Presentations for Mr Skidmore

These are the groups:

Group1: Michaela, Tom W, Ashley, Greg doing HADRONS

Group 2: John, Andrew, Martin W, Tom M doing LEPTONS

Group 3: Naomi, Hannah, Martin F, Oliver doing QUARKS

Your sheets tell you what he wants in the presentations.

I'm saying:

• Powerpoints with white background
• Font size 40 or 44 is always good.
• Do NOT copy and paste things you don't understand. I can always tell!
• Put some whiteboard questions at the end to check that the rest of the class were awake during your presentation.

I'm saying you MUST include a handout!

I've put some worksheets that I've used with classes before on the school network:

Shared area; Read only; Year 12; Science; Physics; Particle physics.

They might help or they might not...

I will ask you in class next week about your progress and make sure that everyone in the group has a job. You can ask me about stuff you don't understand.

I will make you put your powerpoint onto my area of the network on Weds 8 October so that I can check work has been done.

Annihilation and pair production

Everyone has seen this equation. It's not strictly on the syllabus this year: ie you don't need to do calculations with it. However, it explains a lot. It tells us that pure energy (E) can be changed into particles of mass m. Or particles of mass m can be destroyed and turned into energy E.
This is annihilation. A positron and an electron are destroyed and two gamma ray photons are made.

• Positive and negative because charge has to add up to zero. The energy that is made has no charge.
• Equal and opposite motion so that momentum before is zero.
• Two gamma ray photons so that momentum is zero after the annihilation.
  

In reverse, two gamma ray photons can come together to make a positron and an electron. This is pair production.

Cladding

The question on the January exam about curved bit of glass with 3 rays going through (P,Q,R) raised some interesting ideas about cladding. Cladding is a layer of transparent material wrapped around the outside of the central fibre (the core). The purpose of the cladding is to stop the central glass core getting scratched. Scratched cores leak light more easily because it changes the angle of the outside wall of the glass and thus changes the angle of incidence.

Cladding makes it more difficult for the light to stay in the glass:

We will be using this equation. But note that at the critical angle, the second angle is zero.When there is no cladding, there is air around with refractive index of n = 1. So With the cladding, we get this equation:

Simple calculation will show that the critical angle becomes bigger. There are now fewer angles at which the light can hit the glass for TIR, so it is much easier for light to escape into the cladding.

Finally, be careful how you define critical angle.

• The angle of incidence for which the angle of refraction is 90 degrees.
• The minimum angle of incidence for which you get total internal reflection.
  
  
  

Making IV characteristics

In this circuit, a varaible resistor is being used to change the voltage and current through a bulb. This is not a good circuit to use for an IV characteristic. Here's why:

• Suppose the bulb has a resistance of 20 Ohms and we use a 6 Volt battery.
• The variable resistor can go from zero up to a maximum of 20 Ohms.
• When the variable resistor has zero Ohms, then the bulb will take all of the energy and the voltmeter reading will be at a maximum of 6 Volts.
• When the variable resistor has 20 Ohms, then the bulb and the resistor will have the same resistance and they will have equal shares of the energy. Voltmeter will read 3 Volts.
• Hence by using a variable resistor like this, there is a limited range of voltages you can use. In this case it is between 3V and 6V. You can't get down to 0 V.
• I suppose that if you use a variable resistor with a maximum resistance massively bigger than the bulb, you'd get a better range.

It is much better to use this set up, called a potentiometer. All 3 connections on the variable resistor (rheostat) are used and you can have all the voltages from 0V up to 6V.
This is the set up we used in class. (see IV characteristics booklet). We used a version of a potentiometer when we had the wire along the meter stick in the observation room so that every 10 cm represented 1V.

The plum pudding model

Before Geiger and Marsden did the important gold leaf experiment for Ernest Rutherford, the best idea about the inside of an atom was called the "Plum Pudding" model.

• They had just discovered the electron so they knew that there had to be negative particles inside the atom.
• If there were negatives, there had to be positive as well.
• They imagined the positive charge as being thinly smeared all over the atom. There were no definite positive particles in this model.
• The positive is supposed to be like the dough in a Christmas Pudding, and the electrons are like the raisins.

Any alpha particles fired at it should go straight through because

• alpha particles are much bigger and heavier than electrons so they would knock them out of the way.
• the positive charge is so thinly spread that there is no chance of repelling the doubly positive alpha particle.

Then the gold leaf experiment was done and Rutherford invented the Solar System model with the positive nucleus and orbitting electrons. The Plum Pudding model was consigned to the dustbin of history.

Rutherford scattering

In this experiment, a very narrow beam of alpha particles was fired at a very thin piece of gold foil.

• The thick sheets of lead are to stop the beam of alpha spreading out like the beam on a car headlight. They keep the beam narrow. The arrangement is called a collimator.
• The detector is moved around the outside of vacuum chamber and readings taken for different angles.
• The beam has to be narrow because this gives a very definite spot on the foil from which to measure the angle, as shown on the diagram below: The moment of genius was when it was decided to move the detector round onto the same side as the alpha source. No one in their right mind would expect to find alpha particles on this side, but that is excatly what they did discover. It's called backscattering. We can tell that the nucleus is positive because the positive alpha particles are repelled.
• We can tell that the nucleus is heavy by thinking of what happens in snooker when the cue ball hits a coloured ball. The cue ball stops but the coloured ball moves on. Momentum is transferred here because both balls have the same mass. However, the alpha particle bounces back and the nucleus is unmoved. This means that the nucleus is much heavier.
  
  

What happens when a photon carries too little energy?

They seem to be asking a lot of questions about what happens when a photon arrives carrying too little energy for an electron to reach the first level.

The correct answer is that NOTHING HAPPENS.

The electron does not jump up half way and then fall back. It just sits there as if nothing has happened. It does not absorb the photon at all.

Work function

Remember that we model an electron as being at the bottom of the well. In the model, energy is needed to work against gravitational attraction in the same that energy is needed to work against eectrostatic attraction between the positive nucleus and the negative electron.

The work function is the minimum energy required to escape from the atom.

It is possible to give more energy. That just means that the escaping electrons have extra kinetic energy.

That extra kinetic energy is measured in a circuit like this:

Notice the weird extra battery positioned at the top of the circuit. It is the wrong way round. Any electrons that are realeased by the light are pulled back across the gap, stopping the flow so the ammeter reads zero.

The more kinetic energy given to the electons the bigger the backwards voltage V needed to stop the current. This is called the stopping potential.

This is the graph that is produced by this equipment. Change the frequency of the light and measure the stopping potential in eV. Turn that into Joules to give you the kinetic energy. The threshold frequency is obvious.

How to describe the composition of baryons

Baryons are made of 3 quarks. If the question asks for the general composition, it doesn't want specific details of up, down or strange. So you need to write "3 quarks (qqq)", where q stands for any quark.

Clearly, this is how you would have to write a baryon anti-particle:

Links to past papers

You can find past papers via this link to the AQA website:

http://www.aqa.org.uk/qual/gceasa/phya_assess.html


The "brockbankrevision" website deals with Units 1, 2 and 3, which will need revising for the Upper Sixth synoptic module 10.

This blog contains a link to the Upper Sixth site for Units 4 and 5.

IV characteristic for a diode

The mark schemes vary on the points they cover. Here's the most recent picture:

If you put a large negative voltage on a diode, eventually you get breakdown. It conducts almost perfectly ie it's as if nothing is getting in the way of the electricity so the current is very high.

IV characteristic for filament bulb

The most recent mark scheme makes these points:

• For low voltages, the line is straight so current is proportional to voltage
• As the current increases, the filament wire heats up
• This increases the resistance
• Increasing the voltage will push more current through, but because the resistance has increased, the current will not go up as much for an equal increase in voltage
• The same thing happens if you put current through the bulb in the opposite direction so the characteristic is symmetrical.
  

Link to Upper Sixth site

This is for anyone from the Upper Sixth using the Lower Sixth blog to revise for the Synoptic Paper:

www.brockostressline.blogspot.com

Baryon number and lepton number

Baryons are not fundamental particles because they are made of smaller pieces (quarks).

A baryon is made from 3 quarks. Baryon number is conserved in an interaction. In other words, if you have a baryon before the interaction, then you need to have a baryon after the interaction.

The odd bit is that an antiparticle like an antiproton has a baryon number of -1. So if a proton and an antiproton interact, the overall baryon number is +1 + (-1) = 0, so the total baryon number after the interaction must be zero as well.

Leptons ARE fundamental particles. They are not made of smaller particles. Lepton number is conserved in an interaction.

Strangeness is conserved in the strong and electromagnetic interactions. Strangeness is strictly not conserved in the weak interaction but the exam keeps coming up with the SPECIAL CASE where you have zero strangeness before and zero strangeness after, so the mark scheme says strangeness is conserved, even though it's a weak interaction.

EMF

As far as the exam is concerned, you are best to define the EMF as the battery output voltage when there is ZERO current through the battery.





Once the current starts to flow round the circuit, current also has to pass through the INTERNAL RESISTANCE of the battery, heating it up. This wastes energy so the final output p.d. of the battery is reduced.

This equation sums up the paragraph above. It says that the output voltage is equal to the EMF minus the p.d. lost heating up the internal resistance.

This versionof the equation is NOT on the data sheet.

Applying Newton's Laws to terminal velocity

Think about the freee fall parachutist example:

At first, the only force acting on her is her weight. She has no air resistance at first because her initial speed is zero.

Hence she has a resultant force downwards. Newton's Second Law states that rate of change of momentum is proportional to reultant force, so she accelerates.

As her velocity increases in magnitude, her weight remains constant but her air resistance increases. This means that her resultant force is still downwards but decreases in size. She still accelerates, but she does not gain as much speed each time.

Eventually, the air resistance is the same size as the weight. Resultant force is zero. Newton's First law says that objects continue with uniform motion (straight line, constant speed to you) unless an external resultant force acts. No resultant force thus means constant speed.

When she opens her parachute, her weight remains the same but the air resistance increases greatly. This gives her a resultant force upwards. The second part of Newton's Second Law states that the change in momentum takes place in the direction of the resultant force. If you have momentum downwards but the change is upwards, you must slow down.

As she slows down, her air resistance decreases, but her weight stays the same. Hence her resultant force upwards decreases. She continues to slow down but not by as much each time.

Eventually, she has slowed so much that the size of the air resistance is the same as her weight. Once again she goes at constant speed by Newton's First Law. However, this time it is a smaller terminal velocity so she does not get hurt when she hits the ground.

More about the ball on the curve

In the ball on the curve problem, you were asked to calculate the speed of the ball when it reached the bottom of the curve, given the vertical height of the curved section:

The temptation would be to use suvat by saying that the acceleration was 9.81m/s2 and that the distance s was given by h.

This would be the wrong thing to do because 9.81 is only the acceleration at A.

At B, it is the dotted component that points in the direction of motion that causes the acceleration. This is smaller than the weight, so the acceleration has reduced.

And at the end, when the ball is moving horizontally, it no longer accelerates at all.

So to solve the problem, we use an energy argument:

Say that gravitational potential energy at the top = kinetic energy at the bottom

assuming that no energy has been wasted by a transformation into thermal energy by friction.

Thus





And

Falling from the Eiffel Tower

When you are at the top of the Eiffel Tower you have gravitational potential energy.





As you fall, your gravitational potential is transformed into kinetic energy.






Now assuming that there is no air resistance to transform any of the kinetic energy into thermal (heat) energy due to friction, your gravitational potential energy at the top will be equal to your kinetic energy at the bottom. This is because all of the energy will have been transferred.

You would get this equation:





But now we can cancel the mass:






Finally we can rearrange what is left to get an equation for the velocity v when you hit the ground:


Notice that v does not depend on the mass of the object, because that cancelled out. Hence all objects will hit the ground at the same speed if there is no air resistance.

Constant speed up a slope

This is a picture of a ball going up a slope at a constant speed.

If it going at constant speed, then

driving force = counter force

Driving force is F, because it is pointing in the direction of motion, by Newton's Second Law which says "Rate of change of momentum is proortional to the resultant force and that change in momentum is in the direction of the resultant force".

However, although gravity is making it harder for the ball to go up the slope, the weight W is NOT the counter force because it is not in line with the motion.

We need to take components of W:

Notice that this has been done using the method set out in my previous post.

Just in case you were wondering, P stands for parallel to the slope and N stands for normal to the slope.








Lastly we need to look in detail at the components:



Notice that the bottom angle is the same as the angle for the slope.

By Pythagoras,

P = W sin20




P is the counter force because it is in exactly the opposite direction to the motion (and thus the change in momentum)







Finally, since

Driving force = counter force

F = P

F = W sin20


An example of this type of question might say:
"An object of mass 50 kg travels up a slope of angle 20 degrees at a constant speed of 5 m/s. What is the power output needed to maintain the motion?"

You will of course remember that power output P = Fv. It's on the data sheet.

Hence we already know v, we have to find F

F = W sin 20 but weight W = mg = 50 x 9.81 = 490.5 Newtons

F = 490.5 x sin20 = 167.8 Newtons

Hence power output = Fv = 167.8 Newtons x 5 m/s = 839 Watts to 3 sig fig

One final point: If you have had to put your calculator into RADIANS for maths, make sure you put it back into DEGREES.

past papers

This should be a working link to the past papers section of the AQA website. We are physics syllabus A and we are studying module 2 at the moment:

http://www.aqa.org.uk/qual/gceasa/phya_assess.html