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FHSST Physics/Collisions and Explosions/Types of Collisions

=Types of Collisions =

We will consider two types of collisions in this section

 * Elastic collisions.
 * Inelastic collisions.

In both types of collision, total energy and total momentum is always conserved. Kinetic energy is conserved for elastic collisions, but not for inelastic collisions.

Elastic Collisions

This means that the total momentum and the total kinetic energy before an elastic collision is the same as after the collision. For these kinds of collisions, the kinetic energy is not changed into another type of energy.

Before the Collision
In the following diagram, two balls are rolling toward each other, about to collide

Before the balls collide, the total momentum of the system is equal to all the individual momenta added together. The ball on the left has a momentum which we call $$\overrightarrow{p}_1$$ and the ball on the right has a momentum which we call $$\overrightarrow{p}_2$$, it means the total momentum before the collision is

We calculate the total kinetic energy of the system in the same way. The ball on the left has a kinetic energy which we call K1 and the ball on the right has a kinetic energy which we call K2, it means that the total kinetic energy before the collision is

After the Collision
The following diagram shows the balls after they collide

After the balls collide and bounce off each other, they have new momenta and new kinetic energies. Like before, the total momentum of the system is equal to all the individual momenta added together. The ball on the left now has a momentum which we call $$\overrightarrow{p}_3$$ and the ball on the right now has a momentum which we call $$\overrightarrow{p}_4$$, it means the total momentum after the collision is

The ball on the left now has a kinetic energy which we call K3 and the ball on the right now has a kinetic energy which we call K4, it means that the total kinetic energy after the collision is

Since this is an elastic collision, the total momentum before the collision equals the total momentum after the collision and the total kinetic energy before the collision equals the total kinetic energy after the collision

$$\begin{matrix}\mbox{Before}&&\mbox{After}\\\overrightarrow{p}_\mathrm{Before} &=& \overrightarrow{p}_\mathrm{After}\\\overrightarrow{p}_1+\overrightarrow{p}_2 &=&\overrightarrow{p}_3+\overrightarrow{p}_4 \\&\rm {\textbf{and}}&\\K_\mathrm{Before} &=& K_\mathrm{After} \\K_1+K_2 &=& K_3+K_4\end{matrix}$$

Worked Example 46 An Elastic Collision
We will have a look at the collision between two pool balls. Ball 1 is at rest and ball 2 is moving towards it with a speed of 2 m·s−1. The mass of each ball is 0.3 kg. After the balls collide elastically, ball 2 comes to a stop and ball 1 moves off. What is the final velocity of ball 1?

Step 1 : Draw the before diagram

Before the collision, ball 2 is moving we will call it's momentum P2 and it's kinetic energy K2. Ball 1 is at rest, so it has zero kinetic energy and momentum.

Step 2 : Draw the after diagram

After the collision, ball 2 is at rest but ball 1 has a momentum which we call P3 and a kinetic energy which we call K3.

RIAAN Note: second image on page 145 is missing

Because the collision is elastic, we can solve the problem using momentum conservation or kinetic energy conservation. We will do it both ways to show that the answer is the same, whichever method you use.

Step 3 : Show the conservation of momentum

We start by writing down that the momentum before the collision $$\overrightarrow{p}_\mathrm{Before}$$ is equal to the momentum after the collision $$\overrightarrow{p}_\mathrm{After}$$

We know that momentum is just P = mv, and we know the masses of the balls, so we can rewrite the conservation of momentum in terms of the velocities of the balls

So ball 1 exits with the velocity that ball 2 started with!

Step 4 : Show the conservation of kinetic energy

We start by writing down that the kinetic energy before the collision $$K_\mathrm{Before}$$ is equal to the kinetic energy after the collision $$K_\mathrm{After}$$

We know that kinetic energy is just $$K=\frac{mv^2}{2}$$, and we know the masses of the balls, so we can rewrite the conservation of kinetic energy in terms of the velocities of the balls

So ball 1 exits with the velocity that ball 2 started with, which agrees with the answer we got when we used the conservation of momentum.

Worked Example 47 Elastic Collision 2
Question: Now for a slightly more difficult example. We have 2 marbles. Marble 1 has mass 50 g and marble 2 has mass 100 g. I roll marble 2 along the ground towards marble 1 in the positive x-direction. Marble 1 is initially at rest and marble 2 has a velocity of 3 $$\mathrm{m.s^{-1}}$$ in the positive x-direction. After they collide elastically, both marbles are moving. What is the final velocity of each marble?

Answer:

Step 1 :

Firstly, put all the quantities into S.I. units. So:

$$\begin{matrix}m_{1} = 0.05 \rm {kg} &\rm {and} & m_{2} = 0.1\rm {kg}\end{matrix}$$

Step 2 :

Draw the picture:

Before the collision:

After the collision:

Step 3 :

Since the collision is elastic, both momentum and kinetic energy are conserved in the collision. So:

$$\begin{matrix}E_{kBefore} &=& E_{kAfter} \\&\rm {and}& \\\overrightarrow{p_{Before}} &=& \overrightarrow{p_{After}}\end{matrix}$$

There are two unknowns ($$\overrightarrow{v_{1}}$$ and $$\overrightarrow{v_{2}}$$) so we will need two equations to solve for them. We need to use both kinetic energy conservation and momentum conservation in this problem.

Step 4 : Let's start with energy conservation. Then:

$$\begin{matrix}E_{kBefore} &=& E_{kAfter} \\\frac{1}{2}m_{1}{\overrightarrow{u_{1}}}^2 +\frac{1}{2}m_{2}{\overrightarrow{u_{2}}}^2 &=&\frac{1}{2}m_{1}{\overrightarrow{v_{1}}}^2+ \frac{1}{2}m_{2}{\overrightarrow{v_{2}}}^2 \\m_{1}{\overrightarrow{u_{1}}}^2 + m_{2}{\overrightarrow{u_{2}}}^2 &=&m_{1}{\overrightarrow{v_{1}}}^2+ m_{2}{\overrightarrow{v_{2}}}^2 \\\end{matrix}$$

But $$\overrightarrow{u_{1}}$$, and solving for $${\overrightarrow{v_{2}}}^2$$

$$\begin{matrix}{\overrightarrow{v_{2}}}^2 &=&{\overrightarrow{u_{2}}}^2-\frac{m1}{m2}{\overrightarrow{v_{1}}}^2\\{\overrightarrow{v_{2}}}^2 &=& (3)^2 - \frac{(0.05)}{(0.10)}{\overrightarrow{v_{1}}}^2\\{\overrightarrow{v_{2}}}^2 &=& 9-\frac{1}{2}{\overrightarrow{v_{1}}}^2\rm {\ \ \ \ (\textbf{A})}\\\end{matrix}$$

Step 5 : Now we have simplified as far as we can, we move onto momentum conservation:

$$\begin{matrix}\overrightarrow{p_{Before}} &=& \overrightarrow{p_{After}} \\m_{1}\overrightarrow{u_{1}} + m_{2}\overrightarrow{u_{2}} &=&m_{1}\overrightarrow{v_{1}}+m_{2}\overrightarrow{v_{2}} \\\end{matrix}$$

But $$\overrightarrow{u_{1}}$$=0, and solving for $$\overrightarrow{v_{1}}$$

$$\begin{matrix}m_{2}\overrightarrow{u_{2}} &=& m_{1}\overrightarrow{v_{1}}+m_{2}\overrightarrow{v_{2}} \\m_{1}\overrightarrow{v_{1}} &=& m_{2}\overrightarrow{u_{2}} - m_{2}\overrightarrow{v_{2}}\\\overrightarrow{v_{1}} &=& \frac{m_{2}}{m_{1}}\overrightarrow{u_{2}} -\frac{m_{2}}{m_{1}}\overrightarrow{v_{2}}\\\overrightarrow{v_{1}} &=& 2(3) - 2\overrightarrow{v_{2}}\\\overrightarrow{v_{1}} &=& 6 - 2\overrightarrow{v_{2}} \rm {\ \ \ \ (\textbf{B})}\end{matrix}$$

Step 6 :

Now we can substitute (B) into (A) to solve for $$\overrightarrow{v_{2}}$$:

$$\begin{matrix}{\overrightarrow{v_{2}}}^2 &=& 9-\frac{1}{2}{\overrightarrow{v_{1}}}^2\\{\overrightarrow{v_{2}}}^2 &=& 9 - \frac{1}{2}(6-2\overrightarrow{v_{2}})^2 \\{\overrightarrow{v_{2}}}^2 &=& 9 - \frac{1}{2}(36-24\overrightarrow{v_{2}}+4{\overrightarrow{v_{2}}}^2) \\{\overrightarrow{v_{2}}}^2 &=& 9 - 18 + 12\overrightarrow{v_{2}}-2{\overrightarrow{v_{2}}}^2 \\3{\overrightarrow{v_{2}}}^2 &=& -9 + 12\overrightarrow{v_{2}} \\{\overrightarrow{v_{2}}}^2 &=& 4\overrightarrow{v_{2}} -3 \\{\overrightarrow{v_{2}}}^2 -4\overrightarrow{v_{2}} +3 &=& 0 \\(\overrightarrow{v_{2}}-3)(\overrightarrow{v_{2}}-1)&=& 0 \\\overrightarrow{v_{2}}=3 &\rm {\textbf{or}}& \overrightarrow{v_{2}}=1\end{matrix}$$

We were lucky in this question because we could factorise. If you can't factorise, then you can always solve using the formula for solving quadratic equations. Remember:

$$\begin{matrix}x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\end{matrix}$$

So, just to check:

$$\begin{matrix}\overrightarrow{v_{2}} &=& \frac{4 \pm \sqrt{4^2 - 4(1)(3)}}{2(1)} \\\overrightarrow{v_{2}} &=& \frac{4 \pm \sqrt{16 - 12}}{2} \\\overrightarrow{v_{2}} &=& \frac{4 \pm \sqrt{4}}{2} \\\overrightarrow{v_{2}} &=& 2 \pm 1 \\\overrightarrow{v_{2}}=3 &\rm {\textbf{or}}& \overrightarrow{v_{2}}=1\rm {\ \ same \ as \ before}\end{matrix}$$

Step 7 :

So finally, substituting into equation (B) to get $$\overrightarrow{v_{1}}$$:

$$\begin{matrix}\overrightarrow{v_{1}}=6-2\overrightarrow{v_{2}} \\\end{matrix}$$

If $$\overrightarrow{v_{2}}= 3 \ \mathrm{m.s^{-1}}$$ then

$$\begin{matrix}\overrightarrow{v_{1}}=6-2(3) = 0 \ \mathrm{m.s^{-1}}\end{matrix}$$

But, according to the question, marble 1 is moving after the collision. So $$\overrightarrow{v_{1}} \neq 0$$ and $$\overrightarrow{v_{2}} \neq 3$$. Therefore:

$$\begin{matrix}\overrightarrow{v_{2}} &=& \underline{1 \ \mathrm{m.s^{-1}}\rm {\ \ \ in \ the \ positive \ x-direction}}\\&\rm {and}& \\\overrightarrow{v_{1}} &=& \underline{4 \ \mathrm{m.s^{-1}}\rm {\ \ \ in \ the \ positive \ x-direction}}\\\end{matrix}$$

Inelastic Collisions

So the total momentum before an inelastic collisions is the same as after the collision. But the total kinetic energy before and after the inelastic collision is different. Of course this does not mean that total energy has not been conserved, rather the energy has been transformed into another type of energy.

As a rule of thumb, inelastic collisions happen when the colliding objects are distorted in some way. Usually they change their shape. To modify the shape of an object requires energy and this is where the missing kinetic energy goes. A classic example of an inelastic collision is a car crash. The cars change shape and there is a noticeable change in the kinetic energy of the cars before and after the collision. This energy was used to bend the metal and deform the cars. Another example of an inelastic collision is shown in the following picture.

Here an asteroid (the small circle) is moving through space towards the moon (big circle). Before the moon and the asteroid collide, the total momentum of the system is:

$$\begin{matrix}\overrightarrow{p_{Before}} = \overrightarrow{p_{m}}+\overrightarrow{p_{a}}\end{matrix}$$

$$\overrightarrow{p_{m}}$$ stands for $$\overrightarrow{p_{moon}}$$ and $$\overrightarrow{p_{a}}$$ stands for $$\overrightarrow{p_{asteroid}}$$ and the total kinetic energy of the system is:

$$\begin{matrix}E_{Before} = E_{km} + E_{ka}\end{matrix}$$

When the asteroid collides inelastically with the moon, its kinetic energy is transformed mostly into heat energy. If this heat energy is large enough, it can cause the asteroid and the area of the moon's surface that it hit, to melt into liquid rock! From the force of impact of the asteroid, the molten rock flows outwards to form a moon crater.

After the collision, the total momentum of the system will be the same as before. But since this collision is inelastic, (and you can see that a change in the shape of objects has taken place!), total kinetic energy is not the same as before the collision.

So:

$$\begin{matrix}\overrightarrow{p_{Before}} &=& \overrightarrow{p_{After}} \\\overrightarrow{p_{m}}+\overrightarrow{p_{a}} &=&\overrightarrow{p_{After}} \\&\rm {\textbf{but}}& \\E_{kBefore} &\neq& E_{kAfter} \\E_{km} + E_{ka} &\neq& E_{kAfter}\end{matrix}$$

Worked Example 48 Inelastic Collision
Question: Let's consider the collision of two cars. Car 1 is at rest and Car 2 is moving at a speed of $$2\mathrm{m.s^{-1}}$$ in the negative x-direction. Both cars each have a mass of 500 kg. The cars collide inelastically and stick together. What is the resulting velocity of the resulting mass of metal?

Answer:

Step 1 :

Draw the picture: Before the collision:

After the collision:

Step 2 :

We know the collision is inelastic and there was a definite change in shape of the objects involved in the collision - there were two objects to start and after the collision there was one big mass of metal! Therefore, we know that kinetic energy is not conserved in the collision but total momentum is conserved.

So:

$$\begin{matrix}E_{kBefore} & \neq & E_{kAfter} \\& \rm {\textbf{but}} & \\\overrightarrow{p_{TBefore}} & = & \overrightarrow{p_{After}}\end{matrix}$$

Step 3 :

So we must use conservation of momentum to solve this problem. Take the negative x-direction to have a negative sign:

$$\begin{matrix}\overrightarrow{p_{TBefore}} & = & \overrightarrow{p_{After}} \\\overrightarrow{p_{1}} + \overrightarrow{p_{2}} &=&\overrightarrow{p_{After}} \\m_{1}\overrightarrow{u_{1}} + m_{2}\overrightarrow{u_{2}} &=& (m_{1} +m_{2})\overrightarrow{v} \\0 + 500(-2) &=& (500+500)\overrightarrow{v} \\-1000 &=& 1000\overrightarrow{v} \\\overrightarrow{v} &=& -1 \ \mathrm{m.s^{-1}}\end{matrix}$$

Therefore,

$$\begin{matrix}\overrightarrow{v} = \underline{1 \ \mathrm{m.s^{-1}}\rm {\ in \ the \ negative \ x-direction} }\end{matrix}$$


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FHSST Physics/Collisions and Explosions/Explosions

= Explosions =

When an object explodes, it breaks up into more than one piece and it therefore changes its shape. Explosions occur when energy is transformed from one kind e.g. chemical potential energy to another e.g. heat energy or kinetic energy extremely quickly.

So, like in inelastic collisions, total kinetic energy is not conserved in explosions. But total momentum is always conserved. Thus if the momenta of some of the parts of the exploding object are measured, we can use momentum conservation to solve the problem!

In the following picture, a closed can of baked beans is put on a stove or fire:

Before the can heats up and explodes, the total momentum of the system is:

$$\begin{matrix}\overrightarrow{p_{Before}} &=& \overrightarrow{p_{can}} \\&=& 0\end{matrix}$$

and the total kinetic energy of the system is:

$$\begin{matrix}E_{kBefore} &=& E_{kcan} \\&=& 0\end{matrix}$$

since the can isn't moving. Once the mixture of beans, juice and air inside the can reach a certain temperature, the pressure inside the can becomes so great that the can explodes! Beans and sharp pieces of metal can fly out in all directions. Energy in the system has been transformed from heat energy into kinetic energy.

After the explosion, the can is completely destroyed. But momentum is always conserved, so:

$$\begin{matrix}\overrightarrow{p_{Before}} &=& \overrightarrow{p_{After}} \\\overrightarrow{p_{Before}} &=& \overrightarrow{p_{1}} +\overrightarrow{p_{2}} +\overrightarrow{p_{3}} + \overrightarrow{p_{4}}\\0 &=& \overrightarrow{p_{1}} +\overrightarrow{p_{2}} +\overrightarrow{p_{3}} + \overrightarrow{p_{4}}\end{matrix}$$

However, the kinetic energy of the system is not conserved. The can's shape was changed in the explosion. Before the explosion the can was not moving, but after the explosion, the pieces of metal and baked beans were moving when they were flying out in all directions! So:

$$\begin{matrix}E_{kB} \neq E_{kA}\end{matrix}$$

Worked Example 49 Explosions 1
Question: An object with mass mTot = 10 kg is sitting at rest. Suddenly it explodes into two pieces. One piece has a mass of m1 = 5 kg and moves off in the negative x-direction at $$\overrightarrow{v_1}=3{m}{s^{-1}}$$. What is the velocity of the other piece?

Answer:

Step 1 :

Draw the picture. Before the explosion, the object is at rest:

After the explosion, the two pieces move off:

Riaan Note: second image on page 154 is missing

Step 2 :

Now we know that in an explosion, total kinetic energy is not conserved. There is a definite change in shape of the exploding object! But we can always use momentum conservation to solve the problem.

So:

$$\begin{matrix}\overrightarrow{p_{Before}} &=& \overrightarrow{p_{After}} \\\overrightarrow{p_{Before}} &=& \overrightarrow{p_{1}} + \overrightarrow{p_{2}}\end{matrix}$$

But the object was initially at rest so:

$$\begin{matrix}0 &=& \overrightarrow{p_{1}} + \overrightarrow{p_{2}}\\0 &=& m_{1}\overrightarrow{v_{1}} + m_{2}\overrightarrow{v_{2}} \rm {\qquad (\textbf{A})}\end{matrix}$$

Step 3 :

Now we know that m1= 5 kg but we do not know what the mass of m2 is. However, we do know that:

$$\begin{matrix}m_{Tot} &=& m_1+m_2 \\m_2 & =& m_{Tot} -m_1 \\&=&10 \rm {\ kg}-5 \rm {\ kg} \\&=&5 \rm {\ kg}\end{matrix}$$

Step 4 :

Now we can substitute all the values we know into equation (A) and solve for $$\overrightarrow{v_{2}}$$. Let's choose the positive x-direction to have a positive sign and the negative x-direction to have a negative sign:

$$\begin{matrix}0 &=& m_{1}\overrightarrow{v_{1}} + m_{2}\overrightarrow{v_{2}} \rm {\qquad (\textbf{A})} \\0 &=& 5(-3) + 5\overrightarrow{v_{2}} \\0 &=& -15 + 5\overrightarrow{v_{2}} \\5\overrightarrow{v_{2}} &=& 15 \\\overrightarrow{v_{2}} &=& +3 \rm {\ \mathrm{m.s^{-1}}}\end{matrix}$$

Therefore,

$$\begin{matrix}\overrightarrow{v_{2}} = \underline{3 \ \mathrm{m.s^{-1}}\rm {\ in \ the \ positive \ x-direction.} }\end{matrix}$$

Worked Example 50 Explosions 2
Question: An object with mass mTot =15 kg is sitting at rest. Suddenly it explodes into two pieces. One piece has a mass of m1 = 5000 g and moves off in the positive x-direction at $$v_1=7{m}{s^{-1}}$$. What is the final velocity of the other piece?

Answer:

Step 1 :

Draw the picture. Before the collision:

After the collision:

Riaan Note: image on page 156 is missing

Step 2 :

Convert all units into S.I. units:

$$\begin{matrix}m_{1} &=& 5000 \rm {\ g} \\m_{1} &=& 5 \rm {\ kg}\end{matrix}$$

Step 3 :

Now we know that in an explosion, total kinetic energy is not conserved. There is a definite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So:

$$\begin{matrix}\overrightarrow{p_{Before}} &=& \overrightarrow{p_{After}} \\\overrightarrow{p_{Before}} &=& \overrightarrow{p_{1}} + \overrightarrow{p_{2}}\end{matrix}$$

But the object was initially at rest so:

$$\begin{matrix}0 &=& \overrightarrow{p_{1}} + \overrightarrow{p_{2}}\\0 &=& m_{1}\overrightarrow{v_{1}} + m_{2}\overrightarrow{v_{2}} \rm {\qquad (\textbf{A})}\end{matrix}$$

Step 4 :

Now we know that m1= 5 kg but we do not know what the mass of m2 is. However, we do know that:

$$\begin{matrix}m_{Tot} &=& m_1+m_2 \\m_2 & =& m_{Tot} -m_1 \\&=&15 \rm {\ kg}-5 \rm {\ kg} \\&=&10 \rm {\ kg}\end{matrix}$$

Step 5 :

Now we can substitute all the values we know into equation (A) and solve for $$\overrightarrow{v_{2}}$$. Let's choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign:

$$\begin{matrix}0 &=& m_{1}\overrightarrow{v_{1}} + m_{2}\overrightarrow{v_{2}} \rm {\qquad (\textbf{A})} \\0 &=& 5(7) + 10(\overrightarrow{v_{2}}) \\0 &=& 35 + 10(\overrightarrow{v_{2}}) \\10(\overrightarrow{v_{2}}) &=& -35 \\\overrightarrow{v_{2}} &=& -3.5 \rm {\ \mathrm{m.s^{-1}}}\end{matrix}$$

Therefore,

$$\begin{matrix}\overrightarrow{v_{2}} = \underline{3.5 \ \mathrm{m.s^{-1}}\rm {\ in \ the \ negative \ x-direction.} }\end{matrix}$$


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FHSST Physics/Collisions and Explosions/Energy and Heat

= Explosions: Energy and Heat =

In explosions, you have seen that kinetic energy is not conserved. But remember that total energy is always conserved. Let's look at what happens to the energy in some more detail. If a given amount of energy is released in an explosion it is not necessarily all transformed into kinetic energy. Due to the deformation of the exploding object, often a large amount of the energy is used to break chemical bonds and heat up the pieces.

Energy is conserved but some of it is transferred through non-conservative processes like heating. This just means that we cannot get the energy back. It will be radiated into the environment as heat energy but it is all still accounted for.

Now we can start to mix the ideas of momentum conservation with energy transfer to make longer problems. These problems are not more complicated just longer. We will start off short and then combine the different ideas later on.

 Long problems should be treated like a number of smaller problems. Focus on them one at a time.

Worked Example 51 Energy Accounting 1
Question: An object with a mass of mt = 17 kg explodes into two pieces of mass m1 = 7 kg and m2 = 10 kg; m1 has a velocity of 9 m·s−1 in the negative x-direction and m2 has a velocity of 6.3 m·s−1 in the positive x-direction. If the explosion released a total energy of 2000 J, how much was used in a non-conservative way?

Answer:

Step 1 :

Draw the picture. Before the collision:

Riaan Note: first image on page 157 is missing

After the collision:

Step 2 :

We are asked how much energy was used in a non-conservative fashion. This is the difference between how much energy was used in a conservative fashion and how much was used in total. We are lucky because we have everything we need to determine the kinetic energy of both pieces. The kinetic energy of the pieces is energy that was used in a conservative way.

Step 3 :

The sum of the kinetic energy for the two blocks is the total kinetic energy of the pieces. So:

$$\begin{matrix}E_{kTot} &=& E_{k1} + E_{k2} \\&=& \frac{1}{2}m_{1}\overrightarrow{v_{1}}^{2} +\frac{1}{2}m_{2}\overrightarrow{v_{2}}^{2} \\&=& \frac{1}{2}(7)(9)^{2} + \frac{1}{2}(10)(6.3)^{2} \\&=& 283.5 + 198.45 \\E_{kTot} &=& 481.95 \rm {\ J}\end{matrix}$$

Step 4 :

The total energy that was transformed into kinetic energy is 481.95 J. We know that 2000 J of energy were released in total. the question makes no statements about other types of energy so we can assume that the difference was lost in a non-conservative way. Thus the total energy lost in non-conservative work is:

$$\begin{matrix}E - E_{kTot} &=& 2000 - 481.95 \\&=& 1518.05 \rm {\ J}\end{matrix}$$

Worked Example 52 Energy Accounting 2
Question: An object at rest, with mass mTot = 4 kg, explodes into two pieces (m1, m2) with m1 = 2.3 kg. m1 has a velocity of 17 m·s−1 in the negative x-direction. If the explosion released a total energy of 800 J,

 * 1) What is the velocity of m2?
 * 2) How much energy does it carry?
 * 3) And how much energy was used in a non-conservative way?

Answer:

Step 1 :

Draw the picture. Before the collision:

After the collision:

Step 2 :

Now we know that in an explosion, total kinetic energy is not conserved. There is a definite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So:

$$\begin{matrix}\overrightarrow{p_{Before}} &=& \overrightarrow{p_{After}} \\\overrightarrow{p_{Before}} &=& \overrightarrow{p_{1}} + \overrightarrow{p_{2}}\end{matrix}$$

But the object was initially at rest so:

$$\begin{matrix}0 &=& \overrightarrow{p_{1}} + \overrightarrow{p_{2}}\\0 &=& m_{1}\overrightarrow{v_{1}} + m_{2}\overrightarrow{v_{2}} \rm {\qquad (\textbf{A})}\end{matrix}$$

Step 3 :

Now we know that m1= 2.3 kg but we do not know what the mass of m2 is. However, we do know that:

$$\begin{matrix}m_{Tot} &=& m_1+m_2 \\m_2 & =& m_{Tot} -m_1 \\&=&4 \rm {\ kg}-2.3 \rm {\ kg} \\&=&1.7 \rm {\ kg}\end{matrix}$$

Step 4 :

Now we can substitute all the values we know into equation (A) and solve for $$\overrightarrow{v_{2}}$$. Let's choose the positive x-direction to have a positive sign and the negative x-direction to have a negative sign:

$$\begin{matrix}0 &=& m_{1}\overrightarrow{v_{1}} + m_{2}\overrightarrow{v_{2}} \rm {\qquad (\textbf{A})} \\0 &=& (2.3)(-17) + 1.7\overrightarrow{v_{2}}\\0 &=& -39.1 + 1.7\overrightarrow{v_{2}}\\1.7\overrightarrow{v_{2}} &=& 39.1 \\\overrightarrow{v_{2}} &=& 23 \rm {\ \mathrm{m.s^{-1}}}\end{matrix}$$

Step 5 :
 * $$\overrightarrow{v_{2}} = 23 \mathrm{m.s^{-1}}$$ in the positive x-direction

Now we need to calculate the energy that the second piece carries:

$$\begin{matrix}E_{k2} &=& \frac{1}{2}m_{2}\overrightarrow{v_{2}}^{2}\\&=& \frac{1}{2}(1.7)(23)^2 \\&=& 449.65 \rm {\ J}\end{matrix}$$

 * The kinetic energy of the second piece is $$E_{k2}=449.65 J$$

Step 6 :

Now the amount of energy used in a non-conservative way in the explosion, is the difference between the amount of energy released in the explosion and the total kinetic energy of the exploded pieces:

$$\begin{matrix}E - E_{kTot} &=& 800 - E_{kTot} \\\end{matrix}$$

We know that:

$$\begin{matrix}E_{kTot} &=& E_{k1} + E_{k2} \\&=& \frac{1}{2}m_{1}\overrightarrow{v_{1}}^{2} + 449.65 \\&=& \frac{1}{2}(2.3)(17)^{2} + 449.65 \\&=& 332.35 + 449.65 \\&=& 782 \rm {\ J}\end{matrix}$$

Step 7 :

So going back to:

$$\begin{matrix}E - E_{kTot} &=& 800 - E_{kTot} \\&=& 800 - 782 \\&=& 18 \rm {\ J}\end{matrix}$$

 * 18 J of energy was used in a non-conservative way in the explosion


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Cookbook:Chicken Parmigiana

| Italian Cuisine | US Cuisine

Chicken parmigiana is a popular Italian-American dish served in a variety of ways. The same basic recipe works for a number of dishes. The components usually consists of a breaded, tenderized piece of meat (or slice of eggplant), a simple marinara sauce, and cheese (usually mozzarella). A thicker pasta, such as linguini, a toasted loaf of Italian bread, and a light red wine such as Chianti complement this dish well.

This recipe is typical of the parmigiana recipes from New York and New Jersey.

Ingredients

 * 2 chicken breasts
 * 1 teaspoon of salt
 * 1 teaspoon of garlic powder
 * ½ a teaspoon of dried oregano
 * ¼ teaspoon of dried basil
 * ¼ teaspoon of finely-crushed black pepper
 * ½ a cup of all-purpose flour
 * ½ a cup of tomato puree
 * ¼ cup of chicken stock
 * ½ a cup of canola oil (see below if you wish to use olive oil)
 * ½ a cup of shredded mozzarella cheese
 * ¼ a cup of grated parmigiano cheese

Procedure

 * 1) In a heavy skillet (preferably cast iron), heat the oil on medium heat.
 * 2) While the oil is heating, use a tenderizing mallet (or the ball of your hand) to pound the chicken breasts to about ½ inch thick. This step is important to ensure the chicken is tender and more importantly to ensure the chicken cooks thoroughly.
 * 3) Season the chicken breasts with salt and garlic powder.
 * 4) Dredge the breasts in flour and shake off any excess.
 * 5) Place both breasts in the heated oil for 3 minutes, then flip and fry for an additional 2 minutes.
 * 6) Remove the breasts and place on a rack or paper towel.
 * 7) Preheat a broiler.
 * 8) In a large oven-safe saucepan, heat the tomato puree and chicken stock until it begins to simmer.
 * 9) Add the oregano, basil, and pepper to saucepan and mix
 * 10) Place cooked chicken breasts in pan and flip a number of times to coat with sauce.
 * 11) Add mozzarella and parmigiana cheese to the tops of the breasts.
 * 12) Broil for 30–60 seconds. It will burn very quickly, so check it frequently.
 * 13) Remove from oven and serve.

Notes, tips, and variations

 * Use thin-sliced chicken cutlets (available at most supermarkets) to avoid the BIG mess and time of pounding.
 * If pasta is to be served with this meal, it should be timed so that the parmigiana is done right when the pasta is. The pasta should be removed directly from the pot and placed into the pan. The remaining sauce can then be used with the pasta.
 * This recipe works best with breasts that come from slightly older chickens (such as broiler or fryer chickens). These chickens have a stronger taste but are generally tougher. The tenderizing step takes care of the toughness of the older birds though.
 * If you do not live in area where fresh tomatoes are grown, canned tomatoes are usually superior to the low-cost ones you will find in the supermarket.

 * For veal parmigiana, follow the same steps except do not tenderize. Veal is already tender enough.  The veal can also be cooked for 2 minutes on each side.
 * Omit the tenderizing step for eggplant parmigiana.
 * If you want to use fresh garlic instead of powder, add it to the sauce instead of adding it to the chicken breast.
 * You can also try using other types of Italian cheeses such as provolone cheese or fontina cheese to add more of a "tasty kick" compared to a mild cheese like mozzarella
 * You may wish to use olive oil. When frying with olive oil, be sure to use a cheap solvent-extracted grade! The fancy grades can not tolerate heat. They have smoke points as low as 200°F (93°C), making for both a fire hazard and a source of smoke. Save the cold-pressed extra-virgin olive oil for a salad.

title: Cookbook:Chicken Parmigiana


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Teaching Computer Literacy

This is a resource for anyone who is, or will be teaching the basics of computers to any type of beginner. The focus of this book is to provide:
 * General suggestions that can be adapted to your needs.
 * Specific lesson plans for students with varying interests.
 * Links to other relevant information
 * Links to useful software for the purpose of learning or teaching.

Currently we have "Lessons in Paint" which are good for most children, and anyone else who likes to draw. Lessons in Paint is great because many of the Windows features are teachable in Paint.

These lessons are geared toward Windows PCs. The suggestions and lessons could be adapted to work with other operating systems. If you have or make a lesson plan around Mac or Linux, please let us know on the discussion page and we could create a place for it.

Remember, you can edit or add information by simply clicking on the “edit” tab at the top of any page or any section. Don’t worry about mistakes, you will be able to preview before saving. Also other users of can refer to previous versions and fix mistakes. If you are interested in the development of this book click on the “discussion” tab on any page to see what we are working on, and to contribute to the discussion.

= What the Basics Include =

Mouse Familiarity
 * click (left button)
 * drag
 * double click
 * middle click (for instance to open link in a new Window, also available as Ctrl + left click, paste marked text on UNIX-like systems, if held substitute for scrolling)
 * scroll wheel (natural scrolling vs normal)
 * right click (for the context menu)

Keyboard
 * some of the more common letters
 * enter
 * backspace
 * shift
 * tab
 * ctrl
 * alt, Alt-Gr (also right alt key)
 * Windows, command (on Mac) or meta key (other operating systems)
 * arrows

File Management
 * save
 * open
 * locate
 * create a folder
 * create a desktop shortcut
 * name/ rename a file or folder, be aware of special characters, extensions
 * mark files using ctrl and clicking
 * using shift to select more files
 * compress/ decompress

Printing
 * print
 * print preview
 * print to file (PDF, XPS)

GUI Elements
 * use the menu
 * resize a window
 * minimize a window
 * maximize a window
 * move a window
 * locate items in Start Menu
 * using the scrollbar
 * using search

Editing Options
 * copy
 * paste
 * cut
 * undo
 * redo
 * spell check

= Overall Suggestions = Some obvious and hopefully some not so obvious suggestions; It is important to make it interesting for the specific person, while keeping it simple enough to avoid frustration. To achieve both goals at the same time you should;
 * 1) Ask them what they want to do.
 * 2) Find the simplest method of accomplishing the task.
 * 3) Try it a few times to simplify the process and organize your lesson.
 * 4) Then walk them through the lesson.

At first skip the start menu by using a desktop shortcut. Yes they will need to learn it, but getting straight to something useful is key to avoiding frustration.

Use lower screen resolution (for instance 1600x900, 800x600, beware the aspect ratio) or dots per inch (DPI) scaling in newer operating systems and larger icons for improved visibility and mouse accuracy.

Change the mouse sensitivity to be easier to control.

Use only click, and drag on the first lesson.

Choose the save to directory and make sure you are viewing with thumbnails.

Teach the menu from left to right, i.e. start with 'file' on the first day then 'edit' the next. This should prevent having too many students unsure which menu item to click.

Note that similar things can be taught together when they are done in the same way. Like dragging a window and dragging a pasted item. However when two different things could be confusingly similar, teach them at different times, and teach the distinction when teaching the second. Some examples of this; the menu items, click/double click, and save/save-as.

Use consistent language, and find the easiest term to relate too. Students will learn the synonyms later without even realizing they are learning. Here are a few examples

Arrow, pointer, cursor, mouse (I prefer arrow, pointer works well also). Enter, or return (use whatever is written on the key, and stick with it).

In the first few sessions avoid button combinations of any kind. Using Ctrl C to copy is faster, and easier, but not for those new to computers. Even Shift key for capital letters is potentially frustrating. Introduce these combinations after your students are good at computers.

= Lessons in Paint =

In this series of lessons we use Microsoft Paint. This is a good program to use; it is already on the computer, is fun to use, and is great practice for the mouse. You can teach most of the basics using Paint.

Paint Lesson 1
Open MS Paint

First you setup the computer; (This will help students who are new to using the mouse, making it easier to select things.)
 * Change the screen resolution to 800 x 600.
 * Choose the location for saving files.
 * Set to view files as thumbnails.
 * Change the mouse settings to be less sensitive.
 * Next you put an icon of the Paint program on the desktop.

The students will be learning how to click, drag, and point the mouse. To avoid overwhelming the student we will focus on just those 3 things. We will cover double click, and right click in another lesson. We will also want to focus on only things that can be done when pressing one button at a time. An example is shift key for caps; you should consider not using caps, or using the caps lock.

Mini Lesson #1
Watch for anyone having trouble, and be ready to answer any questions.
 * 1) Show how to select the icon and hit enter to open (not double clicking).
 * 2) Have them click on the line tool.
 * 3) Show how to drag to make a line.
 * 4) Demonstrate picking a color from the palette.
 * 5) Show the circle tool (ellipse tool).
 * 6) Let them have fun.

Next we show how to save;
 * 1) Click on file.
 * 2) Click save.
 * 3) Choose a name.
 * 4) Type it in.
 * 5) Click save.
 * 6) Show how to close the Window (using the X).

Congratulate everyone on completing the first (mini) lesson.

Mini lesson #2
Have everyone start again, on their own as much as possible. Show the pencil tool and have the students see what all the tools below it do. Kids can easily remember these tools because the icons have simple representations of what the tool does.
 * spray paint (airbrush)
 * line
 * rectangle
 * oval (ellipse)

Tell the students they can draw a picture for the computer. This will give you some time to help and reassure those who need it.
 * 1) make sure everyone has saved their picture
 * 2) click on 'file'
 * 3) click on 'save as background image centered'

Now their work of art is the center of attention on the computer. This should bring their focus from the paint program to other parts of the computer.

Paint Lesson 2
Check the settings are suitable to your needs, just as in Paint Lesson 1.

You might have them find how many don't have options (pencil, fill tool, eyedropper). Let the students get comfortable using them to see what happens.
 * 1) Have everyone open Paint by double clicking on the icon.
 * 2) Show the Brush and demonstrate the options (shapes and sizes).
 * 3) Have the students draw with the diagonal brush so they can see the ribbon effect.
 * 4) Explain how most tools have options in the same place.
 * 5) Let them look at the tools and options and try them out.

Most of the options are related to size making it easier to understand.

For now we will move on to explain edit options starting with undo.
 * 1) Show how to undo from the edit menu and let everyone practice.
 * 2) Show how it will only go back 3 times.
 * 3) Demonstrate redo, and how it is the opposite of undo.

End the lesson by having everyone save their pictures and make it the background image.

Paint Lesson 3
You should begin by teaching anything you may have skipped in from the past lessons. Next you should cover the text tool

= Reference Links = These sites have great suggestions on how to help people learn computers; Phil Agre Ask MetaFilter

= Software Links =
 * If you want to teach computer programming, kids like . info has an encyclopedia article that discusses of the best educational programming languages and links to their teaching materials.
 * Tux4Kids Has two useful programs that are fun for kids. They are Tux Paint, and Tux Typing.

 * Microsoft Windows Media Encoder is a free program that can record video of all or part of the computer screen in use, including audio recording. This type of tutorial is being called screencasting and it can be a great way to show people how to use computers. People new to computers may not benefit most from learning this way, but this can be used to review lessons more easily.

title: Teaching Computer Literacy


text:
Debate/Topics/Drug Prohibition

Drug prohibition is a set of policies which outlaw the possession, use, manufacture and/or distribution of drugs. In most countries, prohibited drugs include heroin, cocaine, marijuana and LSD.

 * Supporters believe that drug prohibition is a morally acceptable and/or efficacious way to reduce drug abuse
 * Opponents believe drug prohibition is a morally unacceptable and/or ineffective way to reduce drug abuse

Both of these positions rely on the premise that drug abuse is harmful to the individual abusers and/or society in general.

Crime
Drug prohibition reduces the rates of drug addiction and thus reduces drug-related crime such as robbery.

 * This argument relies on the assumption that drug prohibition has reduced drug abuse.
 * This argument relies on the assumption that drug addiction causes crime drug abuse.

Marketing
Legalized drugs would be sold by large corporations that may use irresponsible advertising to encourage risky and addictive behaviors.

Counterpoint
Prohibited drugs are currently sold by black market distributors who do not monitor purity and resort to outright violence in trade disputes, in addition to promoting their product to non-users. Corporations could be more easily held to standards of purity and ethical conduct than a vast network of black market distributors.

Socialist view
Legalized drugs could be produced and managed by the state and done in such a manner as to eliminate advertising and reduce overall use.

Protect individuals
Drug abuse causes harm to the individuals and their family and friends, and society in general. It is the responsibility of the government to prevent that.
 * This argument relies on the premise that drug abuse is harmful to the individual abusers and/or society in general.
 * This argument relies on the premise that the government has the right and responsibility to limit individual rights in this case.
 * This argument relies on the assumption that drug prohibition has reduced drug abuse.

Setting an example
Allowing the legal use of drugs like heroin and cocaine will provide a negative role model for children and thus cause an increase in drug abuse.
 * This argument relies on the factual claim that legalizing drugs will cause an increase in drug abuse.

Black Market
Black Market: the trade in prohibited goods

Much of the harm caused by drug abuse comes from the unregulated market for drugs. There is no legal incentive for manufacturers and distributors to ensure the safety of their product or rely on honest, peaceful transactions with customers and rivals. If drugs were legalized, this black market would disappear, reducing the funding considerably for organized crime including terrorism, Mafia and street gangs.

 * This argument relies on the claim that organized crime is funded by the drug trade.

Drug costs
Legalizing drugs would drastically reduce their cost. Thus, addicts could afford their habit without resorting to violence and robbery.

Efficacy
Because drug prohibition has not proven effective in reducing the rate of drug abuse, it is time to try a different approach, such as a wide variety of harm reduction proposals.

 * This argument relies on the claim that drug prohibition has not reduced drug abuse rates.

Personal freedom
The government does not have the right to limit a person's choices regarding drug use.

Racism
Drug prohibition was founded based on racism, and remains racist in practice.

 * This argument relies on the premise that drug prohibition policies are racist in implementation and design.

Tax revenue
Legalizing drugs would grant a major new stream of tax revenue for the government.

title: Debate/Topics/Drug Prohibition


text:
Debate/Topics/Same-Sex Marriage/United States

Federal Marriage Amendment
Wikipedia article: Federal Marriage Amendment

The Federal Marriage Amendment is a proposed Amendment to the United States Constitution that would legislate the nature of marriage, so that it is limited to occurring between one man and one woman. The Amendment would override any state law that contradicts it, and could be changed only through the passage of another Amendment.

The Federal Marriage Amendment is generally supported more by the Republican Party. Nevertheless, there are Democrats and member of other leftist parties who also support it, as well as Republicans and other rightist parties who do not. Opposition can be inspired either by a belief that same-sex marriage should be allowed, or by a philosophy that the Amendment would needlessly overrule states rights.

Those who believe in states rights on the same-sex marriage issue do so on the principle that the populations of individual states should be allowed to organize their society and culture in whatever way fits their ideals and morals. One potentially divisive issue, however, is whether other states and communities must recognize same-sex marriages performed elsewhere. According to the US Constitution's Full Faith and Credit clause, marriages performed legally in one state must be recognized throughout the nation. The Defense of Marriage Act, a bill signed into law in 1996 by Democratic President Bill Clinton, explicitly grants states the right to refuse to recognize same-sex marriages. Some of the support for the Federal Marriage Amendment comes from those who believe that the Defense of Marriage Act is unconstitutional because it grants an exception to the full faith and credit clause.

Some opponents of the Federal Marriage Amendment have pointed out that the Amendment would be the only binding Amendment to limit the freedom of citizens. The 18th Amendment, which banned alcohol during Prohibition, was also passed, but has been repealed by the 21st Amendment. The 22nd and 16th Amendment could also be seen as curtailing freedoms in establishing a two-term limit for Presidents and instituting a graduated income tax, respectively.

Civil unions
Wikipedia article: Civil Union

The state of Vermont does not allow same-sex marriages, but does allow same-sex "civil unions". These grant most of the same rights as marriage. Opposition to civil unions occurs from both Supporters and Opponents; Supporters believe it is a form of "second-class citizenship" (that is, discrimination) and Opponents believe it is, for all intents and purposes, equivalent to marriage and is opposable on the same grounds. Those who support civil unions include Opponents who object to same-sex marriage on traditional grounds—that marriage is a heterosexual institution by definition, but that same-sex couples should be able to receive the same benefits for their own unions—and Supporters who believe that civil unions are practically identical to marriage.

Discrimination
Supporters: The refusal to extend marriage to homosexual couples is inherently unfairly discriminatory and a violation of the 14th Amendment, which guarantees equal treatment under the law. There are more than 1,000 laws which apply differently depending on an individual's marital status. These extend from trivial facts, such as that a gay president's lover would not be protected by the Secret Service, to complex issues concerning the release of financial and medical information, hospital visitation rights, taxes, wills, child custody, alimony, and Fifth Amendment rights.

Opponents: The 14th Amendment does not apply because a person's homosexual behavior is a choice. The government is not obliged to recognize all choices as of equal validity. (This is part of the nature vs. nurture debate.)

Libertarian View
The United States Libertarian Party, and some persons from Democratic, Republican, other, or no parties, have an opinion different from both Supporters and non-Supporters.

Libertarians: The government has no authority to recognize any marriage. There are over 1,000 legal rights granted to heterosexual couples should not be granted. For example, the tax code should not treat married couples differently than unmarried couples. The government should not compel hospitals to allow any individual access to patients or medical records, except in accordance with that hospital's policy. Individuals can choose to cohabitate, and can declare themselves married, or perform a ceremony. It is up to individuals to choose, according to their own beliefs, whether to recognize any such marriage.

Parties

 * Democratic Party: Opposed to same-sex marriage but in favor of civil unions
 * Republican Party: Opposed to same-sex marriage and civil unions
 * Green Party: Supports same-sex marriage
 * Libertarian Party: Supports the Libertarian View (see above)


text:
Talk:Teaching Computer Literacy

Untitled

 * Teaching tips even basic ones.
 * Lesson plans/lesson outlines.
 * Organizing the information.
 * Changing to multiple pages at some point.
 * Corrections to grammar spelling etc.
 * Links to good software (preferably free).

A little more about me and this project
I'm Eric, and I started this wikibook because I've offered to volunteer teaching computer classes. Classes should start when school is out. They are still cutting through the red tape, and I am not sure if I will get the chance. I didn't realize that non-profit organizations had so much red tape. Anyway I practiced the lessons with my nephews 7 and 4 years old. But they are already familiar with a mouse, hopefully it doesn't skew my test results too much.

I tried to find some public domain material on the subject but came up with nothing, and so this book was started. I started with paint because I know it well. It does keep the interest of kids. As a native windows program you can teach some windows fundamentals. Simplified kids software is great but you can't teach what isn't there.

The organization I've offered to volunteer for says they have Microsoft Office. So far I'm not sure what is going to get teenagers interested in Microsoft Office. Any ideas please click 'edit this page'. Thanks for taking the time to read this, hopefully you find this book useful.

A few comments
1) I think we should either rename the article or perhaps expand its scope to non-windows machines. 2) I hope to make some of my own contributions, and maybe expand this beyond paint. 3) Is this really a worthwhile book? There are other, much more complete books on the same topic. 4) Anyone else still interested? --Jats (talk) 01:55, 27 February 2009 (UTC)

Wordpad
My name is Hall, I'm a computer teacher for both primary and secondary schools. I have found that Wordpad is a great program to teach beginners about word processing, keyboard skills and file management. The interface in wordpad is like a cutdown version of MSWord (without going to the trouble of customising the MS word interface), which helps the student focus on key skills, reduces choice/mistake anxiety, and helps the transition to a "real" Word Pocessor. I have about 3 lessons written up (if I can find them). If Eric wants to set up the chapters I'll paste them in.

Setting up Wordpad lessons
I would be glad to set up the Wordpad lesson. You can leave what you have in this talk page. Thanks, Eric

Timing
A helpful element for this book would be suggested/expected time that the lessons would take. I might offer this myself after trying some of it out today at a primary school in Newtown, Basseterre, St. Kitts today.

Not Paint, but WORD or rtf
The subject is presented well, but I teach Senior Citizens! I therefore prefer a TEXT program to start with, such as WORD since that makes immediate connection to the SC students. Easy to start them off with typing their life history for their grandchildren. Alternately, start them playing card games, which they often already know, but gives them plenty of opportunity to use the mouse, especially at home away from the classroom.

title: Talk:Teaching Computer Literacy


text:
Debate/Topics/Invasion of Iraq

The US-led coalition invaded Iraq and are occupying it. There are two broad areas of dispute:
 * The evidence that an attack was necessary
 * The handling of the invasion and subsequent occupation

International Law

 * According to international law, an attack on another country is illegal unless approved by the security council. The USA was unable to provide this approval.
 * The USA frequently condemn the military actions of other countries when they act without security council approval. Failing to lead by example will diminish the USA's credibility in the eyes of other countries as a law abiding moral state.

Humanitarian principles

 * Saddam Hussein was a tyrant who ruled over Iraq with a reign of terror and murder, committing acts that amount to genocide. Therefore, invading Iraq to install a more peaceful regime was an act of kindness and benevolence.

 * This argument relies on the premise that murder is wrong.
 * This argument relies on the premise that democracy is morally superior to tyrannical regimes.
 * This argument relies on the claim that the government of Saddam Hussein was murderous and dictatorial.
 * This argument relies on the claim that the invasion is not murder itself: See War is murder.

Counterpoint

 * Many governments in the world are dictatorial and sometimes even genocidal. Saddam Hussein's capacity to harm both his own people and his neighbors was so severely harmed by UN sanctions that he posed little threat to anyone, and was in any case less dangerous than many other dictators around the world.  Thus, the invasion was not warranted because other countries represented a bigger threat to world peace and stability.

 * This argument relies on the premise that there are extant governments that routinely violate human rights.
 * This argument relies on the premise that UN sanctions on Iraq worked to reduce Saddam Hussein's military capacity and threat to the world.

Iminent threat

 * The invasion of Iraq was warranted because Iraq posed an imminent threat to the safety and well-being of the United States and other countries.

 * This argument relies on the premise that Saddam Hussein's government represented a threat to world peace.
 * This argument relies on the premise that the safety and well-being of the United States and other countries are important.

Aid to al-Qaeda

 * Saddam Hussein had ties to al-Qaeda, a terrorist organization that threatened the United States and other countries and has masterminded several major terrorist attacks.

 * This argument relies on the claim that Saddam Hussein's government had ties to al-Qaeda.
 * This argument relies on the claim that terrorism, and al-Qaeda specifically, remains a major threat to peace in the United States and other countries.

Weapons of Mass Destruction

 * Saddam Hussein represented a threat because he had Weapons of Mass Destruction, or had the desire and ability to acquire them.

 * This argument relies on the premise that Iraq was in possession of Weapons of Mass Destruction.
 * This argument relies on the premise that Iraq had no right to possess Weapons of Mass Destruction.

title: Debate/Topics/Invasion of Iraq


text:
Debate/Topics/Human Rights

There are many countries in the world, many or most of which could be accused of violating the human rights of citizens and non-citizens.

The following countries may violate human rights:
 * Afghanistan
 * Algeria
 * Armenia
 * Australia
 * Azerbaijan
 * Bangladesh
 * Belarus
 * Brazil
 * Cambodia
 * People's Republic of China
 * Cuba
 * Egypt
 * Georgia
 * Greece
 * India
 * Indonesia
 * Iran
 * Iraq
 * Israel
 * Japan
 * Jordan
 * Kazakhstan
 * Laos
 * Lebanon
 * Libya
 * Malaysia
 * Mexico
 * Morocco
 * Myanmar
 * Nepal
 * North Korea
 * Pakistan
 * Russia
 * Saudi Arabia
 * Syria|
 * Thailand
 * Turkey
 * Turkmenistan
 * Tajikistan
 * Tunisia
 * United Arab Emirates
 * United Kingdom
 * United States
 * Uzbekistan
 * Venezuela
 * Vietnam
 * Yemen
 * Zimbabwe

title: Debate/Topics/Human Rights


text:
FHSST Physics/Collisions and Explosions/Summary

= Important Equations and Quantities =

Momentum:

Kinetic energy:


text:
FHSST Physics/Newtonian Gravitation

= Newtonian Gravitation = All objects on Earth are pulled downward, towards the ground. This phenomenon is called gravity. Every object falls just as fast as any other object (unless the air slows it down like a feather, or pushes it up like a balloon), as first shown by Galileo. In 1687 Isaac Newton stated that gravity is not restricted to the Earth, but instead, there is gravity everywhere in the universe. Newton explained that planets, moons, and comets move in orbits because of the effect of gravity.

title: FHSST Physics/Newtonian Gravitation


text: Template:FHSST Physics/Newtonian Gravitation


text:
FHSST Physics/Newtonian Gravitation/Properties

= Properties =

Gravitational attraction is a force and therefore must be described by a vector - so remember magnitude and direction. The force due to gravity acts between any two objects with mass. To determine the magnitude of the force we use the following equation:

This equation describes the force between two bodies, one of mass m1, the other of mass m2 (both have units of kilograms, or kg for short). The G is Newton's`Gravitational Constant' (6.673×10−11 N·m2·kg−2) and r is the straight line distance between the two bodies in meters.

This means the bigger the masses, the greater the force between them. Simply put, big things matter big with gravity. The 1/r2 factor (or you may prefer to say r−2) tells us that the distance between the two bodies plays a role as well. The closer two bodies are, the stronger the gravitational force between them is. We feel the gravitational attraction of the Earth most at the surface since that is the closest we can get to it, but if we were in outer-space, we would barely even know the Earth's gravity existed!

Remember that

which means that every object on the earth feels the same gravitational acceleration! That means whether you drop a pen or a book (from the same height), they will both take the same length of time to hit the ground... in fact they will be head to head for the entire fall if you drop them at the same time. We can show this easily by using the two equations above (9.1 and 9.2). The force between the Earth (which has the mass me) and an object of mass mo is

and the acceleration of an object of mass mo (in terms of the force acting on it) is

So we substitute equation (9.3) into equation (9.4), and we find that

Since it doesn't matter what mo is, this tells us that the acceleration on a body (due to the Earth's gravity) does not depend on the mass of the body. Thus all objects feel the same gravitational acceleration. The force on different bodies will be different but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it differently, instead of using a we use g which we call the gravitational acceleration.


text:
Talk:IntroAIToday

Merriam-Webster... um, copyright problem here?--Hyperlink 08:09, 26 Mar 2005 (UTC)

title: Talk:IntroAIToday


text:
FHSST Physics/Newtonian Gravitation/Mass and Weight

= Mass and Weight =

Weight is a force which is measured in Newtons, it is the force on an object due to gravity. People are always asking other people "What is your weight?" when a physicist might ask the same question as "What is your mass?".

Mass is measured in kilograms (kg) and is the amount of matter in an object, it doesn't change unless you add or remove matter from the object (if you continue to study physics through to university level, you will find that Einstein's theory of relativity means that mass can change when you travel as fast as light does, but you don't need to worry about that right now). There are 1000 g in 1 kg and 1000 kg in a tonne.

To change mass into weight we use Newton's 2nd Law which is $$F = ma$$. The weight is the force and gravity the acceleration, it can be rewritten as:

W is the weight, measured in newtons. m is the mass, measured in kilograms and g is the acceleration due to gravity, measured in metres per second squared; it is equal to about 10 on the Earth. The exact value of Earths equatorial surface gravity is 9.7801 $$m.s^{-2}$$.

Examples
1. A bag of sugar has a mass of 1 kg, what is its weight? (Acceleration due to gravity = 10 $$m.s^{-2}$$)

Step 1: Always write out the equation, it helps you to understand the question, and you will get marks as well.

Step 2: Fill in all the values you know. (remember to make sure the mass is in kilograms and NOT in grams or tonnes!)

Step 3: Write out the answer remembering to include the units! You will lose marks if you don't

2. A space-man has a mass of 90 kg, what is his weight (a) on the earth? (b) on the moon? (c) in outer space? (The acceleration due to gravity on the earth is 10 $$m.s^{-2}$$, on the moon gravity is about 1/6 of the gravity on earth).

1:

2:

3: Weightless in outer space because g = 0.

So now when somebody asks you your weight, you know to reply ``Anything!! But my mass is a different matter!"


text:
FHSST Physics/Newtonian Gravitation/Normal Forces

= Normal Forces = If you put a book on a table, it does not accelerate; it just lies on the table. We know that gravity is acting on it with a force

but if there is a net force there MUST be an acceleration and there isn't. This means that the gravitational force is being balanced by another force [Newton's third law!].

This force we call the normal force. It is the reaction force between the book and the table. It is equal to the force of gravity on the book. This is also the force we measure when we measure the weight of something.

The most interesting and illustrative normal force question, that is often asked, has to do with a scale in a lift. Using Newton's third law we can solve these problems quite easily.

When you stand on a scale to measure your weight you are pulled down by gravity. There is no acceleration downwards because there is a reaction force we call the normal force acting upwards on you. This is the force that the scale would measure. If the gravitational force were less than the reading on the scale would be less.

Worked Example 53 Normal Forces 1
Question: A man weighing 100kg stands on a scale (measuring newtons). What is the reading on the scale?

Answer:

Step 1 :

We are given the mass of the man. We know the gravitational acceleration that acts on him.

Step 2 :

The scale measures the normal force on the man. This is the force that balances gravity.

Step 3 :

Firstly we determine the net force acting downwards on the man due to gravity.

$$\begin{matrix}F_g&=&mg \\&=& 100kg\times 9.8\frac{m}{s^2} \\&=& 980\frac{kgm}{s^2} \\&=& 980N\ downwards \\\end{matrix}$$

Step 4 :

We now know the gravitational force downwards. We also know that there must be a reactionary force FN upwards. To determine the magnitude of this we use newton's laws again. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 0. if we write out the equation:

$$\begin{matrix}F_r&=&F_g+F_N \\0 &=& -980N+F_N\\F_N = 980N\ upwards\\\end{matrix}$$

The normal force is then 980N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man.

Now we are going to add things to exactly the same problem to show how things change slightly. We will now move to a lift moving at constant velocity. Remember if velocity is constant then acceleration is zero.

Worked Example 55 Normal Forces 2
Question: A man weighing 100kg stands on a scale (measuring newtons) inside a lift accelerating downwards at $$2\frac{m}{s^2}$$. What is the reading on the scale?

Answer:

Step 1 : We are given the mass of the man. We know the gravitational acceleration that acts on him.

Step 2 :

The scale measures the normal force on the man.

Step 3 :

Firstly we determine the net force acting downwards on the man due to gravity.

$$\begin{matrix}F_g&=&mg \\&=& 100kg\times 9.8\frac{m}{s^2} \\&=& 980\frac{kgm}{s^2} \\&=& 980N\ downwards \\\end{matrix}$$

Step 4 :

We now know the gravitational force downwards. We also know that there must be a reactionary force FN upwards. To determine the magnitude of this we use newton's laws again. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is $$2\frac{m}{s^2}$$. if we write out the equation:

$$\begin{matrix}F_r&=&F_g+F_N \\100kg\times (-2)\frac{m}{s^2} &=& -980N+F_N\\-200\frac{kgm}{s^2} &=& -980N+F_N\\-200N &=& -980N+F_N\\F_N = 780N\ upwards\\\end{matrix}$$

The normal force is then 780N upwards. It balances the gravitational force downwards just enough so that the man only accelerates downwards at $$2\frac{m}{s^2}$$.

Worked Example 56 Normal Forces 3
Question: A man weighing 100kg stands on a scale (measuring newtons) inside a lift accelerating upwards at $$4\frac{m}{s^2}$$>. What is the reading on the scale?

Answer:

Step 1 : We are given the mass of the man. We know the gravitational acceleration that acts on him.

Step 2 :

The scale measures the normal force on the man.

Step 3 :

Firstly we determine the net force acting downwards on the man due to gravity.

$$\begin{matrix}F_g&=&mg \\&=& 100kg\times 9.8\frac{m}{s^2} \\&=& 980\frac{kgm}{s^2} \\&=& 980N\ downwards \\\end{matrix}$$

Step 4 :

We now know the gravitational force downwards. We also know that there must be a reactionary force FN upwards. To determine the magnitude of this we use newton's laws again. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is $$4\frac{m}{s^2}$$. if we write out the equation:

$$\begin{matrix}F_r&=&F_g+F_N \\100kg\times (4)\frac{m}{s^2} &=& -980N+F_N\\400\frac{\mbox{kg} \cdot \mbox{m}}{\mbox{s}^2} &=& -980 \mbox{ N}+F_N\\400\mbox{ N} &=& -980 \mbox{ N}+F_N\\F_N = 1380 \mbox{ N}\ upwards\\\end{matrix}$$

The normal force is then 1380 N upwards. It balances the gravitational force and then in addition applies sufficient force to accelerate the man upwards at $$4\frac{m}{s^2}$$.


text:
FHSST Physics/Newtonian Gravitation/Comparative Problems

= Comparative problems =

Here always work with multiplicative factors to find something new in terms of something old.

Worked Example 57 Comparative Problem 1
Question: On Earth a man weighs 700 N. Now if the same man was instantaneously beamed to the planet Zirgon, which has the same size as the Earth but twice the mass, what would he weigh?

Answer:

Step 1 :

We start with the situation on earth:

Step 2 :

Now we consider the provje

but we know that $$m_Z = 2m_E$$ and we know that rZ = r so we could write the equation again and substitute these relationships in:

Step 3 :

Step 4 :

Step 5 :

so on Zirgon he weighs 1400 N.

Principles

 * Write out first case
 * Write out all relationships between variable from first and second case
 * Write out second case
 * Substitute all first case variables into second case
 * Write second case in terms of first case

Worked Example 58 Comparative Problem 2
Question: On Earth a man weighs 70 kg. On the planet Beeble how much will he weigh if Beeble has mass half of that of the Earth and a radius one quarter that of the earth.

Answer:

Step 1 :

We start with the situation on earth:

Step 2 :

Now we consider the provje

but we know that $$m_Z = \frac{1}{2}m_E$$ and we know that $$r_Z = \frac{1}{4}r$$ so we could write the equation again and substitute these relationships in:

Step 3 :

Step 4 :

Step 5 :

$$\begin{matrix}W_Z= 8(W)\end{matrix}$$

But although the man exerts 8 times as much force due to gravity, he still weighs 70 kg on Beeble!


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FHSST Physics/Newtonian Gravitation/Falling Bodies

= Falling bodies =

Objects on the earth fall because there is a gravitation force between them and the earth - which results in an acceleration - as we saw above. So if you hold something in front of you and let it go - it will fall.

It falls because of an acceleration toward the centre of the earth which results from the gravitational force between the two.

These bodies move in a straight line from the point where they start to the centre of the earth. This means we can reuse everything we learnt in rectilinear motion. the only thing that needs thinking about is the directions we are talking about.

We need to choose either up or down as positive just like we had to choose a positive direction in standard rectilinear motion problems. this is the hardest part. If you can do rectilinear motion you can do falling body problems. Just remember the acceleration they feel is constant and because of gravity - but once you have chosen your directions you can forget that gravity has anything to do with the problem - all you have is a rectilinear motion problem with a constant acceleration!!

What is the weight of a free falling body ? Would you be surprised to know that it is zero ?

Weight is measured using a spring balance. (Mass is measured using a physical balance). When a body is falling, the spring balance to which it is attached (the spring balance needs to fall along with the body too !) would show zero weight. When the body stops falling, then it would exert a force on the spring (equivalent to the force exerted by earth on itself) and hence it would show a reading.


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FHSST Physics/Newtonian Gravitation/Terminal Velocity

= Terminal Velocity =

Physics is all about being simple - all we do is look at the world around us and notice how it really works. It is the one thing everyone is qualified to do - we spend most of our time when we are really young experimenting to find out how things work.

Take a book - wave it in the air - change the angle and direction. what happens of course there is resistance. different angles make it greater - the faster the book moves the greater it is. The bigger the area of the book moving in the direction of motion the greater the force.

So we know that air resistance exists! it is a force. So what happens when an object falls? of course there is air resistance - or drag as it is normally called. There is an approximate formula for the drag force as well.

The important thing to realise is that when the drag force and the gravitational force are equal for a falling body there is no net force acting on it - which means no net acceleration. That does not mean it does not move - but it means that its speed does not change.

It falls at a constant velocity! This velocity is called terminal velocity.


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FHSST Physics/Newtonian Gravitation/Drag Force

= Drag force =

The actual force of air resistance is quite complicated. Experiment by moving a book through the air with the face of the book and then the side of the book forward, you will agree that the area of the book makes a difference as to how much you must work in order to move the book at the same speed in both cases. This is why racing cars are slim-lined in design, and not shaped like a big box!

Get a plastic container lid (or anything waterproof) swing it around in air and then try to swing it around under water. The density of the water is much larger than the air, making you have to work harder at swinging the lid in water. This is why boats and submarines are a lot slower than aeroplanes!

So we know that density, area and speed all play a role in the drag force.The expression we use for drag force is

where C is a constant which depends on the object and fluid interactions, $$\rho$$ is the density, A is the area and v is the velocity.


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FHSST Physics/Newtonian Gravitation/Summary

= Important Equations and Quantities =


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Wikijunior talk:Big Cats/Tigons and Ligers

This article seems to contradict much of what is written in the Tigon wikipedia article. - anon

the article on Ligers and Tigons is an informative one and should not be deleted. Hybrids exist in many species, that is a fact and one that most children should have no trouble understanding. Frankly I fail to understand the logic behind deleting a page because most adults have not heard of the content.. isn't that what information books are all about, informing us? Pabsy.

This page is informative and I can see no reason to delete it from the book. Hybrids happen in many species and my children have no trouble understanding that fact. Frankly i fail to understand the logic in deleting a page on the grounds that a reader has never heard of the content or that the content could confuse a child. isn't that what information books are all about.. informing us? oh come on... why stop at deleting a page on hybrid cats?, children are confused with Algebra, logarithms, times tables... lets delete the lot..simple. pabsy 26-03-05

 * Are you sure that lions and tigers can cross-breed? Do they even co-exist anywhere? Visit Uncyclopedia - you will absolutely love it. Charlie123 8 July 2005 12:52 (UTC)

Merger
Content to make an entire book on Tigons and Ligers is quite frankly absurd. This section should be merged into the WikiJunior Book of Lions or Tigers.

Confusing
"Due to conservation efforts, deliberate hybridization is prohibited in most zoos. However, it happens regularly by accident, and some private breeders try to breed ligers for novelty purposes."

I am 12, I find this confusing: deliberate hybridization, novelty purposes, conservation efforts. "Not allowed" is probably more understandable than "prohibited". "Owners" might be easier than "private breeders."

"They are instinctively social animals. Tigers, on the other hand, are instinctively solitary. The offspring of a lion-tiger share the conflicting instincts of both their parents."

I would say this sentence like this-

"Lions naturally like to live together in groups, but tigers like to live by themselves." (The second sentence doesn't make any sense to me.)

"Ligers were originally assumed to be sterile, which is true for most hybrid species such as the mule (a cross between a horse and a donkey). However, females can be fertile."

I thought "sterile" meant clean, but my father told me that it means that they can't have babies. I guess that "fertile" means they can. --Hdavis 00:19, 2 October 2005 (UTC)

Illustrations
This article is crying out for photographs. Unfortunately our options are a bit limited. All the hybrid pictures on commons reside in commons:Category:Pantherinae (commons:Category:Hybrid Cats is just a partial duplicate). Our options look to be:

The last tigon picture is ok, but the rest leave a lot to be desired, especially for a kids book. I'm still looking for more free images, but I'm not holding my breath on this one. What do you folks think? — Laura Scudder | Talk 08:37, 23 November 2005 (UTC)


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FHSST Physics/Pressure

Essay 3: Pressure and Forces

Author: 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.

= Pressure and Forces =

In the mining industry, the roof (hangingwall) tends to drop as the face of the tunnel (stope) is excavated for rock containing gold.

As one can imagine, a roof falling on one's head is not a nice prospect! Therefore the roof needs to be supported.

The roof is not one big uniform chunk of rock. Rather it is broken up into smaller chunks. It is assumed that the biggest chunk of rock in the roof has a mass of less than 20 000 kg therefore each support has to be designed to resist a force related to that mass. The strength of the material (either wood or steel) making up the support is taken into account when working out the minimum required size and thickness of the parts to withstand the force of the roof.

Sometimes the design of the support is such that the support needs to withstand the rock mass without the force breaking the roof..

Therefore hydraulic supports (hydro = water) use the principles of force and pressure such that as a force is exerted on the support, the water pressure increases. A pressure relief valve then squirts out water when the pressure (and thus the force) gets too large. Imagine a very large, modified doctor's syringe.

In the petrochemical industry, there are many vessels and pipes that are under high pressures. A vessel is a containment unit (Imagine a pot without handles, that has the lid welded to the pot that would be a small vessel) where chemicals mix and react to form other chemicals, amongst other uses.

The end product chemicals are sold to companies that use these chemicals to make shampoo, dishwashing liquid, plastic containers, fertilizer, etc. Anyway, some of these chemical reactions require high temperatures and pressures in order to work. These pressures result in forces being applied to the insides of the vessels and pipes. Therefore the minimum thickness of the pipe and vessels walls must be determined using calculations, to withstand these forces. These calculations take into account the strength of the material (typically steel, plastic or composite), the diameter and of course the pressure inside the equipment.

title: FHSST Physics/Pressure


text: Template:FHSST Physics/Pressure

title: Template:FHSST Physics/Pressure


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FHSST Physics/Pressure/Summary

= Important Equations and Quantities =

Non Si Units for pressure
PSI: PSI stands for pounds per square inch. One PSI is equal to 6,894.76 Pascals

Atm: Atm stands for Atmosphere. This is the atmospheric pressure at sea level. I atm is equal to 101,325 Pascals

Hydrostatic pressure (head pressure)
Hydrostatic pressure is the pressure due to the weight of a fluid.

p = \rho g h\, $$ where:

 * ρ (rho) is the density of the fluid (i.e., the practical density of fresh water is 1000 kg/m3);
 * g is the acceleration due to gravity (approximately 9.81 m/s2 on earth's surface);
 * h is the height of the fluid column (in metres). Other units can be used if the rest of the units used in the equation are defined in a consistent way.

It is important to realize that air is a fluid. This equation can be used to find the pressure created by air. This also means that if we are trying to find the pressure of a liquid open to the atmosphere, we must add in the atmospheric pressure to the pressure of the liquid to find the total pressure.

In fluid mechanics the unit used N/m2.

From the unit of density denominator have m3 and acceleration due to gravity have m/s2 one metre from density and other meter from g cancel to each other.then unit will remain as kg/. m2.s2 after that unit of h is meter which is multiply into kg we will got kg.m/s2 =N

Final result is N/m2

1
Pressure is force divided by area, so in a sense it does have a direction as force has a direction. The term pseudovector is sometimes used for such quantities.

title: FHSST Physics/Pressure/Summary


text: FHSST Physics/Heat and Properties of Matter


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text: FHSST Physics/Heat and Properties of Matter/Phases of Matter


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FHSST Physics/Heat and Properties of Matter/Deformation of Solids

= Deformation of solids =


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User talk:Hyperlink

= New =

= Old =

Broken link at US History
Hi Hyperlink, I tried out the links but they seem to work for me. Could you be more specific (on the Talk:US History: Contents page) which was the link that did not work for you? --Andreas 09:47, 29 Mar 2005 (UTC)

Issue Guides
There's some thoughts at Talk:Voter's Guide, but I'm still kind of experimenting and am not really sure about where to take the project and how to format it. (Please feel free to make some suggestions about formatting) So, yes, I will be doing something with organizational top-level pages of the Issue Guides, but I don't really know how to organize it yet, so I haven't done so. TUF-KAT 08:48, 26 Mar 2005 (UTC)

Book Of The Month.
Well... We got there. Thanks for nominating us. Getting BOTM made all of the hard work we put into HTBAC worthwhile. Thanks --Noogz 02:06, 1 May 2005 (UTC)

title: User talk:Hyperlink


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FHSST Physics/Heat and Properties of Matter/Ideal Gasses

= Ideal gases =

Equation of state
An ideal gas has the following equation of state:

$$ pV=\nu RT $$

Where p is pressure, V is volume, $$\nu=m/\mu$$ is number of moles observed, $$R=N_Ak_B=6.022$$ J/(mol·K) is a universal gas constant.

The gas constant has a value of 8.314 J/(mol.K) or 0.0821 (L*atm)/(mol*K)

Pressure of a gas
let us consider n mole of an ideal gas is placed inside the cubical vessel of side l let the mass of each molecule is m and let it moves with velocity c1


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FHSST Physics/Heat and Properties of Matter/Temperature

= Temperature =


text:
FHSST Physics/Heat and Properties of Matter/Thermal Properties

= Thermal Properties of Materials = R

Specific heat capacity
Conversion of macroscopic energy to microscopic kinetic energy thus tends to raise the temperature, while the reverse conversion lowers it. It is easy to show experimentally that the amount of heating needed to change the temperature of a body by some amount is proportional to the amount of matter in the body. Thus, it is natural to write

Riaan Note: not sure about the equation numbers, sticking to those in the PDF released the 1st of March 2005

where $$M$$ is the mass of material, $$\Delta Q$$ is the amount of energy transferred to the material, and $$\Delta T$$ is the change of the material's temperature. The quantity $$C$$ is called the specific heat of the material in question and is the amount of energy needed to raise the temperature of a unit mass of material one degree in temperature. $$C$$ varies with the type of material. Values for common materials are given in table 22.2.

Specific latent heat
When a material changes phases from solid to liquid, or from liquid to gas, a certain amount of energy is involved in this change of space.

First law of thermodynamics
We now address some questions of terminology. The use of the terms heat and quantity of heat to indicate the amount of microscopic kinetic energy inhabiting a body has long been out of favor due to their association with the discredited caloric theory of heat. Instead, we use the term internal energy to describe the amount of microscopic energy in a body. The word heat is most correctly used only as a verb, e. g., to heat the house. Heat thus represents the transfer of internal energy from one body to another or conversion of some other form of energy to internal energy. Taking into account these definitions, we can express the idea of energy conservation in some material body by the equation

where $$\Delta E$$ is the change in internal energy resulting from the addition of heat $$\Delta Q$$ to the body and the work $$\Delta W$$ done by the body on the outside world. This equation expresses the first law of thermodynamics. Note that the sign conventions are inconsistent as to the direction of energy flow. However, these conventions result from thinking about heat engines, i. e., machines which take in heat and put out macroscopic work. Examples of heat engines are steam engines, coal and nuclear power plants, the engine in your automobile, and the engines on jet aircraft.


text:
FHSST Physics/Heat and Properties of Matter/Summary

= Important Equations and Quantities =


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FHSST Physics/Electrostatics

= Electrostatics = The study of the effects of static charges. These charges are produced by too many or too few electrons. Electrons, in turn, are the most movable charge forms, and are found spread around the positive charge in the nucleus of the atom. If there are too few electrons, a positive charge will appear. If there are too many, a negative charge will appear. If the charges are in balance, then no charge is detected. The electron is a fundamental particle, and has interesting characteristics, besides the charge, it spins, and this spin gives rise to a magnetic field, much like an electric motor.

The decision to call some charges positive and others negative was arbitrary. This gives rise to a potential misunderstanding when charges are moving in a circuit. This will be deferred to a later section, on electricity.

The fundamental study of these charges in isolation can best be observed by experiment. If we generate a charge by friction, say fur on hard rubber, or silk on glass, we can transfer the charge to a small ball coated with aluminum foil. This ball is held up by a thin thread. Nylon or silk work well.

If however the air is damp, the charge will leak off the metal balls.

Two such balls, charged by the same static source will repel each other. If we use two different sources, they will attract each other. Maybe.

We can measure the amount of attraction or repulsion by a simple means. If we measure the weight of these balls, then we know that they are attracted by the earth. Other forces on these balls will cause them to swing away from hanging straight down.

By measuring the movement, or the angle, we can figure out what tiny force is pushing the ball away from the vertical. The tangent function is the perfect candidate to resolve this problem. ((add more text and diagrams here))

It would be nice to find out what happens when some charge is removed. It turns out, that if we use a third ball, of the same size as the other two, we can exactly halve the charge.

From this we can see that we can measure the effects of increasing distance, and decreasing charge.

title: FHSST Physics/Electrostatics


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FHSST Physics/Electrostatics/Definition

= What is Electrostatics? = Electrostatics is the study of electric charge which is not moving i.e. is static.

title: FHSST Physics/Electrostatics/Definition


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FHSST Physics/Electrostatics/Charge

= Charge =

All objects surrounding us (including people!) contain large amounts of electric charge. Charge can be negative or positive and is measured in units called coulombs (C). Usually, objects contain the same amount of positive and negative charge so its effect is not noticeable and the object is called electrically neutral. However, if a small imbalance is created (i.e. there is a little bit more of one type of charge than the other on the object) then the object is said to be electrically charged.

Some rather amusing examples of what happens when a person becomes charged are for example when you charge your hair by combing it with a plastic comb and it stands right up on end! Another example is when you walk fast over a nylon carpet and then touch a metal doorknob and give yourself a small shock (alternatively you can touch your friend and shock them!)

Charge has 3 further important properties:

 * Charge is always conserved. Charge, just like energy, cannot be created or destroyed.
 * Charge comes in discrete packets. The smallest unit of charge is that carried by one electron called the elementary charge, e, and by convention, it has a negative sign ( $$e = -1.6 \times 10^{-19} \rm {C} $$ ).
 * Charged objects exert electrostatic forces on each other. Like charges repel and unlike charges attract each other.

You can easily test the fact that like charges repel and unlike charges attract by doing a very simple experiment. Take a glass rod and rub it with a piece of silk, then hang it from its middle with a piece string so that it is free to move. If you then bring another glass rod which you have also charged in the same way next to it, you will see the rod on the string turn away from the rod in your hand i.e. it is repelled. If, however, you take a plastic rod, rub it with a piece of fur and then bring it close to the rod on the string, you will see the rod on the string turn towards the rod in your hand i.e. it is attracted!

What actually happens is that when you rub the glass with silk, tiny amounts of negative charge are transferred from the glass onto the silk, which causes the glass to have less negative charge than positive charge, making it positively charged. When you rub the plastic rod with the fur, you transfer tiny amounts of negative charge onto the rod and so it has more negative charge than positive charge on it, making it negatively charged.

Conductors and Insulators
Some materials allow charge carriers to move relatively freely through them (e.g. most metals, tap water, the human body) and these materials are called conductors. Other materials, which do not allow the charge carriers to move through them (e.g. plastic, glass), are called non-conductors or insulators.

 Aside: As mentioned above, the basic unit of charge, namely the elementary charge, e, is carried by the electron. In a conducting material (e.g. copper), when the atoms bond to form the material, some of the outermost, loosely bound electrons become detached from the individual atoms and so become free to move around. The charge carried by these electrons can move around in the material! In insulators, there are very few, if any, free electrons and so the charge cannot move around in the material.

If an excess of charge is put onto an insulator, it will stay where it is put and there will be a concentration of charge in that area on the object. However, if an excess of charge is put onto a conductor, the charges of like sign will repel each other and spread out over the surface of the object. When two conductors are made to touch, the total charge on them is shared between the two. If the two conductors are identical, then each conductor will be left with half of the total charge.

title: FHSST Physics/Electrostatics/Charge


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FHSST Physics/Electrostatics/Electrostatic Force

= Electrostatic Force =

As we now know, charged objects exert a force on one another. If the charges are at rest then this force between them is known as the electrostatic force. An interesting characteristic of the electrostatic force is that it can be either attractive or repulsive, unlike the gravitational force which is only ever attractive. The relative charges on the two objects is what determines whether the force between the charged objects is attractive or repulsive. If the objects have opposite charges they attract each other, while if their charges are similar they repel each other (e.g. two metal balls which are negatively charged will repel each other, while a positively charged ball and negatively charged ball will attract one another).

It is this force that determines the arrangement of charge on the surface of conductors. When we place a charge on a spherical conductor the repulsive forces between the individual like charges cause them to spread uniformly over the surface of the sphere. However, for conductors with non-regular shape there is a concentration of charge near the point or points of the object.

This collection of charge can actually allow charge to leak off the conductor if the point is sharp enough. It is for this reason that buildings often have a lightning rod on the roof to remove any charge the building has collected. This minimises the possibility of the building being struck by lightning.

This spreading out of charge would not occur if we were to place the charge on an insulator since charge cannot move in insulators.

Coulomb's Law
The behaviour of the electrostatic force was studied in detail by Charles Coulomb around 1784. Through his observations he was able to show that the electrostatic force between any point two point charges is directly proportional to their magnitude and inversely proportional to the square of the distance between the charges. Thus if F is the force, Q1 and Q2 are the charges and R is their distance apart. then the law is summarized as                    F=(Q1Q2)/(R^2) where Q1 is the charge on the one point-like object, Q2 is the charge on the second, and r is the distance between the two. Remember her F is the mutual force that acts on both the charges exerted by one another as in Newtons third law of motion.

The magnitude of the electrostatic force between two point-like charges is given by Coulomb's Law:

and the proportionality constant k is called the electrostatic constant. We will use the value

$$\begin{matrix}k=8.99\times10^9 {\rm N\cdot m^2/C^2.}\end{matrix}$$

The value of the electrostatic constant is known to a very high precision (9 decimal places). Not many physical constants are known to as high a degree of accuracy as k.

 Aside: Notice how similar Coulomb's Law is to the form of Newton's Universal Law of Gravitation between two point-like particles:

$$\begin{matrix}F_G=G\frac{m_1m_2}{r^2},\end{matrix}$$

where m1 and m2 are the masses of the two particles, r is the distance between them, and G is the gravitational constant. It is very interesting that Coulomb's Law has been shown to be correct no matter how small the distance, nor how large the charge: for example it still applies inside the atom (over distances smaller than $$10^{-10}\rm {m}$$

Let's run through a simple example of electrostatic forces.

Worked Example 59 Coulomb's Law I
Question: Two point-like charges carrying charges of +3×10−9 C and −5×10−9 C are 2 m apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.

Answer:

Step 1 :

First draw the situation:

Step 2 :

Is everything in the correct units? Yes, charges are in coulombs [C] and distances in meters [m].

Step 3 :

Determine the magnitude of the force: Using Coulomb's Law we have

$$\begin{matrix}F &=& k\frac{Q_1Q_2}{r^2}\\&=& (8.99\times10^{9} {\rm N\cdot m^2/C^2}) \frac{(+3\times10^{-9}{\rm C})(-5\times 10^{-9}{\rm C})}{(2 {\rm m})^2}\\ &=&-\frac{(8.99)(3)(5)}{4}(10^{-9})\frac{(N\cdot m^2/C^2\!\!\!/)(C^2)}{m^2}\\&=& -3.37\times10^{-8} \rm N\end{matrix}$$

Thus the magnitude of the force is $$3.37\times10^{-8}\rm {N}$$. The minus sign is a result of the two point charges having opposite signs.

Step 4 :

Is the force attractive or repulsive? Well, since the two charges are oppositely charged, the force is attractive. We can also conclude this from the fact that Coulomb's Law gives a negative value for the force.

Next is another example that demonstrates the difference in magnitude between the gravitational force and the electrostatic force.

Worked Example 60 Coulomb's Law II
Question: Determine the electrostatic force and gravitational force between two electrons 1Aring apart (i.e. the forces felt inside an atom) Answer: Å Step 1 :

First draw the situation:

Riaan Note: this image is faulty, recreate

Step 2 :

Get everything into S.I. units: The charge on an electron is $$-1.60\times10^{-19}\rm C$$, the mass of an electron is $$9.11\times10^{-31}\rm kg$$, and 1 Å = $$1\times10^{-10}\rm m$$

Step 3 :

Calculate the electrostatic force using Coulomb's Law:

$$\begin{matrix}F_E&=&k\frac{Q_1 Q_2}{r^2}=k\frac{e\cdot e}{1\AA^2}\\&=&(8.99\times10^{9}\rm N\cdot m^2/C^2)\frac{(-1.60\times10^{-19}{\rm C})(-1.60\times10^{-19}\rm C)}{(10^{-10}\rm m)^2}\\&=&2.30\times10^{-8}\rm N\end{matrix}$$

Hence the magnitude of the electrostatic force between the electrons is $$2.30\times10^{-8}\rm N$$. (Note that the electrons carry like charge and from this we know the force must be repulsive. Another way to see this is that the force is positive and thus repulsive.)

Step 4 :

Calculate the gravitational force:

$$\begin{matrix}F_E&=&G\frac{m_1m_2}{r^2}=G\frac{m_e\cdot m_e}{(1\AA)^2}\\&=&(6.67\times10^{-11}{\rm N\cdot m^2/kg^2})\frac{(9.11\times10^{-31}{\rm C})(9.11\times10^{-31}{\rm kg})}{(10^{-10}{\rm m})^2}\\&=&5.54\times10^{-51}{\rm N}\end{matrix}$$

The magnitude of the gravitational force between the electrons is $$5.54\times10^{-51}\rm N$$

Notice that the gravitational force between the electrons is much smaller than the electrostatic force. For this reason, the gravitational force is usually neglected when determining the force between two charged objects.

We mentioned above that charge placed on a spherical conductor spreads evenly along the surface. As a result, if we are far enough from the charged sphere, electrostatically, it behaves as a point-like charge. Thus we can treat spherical conductors (e.g. metallic balls) as point-like charges, with all the charge acting at the centre.

Worked Example 61 Coulomb's Law: Challenge Question
Question: In the picture below, X is a small negatively charged sphere with a mass of 10 kg. It is suspended from the roof by an insulating rope which makes an angle of 60o with the roof. Y is a small positively charged sphere which has the same magnitude of charge as X. Y is fixed to the wall by means of an insulating bracket. Assuming the system is in equilibrium, what is the magnitude of the charge on X?

Answer: How are we going to determine the charge on X? Well, if we know the force between X and Y we can use Coulomb's Law to determine their charges as we know the distance between them. So, firstly, we need to determine the magnitude of the electrostatic force between X and Y.

Step 1 :

Is everything in S.I. units? The distance between X and Y is $$50{\rm cm}=0.5\rm m$$, and the mass of X is 10 kg.

Step 2 :

Draw the forces on X (with directions) and label.

Step 3 :

Determine the magnitude of the electrostatic force (FE). Since nothing is moving (system is in equilibrium) the vertical and horizontal components of the forces must cancel. Thus

$$\begin{matrix}F_E=T\sin(60^o) \qquad\qquad F_g=T\sin(60^o).\end{matrix}$$

The only force we know is the gravitational force Fg = mg. Now we can calculate the magnitude of T from above:

$$\begin{matrix}T=\frac{F_g}{\sin(60^o)}=\frac{(10{\rm kg})(10\rm m/s^2)}{\sin(60^o)}=1155{\rm N}.\end{matrix}$$

Which means that FE is:

$$\begin{matrix}F_E=T\cos(60^o)=1154\rm N\cdot \cos(60^o)=577.5\rm N\end{matrix}$$

Step 4 :

Now that we know the magnitude of the electrostatic force between X and Y, we can calculate their charges using Coulomb's Law. Don't forget that the magnitudes of the charges on X and Y are the same: $$|Q_{\rm X}|=|Q_{\rm Y}|$$. The magnitude of the electrostatic force is

$$\begin{matrix}F_E &=& k\frac{|Q_{\rm X}Q_{\rm Y}|}{r^2}= k\frac{Q_{\rm X}^2}{r^2}\\ |Q_{\rm X}|&=&\sqrt{\frac{F_E r^2}{k}}\\&=&\sqrt{\frac{(577.5{\rm N})(0.5{\rm m})^2}{8.99\times10^{9}{\rm N\cdot m^2/C^2}}}\\&=&1.27\times10^{-4}{\rm C}\end{matrix}$$

Thus the charge on X is $$-1.27\times10^{-4}{\rm C}$$