a2 Physics (Updated)

Momentum is the product of mass and velocity. It’s represented by the symbol P. P = mv It’s a vector quantity since velocity is vector. If we look at the equation representing Newton’s 2nd law: F = ma Since a = ∆v/t F = m∆v/t F = ∆P/t So Newton’s 2nd law can also by defined as: the resultant force is directly proportional to rate of change of momentum and the momentum is in the direction of the resultant force. One more equation can be derived which is: ∆P = F x t Impulse = F x t A change in momentum will cause an impulse.

PRINCIPLE OF CONSERVATION OF MOMENTUM

The momentum before collision is equal to the momentum after collision provided that there is no external resultant force acting on the system.

HOW TO INVESTIGATE THE PRINCIPLE?

Two trolleys A and B are kept on a friction compensated slope. A is kept at the start of the slope with a card. It has a cork at the front with a pin. B is kept in between the two light gates (connected to a computer) without a card. It has a cork on its front. A is given a push and the first light gate will record the time for it to cut it. Velocity = length of card/time When A attaches to B, they will both start moving in the same direction with a common final velocity. This velocity is recorded by the second light gate. The principle can be verified by: Mau = v(ma + mb)

The recorded and the calculated velocity will be same.

 When the masses are added to the end of the pulley, the trolley will accelerate.

 The motion sensor will record the displacement of the trolley at regular intervals.

 The different velocities will be calculated by the computer.  The mass is kept constant.  Calculate P = mv for all the velocities.  Graph will be:

Gradient is F since ∆P/t = F F is constant because mass is constant.

Ek = ½ mv2

P = mv

To remove v: V2 = 2Ek/m

P2 = m2 v2

Therefore: P2 = m2 x 2Ek/m P2 = m22Ek/m P2 = m2Ek

Ek = P2/2m  This equation works for problems in everyday life.  This equation can only be used to fast moving particles when they are non-relativistic (particles that have a speed less than 10 % of the speed of light.)

CONSERVATION OF MOMENTUM IN 2D When a particles of mass m collide with another particle of same mass at rest, in an off-centered collision they move away from each other at 90o

When a particle of mass m collides with another particle of same mass at rest in a centered collision, the one moving comes to rest and the other starts moving.

When a bigger particle collides with a smaller particle at rest, the smaller particle will move at an acute angle and bigger particle will move in the same direction as it was moving.

When a smaller particle collides with a bigger particle at rest, the smaller particle will move at an obtuse angle and the bigger particle moves forward.

 All of these only apply if the collision is elastic.

ELASTIC COLLISION  Kinetic energy before collision is equal to kinetic energy after collision.  Kinetic energy is conserved.  This type of collision is called elastic collision. INELASTIC COLLISION  Kinetic energy before collision is not equal to kinetic energy after collision.  Kinetic energy not conserved.  This type of collision is called inelastic collision.

 When a body is moving in a circular motion, the speed remains constant.  But since the direction is continuously changing, the velocity changes.  Hence there is an acceleration.  According to Newton’s 1st law if there is a motion there must be a resultant force.

 This resultant force needed for circular motion is called centripetal force.(Fc)  Centripetal force is directed towards the center.  According to Newton’s 2nd law if there is a resultant force there must be an acceleration.  This acceleration is called centripetal acceleration (ac)  If the resultant force is not there, there won’t be an acceleration. Hence there won’t be a change in velocity so no circular motion.

 Angular displacement is the vector measurement of angle in radians. 1 degree = 180/pi

 For a body travelling from A to B: Angle = s/r  For a body completing a whole circle: Angle in radian = 2pir/r

= 2pi

 To calculate velocity for a body moving in a circle, we need to know the time it takes to complete one revolution.  The quantity is called angular velocity (w), unit: rads-1 Angular velocity = angular displacement/ time  For a body completing a whole circle: W = 2π/T T= 2π/ω Since f = 1/T W = 2πf And since instantaneous v = s/t And s = rtheta/t Theta/t = W Therefore v = rw

A = v2/r Therefore: F = ma = mv2/r And since: v = rw = m(rw)2/r = mr2w2/r = m rw2 F=ma a= rw2

The mass ‘m’ can be a rubber bung. Rotate the rubber bung making sure that the radius ‘r’ remains constant. Start a stopwatch as soon as you start rotating. Use the stopwatch to record the time taken to make 10 revolutions. Repeat with different masses added to the weight hanger. Each time measure the radius ‘r’ and record it too. Draw a graph of F against rw2 where F= mrw2 The graph will be a straight line through origin and its gradient will be mass ‘m’. Assumptions: The bung rotates in a horizontal circle. There is no friction between glass tubing and string. Safety precautions: Wear goggles to protect the eyes. Carry out investigation in a wide space.

UNIFORM:

 Electric field is an area where an electric force can be experienced. Q

q1

q2

 Force from Q experienced by q1 is greater than the force experienced by q2. Electric Field Strength  It’s a vector quantity.  Magnitude is E. E = F/Q  Unit is NC-1

Move the probe across the paper and mark a position with some voltage for example 6V. Mark another position with the voltage 6V. Repeat this to get 5-8 dots. Join all these points and you will get equipotential lines. Those lines are perpendicular to electric field lines. There is no p.d on equipotential lines. Hence the work done to move a charge along an equipotential line is zero W = VQ V = 0 W=0

Applying a potential difference to two parallel plates produces a uniform electric field between them. We can investigate this using the following procedure. Attach an aluminum foil to a ruler. Keep the ruler at different positions between the plates. The foil will deflect. The angle between the ruler and the foil will be constant. By this we know that F is constant. Since q is also constant, we conclude that E is constant. There is another equation which is specific for uniform electric fields which is:

E = V/s

E = 12000/2 = 6000 Vm-1 Vm-1 is actually the same as NC-1 since: Vm-1 V= J/C = NmC-1 NmC-1 x m-1 = NC-1

 Coulumb’s Law states that the force acting on two masses is directly proportional to the product of their charges divided by the square of their separation. F directly proportional to Q1Q2/r2 F = kQ1Q2/r2 K = 1/

0

And for a point charge: E = kQ1/r2

 E0 is epsilon zero and this gives an idea of how easily an electric field can pass through space.

 When ‘r’ decreases by 2x reading increases by 4x.  We can measure ‘r’ using a ruler kept between the 2 spheres.  The balance reading will give an idea of the magnitude of the force of attraction between the spheres.  It can be seen that F is inversely proportional to ‘r’.

 When a capacitor is connected to a power supply, electrons move from the negative terminal towards the capacitor but they cant pass to the next plate due to the insulating medium between the plates.  The other plate gets positively charged by induction.  Electrons accumulate in the capacitor and charge builds up. Charge build up is directly proportional to potential difference Q directly proportional to V Q = CV Where ‘C’ is capacitance C = Q/V = CV-1 SI unit: Farad (F)  Capacitance is the capacity of the capacitor to store charge.

Experiment to measure capacitance

 Record current and voltage at regular intervals using a stopwatch so that we can find Q = It and hence C = Q/V  Instead of a stopwatch and ammeter we can use coulometer but that is used for very small charge.

Charge is directly proportional to voltage Q = CV V = W/Q W = VQ W = V x CV W = ½ CV2 Or W = VQ

W = Q/C x Q W= ½ Q2/C Note: ½ because ½ of the energy is used to charge the capacitor and ½ is stored.

 You can investigate how the energy stored in a capacitor can be altered in order to make it suitable for a particular combination of bulbs.

 For the circuit above, the total resistance is 5 ohm.

 The 100 F capacitor was charged with 6V and then connected to this circuit.

 The bulb lighted up with the highest brightness.  Now use a different combination of bulbs.

 Each individual bulb has a resistance of 5 ohm.  Calculating the total resistance will give 5 ohms.  This is the same total resistance as the previous circuit.  Now with the capacitor charged to 6V, connect it to this circuit.  This time all bulbs wont light up with the same brightness as before since the voltage is split between bulbs in series.  Therefore we can remove the capacitor and charge it to a higher voltage say 8V.  Connect the capacitor back to the circuit above and see whether the bulbs light up with the highest brightness.  Keep increasing the voltage of the capacitor until all bulbs light up to the highest brightness.  And then calculate E = ½ CV2 with that final voltage used and the capacitance on the capacitor. Note: in this investigation we must make combinations of bulbs such that for all the combinations, the total resistance is constant.

Discharging graphs

 Q decreases with time since the capacitor is discharging.  V decreases with time since there is nothing else in the circuit besides the capacitor, and the potential difference across the capacitor is decreases.

 Current with which the capacitor is discharged decreases since the voltage across it decreases. Charging Curves

 Charge on capacitor increases since the capacitor is being charged.  Voltage across capacitor increases since it is gaining the voltage from the cell.

 Current decreases since the potential difference across the cell and capacitor decreases.

Time Constant It is the time required to decrease the charge/voltage/current on the capacitor by 37% of its initial charge/voltage/current. It is the time required to increase the charge/voltage/current on the capacitor by 63% of its final charge/voltage/current Time constant = RC

Q = Q0e-t/RC Since Q = CV And Q0 = CV0 CV = CV0e-t/RC C is cancelled Therefore: V = V0e-t/RC since V = IR and V0 = I0R IR = I0Re-t/RC

I is cancelled. Therefore: I = I0e-t/RC

magnetic flux density: the strength of a magnetic field symbol: B unit: Tesla (T) flux: the strength of magnetic field in a whole area symbol: Φ formula: phi = BA unit: Weber (Wb) flux linkage: strength of magnetic field in whole area through a coil symbol: N Φ formula: Nphi = BAN unit: Weber turns (Wb turns)

N= number of turns

 F = BIlsin θ is used to find the force acting on a current carrying conductor placed in a magnetic field

 The theta is the angle given by the vertical component of conductor.

 Blue line is the conductor.  The angle between the pink and blue line is taken.  If the conductor is perpendicular to magnetic field, the angle would be 90 degrees. Therefore: F = BIl sin 90 = BIl x 1 =BIl Investigating the Equation  Keep a current carrying conductor in a magnetic field (U- magnet)  The U magnet must be kept on a balance reading zero.  The circuit consisting of the conductor must have an ammeter.  Switch on the current and start taking readings from the balance.  For each reading of balance, take the current reading as well.  Applying Newton’s third law, the magnetic force on wire must be equal and opposite to the magnetic force on magnet.  Magnetic force on magnet is given by the balance reading.  Calculate F = mg  Draw a graph of F against I

 It will be a straight line with gradient Bl.

 Electron beams would follow a circular path when passing through a magnetic field.

 This is force F = BeV  Force F is perpendicular to both magnetic field and the direction the electron is moving

 The electron tries to move in a straight line but that force continuously pulls the electron to the center of the curvature.

 Hence the force is known as the centripetal force.  Mv2/r = BeV  Since BeV is perpendicular to the path of the electrons, it does no work on them.

 Therefore the speed and KE of electrons remain constant.

 A relative motion between a coil and permanent magnet will produce an emf.

 This is because a change in magnetic flux will produce a voltage. dN/dt BAN/s NA-1m-1m2/s NmA-1/s J/As E/It V = E/Q  When there is a change of current in the coil linked with the magnet, then also an emf will be induced.  Factors that increase induced emf:  Increasing number of turns on coil.  Increasing magnetic field strength. Investigating Induced EMF  Prepare a circuit with a coil of wire and a resistor.  Connect the circuit to a data logger and the data logger to a computer.  Set the computer so that it draws a graph of V against t  Drop a short bar magnet through the coil.  The graph will be:

 At first the emf is increasing as the magnet first enters the coil and there is a change in magnetic flux.  Emf starts decreasing since there is less and less of change in magnetic flux as the bar moves into the coil.  Emf becomes zero since there is no more change in magnetic flux as the bar is fully in the coil  There is a magnetic flux but not a change in magnetic flux.  Beyond zero, graph goes below the x axis since an emf is again produced as the bar moves out of the coil.  Its negative since this time the emf is in opposite direction.

Nucleon number: total number of protons and neutron in the nucleus of an atom. Proton number: the number of protons in the nucleus of an atom.

Observations and Conclusions Drawn from Rutherford’s Alpha Scattering Experiment It was seen that most of the alpha particles pass the gold foil without any deflection. This means that most of the atom is empty space. Some alpha particles get deflected. This means that the atom has a nucleus with a positive charge. Only a few alpha particles bounce back. This means there is a small nucleus concentrated at the center.

 The escape of electrons from a metal surface when it is heated to a very high temperature is called thermionic emission. How are electrons accelerated by:

1. Electric Field  Force on an electron in an electric field is parallel to the field.  Electrons are negative therefore there is a force of attraction on them towards the positive plate.  If there is a force there must be acceleration. F = ma E = F/Q F = EQ EQ = ma a= EQ/m

2. Magnetic Field  Force on an electron in a magnetic field is perpendicular to the field.  When an electron is in a magnetic field, it moves in a circular path.  This is because the magnetic force on the electron is perpendicular to both magnetic field and electron path.  Applying the Fleming’s left hand rule, we would find that the force F is always towards the center.  This resultant force F is the centripetal force.  Centripetal force produces a centripetal acceleration.  Momentum can be found from radius. r= P/BQ

Particles need to be accelerated so that they can collide with one another, break into pieces thereby allowing us to study them. If particles collide with slow speeds they would just bounce back. There are 2 types of particle accelerators. 1. Linear accelerators 2. Cyclotron Linear Accelerators (Linac)

Linear accelerators are a series of metal tubes connected to an ac power supply. The proton is fired when the left of tube one is negatively charged. When the proton is midway along tube 1, the ac changes so that now the right of that tube is negative. Proton is accelerated towards the end of tubes 1 As it is just about to leave, ac changes so that now the start of tube 2 is negative. So the proton is further accelerated towards tube 2 and the process repeats. Every successive tube is made longer since at each succession the particle is moving with a higher speed. And also because the frequency at which the ac changes is constant. If all tubes in the diagram above were of the same size, the proton might reach the end of tube 2 while it’s still positive. This will cause the proton to repel and decelerate. Disadvantage of the linear accelerator would be that it takes a lot of space since it needs to be made longer for greater accelerations. Cyclotron

Cyclotron makes the particles move in a circle. When a charged particle, for instance an electron, moves in the first hollow chamber (dee A) it will move in a circular path since there is a magnetic field. Magnetic field will provide the centripetal force to maintain circular motion. In dee A, the electron will complete a semicircle and just as it does, ac changes so that there is a positive charge on dee B. So the electron gets accelerated towards dee B. Electron will make another semicircle in dee B and just as it does, ac changes making dee A positive. Electron gets accelerated towards dee A and moves in a semicircle with a greater radius. It moves in a greater radius since it has gained kinetic energy.

 In a GM tube, the particle to be detected passes through it ionizing the argon gas present in it.  The electron released while ionizing together with the ions are accelerated by an electric field.  The electrons are discharged when they reach the electrodes that produce the electric field.  This produces a pulse of electricity which is counted by counter connected to a tube.  In a mass spectrometer, the sample is vaporized then ionized by bombardment of electrons.  The ions are accelerated by an electric field.  The ions are deflected by a magnetic field.  The amount of deflection will be proportional to m/e ratio.  Hence ion can be identified by the amount of deflection.

 Force acting on a charged particle moving in a magnetic field is given by: F = Bqv  The charged particle moves in a circle in a magnetic field.  For any object moving in a circle the force acting on it is given by: F = mv2/r Therefore: mv2/r = Bqv Mv2/v = Bqr Mv = Bqr P = Bqr R = p/Bq

 Hence for a given magnetic field, the radius of the path of the particle is proportional to momentum.

 For a particle interaction to occur, the charge, energy and momentum should be conserved.  Consider the following collision e+ + e- -------  +   When a positron and an electron collided, massless particles of gamma are produced. How is energy conserved? The energy is conserved since massless particles with high velocity are produced from particles that have a greater mass and lower velocity. E = ½ mv2 How is charge conserved? Charge is conserved since the charge is zero even at the start and the end. How is momentum conserved? Conserved since massless particles with high velocity are produced from particles that have a greater mass and lower velocity.

Initial momentum

=

P = high mass x low velocity

=

final momentum p = low mass x high velocity

 The length of the tracks depends on the energy of the particles.  Magnetic fields deflect the charged particles and hence the tracks are bent.  The curvature depends on momentum, charge and strength of magnetic field.  We can tell whether the particle is positive or negative using Fleming’s left hand rule. The direction of current will be the direction of positive charge and for negative charge, the direction is just opposite.  If the radius is decreasing, it means the particles are slowing down as they are releasing energy.

 If the particles collide with a low energy, they would just bounce back and we wont be able to observe the smaller particles within them.  When particles collide with a higher energy, they split into the particles they are made up of. So we can study the fine structure.

E = energy M = mass C = speed of light (3 x 108)  Light is the fastest moving particle.  If any other object tries to move faster than light, its energy gets interchanged to mass.  This is called Einstein’s Special Relativity.

o According to the equation, matter can appear out of nowhere. o It is converted from energy. Equation applied to Annihilation o Annihilation is the process by which matter and antimatter collide to form a high energy, massless particle. (usually electromagnetic radiation) o Just as energy can be interchanged to mass, mass can be interchanged to energy. o The matter and antimatter vanish from existence to produce equivalent energy in the form of electromagnetic radiation.

To convert MeV to joule: Energy value x 1.6 x 10-19 x 106 To convert GeV to joule: Energy value x 1.6 x 10-19 x 109 To convert MeV/c2 to kg: Mass value x 1.6 x 10-19 x 106/ (3 x 108)2 To convert GeV/c2 to kg: Mass value x 1.6 x 10-19 x 109/ (3 x 108)2 To convert kg to eV Mass value / (1.6x10-19)/(3x108)

When an object tries to move faster than light, some of its energy is converted to mass so that it does not reach that speed. This fact needs to be taken into account when dealing with speeds near to that of light. (for particles that have a speed of 10% less than that of the speed of light.)

 There are 12 fundamental particles.  The fundamental particles are divided into 2 families which are: 1. Quarks (6 members) 2. Leptons (6 members)  Each quark and each lepton has their antiparticle of the same mass and opposite charge.

QUARKS QUARK Up (u) Down (d) Charm (c)

CHARGE +2/3 e -1/3 e +2/3 e

Strange (s)

-1/3 e

Top (t) Bottom (b)

+2/3 e -1/3 e

ANTIQUARK Anti- up quark Anti-down quark Anti-charm quark Anti-strange quark Anti-top quark Anti-bottom quark

CHARGE -2/3 e +1/3 e -2/3 e +1/3 e -2/3 e +1/3 e

First generation: up and down quarks Second generation: charm and strange quarks Third generation: top and bottom quarks. Strong bonds form between the quarks. LEPTONS LEPTON Electron (e-) Electron neutrino (Ve-) Muon (-) Muon neutrino (V-) Tau (-) Tau neutrino (V-)

CHARGE -e 0 -e 0 -e 0

ANTILEPTON Antielectron Antielectron neutrino Antimuon Antimuon neutrino Antitau Antitau neutrino

CHARGE +e 0 +e 0 +e 0

All antileptons have the same symbol as their lepton except that they have + instead of –

When a proton was bombarded with high electrons, the proton splits into three particles The particles are 2 up quarks and 1 down quark. (uud) Therefore the charge would be: 2/3e + 2/3e – 1/3e = +e A neutron has 0 charge.

Therefore it must have 2 down quarks and 1 up quark. -1/3e + -1/3e + 2/3e = 0 Particles like neutrons and protons are called baryons. Baryons are particles having 3 quarks. Mesons are particles having a quark and an antiquark.  It was previously known that there are 6 leptons before all the quarks were discovered.  Then when up, down, strange and charm quarks were discovered, the quarks were still lagging behind leptons.  The scientists like to believe that the universe is balanced.  Therefore they came to the conclusion that there must be 6 quarks with the 6 leptons.  Hence the top and bottom quarks were predicted.  Later they were even discovered for real.

Write and interpret equations for the following reaction Neutral pion decay in which the pion becomes an electron, a positron and a gamma photon. 0 ---- e- + e+ + 

This reaction will occur since the charge is conserved. The momentum, energy/mass should also be conserved for this reaction to occur.

 - Wavelength h- Plank’s constant – 6.63 x 10-34 p = momentum p can also be written as mv when asked to find mass or velocity.