Cheney Y12 Physics: May 2013

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	E_k	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

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

[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

[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

[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]

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

Cheney Y12 Physics

Saturday 25 May 2013

Revision for Unit 2 Physics Mechanics Materials and Waves

Saturday 18 May 2013

Revision for Unit 1 this Sunday 19th May in school

Unit prefixes, symbols and units for unit 1 exam

Thursday 9 May 2013

Physics revision and extra help 2013

Wednesday 1 May 2013

PHYA2 answers to booklet vectors moments etc