J6 Lessons
RM 10/09/04 J6
Distance, displacement, speed and velocity were discussed. Vector quantities have a direction aswell as a size (magnitude).
Accleration = Change in velocity/Time taken for change
It is the rate of change of velocity. You can accelerate without changing speed, so long as you are changing direction (as in constant circular motion).
HW Do the practise questions for chapter 6 on P80 of the text book.
RM 14/09/04
Acceleration was discussed and notes taken. Acceleration is a vector and therefore has a direction. Even for a one dimensional calculation you must decide which direction is positive and which negative. a = (v-u)/t gives you an average acceleration. Most situations you will come across will involve a constant acceleration.
Motion graphs were discussed. The gradient of a displacement vs time graph gives you the velocity. The gradient of a velocity vs time graph gives you the acceleration. The displacement vs time graph contains all the infromation about an object's motion.
It is possible to reconstruct the displacement vs time graph from the speed vs time or even the accel vs time graph, but extra information is required and it's a deal tougher.
HW Look through the bouncing ball chapter and ensure that you are happy with the motion graphs as drawn (badly) by me on the board.
RM 17/09/04
We were struck slightly by the Jewish holiday, but soldiered on nevertheless. Equations of motion were derived. This is in the text book, make sure you are happy with it absentees. We practised doing questions based on motion graphs. The important thing is to remember the link between the gradient of a displacement/time graph and the values on a velocity/time graph. Similarly, the gradient of a velocity/time graph and the values on an acceleration/time graph.
We looked at the idea of horizontal projection. Just because something is moving horizontally, doesn't have any effect on its motion in a vertical plane. Thus a ball which is projected horizontally as another is dropped vertically will hit the ground at the same time. The equations of motion can be used for horizontal and vertical motion, but the 2 must be kept seperate in your calculations. More of this, next time.
HW Do questions 9.1-9.5 practising the use of the equations of motion.
RM 21/09/04
A very hard problem to start with. Introducing the idea of resolving vectors (that just means seperating out the horizontal and vertical bits). In this case, the horizontal and vertical bits of the velocity of an object. All projectile problems can be solved by keeping horizontal and vertical motion seperate and using seperate x,u,v,a,t equations to deal with them. The time is the only thing which is the same for both sets of equations.
To calculate how long a projectile takes to hit the floor again, you must know how fast it was projected in an upward direction. When it hits the floor again, its displacement is zero. You therefore know u and x. t can be calculated by solving the relevant equation of motion. HW was taken in and will be marked. Some claimed to have struggled, we may do some more practise on this.
RM 24/09/04
Some more practise on motion equations. Answers must be given to the same degree of accuracy as given information. Units must be included. There are often different ways to solve a given problem, so make sure that your method is shown clearly.
Here's that Powerpoint file that I never printed out. Have a go, if still in need of practise.
We looked very quickly at Newton's 1st law which says that when forces acting on an object are balanced, it's velocity remains the same.
HW Questions from Muncaster A level big text book P25 Qs 1-6.
RM 28/09/04
Apparently, the homework was tricky. It turned out to be the resolving of vectors again, done by trigonometry. Once you know the vertical and horizontal components of the initial velocity of an object, you keep all horizontal and vertical equations seperate. Commonly, you work out the time an object spends in the air using the vertical equations. You can then work out how far it has gone within that time horizontally.
We performed an experiment to see how the acceleration of a body varies with the force you apply to it. The experiment showed (roughly) that the acceleration of a body is proportional to the force that you apply to it. We'll look at varying the mass next time.
HW: Revise for a test which will consist of Assessment Qs 3,4,5,6 from the back of the book. You are going to have to answer them in exam type conditions. Those away on Fri, complete for HW.
RM 01/10/04
We sat the test consisting of assessment Qs 3,4,5,6. This took a while. Then we went on to do an air track experiment which varied the mass of an object and kept the force on it constant. This caused the acceleration of the object to change.
HW Complete graphs of acceleration against mass, and acceleration against (1/mass) for next time.
RM 05/10/04
We started a large set of practise motion questions to try and get you a bit more used to them after the test results weren't spectacular. We'll finally go through the test next Tues, as we will be incomplete again on Friday. This Friday, perhaps a bit more practise on motion Qs, and onto free body force diagrams.
We went into Newton's 3rd law. When body A exerts a force on body B, body B always exerts a force of the same type, in the opposite direction on body A. There is a common misconception that some pairs of forces are Newton's 3rd law pairs. (e.g. a person's weight and the normal reaction force exerted on a person by the ground.) They are not! A person's weight is a force exerted on the person by the Earth (gravitational).The equal and opposite force exerted on the Earth by the person is also gravitational and equal in size to the person's weight. Whilst such a force has a large effect on the small mass of a person, the Earth is largely unaffected.
RM 8/10/04
We were scuppered by a security alert in the latter part of the lesson, but not before some more practise at motion questions. The main faults are getting confused as to what equations to use, muddling vertical and horizontal motion and not showing working. We'll go through the test in detail on Tuesday when everyone is back and then we'll get on with the syllabus.
HW Hand in the first 9 practise motion Qs next lesson.
RM 12/10/04
Oh dear. Went through the little test on motion Qs. I hope we have them largely covered now which I will check by marking Qs 1-9 which were handed in today.Then onto free body force diagrams. The main trouble with these is that they are so boring they make you want to gnaw your own limbs off. The basic point behind one of these diagrams is that it must be drawn showing the forces on one body only, making clear their directions and types. Do not confuse balanced forces acting on a single body (making it a Newton 1 body staying at constant velocity) with a Newton's third law pair. These pairs are of the same size, but act on different bodies in opposite directions and are of the same type.
We started to look at resolving vectors - this means working out the resultant size and direction of a number of different vectors working at different angles to each other. This can be done with any vector quantity.
HW Do free body force diagram Qs 17.1-5.
RM 15/10/04
We looked more at resolving vectors. If you need to combine 2 or more vectors, the resultant effect can always be found by joining the vectors nose to tail one after another. This is often relevant when combining the different forces acting on a body which you have identified using a free body force diagram.
In general, chose a direction to be positive in each dimension. Then add the components of all the forces together in each dimension. The resultant vector will give you the resultant force on the body. Its size can always be found using Pythagoras, and its direction using trig. (be sure to quote where the angle you give for the direction is measured from - on maps etc. they tend to be given as bearings from North).
HW Example questions on vectors on stapled sheets.
RM 19/10/04
You were confused by the more complex free body force diagrams. Make sure that only the forces involved on one body at a time are drawn on each force diagram. A situation diagram can help show what is going on. The forces on each individual body may be equal and opposite, but can never be a Newton 3 pair. We did some more practice at resolving vectors with Chap 19 questions. We touched briefly on fluid forces, looking first of all at upthrust.
HW None!
RM 22/10/04
We finished off fluid forces. The other 2 are: drag - force opposite to direction of travel which increases as you go faster, and lift - force due to aeroplane wing shape in which fluid has to travel faster over one side than the other and a pressure difference is caused. It is perpendicular to the direction of travel.
Static friction is like a contact force which acts tangentially along surfaces and must be taken into account when an object is sitting on a slope for example.
HW Qs 1B from P10 of Muncaster.
RM 02/11/04
RM absent. You looked at questions from chapters 22-24.
RM 05/11/04
We looked at the beam balance to emphasise the idea that: An object in equilibrium has no resultant moment on it about any point, and no resultant force either.
Most problems involving static, solid structures can by solved by these 2 facts.
HW Set of Qs on moments, couples and statics.
P.S. Those with missing bits get them to me on Monday so I can say nice things about you to you parents.
RM 09/11/04
Looking to finish off statics we went through some of the questions set for HW. They were on the fiendishly hard end of the scale and quite hard work. However, the principals were the same as for all the most simple questions. You know 2 things for a body that is still. 1. The forces up/down and left/right on it must be balanced 2. The twisting forces clockwise/anticlockwise must be balanced.
Using these 2 facts in one order or another will always allow you to solve your statics problem.
To practise for the test, I suggest looking at Muncaster chapter 4. Some good worked examples here and Questions 4A are all suitable to look at. Some other suitable questions can be found here.
HW Revise for a smallish test using past paper questions on statics and resolving vectors. Work hard for this please, I classify this as important. Also, anyone with continuing gaps will be yellow carded at the end of this week. That is all.
RM 12/11/04
We sat the small test on vectors and moments.
Then we started momentum as a topic. Momentum = Mass times velocity. Collisions were observed on the air track using light gates to measure speeds.
Calculate the momentum before and after each of the first 2 collisions. Comment on the results explaining any problems with the experiment.
RM 16/11/04
We looked at a third case of 2 bodies exploding away from each other after initially being at rest. When the 2 bodies were of the same mass, they exploded away at the same speed. When one body was more massive than the other, it carried less speed. This is because the initial total momentum was zero. The final momentum of the 2 bodies must be equal and opposite so the final total momentum is zero too.
The force experienced by each body during an explosion or a collision must be equal and opposite (Body A exerts a force on Body B, then body B exerts an equal and opposite force on body A). The force would only be exerted for a short space of time and may well vary over that time. The longer the force is exerted for, the larger an effect it will have on the bodies it is acting on. (and the larger the force the greater effect obviously.)
Newton's 2nd law can be stated: F = ma
but: a = (v-u)/t
which means: F = m(v-u)/t
and: F = (mv-mu)/t mv-mu = change in momentum of a body
so: Force = Rate of change of momentum
also: Ft = mv - mu
The force on an object multiplied by the time the force is exerted for equals the change in momentum of the object.
Force times time, or Ft is a quantity called impulse.
Impulse = Change in momentum - this is another way of stating Newton's 2nd
The units of impulse are Newton seconds (Ns). They are totally equivalent to kilogram metres per second (kgm/s).
HW Do chapter 27 practise questions at the back of the book. Cover work on Friday will consist of kinetic energy calculations and looking at elastic and inelastic collisions.
RM 19/11/04
RM absent at CCF. Those present worked out the kinetic energy present before and after in each of the collisions observed before to test the conservation of momentum. Energy is lost in almost all collisions but momentum is always conserved. Energy cannot be gained in a collision unless there is an external power source involved somehow. A loss free collision is called an elastic collision. If some KE is lost during the collision then it is called inelastic. Sheet of calculations on elastic and inelastic collisions was completed.
RM 23/11/04
We did some exam style questions on momentum, and then on elastic and inelestic collisions.
Remember: Momentum is always conserved incollisions. Add the total momentum beforehand (taking into account directions) and it will sum to the total momentum afterwards.
Kinetic energy is not always conserved. Add total KE beforehand (bearing in mind that energy is a scalar quantity and so direction is irrelevant) and then add total KE after. If they are equal, you have an elastic collision, if not, it is inelastic. If you gain in KE, there must be an external power source (i.e. the system is not isolated.)
HW Complete question 3 (arrows and apples). Also complete the KE calculations for our lab collisions to see if they are elastic or not in your neat books. I'll take them in on Friday.
RM 26/11/04
When work is done, energy is converted from one form to another. Forces often do work.
Work done by force = Force (N) times Distance force is applied for (m)
If the force doesn't move (i.e. the distance is zero, then no work is done.) This can be counter intuitive. For example, humans appear to be expending effort when holding up heavy masses, (just try to hold up your bag at arms length for a minute). However, place your bag on a table and it will continue to sit there ad nauseum. The table never gets tired and runs out of energy to support it. Yet the table certainly doesn't have an endless supply of fuel providing energy to keep the bag in place. The fact is that the table is simply doing no work, there is no energy transfer going on. Humans are doing work when they hold a mass at arms length due to their muscle configuration (different sets of muscles alternately work against each other to hold the arm in place). Energy is transferred from chemical potential energy stored in the human's cells and converted to heat energy.
Forces can do work by giving a mass some kinetic energy. They can do some work against a frictional force acting against them in which case energy is converted to heat. They can do some work against another kind of opposing force such as a gravitational, electric or magnetic field (or a field associated with a nuclear force). In this case potential energy associated with the particular field is being converted by the force.
Force * distance = Work done
Power = Work done/time (it is the rate of energy conversion)
Power = Force * distance/time
Distance/time = Velocity
Power = Force * Velocity
The above formula can be useful for working out the power of devices which are moving at a constant speed against a frictional force.
HW Do the double sided sheet with the man doing press ups and the car moving up a slope for Tuesday. Yellow cards have been instigated from now on in - you have been warned.
RM 30/11/04
We looked at the idea of using a ballistic pendulum to try and measure the speed of an air rifle pellet. The pellet hit the pendulum and became embedded in it and caused it to swing away and up. We measured the length of the pendulum. We took measurements of how far the pendulum swang horizontally and used an approximation to turn this into the angle that the pendulum moved through. This in turn allowed us to calculate the height by which the pendulum had raised.
An initial calculation assumed that the collision was elastic. (all the energy from the pellet passed to the pendulum bob and went into raising its height, giving it GPE). We measured the mass of a pellet and the mass of the pendulum bob.
GPEgained by pendulum = KEpellet
Mbob*g*h = 1/2mpelletvpellet2
This was clearly nonsense as loads of energy would have been lost as the bullet dug into the plasticine bob. Our calculation gave a figure of 16m/s as the pellet's speed. Nowhere near high enough. Whilst energy is rarely conserved, momentum is always conserved. This allowed us to try a different tack.
We assumed that the GPE gained by the pendulum bob + embedded pellet was equal to the KE of the bob + pellet just after the collision.
mbob+pelletgh = 1/2mbob+pelletv2
This allowed us to calculate the velocity of the bob just after the collision. We were then able to use this information to do a conservation of momentum calculation.
mpellet*vpellet = mbob+pellet*vbob+pellet
This calculation gave us a much more sensible pellet speed of about 250m/s. You must be aware of this method of combining an energy calculation with a momentum calculation to get an answer.
The law of energy conservation states that energy cannot be lost or gained in an isolated system. The largest possible isolated system is the whole universe (the amount of energy in the universe as a whole is constant). Smaller, near isolated experimental situations can be set up.
HW An entire past module paper to be sat in exam conditions on Friday. Revise and get a good mark please.
RM 03/12/04
We sat the module past paper Jan '02.
RM 07/12/04
Test paper was gone through. I want to see corrections from those who scored below 60%.
RM 10/12/04
We looked at charge. A fundamental property of matter. Comes in 2 sorts positive and negative. Carried by some particles, electrons which are negative and protons which are positive. Electrons can be persuaded to move from orbiting their nucleus at the surface of a material by a frictional force. Duster rubbing polyethene leaves the polyethene negatively charged overall - it has collected extra electrons from the duster. Electrons being light are much more mobile than the positive nuclei of atoms and tend to be the things that move.
When charge moves - you have a current. Charge Q is measured in Coulombs. The Coulomb is defined as the amount of charge that will move past a point if 1 Amp flows for 1 second. Coulombs are therefore, surprisingly, a derived quantity.
Q = I t
The definition of the Amp is based on electromagnetism.
Coulombs are more convenient to use than whole numbers of electrons which would get very large.
Van der Graff generator can create a large stored charge by friction.
HW 1.1, 2.1,2.3,2.4,2.5,2.6 practise questions from book.
RM 14/12/04
You measured the amount of charge which you were able to put on a polyethene rod using a duster with a digital Coulomb meter. You calculated the number of electrons which had been transferred. You used the formula Q=It to estimate the current which flowed. Current is the rate of flow of charge. It is therefore wrong to say "flow of current", as the term "current" implies a flow of charge in itself. The charge displaced in static charging by friction is small compared to the number of electrons pushed around a conducting circuit by a battery. Approximately 100 billion electrons were passed from the polyethene to the Coulomb meter. A much larger number of electrons were flowing around the simple circuit containing a light bulb.
Charge cannot be created or destroyed, the overall charge in any electrostatic interaction is always conserved. It directly follows from this that when charge is moving, (current) no charge can "disappear", so the current flowing into a junction must always equal the current flowing out of a junction. This is Kirchoff's law and is obvious really.
HW 1.1, 2.1,2.3,2.4,2.5,2.6 practise questions from book. Lets see these all in on Fri please, these are for handing in and marking.