ISA 5 mark experiment question - do for HW

It is suggested that the power supplied to an external resistor, R, is maximum when its
resistance is equal to the internal resistance of the supply. Describe how this hypothesis
could be tested experimentally. (5 marks)

For paper we did - anyone who didn't answer that question, here it is again

The temperature of the 10 cm diameter ball referred to in Question 2 may affect the
height to which it bounces. Describe an experiment, based on your experiment in
Stage 1, which would enable you to investigate whether or not temperature does affect
the height of the bounce. You should consider safety issues and the reduction of
random errors in this experiment. (5 marks)

12d homework

Exam question - answer in full for 6 marks

When monochromatic light is incident on a metal plate, electrons are emitted only when the frequency of light exceeds a certain frequency.
Explain in terms of energy, why this threshold frequency exists and why a photon theory of light provides a better explanation of the photoelectric effect than a wave theory of light.
The quality of your written answer will be assessed in this question.

Upcoming FREE events at Oxford Physics

Free lectures and events at Oxford University Department of Physics - links below 

NEW Tickets available for our alumni talk, ‘Space Lab’. More information is presented below and in the attached poster. Please contact the Alumni Office directly to book places.

Space Lab: Inspiring the next generation of scientists

Thursday 28^th November 2013, 4pm-5.30pm

Oxford Physics alumni and Google employee talks about his life and inspiring idea to run a global competition called YouTube Space Lab where school students designed an experiment that was sent into space. More information and booking: http://www2.physics.ox.ac.uk/events/2013/11/28/space-lab-inspiring-the-next-generation-of-scientists.

NEW, next month our second instalment of Flash Talk Physics:

Flash Talk Physics

Wednesday 4^th December 2013, 6.30pm-8.30pm.

Our research students present their work, look out for interesting props and demonstrations and vote for your favourite talks: http://www2.physics.ox.ac.uk/events/2013/12/04/flash-talk-physics. 

Physics A-level Club Tuesdays in W8

Physics drop-in
At least one Physics teacher will be available every Tuesday after school from 3.00- 4.00pm to help with your Physics needs.  Bring some work to do, e.g. questions, homework you are stuck with, or come with things to ask about.
We will be operating a loyalty card system with an end of year reward for the person with the most points!
Thanks and hope to see you there
Ms H

Revision for Unit 2 Physics Mechanics Materials and Waves

There will be a revision session for the Unit 2 paper on Monday 3rd June from 10.30 until about 1pm in W8,   lots of questions, papers and markschemes available as well as at least 2 Physics teachers

For those not doing Chemistry there will also be the usual revision session running on Tuesday at 3pm.

Good luck with your revision and happy half term!

Ms Hamnett

Revision for Unit 1 this Sunday 19th May in school

Revision for unit 1, 12.00pm till 2.30pm in school, W8, bring some food and some work, there will be tea, lots of papers, markschemes, questions, answers and Physics teachers available
See you then!
Ms Hamnett

Unit prefixes, symbols and units for unit 1 exam

Physics Alphabet AS Unit 1

Quantities, symbols and units for key quantities in Unit 1

Quantity	Symbol	Unit	Unit symbol
Energy	E	Joule	J
Energy	E	Electron volts	eV  (1eV= 1.6 x 10^-19 J
Kinetic energy	Ek	Joule	J
Frequency	f	Hertz	Hz
Wavelength	λ	metre	m
Momentum	p  (or mv)	Kilogram metres pe second	Kgm/s
Charge	Q	Coulomb	C
Current	I	Amp	A
Potential Difference	V	Volt  (or Joule per coulomb)	V
Resistance	R	Ohm	Ω
Power	P	Watt ( or joule per second)	W  (or J/s)
Resistivity	ρ	Ohm metre	Ωm

Here are the unit prefixes you will actually need to know, have deleted the irrelevant ones.

Factor Name  Symbol 1012 tera T 109 giga G 106 mega M 103 kilo k

Factor Name  Symbol 10-2 centi c 10-3 milli m 10-6 micro µ 10-9 nano n 10-12 pico p 10-15 femto f

Physics revision and extra help 2013

The following is a list of where and when a Physicist will be available to help you with revision - past papers will be available and you can come and work quietly or ask for help with particular areas.

There will also be a Unit 1 revision session in school (if enough people want one!) on Sunday 19th May from 12.00 till 2.30pm in W8.

Day	Room	Time	Period	Teacher
Monday	W8	12.20	3	MPO
Wednesday	W8	9.30	2	MPO/CHA
Thursday	W8 or W9	2pm	5	MPO/CHA
Friday	W8	9.30	2	MPO/CHA
	W13	12.20	4	CHA

PHYA2 answers to booklet vectors moments etc

PHYA2 answers vectors, moments, graphing motion

1.         (a)        (i)         resultant force acting on tray is zero [or P + W = Q] (1)resultant torque is zero
[or correct moments equation
or anticlockwise moments = clockwise moments] (1)

(ii)        W= 0.12 × 9.81 = 1.2N (1) (1.18N)

(iii)       (taking moments about P gives)
Q × 0.1 = 0.12 × 9.81 × 0.25 (1)Q = 2.9 N (2.94 N) (1)P = 2.9 – 1.2 = 1.7 N (1) (or 2.94 – 1.18 = 1.76 N)
                (allow C.E. for values of W and Q)                                                                       6

(b)        placed at Q (1)no additional turning moment about Q (1)                                                                                      2

[8]

2.         (a)        (i)         acceleration  (1)

(ii)        both represent acceleration of free fall
[or same acceleration]  (1)

(iii)       height/distance ball is dropped from above the ground
[or displacement]  (1)

(iv)      moving in the opposite direction  (1)

(v)       kinetic energy is lost in the collision
[or inelastic collision]  (1)                                                                                                        5

(b)        (i)         v2 = 2 × 9.81 × 1.2  (1)
v = 4.9 m s–1  (1)    (4.85 m s–1)

(ii)        u2 = 2 × 9.81 ×0.75  (1)
u = 3.8 m s–1  (1)    (3.84 m s–1)

(iii)       change in momentum = 0.15 × 3.84 – 0.15 × 4.85  (1)
= –1.3 kg m s–1  (1)            (1.25 kg m s–1)

             (allow C.E. from (b) (i) and (b)(ii))

(iv)      F =   (1)

            = 13 N  (1)

             (allow C.E. from (b)(iii))                                                                                              8

[13]

3.         (a)        (i)         component velocity North = 20 cos68° (1)
= 7.5 m s–1
which is supplied by wind (1)
by triangle of velocities [or by components] (aircraft must
point East) (1)

            alternative (a)(i)
triangle or parallelogram of velocities (1)
find angle between aircraft component and wind using sine and cosine
formulae – prove 90° (1) (1)

(ii)        work done = Fs cosq [or force × distance moved in direction
of force or 2.0 × 103 × 10 × 103 cos22°] (1)
= 1.8(5) × 107 J (1)

(iii)       power =  = 1.8(5) × 107 ÷ (1)
= 3.6 × 104 W (1)

            alternative (iii)
power = force × vel. component East = 2.0 × 103 × 20 cos22° (1)
              = 3.6 × 104 W (1)                                                                                               max 6

(b)        return time =  = 714 s \ total time = 1214 s (1)
average speed =  = 16[16.5]m s–1 (1)                                                                                 2

[8]

4.         (a)        (i)         a force multiplied by a distance
perpendicular distance from line of action of the force to the
point P (1)
                         (stated or from diagram)

(ii)        N m (1)                                                                                                                                       3

(b)        (i)         force up at pivot (1)

two downward forces at correct points (1)

(ii)        weight of tube ( = mg) = 12.0 × 9.81 = 118 N (1)

(iii)       moments about pivot equated (1)118 × 1.6 = W × 0.3 gives W = 629 (N) (1)(allow e.c.f. for weight in (ii))
mass =  = 64.1 kg (1)               (allow e c f for W)                                                      5

[8]

5.         (a)        (i)        

two forces opposing (1)forces parallel (1)s correct (1)

(ii)        N m (1)                                                                                                                                       4

(b)        (i)         anticlockwise moments = clockwise moments (1)

(ii)        weight of beam acts at centre (1)this is through the pivot (1)                                                                                                     3

(c)        (equating moments gives) 400 × 1.0 = 200 × 0.50 + 250 × d (1)\400 – 100 = 250 × d and d = 1.2 m (1)                                                                                         2

[9]

6.         (a)        (i)         rate of change of velocity
[or a = ] (1)

(ii)        (acceleration) has (magnitude and) direction (1)                                                               2

(b)        (i)         (acceleration) is the gradient (or slope) of the graph (1)

(ii)        (displacement) is the area (under the graph)                                                                       2

(c)       
    4

[8]

7.         (a)        (i)        

n.b. B must make an appreciable angle with wall and bar

(ii)        A          weight of sign and bar (accept gravity) (1)
B          reaction of wall (1)
C          tension in wire (1)                                                                                               max 5

(b)       
use of mg (1)
clockwise moments 118 × 0.375 (1)

             = anticlockwise moments (Tcos40° (1)) × 0.750 (1)
T = 77 N (1)                                                                                                                                   max 4

[9]

8.         (a)        (i)         gradient =  = 3.0 ms–2 (1)

(ii)        distance is area under graph (to t = 0.1 s)
or  × 0.7 × 2.1 0.3  (1) = 1.4(2) m (1)                                                        3

(b)        (i)         T – mg = ma [or T = 1500(9.8+3.0)] (1)
= 1.9 × 104 N (1)

T = mg = l.5 × 104 N (1)

(ii)        EF (1)                                                                                                                                          4

(c)        power = Fu or l.5 × 104 × 2.5 (1)
= 3.7[3.8] × 104 W (1)                                                                                                                          2

[9]

9.         (i)         a =  = 11 ms–2 (1)

F = ma =1.1 × 105 N (1)

(ii)        D  = 236 m s–1

a =  = 29.5 ms–2 (1)

(iii)       sone =  × t = × 4.0 = 88m (1)

stwo =  × t =  × 8.0 (1) = 1296(m) (1)

total distance = 1384 m (1)

[6]

10.       (a)        (i)         region A: uniform acceleration

                                                                     (or (free-fall) acceleration = g( = 9.8(i) m s–2))

force acting on parachutist is entirely his weight

             (or other forces are very small) (1)

(ii)        region B: speed is still increasing

acceleration is decreasing (2)                  (any two)

because frictional (drag) forces become significant
(at higher speeds)

(iii)       region C: uniform speed (50 m s–1)

because resultant force on parachutist is zero (2) (any two)

weight balanced exactly by resistive force upwards                                                          6
                                                                                                                                 QWC

(b)        deceleration is gradient of the graph (at t = 13s) (1)

(e.g. 20/1 or 40/2) = 20 m s–2 (1)                                                                                                        2

(c)        distance = area under graph (1)suitable method used to determine area (e.g. counting squares) (1)with a suitable scaling factor (e.g. area of each square = 5 m2) (1)distance=335m (±15m) (1)                                                                                                                 4

(d)        (i)         speed = Ö(5.02 + 3.02) = 5.8 m s–1 (1)

(ii)        tan q =     gives q = 31°(1)                                                                                                  2

[14]