correct units.

Step 2 :

We are asked to find the magnitude of the force applied to the block. Newton's Second Law tells us the relationship between acceleration and force for an object. Since we are only asked for the magnitude we do not need to worry about the directions of the vectors:

$\begin{matrix}F_{Net} &=& ma \\&=& 10\ \mbox{kg} \times 2 \mbox{ m} \cdot \mbox{s}^{-2} \\&=& 20\ \mbox{N}\end{matrix}$

Thus, there must be a net force of 20 N acting on the box.

Worked Example 17 Newton's Second Law 2
Question: A 12 N force is applied in the positive x-direction to a block of mass 100 mg resting on a frictionless flat surface. What is the resulting acceleration of the block?

Answer:

Step 1 :

We are given

 * the block's mass
 * the applied force

but the mass is not in the correct units.

Step 2 :

Let us begin by converting the mass:

$\begin{matrix}100 \mbox{ mg} &=& 100 \times 10^{-3} \mbox{ g} = 0.1 \mbox{ g} \\1000 \mbox{ g} &=& 1 \ \mbox{kg} \\1 &=& 1kg\times \frac{1}{1000g} \\&=& \frac{1kg}{1000g} \\0.1g &=& 0.1g \times 1 \\&=& 0.1g \times \frac{1kg}{1000g} \\&=& 0.0001\ kg \\\end{matrix}$

Step 3 :

We know that net force results in acceleration. Since there is no friction the applied force is the resultant or net force on the block (refer to the earlier example of the block pushed on the surface of the table). The block will then accelerate in the direction of this force according to Newton's Second Law.

Step 4 :

To determine the magnitude of the acceleration:

$\begin{matrix}F_{Res} &=& ma \\12 \mbox{ N} &=& (0.0001 \mbox{ kg})a\\a &=& \frac{12 \mbox{ N}}{0.0001 \mbox{kg}}\\&=& 120000\frac{\mbox{N}}{\mbox{kg}} \\&=& 120000\frac{\mbox{kg} \cdot \mbox{m}}{\mbox{s}^2 \cdot \mbox{kg}} \\&=& 120000\frac{\mbox{m}}{\mbox{s}^2} \\&=& 1.2\times 10^{5}\ \mbox{ m} \cdot \mbox{s}^{-2}\end{matrix}$|undefined

From Newton's Second Law the direction of the acceleration is the same as that of the resultant force. The final result is then that the block accelerates at $$1.2\times10^{5}\ \mbox{ m} \cdot \mbox{s}^{-2}$$ in the positive x-direction.

Weight and Mass
You must have heard people saying My weight is 60 kg. This is actually correct, though physicists often have difficulty understanding it because it is mass that is measured in kilograms. This is a different meaning of the word weight from the one used in mechanics, where weight is the force of gravity exerted by the earth on an object with mass:

$\begin{matrix}F_{weight} = mg\end{matrix}$

As such, weight is measured in newtons.

If you compare this equation to Newton's Second Law you will see that it looks exactly the same with the a replaced by g. Thus, when weight is the only force acting on an object (i.e. when $$F_{weight}$$

Fweight is the resultant force acting on the object) the object has an acceleration g. Such an object is said to be in free fall. The value for g is the same for all objects (i.e. it is independent of the objects mass):

$$\begin{matrix}g=9.8 \mbox{m} \cdot \mbox{s}^{-2} \approx 10 \mbox{m} \cdot \mbox{s}^{-2}\end{matrix}$$

You will learn how to calculate this value from the mass and radius of the earth in Chapter ???. Actually the value of g varies slightly from place to place on the Earth's surface.

The reason that we often get confused between weight and mass, is that scales measure your weight (in newtons) and then display your mass using the equation above.

Worked Example 18 Calculating the resultant and then the acceleration
Question: A block (mass 20 kg) on a frictionless flat surface has a 45 N force applied to it in the positive x-direction. In addition a 25 N force is applied in the negative x-direction. What is the resultant force acting on the block and the acceleration of the block?

Answer:

Step 1 :

We are given

 * the block's mass
 * Force F1 = 45 N in the positive x-direction
 * Force F2 = 25 N in the negative x-direction

all in the correct units.

Step 2 :

We are asked to determine what happens to the block. We know that net force results in an acceleration. We need to determine the net force acting on the block.

Step 3 :

Since F1 and F2 act along the same straight line, we can apply the algebraic technique of vector addition discussed in the Vectors chapter to determine the resultant force. Choosing the positive x-direction as our positive direction:

Positive x-direction is the positive direction:

$\begin{matrix}F_{Res}&=& (+45 \ \mbox{N}) + (-25 \mbox{ N}) \\& = & +20 \ \mbox{N} \\&=& 20 \mbox{N}\ in\ the\ positive\ x-direction\end{matrix}$

where we remembered in the last step to include the direction of the resultant force in words. By Newton's Second Law the block will accelerate in the direction of this resultant force.

Step 4 :

Next we determine the magnitude of the acceleration:

$\begin{matrix}F_{Res} & =& ma \\20 \mbox{N} &=& (20 \mbox{kg})a\\a &=& \frac{20 \mbox{N}}{20 \mbox{kg}} \\&=& 1\frac{ \mbox{N}}{ \mbox{kg}} \\&=& 1\frac{ \mbox{kg} \cdot \mbox{m}}{ \mbox{s}^2 \cdot \mbox{kg}} \\&=& 1\ \mbox{ m} \cdot \mbox{s}^{-2} \\\end{matrix}$|undefined

The final result is then that the block accelerates at $$1\ \mbox{ m} \cdot \mbox{s}^{-2}$$ in the positive x-direction (the same direction as the resultant force).

Worked Example 19 Block on incline
Question: A block (mass 10 kg) is released on an inclined plane. What is the resulting rate of acceleration of the block if the angle theta is 25 degrees and the coefficient of friction between the block and the plane is 0.25?

Solution:

Step 1: The acceleration of the block will be parallel to the plane, so we break the problem into two parts - forces perpendicular to the plane (which must cancel out since there is no acceleration perpendicular to the plane) and parallel to the plane.

Step 2: We know that there is no acceleration in the direction perpendicular to the surface of the inclined plane, or in other words $$a_{\perp}=0$$. Therefore we know that:

 * $$\begin{matrix} \\ \sum{F_{\perp}} &=& 0

\\ N - W_{\perp} &=& 0 \\ N - m \   g \  {\cos \theta} &=& 0 \\ N &=& m \ g \  {\cos \theta} \\ &=& (10 \ kg)(9.8 \  {m \over {s^2}}) \  {\cos 25^\circ} \\ &=& 88.82 \ N \end{matrix}$$

Step 3: Summing forces parallel to the plane yields:
 * $$\begin{matrix} \\ \sum{F_{\|}} &=& m \ g \  {\sin \theta} - \mu \  N

\\ m a_{\|} &=& m \  g \  {\sin \theta} - \mu \  N \\  (10 kg) a_{\|} &=& (10 kg) (9.8 \  {m \over s^2}) \sin 25^\circ - (0.25) \  (88.82 \  N) \\ a_{\|} &=& (9.8 \  {m \over s^2}) \  \sin 25^\circ - (0.25) \  {{88.82 \ N} \over {10 \ kg}} \\ &=& 5.5 \ {m \over {s^2}} \end{matrix}$$

The final result is that the block accelerates down the inclined plane with an acceleration of $$5.5 \ m \cdot s^{-2}$$.

Third Law
This law is a direct consequence of the Principle of Conservation of Linear Momentum.

Newton's Third Law is easy to understand but it can get quite difficult to apply it. An important thing to realise is that the action and reaction forces can never act on the same object and hence cannot contribute to the same resultant.

Worked Example 20 Identifying action-reaction Pairs
Question: Consider pushing a box on the surface of a rough table.

 * 1) Draw a force diagram indicating all of the forces acting  on the box.
 * 2) Identify the reaction force for each of the forces acting on the box.

Answer:

 * 1) The following force diagram shows all of the forces acting on the box

There is an important thing to realise which is related to Newton's Third Law. Think about dropping a stone off a cliff. It falls because the earth exerts a force on it (see Chapter ???) and it doesn't seem like there are any other forces acting. So is Newton's Third Law wrong? No, the reactionary force to the weight of the stone is the force exerted by the stone on the earth. This is illustrated in detail in the next worked example.

Worked Example 21 Newton's Third Law
Question: A stone of mass 0.5 kg is accelerating at 10 m·s−2 towards the earth.

 * 1) What is the force exerted by the earth on the stone?
 * 2) What is the force exerted by the stone on the earth?
 * 3) What is the acceleration of the earth, given that its mass is 5.97 × 1027 kg?

Answer:

 * 1) Step 1 :  We are given

 * the stone's mass
 * the stone's acceleration (g)

 * 1) By Newton's Third Law the stone must exert an equal but opposite force on the earth. Hence the stone exerts a force of 5 N towards the stone on the earth.
 * 2) We have

 * the force acting on the earth
 * the Earth's mass

The next two worked examples are quite long and involved but it is very important that you understand the discussion as they illustrate the importance of Newton's Laws.

Worked Example 22 Rockets
Question: How do rockets accelerate in space?

Answer:

 * Gas explodes inside the rocket.
 * This exploding gas exerts a force on each side of the rocket (as shown in the picture below of the explosion chamber inside the rocket).

''Note that the forces shown in this picture are representative. With an explosion there will be forces in all directions.''

 * Due to the symmetry of the situation, all the forces exerted on the rocket are balanced by forces on the opposite side, except for the force opposite the open side. This force on the upper surface is unbalanced.
 * This is therefore the resultant force acting on the rocket and it makes the rocket accelerate forwards.

Systems and External Forces
The concepts of a system and an external forces are very important in physics. A system is any collection of objects. If one draws an imaginary box around such a system then an external force is one that is applied by an object or person outside the box. Imagine for example a car pulling a trailer.

Concept of system is extremely important. Consider an example of a person holding a load (box) on his head and standing on a platform.

Now count the forces :

(i) Box presses the person down (ii) The person pushes the box up (reaction to box's weight) (iii) Box is pulled down by the earth (iv) Earth is pulled up by the box (v) Person presses the floor (vi) Floor pushes the person up (vii) Person is pulled down by earth and (viii) Earth is pulled up by the person.

Quite a mess. Which ones are internal forces and which ones are external forces? Identification of system comes handy here. We can select the box or the person as a system for analyzing forces or we may even consider box and person together as a single system. Only condition to watch while selecting a system is that all parts of the system should have same acceleration. If two parts are at different accelerations then each of them should be treated as a separate system.


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FHSST Physics/Forces/Examples

= Examples of Forces Studied Later =

Most of physics revolves around forces. Although there are many different forces we deal with them all in the same way. The methods to find resultants and acceleration do not depend on the type of force we are considering.

At first glance, the number of different forces may seem overwhelming - gravity, drag, electrical forces, friction and many others. However, physicists have found that all these forces can be classified into four groups. These are gravitational forces, electromagnetic forces, strong nuclear force and weak nuclear force. Even better, all the forces that you will come across at school are either gravitational or electromagnetic. Doesn't that make life easy?

Newtonian Gravity
Gravity is the attractive force between two objects due to the mass of the objects. When you throw a ball in the air, its mass and the Earth's mass attract each other, which leads to a force between them. The ball falls back towards the Earth, and the Earth accelerates towards the ball. The movement of the Earth toward the ball is, however, so small that you couldn't possibly measure it.

Electromagnetic Forces
Almost all of the forces that we experience in everyday life are electromagnetic in origin. They have this unusual name because long ago people thought that electric forces and magnetic forces were different things. After much work and experimentation, it has been realised that they are actually different manifestations of the same underlying theory.

The Electric Force
If we have objects carrying electrical charge, which are not moving, then we are dealing with electrostatic forces (Coulomb's Law). This force is actually much stronger than gravity. This may seem strange, since gravity is obviously very powerful, and holding a balloon to the wall seems to be the most impressive thing electrostatic forces have done, but think about it: for gravity to be detectable, we need to have a very large mass nearby. But a balloon rubbed in someone's hair can stick to a wall with a force so strong that it overcomes the force of gravity - with just the charges in the balloon and the wall!

Magnetic force
The magnetic force is a different manifestation of the electromagnetic force. It stems from the interaction between moving charges as opposed to the fixed charges involved in Coulomb's Law.

Examples of the magnetic force in action include magnets, compasses, car engines, computer data storage and your hair standing on end. Magnets are also used in the wrecking industry to pick up cars and move them around sites.

Friction
Newton's First Law states that an object moving without a force acting on it will keep moving. Then why does a box sliding on a table stop? The answer is friction. Friction arises from the interaction between the molecules on the bottom of a box with the molecules on a table. This interaction is electromagnetic in origin, hence friction is just another view of the electromagnetic force. The great part about school physics is that most of the time we are told to neglect friction but it is good to be aware that there is friction in the real world.

Friction is also useful. If there was no friction and you tried to prop a ladder up against a wall, it would simply slide to the ground. Rock climbers use friction to maintain their grip on cliffs.

Friction is a force that impedes motion. It is in parallel to the contact surface and acts against the motion of the body.

Drag Force
This is the force an object experiences while travelling through a medium. When something travels through the air it needs to displace air as it travels and because of this the air exerts a force on the object. This becomes an important force when you move fast and a lot of thought is taken to try and reduce the amount of drag force a sports car experiences.

The drag force is very useful for parachutists. They jump from high altitudes and if there was no drag force, then they would continue accelerating all the way to the ground. Parachutes are wide because the more surface area you show, the greater the drag force and hence the slower you hit the ground.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: FHSST Physics/Forces/Examples


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FHSST Physics/Forces/Summary

= Summary of Important Quantities, Equations and Concepts =

Equilibrium: Objects at rest or moving with constant velocity are in equilibrium and have a zero resultant force.

Equilibrant: The equilibrant of any number of forces is the single force required to produce equilibrium.

Triangle Law for Forces in Equilibrium: Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order.

Newton's First Law: Every object will remain at rest or in uniform motion in a straight line unless it is made to change its state by the action of an external force.

Newton's Second Law: The resultant force acting on a body results in an acceleration which is in the same direction as the resultant force and is directly proportional to the magnitude of this force and inversely proportional to the mass of the object.

Newton's Third Law: For every force or action there is an equal but opposite force or reaction.

title: FHSST Physics/Forces/Summary


text: Template:FHSST Physics/Rectilinear Motion


text: FHSST Physics/Rectilinear Motion

title: FHSST Physics/Rectilinear Motion


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FHSST Physics/Rectilinear Motion/Definition

= What is rectilinear motion? =

Rectilinear motion means motion along a straight line. This is a useful topic to study for learning how to describe the movement of cars along a straight road or of trains along straight railway tracks. In this section you have only two directions to worry about: (1) along the direction of motion, and (2) opposite to the direction of motion.

To illustrate this imagine a train heading east.

If it is accelerating away from the station platform (P), the direction of acceleration is the same as the direction of the train's velocity — east. If it is braking the direction of acceleration is opposite to the direction of its motion, i.e. west.


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FHSST Physics/Rectilinear Motion/Speed and Velocity

= Speed and Velocity =

Let's take a moment to review our definitions of velocity and speed by looking at the worked example below:

Worked Example 23 Speed and Velocity

Question: A cyclist moves from A through B to C in 10 seconds. Calculate both his speed and his velocity.

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

 * the distance between A and B
 * the distance between B and C
 * the total time for the cyclist to go from A through B to C

all in the correct units!

Step 2 :

What is being asked? We are asked to calculate the average speed and the average velocity of the cyclist.

His speed - a scalar - will be

$$\begin{matrix}v&=&\frac{s}{t}\\&=&\frac{30m+40m}{10s}\\&=&7\frac{m}{s}\end{matrix}$$

Since velocity is a vector we will first need to find the resultant displacement of the cyclist. His velocity will be

$$\begin{matrix}\overrightarrow{v}=\frac{\overrightarrow{s}}{t}\end{matrix}$$

The total displacement is the vector from A to C, and this is just the resultant of the two displacement vectors, i.e.

$$\begin{matrix}\overrightarrow{s} = \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}\end{matrix}$$

Using the rule of Pythagoras:

$$\begin{matrix}\overrightarrow{s} &=& \sqrt{{(30m)}^2+{(40m)}^2}\\&=& 50m \ in \ the \ direction \ from \ A \ to \ C\end{matrix}$$

∴

$$\begin{matrix}\overrightarrow{v} &=&\frac{50\ \mbox{m}}{10\ \mbox{s}}\\&=& 5\ \frac{\mbox{m}}{\mbox{s}} \ in\ the \ direction \ from \ A \ to \ C\end{matrix}$$

For this cyclist, his velocity is not the same as his speed because there has been a change in the direction of his motion. If the cyclist traveled directly from A to C without passing through B his speed would be

$$\begin{matrix}v &=& \frac{50m}{10s}\\&=& 5 \frac{m}{s}\end{matrix}$$

and his velocity would be

$$\begin{matrix}\overrightarrow{v} &=& \frac{50m}{10s}\\&=& 5 \frac{m}{s} \ in \ the\ direction \ from \ A \ to \ C\end{matrix}$$

In this case where the cyclist is not undergoing any change of direction (i.e. he is traveling in a straight line) the magnitudes of the speed and the velocity are the same. This is the defining principle of rectilinear motion.


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FHSST Physics/Rectilinear Motion/Graphs

= Graphs =

In physics we often use graphs as important tools for picturing certain concepts. Below are some graphs that help us picture the concepts of displacement, velocity and acceleration.

Displacement-Time Graphs
Below is a graph showing the displacement of the cyclist from A to C:

This graphs shows us how, in 10 seconds time, the cyclist has moved from A to C. We know the gradient (slope) of a graph is defined as the change in y divided by the change in x, i.e. $$\frac{\Delta y}{\Delta x}$$. In this graph the gradient of the graph is just $$\frac{\Delta \overrightarrow{s}}{\Delta t}$$ - and this is just the expression for velocity.

The slope is the same all the way from A to C, so the cyclist's velocity is constant over the entire displacement he travels. In figure 5.1 are examples of the displacement-time graphs you will encounter.

a) shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity.

b) shows the graph for an object moving at a constant velocity. You can see that the displacement is increasing as time goes on. The gradient, however, stays constant (remember: its the slope of a straight line), so the velocity is constant. Here the gradient is positive, so the object is moving in the direction we have defined as positive.

c) shows the graph for an object moving at a constant acceleration. You can see that both the displacement and the velocity (gradient of the graph) increase with time. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating.

Uniform acceleration and gradient (slope)
Look at the velocity-time graph below:

This is the velocity-time graph of a cyclist traveling from A to B at a constant acceleration, i.e. with steadily increasing velocity. The gradient (slope) of this graph is just $$\frac{\Delta \overrightarrow{v}}{\Delta t}$$ - and this is just the expression for acceleration. Because the slope is the same at all points on this graph, the acceleration of the cyclist is constant. And the constant acceleration is $$2 / m/s^2$$, or 2 meters per second per second (or 2 meters per second squared)

$$\begin{matrix}\mathrm{slope \ of \ line} &=& \Delta v / \Delta t \\&=& \frac{10 m/s}{5 s} \\&=& 2 \frac{m}{s^2} \\\end{matrix}$$

Distance travelled
Not only can we get the acceleration of an object from its velocity-time graph, but we can also get some idea of the displacement traveled. Look at the graph below:

This graph shows an object moving at a constant velocity of 10m/s for a duration of 5s. The area between the graph and the time axis (the shaded area) of the above plot will give us the displacement of the object during this time. In this case we just need to calculate the area of a rectangle with width 5s and height 10 m/s

$$\begin{matrix}\mathrm{area \ of \ rectangle} &=& \mathrm{height} \times \mathrm{width}\\&=& \overrightarrow{v} \times t\\&=& 10 \frac{m}{s} \times 5 s\\&=& 50 m\\&=& \overrightarrow{s}= \mathrm{displacement}\\\end{matrix}$$

So, here we've shown that an object traveling at 10 m/s for 5s has undergone a displacement of 50m.

Here are a couple more velocity-time graphs to get used to:

In figure 5.2 are examples of the displacement-time graphs you may encounter.

a) shows the graph for an object moving at a constant velocity over a period of time. The gradient is zero, so the object is not accelerating.

b) shows the graph for an object which is decelerating. You can see that the velocity is decreasing with time. The gradient, however, stays constant (remember: its the slope of a straight line), so the acceleration is constant. Here the gradient is negative, so the object is accelerating in the opposite direction to its motion, hence it is decelerating.

Acceleration-Time Graphs
In this chapter on rectilinear motion we will only deal with objects moving at a constant acceleration, thus all acceleration-time graphs will look like these two:

Here is a description of the graphs below:

a) shows the graph for an object which is either stationary or traveling at a constant velocity. Either way, the acceleration is zero over time.

b) shows the graph for an object moving at a constant acceleration. In this case the acceleration is positive - remember that it can also be negative.

We can obtain the velocity of a particle at some given time from an acceleration time graph - it is just given by the area between the graph and the time-axis. In the graph below, showing an object at a constant positive acceleration, the increase in velocity of the object after 2 seconds corresponds to the shaded portion.

$$\begin{matrix}\mathrm{area \ of \ rectangle} &=& \overrightarrow{a} \times t \\&=& 5 \frac{m}{s^2} \times 2s\\&=& 10 \frac{m}{s}\\&=& \overrightarrow{v}\\\end{matrix}$$

Its useful to remember the set of graphs below when working on problems. Figure 5.3 shows how displacement, velocity and time relate to each other. Given a displacement-time graph like the one on the left, we can plot the corresponding velocity-time graph by remembering that the slope of a displacement-time graph gives the velocity. Similarly, we can plot an acceleration-time graph from the gradient of the velocity-time graph.

Worked Example 24 Relating displacement-, velocity-, and acceleration-time graphs
Question: Given the displacement-time graph below, draw the corresponding velocity-time and acceleration-time graphs, and then describe the motion of the object.

Answer:

Step 1 : Analyse the question to determine what is given. The question explicitly gives a displacement-time graph.

Step 2 : What is asked?

3 things:

 * 1) Draw a velocity-time graph
 * 2) Draw an acceleration-time graph
 * 3) Describe the behaviour of the object

For the first 2 seconds we can see that the displacement remains constant - so the object is not moving, thus it has zero velocity during this time. We can reach this conclusion by another path too: remember that the gradient of a displacement-time graph is the velocity. For the first 2 seconds we can see that the displacement-time graph is a horizontal line, i.e. it has a gradient of zero. Thus the velocity during this time is zero and the object is stationary.

For the next 2 seconds, displacement is increasing with time so the object is moving. Looking at the gradient of the displacement graph we can see that it is not constant. In fact, the slope is getting steeper (the gradient is increasing) as time goes on. Thus, remembering that the gradient of a displacement-time graph is the velocity, the velocity must be increasing with time during this phase.

For the final 2 seconds we see that displacement is still increasing with time, but this time the gradient is constant, so we know that the object is now travelling at a constant velocity, thus the velocity-time graph will be a horizontal line during this stage.

So our velocity-time graph looks like this one below. Because we haven't been given any values on the vertical axis of the displacement-time graph, we cannot figure out what the exact gradients are and hence what the values of the velocity are. In this type of question it is just important to show whether velocities are positive or negative, increasing, decreasing or constant.

Once we have the velocity-time graph its much easier to get the acceleration-time graph as we know that the gradient of a velocity-time graph is the just the acceleration.

For the first 2 seconds the velocity-time graph is horizontal at zero, thus it has a gradient of zero and there is no acceleration during this time. (This makes sense because we know from the displacement time graph that the object is stationary during this time, so it can't be accelerating).

For the next 2 seconds the velocity-time graph has a positive gradient. This gradient is not changing (i.e. its constant) throughout these 2 seconds so there must be a constant positive acceleration.

For the final 2 seconds the object is traveling with a constant velocity. During this time the gradient of the velocity-time graph is once again zero, and thus the object is not accelerating.

The acceleration-time graph looks like this:

A brief description of the motion of the object could read something like this: At t = 0s and object is stationary at some position and remains stationary until t = 2s when it begins accelerating. It accelerates in a positive direction for 2 seconds until t = 4s and then travels at a constant velocity for a further 2 seconds.

Worked Example 25 Calculating distance from a velocity-time graph
Question: The velocity-time graph of a car is plotted below. Calculate the displacement of the car has after 15 seconds.

Answer: We are asked to calculate the displacement of the car. All we need to remember here is that the area between the velocity-time graph and the time axis gives us the displacement.

For t = 0s to t = 5s this is the triangle on the left:

$$\begin{matrix}Area \triangle &=& \frac{1}{2}b \times h\\&=& \frac{1}{2}5s\times 4m/s\\&=&10m\end{matrix}$$

For t = 5s to t = 12s the displacement is equal to the area of the rectangle

$$\begin{matrix}Area \Box &=& w \times h\\&=&7s \times 4m/s\\&=&28m\end{matrix}$$

For t = 12s to t = 14s the displacement is equal to the area of the triangle above the time axis on the right

$$\begin{matrix}Area \triangle &=& \frac{1}{2}b \times h\\&=& \frac{1}{2}2s \times 4m/s\\&=&4m\end{matrix}$$

For t = 14s to t = 15s the displacement is equal to the area of the triangle below the time axis

$$\begin{matrix}Area \triangle &=& \frac{1}{2}b \times h\\&=& \frac{1}{2}1s \times 2m/s\\&=&1m\end{matrix}$$

Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from t = 14s to t = 15s) the velocity of the car is negative, it means that the car was going in the opposite direction, i.e. back where it came from! So, to get the total displacement, we have to add the first 3 areas (those with positive displacements) and subtract the last one (because it signifies a displacement in the opposite direction).

$$\begin{matrix}\overrightarrow{s}&=&10 +28 +4 -1\\&=&41m\ in\ the\ positive\ direction\end{matrix}$$

Worked Example 26 Velocity from a displacement-time graph
Question: Given the diplacement-time graph below,

 * 1) what is the velocity of the object during the first 4 seconds?
 * 2) what is the velocity of the object from t = 4s to t = 7s?

Answer:
 * 1) for the first 4sec velocity is slope of the curve i.e. 2/4=0.5 m/s

 * 1) For the last 3 seconds we can see that the displacement stays constant, and that the gradient is zero. Thus $$\overrightarrow{v}=0m/s$$.

Worked Example 27 From an acceleration- to a velocity-time graph
Question: Given the acceleration-time graph below, assume that the object starts from rest and draw its velocity-time graph.

Answer:

title: FHSST Physics/Rectilinear Motion/Graphs


text:
FHSST Physics/Rectilinear Motion/Equations of Motion

= Equations of Motion =

This section is about solving problems relating to uniformly accelerated motion. We'll first introduce the variables and the equations, then we'll show you how to derive them, and after that we'll do a couple of examples.

 * u = starting velocity (m/s) at t = 0
 * v = final velocity (m/s) at time t
 * s = displacement (m)
 * t = time (s)
 * a = acceleration (m/s²)

$$\Delta x = 1/2*a*(\Delta t)^2 + u(\Delta t)$$

Make sure you can rhyme these off, they are very important! There are so many different types of questions for these equations. Basically when you are answering a question like this:

 * 1) Find out what values you have and write them down.
 * 2) Figure out which equation you need.
 * 3) Write it down!!!
 * 4) Fill in all the values you have and get the answer.

Equation 5.1
By the definition of acceleration $$a = \frac{\Delta v}{t}$$ where $$\Delta$$v is the change in velocity, i.e. $$\Delta v = v - u$$. Thus we have

$$\begin{matrix}a &=& \frac{v-u}{t} \\v &=& u +at\end{matrix}$$

Equation 5.2
In the previous section we saw that displacement can be calculated from the area between a velocity-time graph and the time-axis. For uniformly accelerated motion the most complicated velocity-time graph we can have is a straight line. Look at the graph below — it represents an object with a starting velocity of u, accelerating to a final velocity v over a total time t.

To calculate the final displacement we must calculate the area under the graph — this is just the area of the rectangle added to the area of the triangle.

$$\begin{matrix}Area \triangle &=& \frac{1}{2}b \times h\\&=& \frac{1}{2}t \times (v-u)\\&=& \frac{1}{2}vt - \frac{1}{2}ut\end{matrix}$$

$$\begin{matrix}Area \Box &=& w \times h\\&=& t \times u\\&=& ut\end{matrix}$$

$$\begin{matrix}Displacement &=& Area \Box + Area \triangle\\s &=& ut + \frac{1}{2}vt -\frac{1}{2}ut\\&=& \frac{(u+v)}{2}t\end{matrix}$$

Equation 5.3
This equation is simply derived by eliminating the final velocity v in equation 5.2. Remembering from equation 5.1 that

$$v=u+at$$

then equation 5.2 becomes

$$\begin{matrix}s &=& \frac{u+u+at}{2}t\\&=&\frac{2ut+at^2}{2}\\&=&ut+\frac{1}{2}at^2\end{matrix}$$

Equation 5.4
This equation is just derived by eliminating the time variable in the above equation. From Equation 5.1 we know

$$t=\frac{v-u}{a}$$

Substituting this into Equation 5.3 gives

This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.

Worked Example 28
Question: A racing car has an initial velocity of 100 m/s and it covers a displacement of 725 m in 10 s. Find its acceleration.

Answer:

Step 1 :

We are given the quantities u, s and t — all in the correct units. We need to find a.

Step 2 :

We can use equation 5.3 $$s=ut +\frac{1}{2}at^2$$

Step 3 :

Rearranging equation 5.3 we have $$a=\frac{2(s-ut)}{t^2}$$

Substituting in the values of the known quantities this becomes

$$\begin{matrix}a &=& \frac{2(725 \mbox{ m}-100\frac{\mbox{m}}{\mbox{s}}\cdot10\mbox{ s})}{10^2 \mbox{ s}^2}\\&=&\frac{2(-275 \mbox{ m})}{100 \mbox{ s}^2}\\&=&-5.5\frac{\mbox{m}}{\mbox{s}^2}\end{matrix}$$

The racing car is accelerating at -5.5 m/s², or we could say it is decelerating at 5.5 m/s². this ans is wrong

its better to use a=v-u/t then a=(725-100)/10s you will get a=625/10 which will be equal to 62.5m/s2

Worked Example 29
Question: An object starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4 s. Calculate

 * its acceleration
 * its final velocity
 * at what time the object had covered half the total distance
 * what distance the object had covered in half the total time.

Answer:

Step 1 : We are given the quantities u, s and t in the correct units.

Step 2 : To calculate the acceleration we can use equation [#eq:eq3 5.3]. Rearranging it we have:

$$a=\frac{2(s-ut)}{t^2}$$

Substituting in the values of the known quantities this becomes

$$\begin{matrix}a &=& \frac{2(64m-0\frac{m}{s}4s}{4^2s^2}\\&=&\frac{128m}{16s^2}\\&=&8\frac{m}{s^2}\end{matrix}$$

Step 3 : To calculate its final velocity we can use equation 5.1 - remember we now also know the acceleration of the object.

$$\begin{matrix}v&=&u+at\\&=&0\frac{m}{s}+(8\frac{m}{s^2})(4s)\\&=&32\frac{m}{s}\end{matrix}$$

Step 4 : The time at which the object had covered half the total distance. Half the distance is 32m. Here we have the quantities s, u and a so we first use equation 5.4 to calculate the velocity at this distance:

$$\begin{matrix}v^2&=&u^2+2as\\&=&(0m)^2+2(8m/s^2)(32m)\\&=&512m^2/s^2\\v&=&22.6m/s\end{matrix}$$

Now we can use equation 5.2 to calculate the time:

$$\begin{matrix}t&=&\frac{2s}{u+v}\\&=&\frac{(2)(32m)}{0m/s+22.6m/s}\\&=&2.8s\end{matrix}$$

Step 5 : The distance the object had covered in half the time. Half the time is 2s. Thus we have u, a and t - all in the correct units. We can use equation 5.3 to get the distance:

$$\begin{matrix}s&=&ut+\frac{1}{2}at^2\\&=&(0m/s)(2s)+\frac{1}{2}(8\frac{m}{s^2})(2s)^2\\&=&16m\end{matrix}$$

Worked Example 30
Question: A ball is thrown vertically upwards with a velocity of 10 m/s from the balcony of a tall building. The balcony is 15 m above the ground and gravitational acceleration is 10 m/s2. Find a) the time required for the ball to hit the ground, and b) the velocity with which it hits the ground.

Answer:1
Step 1 : In this case it often helps to make the problem easier to understand if we draw ourselves a picture like the one below:

First the ball goes upwards with gravitational acceleration slowing it until it reaches its highest point — here its speed is 0 m/s — then it begins descending with gravitational acceleration causing it to increase its speed on the way down. We can separate the motion into 2 stages:

Stage 1 - the upward motion of the ball

Stage 2 - the downward motion of the ball.

We'll choose the upward direction as positive - this means that gravitation acceleration is negative - and we'll begin by solving for all the variables of Stage 1. So far we have these quantities:

$$\begin{matrix}u_1&=&10m/s\\v_1&=&0m/s\\a_1&=&-10m/s^2\\t_1&=&?\\s_1&=&?\end{matrix}$$

Using equation 5.1 to find t1:

$$\begin{matrix}v_1&=&u_1+a_1t_1\\t_1&=&\frac{v_1-u_1}{a_1}\\&=&\frac{0m/s-10m/s}{-10m/s^2}\\&=&1s\end{matrix}$$

We can find s1 by using equation 5.4

$$\begin{matrix}v_1^2&=&u_1^2+2a_1s_1\\s_1&=&\frac{v_1^2-u_1^2}{2a}\\&=&\frac{(0m/s)^2-(10m/s)^2}{2(-10m/s^2)}\\&=&5m\end{matrix}$$

For Stage 2 we have the following quantities:

$$\begin{matrix}u_2&=&0m/s\\v_2&=&?\\a_2&=&-10m/s^2\\t_2&=&?\\s_2&=&-15m-5m=20m\end{matrix}$$

We can determine the final velocity v2 using equation 5.4:

$$\begin{matrix}v_2^2&=&u_2^2+2a_2s_2\\&=&(0m/s)^2+2(-10m/s^2)(-20m)\\&=&400(m/s)^2\\v_2&=&20m/s \mbox{ downwards}\end{matrix}$$

Now we can determine the time for Stage 2, t2, from equation 5.1:

$$\begin{matrix}v_2&=&u_2+a_2t_2\\t_2&=&\frac{v_2-u_2}{a_2}\\&=&\frac{-20m/s-0m/s}{-10m/s^2}\\&=&2s\end{matrix}$$

Finally,

a) the time required for the stone to hit the ground is

$$t=t_1+t_2=1 \mbox{ s}+2 \mbox{ s}=3 \mbox{ s}$$

b) the velocity with which it hits the ground is just

$$v_2=-20 \ \mbox{m}/\mbox{s}$$

Answer:2
applying

$$\vec s(t) = \vec S_i + \vec U_i \ (t- T_i) + \frac {\vec A}{2} (t - T_i)^2 $$

$$ 0 = 15 \hat j + 10 \hat j (T_{total} - 0 ) + \frac {-10 \hat j}{2} (T_{total} - 0)^2 $$

$$ T_{total}^2 - 2 \ T_{total} - 3 = 0 $$

solving we will get

$$ T_{total} = 3, -1 $$

so time to reach is 3 second, now applying

$$ \vec v (t) = \vec U_i + \vec A (t- T_i)$$

$$ \vec v (3) = 10 \hat j + (-10 \hat j) (3- 0)$$

$$ \vec v (3) = -20 \hat j $$

These questions do not have the working out in them, but they are all done in the manner described on the previous page.

Question: A car starts off at 10 m/s and accelerates at 1 m/s2 for 10 seconds. What is its final velocity?

Answer: 20 m/s

Question: A car starts from rest, and accelerates at 1 m/s2 for 10 seconds. How far does it move?

Answer: 50 m

Question: A car is going 30 m/s and stops in 2 seconds. What is its stopping distance for this speed?

Answer: 30 m

Question: A car going at 20 m/s stops in a distance of 20 m.

 * 1) What is its deceleration?
 * 2) If the car is 1 tonne (1000 kg, or 1 Mg) how much force do the brakes exert?


text:
FHSST Physics/Rectilinear Motion/Summary

= Important Equations and Quantities =


text: Template:FHSST Physics/Momentum

title: Template:FHSST Physics/Momentum


text:
FHSST Physics/Momentum/Definition

= What is Momentum? =

Momentum is a physical quantity which is closely related to forces. We will learn about this connection a little later. Remarkably momentum is a conserved quantity. This makes momentum extremely useful in solving a great variety of real-world problems. Firstly we must consider the definition of momentum.

Mathematically,

Thus, momentum is a property of a moving object and is determined by its velocity and mass. A large truck travelling slowly can have the same momentum as a much smaller car travelling relatively fast.

Note the arrows in the equation defining momentum- momentum is a vector with the same direction as the velocity of the object.

Since the direction of an object's momentum is given by the direction of its motion, one can calculate an object's momentum in two steps:

 * include in the final answer the direction of the object's motion

Worked Example 31 Calculating Momentum 1
Question: A ball of mass 3kg moves at 2m.s−1 to the right. Calculate the ball's momentum.

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

 * the ball's mass, and
 * the ball's velocity

in the correct units!

Step 2 :

What is being asked? We are asked to calculate the ball's momentum. From the definition of momentum,

$$\begin{matrix}\overrightarrow{p} = m\overrightarrow{v},\end{matrix}$$

we see that we need the mass and velocity of the ball, which we are given.

Step 3 :

Firstly we calculate the magnitude of the ball's momentum,

$$\begin{matrix}p &=& mv \\&=& (3kg)(2m.s^{-1}) = 6\ kg.m.s^{-1}.\end{matrix}$$

Finally we quote the answer with the direction of the ball's motion included,

$$\begin{matrix}\overrightarrow{p} = 6\ kg.m.s^{-1}\textbf{\ to\ the\ right}\end{matrix}$$

Worked Example 32 Calculating Momentum 2
Question: A ball of mass 500g is thrown at 2m.s−1. Calculate the ball's momentum.

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

 * the ball's mass, and
 * the magnitude of the ball's velocity

but with the ball's mass in the incorrect units!

Step 2 :

What is being asked? We are asked to calculate the momentum which is defined as

$$\begin{matrix}\overrightarrow{p} = m\overrightarrow{v}.\end{matrix}$$

Thus, we need the mass and velocity of the ball but we have only its mass and the magnitude of its velocity.

Step 3 :

In order to determine the velocity of the ball we need the direction of the ball's motion. If the problem does not give an explicit direction we are forced to be general. In a case like this we could say that the direction of the velocity is in the direction of motion of the ball. This might sound silly but the lack of information in the question has forced us to be, and we are certainly not wrong! The ball's velocity is then m.s−1 in the direction of motion.

Step 4 :

Next we convert the mass to the correct units,

$$\begin{matrix}1000g&=&1kg \\1 & = & \frac{1kg}{1000g}\\500g \times 1 &=& 500g \times \frac{1kg}{1000g}\\& = & 0.500kg\end{matrix}$$

Step 5 :

Now, let us find the magnitude of the ball's momentum,

$$\begin{matrix}p &=& mv\\&=& (0.500kg)(2m.s^{-1}) = 1\ kg.m.s^{-1}\end{matrix}$$

Step 6 :

Finally, we quote the answer with the direction of the momentum included,

$$\begin{matrix}\overrightarrow{p} &=& 1\ kg.m.s^{-1}\textbf{\ in\ the\ direction\ of\ motion\ of\ the\ ball} \end{matrix}$$

Worked Example 33 Calculating the Momentum of the Moon
Question: The moon is $$384\ 400km$$ away from the earth and orbits the earth in 27.3 days. If the moon has a mass of $$7.35\times10^{22}kg$$ [footnode.html#foot11743 6.1] what is the magnitude of its momentum if we assume a circular orbit?

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

 * the moon's mass,
 * the distance to the moon, and
 * the time for one orbit of the moon

with mass in the correct units but all other quantities in the incorrect units.

The units we require are

 * seconds (s) for time, and
 * metres (m) for distance pesteng physics→→→→→→

Step 2 :

What is being asked? We are asked to calculate only the magnitude of the moon's momentum (i.e. we do not need to specify a direction). In order to do this we require the moon's mass and the magnitude of its velocity, since $$\begin{matrix}p = mv.\end{matrix}$$

Step 3 :

How do we find the speed or magnitude of the moon's velocity? Speed is defined as,

$$\begin{matrix}speed &=& \frac{Distance}{time}\end{matrix}$$

We are given the time the moon takes for one orbit but not how far it travels in that time. However, we can work this out from the distance to the moon and the fact that the moon's orbit is circular. Firstly let us convert the distance to the moon to the correct units,

$$\begin{matrix}1km &=& 1000m\\1 &=& \frac{1000m}{1km}\\384\ 400km \times 1 &=& 384\ 400km\times\frac{1000m}{1km}\\&=& 384\ 400\ 000 m\\&=& 3.844\times10^8\ m\end{matrix}$$

Using the equation for the circumference, C, of a circle in terms of its radius, we can determine the distance travelled by the moon in one orbit:

$$\begin{matrix}C&=&2\pi r\\&=&2\pi (3.844\times10^8\ m)\\&=& 2.42\times10^{9}\ m.\\\end{matrix}$$

Next we must convert the orbit time, T, into the correct units. Using the fact that a day contains 24 hours, an hour consists of 60 minutes, and a minute is 60 seconds long,

$$\begin{matrix}1 day&=&(24)(60)(60) seconds \\1 &=&\frac{(24)(60)(60)s}{1 day}\\27.3days\times 1 &=& 27.3 days \frac{(24)(60)(60)s}{1day}\\& = & 2.36\times10^{6}s\end{matrix}$$

Therefore, $$\begin{matrix}T = 2.36\times10^{6}s. \end{matrix}$$

Combining the distance travelled by the moon in an orbit and the time taken by the moon to complete one orbit, we can determine the magnitude of the moon's velocity or speed,

$$\begin{matrix}v &=& \frac{Distance}{time}\\&=& \frac{C}{T}\\&=& 1.02\times10^{3}\ m.s^{-1}.\end{matrix}$$

Step 4 :

Finally we can calculate the magnitude of the moon's momentum,

$$\begin{matrix}p &=& mv\\&=& (7.35\times10^{22}kg)(1.02\times10^{3}\ m.s^{-1})\\&=& 7.50 \times10^{25}\ kg.m.s^{-1}.\end{matrix}$$

→33→

title: FHSST Physics/Momentum/Definition


text:
FHSST Physics/Momentum/System

= The Momentum of a System =

In Chapter 4 Forces the concept of a system was introduced. The bodies that make up a system can have different masses and can be moving with different velocities. In other words they can have different momenta.

Since momentum is a vector, the techniques of vector addition discussed in This chapter must be used to calculate the total momentum of a system. Let us consider an example.

Worked Example 34 Calculating the Total Momentum of a System
Question: Two billiard balls roll towards each other. They each have a mass of 0.3kg. Ball 1 is moving at $$v_1=1\ m.s^{-1}$$ to the right, while ball 2 is moving at $$v_2=0.8\ m.s^{-1}$$ to the left.

Calculate the total momentum of the system.

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

all in the correct units!
 * the mass of each ball,
 * the velocity of ball 1, $$\overrightarrow{v_1}$$ and
 * the velocity of ball 2, $$\overrightarrow{v_2}$$,

Step 2 :

What is being asked? We are asked to calculate the total momentum of the system. In this example our system consists of two balls. To find the total momentum we must sum the momenta of the balls,

$$\begin{matrix}\overrightarrow{p}_{total} &=& \overrightarrow{p_1}+\overrightarrow{p_2}\end{matrix}$$

Since ball 1 is moving to the right, its momentum is in this direction, while the second ball's momentum is directed towards the left.

Thus, we are required to find the sum of two vectors acting along the same straight line. The algebraic method of vector addition introduced in this chapter can thus be used.

Step 3 :

Firstly we choose a positive direction. Let us choose right as the positive direction, then obviously left is negative.

Step 4 :

The total momentum of the system is then the sum of the two momenta taking the directions of the velocities into account. Ball 1 is travelling at $$1\ m.s^{-1}\ to\ the\ right$$ or $$+1\ m.s^{-1}$$. Ball 2 is travelling at $$0.8\ m.s^{-1}\ to\ the\ left$$ or $$-0.8\ m.s^{-1}$$. Thus,

Right is the positive direction

$$\begin{matrix}\overrightarrow{p}_{total} &=& m_1\overrightarrow{v_1} +m_2\overrightarrow{v_2}\\&=& (0.3kg)(+1\ m.s^{-1}) + (0.3kg)(-0.8\ m.s^{-1})\\&=& (+0.3\ kg.m.s^{-1}) + (-0.24\ kg.m.s^{-1}) \\&=& +0.06\ kg.m.s^{-1}\\&=& 0.06\ kg.m.s^{-1}\textbf{\ to\ the\ right}\end{matrix}$$

In the last step the direction was added in words. Since the result in the second last line is positive, the total momentum of the system is in the positive direction (i.e. to the right).

title: FHSST Physics/Momentum/System


text:
FHSST Physics/Momentum/Change

= Change in Momentum =

If either an object's mass or velocity changes then its momentum too will change. If an object has an initial velocity $$\overrightarrow{u}$$ and a final velocity $$\overrightarrow{v}$$, then its change in momentum, $$\Delta \overrightarrow{p}$$, is

Worked Example 35 Change in Momentum
Question: A rubber ball of mass 0.8kg is dropped and strikes the floor at a velocity of $$6\ m.s^{-1}$$. It bounces back with an initial velocity of $$4\ m.s^{-1}$$. Calculate the change in momentum of the rubber ball caused by the floor.

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

 * the ball's mass,
 * the ball's initial velocity, and
 * the ball's final velocity

all in the correct units.

Do not be confused by the question referring to the ball bouncing back with an ``initial velocity of $$4\ m.s^{-1}$$. The word initial is included here since the ball will obviously slow down with time and $$4\ m.s^{-1}$$ is the speed immediately after bouncing from the floor.

Step 2 :

What is being asked? We are asked to calculate the change in momentum of the ball,

$$\begin{matrix}\Delta\overrightarrow{p} &=& m\overrightarrow{v} - m\overrightarrow{u}.\end{matrix}$$

We have everything we need to find $$\Delta\overrightarrow{p}$$. Since the initial momentum is directed downwards and the final momentum is in the upward direction, we can use the algebraic method of subtraction discussed in the vectors chapter.

Step 3 : Firstly, we choose a positive direction. Let us choose down as the positive direction. Then substituting,

Down is the positive direction

$$\begin{matrix}\Delta\overrightarrow{p} &=& m\overrightarrow{v} -m\overrightarrow{u}\\&=& (0.8kg)(-4\ m.s^{-1})-(0.8kg)(+6\ m.s^{-1})\\&=& (0.8kg)(-10\ m.s^{-1})\\&=& -8\ kg.m.s^{-1}\\&=& 8\ kg.m.s^{-1}\textbf{\ up}\end{matrix}$$

where we remembered in the last step to include the direction of the change in momentum in words.

title: FHSST Physics/Momentum/Change


text:
FHSST Physics/Momentum/Properties

= What properties does momentum have? =

You may at this stage be wondering why there is a need for introducing momentum. Remarkably momentum is a conserved quantity. Within an isolated system the total momentum is constant. No matter what happens to the individual bodies within an isolated system, the total momentum of the system never changes! Since momentum is a vector, its conservation implies that both its magnitude and its direction remains the same.

Momentum is conserved in isolated systems!

This Principle of Conservation of Linear Momentum is one of the most fundamental principles of physics and it alone justifies the definition of momentum. Since momentum is related to the motion of objects, we can use its conservation to make predictions about what happens in collisions and explosions. If we bang two objects together, by conservation of momentum, the total momentum of the objects before the collision is equal to their total momentum after the collision.

Let us consider a simple collision of two pool or billiard balls. Consider the first ball (mass m1) to have an initial velocity ($$\overrightarrow{u_1}$$). The second ball (mass m2) moves towards the first ball with an initial velocity $$\overrightarrow{u_2}$$. This situation is shown in Figure 6.1. If we add the momenta of each ball we get a total momentum for the system. This total momentum is then $$\begin{matrix}\overrightarrow{p}_{total\ before}=m_1\overrightarrow{u_1}+m_2\overrightarrow{u_2},\end{matrix}$$

After the two balls collide and move away they each have a different momentum. If we call the final velocity of ball 1 $$\overrightarrow{v_1}$$ and the final velocity of ball 2 $$\overrightarrow{v_2}$$ (see Figure 6.2), then the total momentum of the system after the collision is

$$\begin{matrix}\overrightarrow{p}_{total\ after}=m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2},\end{matrix}$$

This system of two balls is isolated since there are no external forces acting on the balls. Therefore, by the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision. This gives the equation for the conservation of momentum in a collision of two objects,

This equation is always true - momentum is always conserved in collisions.

The chapter `Collisions and Explosions' deals with applications of momentum conservation.

title: FHSST Physics/Momentum/Properties


text:
FHSST Physics/Momentum/Impulse

At the beginning of this chapter it was mentioned that momentum is closely related to force. We will now explain the nature of this connection.

Consider an object of mass m moving with constant acceleration $$\overrightarrow{a}$$. During a time $$\Delta$$t the object's velocity changes from an initial velocity $$\overrightarrow{u}$$ to a final velocity $$\overrightarrow{v}$$ (refer to Figure 6.3). We know from Newton's First Law that there must be a resultant force $$\overrightarrow{F}_{Res}$$ acting on the object.

Starting from Newton's Second Law, $$\begin{matrix}\overrightarrow{F}_{Res}&=&m\overrightarrow{a}\\&=&m(\frac{\overrightarrow{v}-\overrightarrow{u}}{\Delta t})\qquad\qquad\rm {since}\qquad\overrightarrow{a}=\frac{\overrightarrow{v}-\overrightarrow{u}}{\Delta t}\\&=&\frac{m\overrightarrow{v}-m\overrightarrow{u}}{\Delta t}\\&=&\frac{\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}}{\Delta t}\\&=&\frac{\Delta \overrightarrow{p}}{\Delta t}\end{matrix}$$

This alternative form of Newton's Second Law is called the Law of Momentum.

Mathematically,

Rearranging the Law of Momentum,

$$\begin{matrix}\overrightarrow{F}_{Res}\Delta t &=& \Delta \overrightarrow{p}.\end{matrix}$$

The product $$\overrightarrow{F}_{Res}\Delta t$$ is called impulse,

From this equation we see, that for a given change in momentum,$$\overrightarrow{F}_{Res}\Delta t$$ is fixed. Thus, if FRes is reduced, $$\Delta t$$ must be increased (i.e. the resultant force must be applied for longer). Alternatively if $$\Delta t$$ is reduced (i.e. the resultant force is applied for a shorter period) then the resultant force must be increased to bring about the same change in momentum.

Worked Example 36 Impulse and Change in momentum
Question: A 150  N resultant force acts on a 300   kg object. Calculate how long it takes this force to change the object's velocity from $$2\ m.s^{-1}\ to\ the\ right$$ to $$6\ m.s^{-1}\ to\ the\ right$$

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

 * the object's mass,
 * the object's initial velocity,
 * the object's final velocity, and
 * the resultant force acting on the object

all in the correct units!

Step 2 :

What is being asked? We are asked to calculate the time taken $$\Delta t$$ to accelerate the object from the given initial velocity to final velocity. From the Law of Momentum,

$$\begin{matrix}\overrightarrow{F}_{Res}\Delta t &=& \Delta \overrightarrow{p} \\&=& m\overrightarrow{v}-m\overrightarrow{u}\\&=& m(\overrightarrow{v}-\overrightarrow{u}).\end{matrix}$$

Thus we have everything we need to find $$\Delta t$$!

Step 3 :

First we choose a positive direction. Although not explicitly stated, the resultant force acts to the right. This follows from the fact that the object's velocity increases in this direction. Let us then choose right as the positive direction.

Step 4 :

Substituting,

Right is the positive direction

$$\begin{matrix}\overrightarrow{F}_{Res}\Delta t &=& m(\overrightarrow{v}-\overrightarrow{u})\\(+150N)\Delta t &=& (300kg)((+6\frac{m}{s})-(+2\frac{m}{s}))\\(+150N)\Delta t &=& (300kg)(+4\frac{m}{s})\\\Delta t &=& \frac{(300kg)(+4\frac{m}{s})}{+150N}\\\Delta t &=& 8s\end{matrix}$$

Worked Example 37 Calculating Impulse
Question: A cricket ball weighing 156 g is moving at 54 km/h towards a batsman. It is hit by the batsman back towards the bowler at 36\ km/h. Calculate i) the ball's impulse, and ii) the average force exerted by the bat if the ball is in contact with the bat for 0.13 s.

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

 * the ball's mass,
 * the ball's initial velocity,
 * the ball's final velocity, and
 * the time of contact between bat and ball

all except the time in the wrong units!

Answer to (i):

Step 2 :

What is being asked? We are asked to calculate the impulse

$$\begin{matrix}\mathrm{Impulse} = \Delta\overrightarrow{p} =\overrightarrow{F}_{Res}\Delta t. \end{matrix}$$

Since we do not have the force exerted by the bat on the ball ($$\overrightarrow{F}_{Res}$$), we have to calculate the impulse from the change in momentum of the ball. Now, since

$$\begin{matrix}\Delta\overrightarrow{p} &=&\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}\\&=& m\overrightarrow{v} - m\overrightarrow{u},\end{matrix}$$

we need the ball's mass, initial velocity and final velocity, which we are given.

Step 3 : Firstly let us change units for the mass

$$\begin{matrix}1000g &=& 1kg\\1 &=& \frac{1kg}{1000g}\\156g \times 1 &=& 156g \times \frac{1kg}{1000g}\\&=& 0.156kg\end{matrix}$$

Step 4 :

Next we change units for the velocity

$$\begin{matrix}1km &=& 1000 m\\1 &=& \frac{1000m}{1km}\end{matrix}$$

$$\begin{matrix}3600s &=& 1hr\\1 &=& \frac{1hr}{3600 s}\end{matrix}$$

$$\begin{matrix}54\frac{km}{hr}\times 1\times1 &=&54\frac{km}{hr}\times\frac{1000m}{1km}\times\frac{1hr}{3600 s}\\&=& 15\frac{m}{s}\end{matrix}$$

$$\begin{matrix}36\frac{km}{hr}\times 1\times1 &=&36\frac{km}{hr}\times\frac{1000m}{1km}\times\frac{1hr}{3600 s}\\&=& 10\frac{m}{s}\end{matrix}$$

Step 5 :

Next we must choose a positive direction. Let us choose the direction from the batsman to the bowler as the positive direction. Then the initial velocity of the ball is $$\overrightarrow{u}=-15\ m.s^{-1}$$, while the final velocity of the ball is $$\overrightarrow{v}=+10\ m.s^{-1}$$

Step 6 :

Now we calculate the change in momentum,

Direction from batsman to bowler is the positive direction

$$\begin{matrix}\Delta\overrightarrow{p} &=&\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}\\&=& m\overrightarrow{v}-m\overrightarrow{u}\\&=& m(\overrightarrow{v}-\overrightarrow{u})\\&=&(0.156kg)((+10\ m.s^{-1})-(-15\ m.s^{-1}))\\&=& +3.9\ kg.m.s^{-1}\\&=& 3.9\ kg.m.s^{-1}\ \textbf{ in\ the\ direction\ from\ batsman\ to\ bowler}\end{matrix}$$

where we remembered in the last step to include the direction of the change in momentum in words.

Step 7 :

Finally since impulse is just the change in momentum of the ball,

$$\begin{matrix}\mathrm{Impulse} &=& \Delta\overrightarrow{p}\\&=& 3.9\ \mbox{kg} \cdot \mbox{m} \cdot \mbox{s}^{-1}\\&=& 3.9\ \mbox{N} \cdot \mbox{s} \ \textbf{ in\ the\ direction\ from\ batsman\ to\ bowler}\end{matrix}$$

Answer to (ii):

Step 8 :

What is being asked? We are asked to calculate the average force exerted by the bat on the ball, $$\overrightarrow{F}_{Res}$$. Now,

$$\begin{matrix}\mathrm{Impulse} = \overrightarrow{F}_{Res}\Delta t = \Delta\overrightarrow{p}.\end{matrix}$$

We are given $$\Delta t$$ and we have calculated the change in momentum or impulse of the ball in part (i)!

Step 9 :

Next we choose a positive direction. Let us choose the direction from the batsman to the bowler as the positive direction. Then substituting,

Direction from batsman to bowler is the positive direction

$$\begin{matrix}\overrightarrow{F}_{Res}\Delta t &=& \mathrm{Impulse}\\\overrightarrow{F}_{Res}(0.13s) &=& +3.9\frac{kg.m}{s}\\\overrightarrow{F}_{Res} &=& \frac{+3.9\frac{\mbox{kg} \cdot \mbox{m}}{\mbox{s}}}{0.13 \mbox{ s}}\\&=& 30 \mbox{ N} \ \textbf{ in\ the\ direction\ from\ batsman\ to\ bowler}\end{matrix}$$

where we remembered in the final step to include the direction of the force in words.

title: FHSST Physics/Momentum/Impulse


text:
FHSST Physics/Momentum/Summary

= Summary of Important Quantities, Equations and Concepts =

Momentum: The momentum of an object is defined as its mass multiplied by its velocity.

Momentum of a System: The total momentum of a system is the sum of the momenta of each of the objects in the system.

Principle of Conservation of Linear Momentum:: `The total linear momentum of an isolated system is constant' or `In an isolated system the total momentum before a collision (or explosion) is equal to the total momentum after the collision (or explosion)'.

Law of Momentum:: The applied resultant force acting on an object is equal to the rate of change of the object's momentum and this force is in the direction of the change in momentum.

title: FHSST Physics/Momentum/Summary


text: Template:FHSST Physics/Work and Energy

title: Template:FHSST Physics/Work and Energy


text:
FHSST Physics/Work and Energy

= Work and Energy =

title: FHSST Physics/Work and Energy


text:
FHSST Physics/Work and Energy/Definition

= What are Work and Energy? =

The energy is the ability to do a work.

Energy is measured by the result of applied force/power over a period of time to make changes/work.

Work is defined by a force applied to an object to change this object’s physical properties.

Power = Energy / Time No force used no change and none work was done.

Power and energy are closely related, although, they are not the same.

Power is the rate at which energy is delivered, not an amount of energy itself.

Electrical unit of power is Watt (named after the scientist James Watt):

1 Watt = 1 Joule / Second.

Energy = Power × Time.

Energy exist in various forms: kinetic [Ek = ½ m, v2 ], potential [Ep = m g h], thermal (heat), chemical, electrical, electrochemical, magnetic, sound, light, and nuclear.

Energy can be converted from one form into another in many ways:

Through the gravitation forces When gravity accelerates a falling object, it converts its potential energy to kinetic energy, or when an object is lifted, the gravitational field stores the energy exerted by the lifter as potential energy in the earth-object system.

Through the electric and magnetic fields forces Electrically charged particles in the presence of an electric field possess potential energy. The fields’ forces can accelerate particles and convert this particle's potential energy into kinetic energy. Charged particles can interact via electric and magnetic fields to transfer energy between them, i.e.: an electrical current in a conductor transforms electrical energy into heat.

Frictional Forces A mass object with its potential and kinetic energy associated with the position, orientation, and the object’s motion can be converted into thermal energy (heat), whenever the object slides against another object. The sliding causes the molecules on the surfaces of contact to interact via electromagnetic fields with one another and start vibrating.

Through emitting or absorbing photons of light When photons of lights fall on an object, a photon may pass through the object, be reflected by the object, or be absorbed by the atoms making up the object. Depending on the smoothness of the surface, the scale of the photon's wavelength, the reflection may be either diffuse (rough surface) or coherent (smooth surface). If the photon is absorbed, the photon's energy may also be split and converted in one of these ways: •	Photo-thermal effect The energy absorbed may simply produce thermal energy, or heat in the object. In this case, the photon's energy is converted into vibrations of the molecules called phonons, which is actually heat energy. •	Photoelectric effect The energy absorbed may be converted into kinetic energy of conduction electrons, and hence electrical energy. •	Photochemical effect The energy may bring about chemical changes that effectively store the energy. Nuclear reactions occur when the nuclei of particles combine [fusion reaction] or when nuclei split apart [fission reaction].

In the International System of Units [SI system], the electrical energy unit is 1 Joule (named after the English physicist James Prescott Joule).

A one Joule is the amount of energy we expend as work if we exert a force of a one Newton of force over a distance of one meter.

It takes a one Joule of energy to lift 1 lb. about 9 inches.

The unit of force in the International System of Units is 1 Newton (named after the English physicist Isaac Newton).

One Newton of force is the force that can accelerate a mass of 1 kilogram (about 2.205 lbs), such that it picks up 1 meter per second of velocity during each second that the force is exerted.

Appliances rating is how much energy per unit time these appliances draw. This quantity is called the "power": Energy = Power x Time = (100 Joules/Second) × (3600 Seconds) = 360,000 Joules


text:
FHSST Physics/Work and Energy/Work

= Work = To do work on an object, one must move the object by applying a force with at least a component in the direction of motion. The work done is given by

It is very important to note that for work to be done there must be a component of the applied force in the direction of motion. Forces perpendicular to the direction of motion do no work.

As with all physical quantities, work must have units. As follows from the definition, work is measured in N.m. The name given to this combination of S.I. units is the joule (J).

The work done by an object can be positive or negative. Since force (Fundefined) and displacement (s) are both vectors, the result of the above equation depends on their directions:

 * If Fundefined acts in the same direction as the motion then positive work is being done. In this case the object on which the force is applied gains energy.
 * If the direction of motion and Fundefined are opposite, then negative work is being done. This means that energy is transferred in the opposite direction. For example, if you try to push a car uphill by applying a force up the slope and instead the car rolls down the hill you are doing negative work on the car. Alternatively, the car is doing positive work on you!

Worked Example 38 Calculating Work Done I
Question: If you push a box 20m forward by applying a force of 15N in the forward direction, what is the work you have done on the box?

Answer:

Step 1 : Analyse the question to determine what information is provided

These quantities are all in the correct units, so no unit conversions are required.
 * The force applied is F = 15N.
 * The distance moved is s = 20m.
 * The applied force and distance moved are in the same direction. Therefore, $$F_{\|}=15N$$

Step 2 : Analyse the question to determine what is being asked

 * We are asked to find the work done on the box. We know from the definition that work done is W = Fundefineds

Step 3 : Next we substitute the values and calculate the work done

$$\begin{matrix}W&=&F_{\|} s\\&=& (15N)(20m)\\&=& 300\ N\cdot m\\&=& 300\ J\end{matrix}$$

Remember that the answer must be positive as the applied force and the motion are in the same direction (forwards). In this case, you (the pusher) lose energy, while the box gains energy.

Worked Example 39 Calculating Work Done II
Question: What is the work done by you on a car, if you try to push the car up a hill by applying a force of 40N directed up the slope, but it slides downhill 30cm?

Answer:

Step 1 : Analyse the question to determine what information is provided

 * The force applied is F = 40N
 * The applied force and distance moved are in opposite directions. Therefore, if we take s = 0.3m, then $$F_{\|}=-40N$$.

Step 2 : Analyse the question to determine what is being asked

 * We are asked to find the work done on the car by you. We know that work done is W = Fundefineds

Step 3 : Substitute the values and calculate the work done

Again we have the applied force and the distance moved so we can proceed with calculating the work done:

$$\begin{matrix}W&=&F_{\|} s\\&=& (- 40N)(0.3m)\\&=& -12N\cdot m\\&=& -12\ J\end{matrix}$$

Note that the answer must be negative as the applied force and the motion are in opposite directions. In this case the car does work on the person trying to push.

What happens when the applied force and the motion are not parallel? If there is an angle between the direction of motion and the applied force then to determine the work done we have to calculate the component of the applied force parallel to the direction of motion. Note that this means a force perpendicular to the direction of motion can do no work.

Worked Example 40 Calculating Work Done III
Question: Calculate the work done on a box, if it is pulled 5m along the ground by applying a force of F = 10N at an angle of 60o to the horizontal.

Answer:

Step 1 :

Analyse the question to determine what information is provided

 * The force applied is F = 10N
 * The distance moved is s = 5m along the ground
 * The angle between the applied force and the motion is 60o

These quantities are in the correct units so we do not need to perform any unit conversions.

Step 2 : Analyse the question to determine what is being asked

 * We are asked to find the work done on the box.

Step 3 : Calculate the component of the applied force in the direction of motion

Since the force and the motion are not in the same direction, we must first calculate the component of the force in the direction of the motion.

From the force diagram we see that the component of the applied force parallel to the ground is

$$\begin{matrix}F_{||}&=&F\cdot \cos(60^o)\\&=& 10N\cdot \cos(60^o)\\&=& 5\ N\end{matrix}$$

Step 4 : Substitute and calculate the work done

Now we can calculate the work done on the box:

$$\begin{matrix}W&=&F_{\|} s\\&=& (5N)(5m) \\&=& 25\ J\end{matrix}$$

Note that the answer is positive as the component of the force Fundefined is in the same direction as the motion.

We will now discuss energy in greater detail.

title: FHSST Physics/Work and Energy/Work


text:
FHSST Physics/Work and Energy/Energy

= Energy =

As we mentioned earlier, energy is the capacity to do work. When positive work is done on an object, the system doing the work loses energy. In fact, the energy lost by a system is exactly equal to the work done by the system.

Like work (W) the unit of energy (E) is the joule (J). This follows as work is just the transfer of energy.

A very important property of our universe which was discovered around 1890 is that energy is conserved.

Energy conservation and the conservation of matter are the principles on which classical mechanics is built.

 Thermal energy (heat) is the disorganized movement of microscopic particles. Once energy is converted to this form, it has limited usefulness for doing further work in the system. Friction is the general name for forces that converts energy to heat.

Types of Energy
So what different types of energy exist? Kinetic, mechanical, thermal, chemical, electrical, radiant, and atomic energy are just some of the types that exist. By the principle of conservation of energy, whenwork is done energy is merely transferred from one object to another and from one type of energy to another.

Kinetic Energy
Kinetic energy is the energy of motion that an object has. Objects moving in straight lines possess translational kinetic energy, which we often abbreviate as Ek.

The translational kinetic energy of an object is given by

Note the dependence of the kinetic energy on the speed of the object- kinetic energy is related to motion. The faster an object is moving the greater its kinetic energy.

Worked Example 41 Calculation of Kinetic Energy
Question: If a rock has a mass of 1kg and is thrown at 5m/s, what is its kinetic energy?

Answer:

Step 1 : Analyse the question to determine what information is provided

 * The mass of the rock m = 1kg
 * The speed of the rock v = 5m/s

These are both in the correct units so we do not have to worry about unit conversions.

Step 2 : Analyse the question to determine what is being asked

 * We are asked to find the kinetic energy. From the definition we know that to work out Ek, we need to know the mass and the velocity of the object and we are given both of these values.

Step 3 : Substitute and calculate the kinetic energy

$$\begin{matrix}E_k & = & \frac{1}{2}mv^2 \\&= & \frac{1}{2}(1kg)(5\frac{m}{s})^2 \\&= & 12.5\frac{kg\cdot m^2}{s^2} \\&=& 12.5\ J\end{matrix}$$

To check that the units in the above example are in fact correct:

$$\begin{matrix}\frac{kg\cdot m^2}{s^2}&=& \left(\frac{kg\cdot m}{s^2}\right)\cdot m =N\cdot m\\&=& J\end{matrix}$$

The units are indeed correct!

Worked Example 42 Mixing Units and Kinetic Energy Calculations 1
Question: If a car has a mass of 900kg and is driving at 60km/hr, what is its kinetic energy?

Answer:

Step 1 : Analyse the question to determine what information is provided

 * The mass of the car m = 900kg

Step 2 : Analyse the question to determine what is being asked

 * We are asked to find the kinetic energy.

Step 3 : Substitute and calculate

We know we need the mass and the speed to work out Ek and we are given both of these quantities. We thus simply substitute them into the equation for Ek:

$$\begin{matrix}E_k & = & \frac{1}{2}mv^2 \\&= & \frac{1}{2}(900kg)(16.67\frac{m}{s})^2 \\&= & 125\ 000\frac{kgm^2}{s^2} \\&=& 125\ 000\ J\end{matrix}$$

Worked Example 43 Mixing Units and Kinetic Energy Calculations 2
Question: If a bullet has a mass of 150kg and is shot at a muzzle velocity of 960m/s, what is its kinetic energy?

Answer:

Step 1 : Analyse the question to determine what information is provided

 * We are given the muzzle velocity which is just how fast the bullet leaves the barrel and it is v = 960m/s.

Step 2 : Analyse the question to determine what is being asked

 * We are asked to find the kinetic energy.

Step 3 : Substitute and calculate

We just substitute the mass and velocity (which are known) into the equation for Ek:

$$\begin{matrix}E_k & = & \frac{1}{2}mv^2 \\&= & \frac{1}{2}(150kg)(960\frac{m}{s})^2 \\&= & 69\ 120 \frac{kgm^2}{s^2} \\&=& 69\ 120\ J\end{matrix}$$

Potential Energy
If you lift an object you have to do work on it. This means that energy is transferred to the object. But where is this energy? This energy is stored in the object and is called potential energy. The reason it is called potential energy is because if we let go of the object it would move.

As an object raised above the ground falls, its  potential energy is released and transformed into kinetic energy. The further it falls the faster it moves as more of the stored potential energy is transferred into kinetic energy. Remember, energy is never created nor destroyed, but merely transformed from one type to another. In this case potential energy is lost but an equal amount of kinetic energy is gained.

In the example of a falling mass the potential energy is known as gravitational potential energy as it is the gravitational force exerted by the earth which causes the mass to accelerate towards the ground. The gravitational field of the earth is what does the work in this case.

Another example is a rubber-band. In order to stretch a rubber-band we have to do work on it. This means we transfer energy to the rubber-band and it gains potential energy. This potential energy is called  elastic potential energy. Once released, the rubber-band begins to move and elastic potential energy is transferred into kinetic energy.

Gravitational Potential Energy
As we have mentioned, when lifting an object it gains gravitational potential energy. One is free to define any level as corresponding to zero gravitational potential energy. Objects above this level then possess positive potential energy, while those below it have negative potential energy. To avoid negative numbers in a problem, always choose the lowest level as the zero potential mark. The change in gravitational potential energy of an object is given by:

When an object is lifted it gains gravitational potential energy, while it loses gravitational potential energy as it falls.

Worked Example 44 Gravitational potential energy
Question: How much potential energy does a brick with a mass of 1kg gain if it is lifted 4m.

Answer:

Step 1 : Analyse the question to determine what information is provided

 * The mass of the brick is m = 1kg
 * The height lifted is $$\Delta h=4m$$

These are in the correct units so we do not have to worry about unit conversions.

Step 2 : Analyse the question to determine what is being asked

 * We are asked to find the gain in potential energy of the object.

Step 3 : Identify the type of potential energy involved

Since the block is being lifted we are dealing with gravitational potential energy. To work out $$\Delta E_P$$, we need to know the mass of the object and the height lifted. As both of these are given, we just substitute them into the equation for $$\Delta E_P$$.

Step 4 : Substitute and calculate

$$\begin{matrix}\Delta E_P & = & mg\Delta h \\&= & (1kg)\left(10\frac{m}{s^2}\right)(4m) \\&= & 40\frac{kg\cdot m^2}{s^2} \\&=& 40\ J\end{matrix}$$

title: FHSST Physics/Work and Energy/Energy


text:
FHSST Physics/Work and Energy/Mechanical Energy and Energy Conservation

=Mechanical Energy and Energy Conservation=

Kinetic energy and potential energy are together referred to as mechanical energy. The total mechanical energy (U) of an object is then the sum of its kinetic and potential energies:

Now,

This principle of conservation of mechanical energy can be a very powerful tool for solving physics problems. However, in the presence of friction some of the mechanical energy is lost:

Worked Example 45 Using Mechanical Energy Conservation
Question: A 2kg metal ball is suspended from a rope. If it is released from point A and swings down to the point B (the bottom of its arc) what is its velocity at point B?

Answer:

Step 1 : Analyse the question to determine what information is provided

 * The mass of the metal ball is m = 2kg
 * The change in height going from point A to point B is h = 0.5m
 * The ball is released from point A so the velocity at point A is zero (vA = 0m/s).

These are in the correct units so we do not have to worry about unit conversions.

Step 2 : Analyse the question to determine what is being asked

 * Find the velocity of the metal ball at point B.

Step 3 : Determine the Mechanical Energy at A and B

To solve this problem we use conservation of mechanical energy as there is no friction. Since mechanical energy is conserved,

$$\begin{matrix}U_A=U_B\end{matrix}$$

Therefore we need to know the mechanical energy of the ball at point A (UA) and at point B (UB). The mechanical energy at point A is

$$\begin{matrix}U_A = mgh_A + \frac{1}{2}m(v_A)^2\end{matrix}$$

We already know m, g and vA, but what is hA? Note that if we let hB = 0 then hA = 0.5m as A is 0.5m above B. In problems you are always free to choose a line corresponding to h = 0. In this example the most obvious choice is to make point B correspond to h = 0.

Now we have,

$$\begin{matrix}U_A &=& (2kg)\left(10\frac{m}{s^2}\right)(0.5m)+\frac{1}{2}(2kg)(0)^2\\&=& 10\ J\end{matrix}$$

As already stated UB = UA. Therefore UB = 10J, but using the definition of mechanical energy

$$\begin{matrix}U_B &=& mgh_B + \frac{1}{2}m(v_B)^2\\&=& \frac{1}{2}m(v_B)^2\end{matrix}$$

because hB = 0. This means that

$$\begin{matrix}10J&=&\frac{1}{2}(2kg)(v_B)^2\\(v_B)^2 &=& 10\frac{J}{kg}\\v_B &=& \sqrt{10}\frac{m}{s}\end{matrix}$$

authors: יהודה שמחה ולדמן


text:
FHSST Physics/Work and Energy/Summary

= Summary of Important Quantities, Equations and Concepts =

Principle of Conservation of Energy: Energy is never created nor destroyed, but merely transformed from one form to another.

Conservation of Mechanical Energy: In the absence of friction, the total mechanical energy of an object is conserved.

title: FHSST Physics/Work and Energy/Summary


text:
FHSST Physics/Work and Energy/Sources

=Energy=

Asogan Moodaly
Asogan Moodaly received his Bachelor of Science degree (with honours) in Mechanical Engineering from the University of Natal, Durban in South Africa. For his final year design project he worked on a 3-axis filament winding machine for composite (Glass re-enforced plastic in this case) piping. He worked in Vereeniging, Gauteng at Mine Support Products (a subsidiary of Dorbyl Heavy Engineering) as the design engineer once he graduated. He currently lives in the Vaal Triangle area and is working for Sasol Technology Engineering as a mechanical engineer, ensuring the safety and integrity of equipment installed during projects.

Energy and electricity. Why the fuss?
Disclaimer: The details furnished below are very basic and for illustration purposes only.

Why do we need energy? Note that I use the word energy and not electricity. On a broad scale it stimulates economic growth, etc., etc. but on a personal level it allows us to lead a comfortable lifestyle.

e.g. Flick a switch and Heat for cooking Entertainment such as television and radio Heat for water and interior of house Ironing Electronic and electrical devices such as alarms, garage doors, etc.

In a modern household this energy is provided in the form of electricity which is powered via fossil fuels or nuclear.

How is electricity made? In a nutshell: By moving a magnet through or near a set of conducting coils.

Most power stations produce steam through heat (nuclear reaction or burning fossil fuels), the steam drives a turbine which moves a magnet relative to a coil (the generator like the above but on a much larger scale i.e. bigger magnets, bigger coils, etc.), which produces electricity that is transmitted via a power network to our homes. Gas fired plants burn gas directly in a gas turbine to produce the same desired relative motion between permanent magnet and coil.

Coal, oil and gas are fossil fuels. Fossil fuels were created by decomposing organic (plant and animal) matter a long, long time ago and are typically found underground. Different temperatures and pressures resulted in the organic matter transforming into coal, oil or gas.

Why the fuss about fossil fuels?

 * 1) Fossil fuel power is bad news in the long run. It pollutes and contributes to the greenhouse effect (global warming resulting in melting polar ice caps, floods, droughts, disease, etc.).
 * 2) It s not going to last forever.
 * 3) Nuclear power is  cleaner  in terms of emissions but there s no proven way of disposing of the nuclear waste. Oh, and it won t last forever either!

Renewable Energy As the name suggests renewable energy lasts forever. Solar (sun), wind, geothermal, wave, hydro and biomass (organic) are all sources of energy that will last until the sun eventually explodes many billions of years from now. Hopefully the human race will have moved from the earth by then! Generally the principal of renewable electricity generation is similar to fossil fuel electricity generation in that electricity is generated by moving a magnet relative to a conducting coil. What is different is the way energy is supplied to cause that motion.

The below are a few different types of available renewable energy technologies.

Wind
Wind turbines catch wind that spins the blades. The blades are connected to a shaft that spins because of the wind. This spinning shaft spins another shaft that turns a permanent magnet relative to conducting coils.

Note that the gears are used to convert the slow spinning of the 1st shaft to a faster spin on the 2nd shaft. The generator shaft needs to spin at the correct speed to produce the right amount and quality of electricity. Some generators are now being modified to run at slower speeds. This saves money as gears are not needed.

Biomass
Biomass is anything organic i.e. plant or animal matter. It can be used in the place of coal as per a normal coal fired plant and is renewable as long as the biomass e.g. wood is handled in a sustainable manner. By sustainable I mean that suitable farming practices are used so that the land is not over farmed which will result in the soil becoming barren and nothing growing there again.

Biomass can also be processed using anaerobic digestion to produce a gas that can be burned for heat or electricity. This biogas  is made up of a number of other gases that are similar to those found in fossil fuel natural gas   Except the amount of the gases are different. E.g. Natural gas has about 94 Anaerobic digestion: Anaerobic means "No air". Therefore anaerobic digestion  means to digest in the absence of air. Bacteria that naturally exist in organic matter will convert organic matter to biogas and fertilizer when all the air is removed.

Thousands of anaerobic digesters have been installed in rural India, Nepal and China in rural area s where cow dung, human waste and chicken litter (faeces) are all processed using anaerobic digestion to produce gas that can be burned in the home for cooking and heating. The leftover is used as fertilizer.

Geothermal Energy
In some places on earth, the Earth's crust is thinner than others. As a result the heat from the Earth's core escapes. The heat can be captured by converting water to steam, and using the steam to drive a steam generator as discussed above.

Hydroelectric power Water from a river is diverted to turn a water turbine to create electricity similar to the principles of steam generation. The water is returned to the river after driving the turbine.

Wave Energy
Some wave energy generators work similarly to wind turbines except that underwater ocean currents turns the blades instead of wind and of course most of the structure is under water!

Another concept uses the rising and falling of the tides to suck air in using a one way valve. As a result air becomes compressed in a chamber and the compressed air is let out to drive a turbine which in turn drives a generator

These are relatively new technologies. Liquid fuels are used mainly for transportation. Petrol and diesel are the most common liquid fuels and are obtained from oil.

Sasol is the only company in the world that makes liquid fuels from coal and will be one of the leading companies in the world to make liquid fuels from natural gas! The Sasol petro-chemical plants are based in Sasolburg on the border of the Free State and in Secunda in Mpumalanga.

However, as discussed above coal, gas and oil are fossil fuels and are not renewable. Petrol and diesel are obtained from fossil fuels and therefore pollute and contribute to the green house effect (global warming).

Biodiesel
Oil can be extracted from plants such as the soya bean, sunflower and rapeseed by pressing it through a filter. This oil if mixed correctly with either methanol or dry ethanol and sodium hydroxide will separate the plant oil into biodiesel, glycerol and fertilizer.

The biodiesel can be used as produced in a conventional diesel engine with little or no modifications required.

The glycerol can be refined a bit further for pharmaceutical companies to use, or can be used to make soap.

This information is not enough.

Ethanol
Corn, maize and sugar cane can be used to make ethanol as a fuel substitute for petrol. It's made by the same fermentation process used to make alcohol. Enzymes are often used to speed up the process.

In ethanol from sugar cane production, the leftover bagasse (the fibre part of the sugar cane) can be burned in a biomass power station to produce electricity.

Hydrogen
Through the process of electrolysis electricity (hopefully clean, renewable electricity!) can split water into hydrogen and oxygen. The stored hydrogen can be used in a fuel cell to create electricity in a process that is opposite to electrolysis  to drive electric motors in a car. The hydrogen can also be burned directly in a modified internal combustion engine.

In both cases the waste product is water.

title: FHSST Physics/Work and Energy/Sources


text: Template:FHSST Physics/Collisions and Explosions


text:
FHSST Physics/Collisions and Explosions/Tiny, Violent Collisions

Essay 2: Tiny, Violent Collisions

Author: Thomas D. Gutierrez

Tom Gutierrez received his Bachelor of Science and Master degrees in physics from San Jose State University in his home town of San Jose, California. As a Master's student he helped work on a laser spectrometer at NASA Ames Research Centre. The instrument measured the ratio of different isotopes of carbon in CO2 gas and could be used for such diverse applications as medical diagnostics and space exploration. Later, he received his Ph.D. in physics from the University of California, Davis where he performed calculations for various reactions in high energy physics collisions. He currently lives in Berkeley, California where he studies proton-proton collisions seen at the STAR experiment at Brookhaven National Laboratory on Long Island, New York.

=High Energy Collisions= Take an orange and expanded it to the size of the earth. The atoms of the earth-sized orange would themselves be about the size of regular oranges and would fill the entire earth-orange. Now, take an atom and expand it to the size of a football field. The nucleus of that atom would be about the size of a tiny seed in the middle of the field. From this analogy, you can see that atomic nuclei are very small objects by human standards. They are roughly 10−15 meters in diameter - one-hundred thousand times smaller than a typical atom. These nuclei cannot be seen or studied via any conventional means such as the naked eye or microscopes. So how do scientists study the structure of very small objects like atomic nuclei?

The simplest nucleus, that of hydrogen, is called the proton. Faced with the inability to isolate a single proton, open it up, and directly examine what is inside, scientists must resort to a brute-force and somewhat indirect means of exploration: high energy collisions. By colliding protons with other particles (such as other protons or electrons) at very high energies, one hopes to learn about what they are made of and how they work. The American physicist Richard Feynman once compared this process to slamming delicate watches together and figuring out how they work by only examining the broken debris. While this analogy may seem pessimistic, with sufficient mathematical models and experimental precision, considerable information can be extracted from the debris of such high energy subatomic collisions. One can learn about both the nature of the forces at work and also about the sub-structure of such systems.

The experiments are in the category of high energy physics (also known as subatomic physics). The primary tool of scientific exploration in these experiments is an extremely violent collision between two very, very small subatomic objects such as nuclei. As a general rule, the higher the energy of the collisions, the more detail of the original system you are able to resolve. These experiments are operated at laboratories such as CERN, SLAC, BNL, and Fermilab, just to name a few. The giant machines that perform the collisions are roughly the size of towns. For example, the RHIC collider at BNL is a ring about 1 km in diameter and can be seen from space. The newest machine currently being built, the LHC at CERN, is a ring 9 km in diameter!

Let's examine the kinematics of such a collisions in some detail...


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FHSST Physics/Collisions and Explosions

=Collisions and Explosions =

In most physics courses questions about collisions and explosions occur and to solve these we must use the ideas of momentum and energy with a bit of mathematics of course! This section allows you to pull the momentum and energy ideas together easily with some specific problems.

title: FHSST Physics/Collisions and Explosions