ACTIVITY MEMORANDA

Mechanics

Forces

QUESTION 2 STUDYING LIKE A BOSS

2.1

OR

Triangle correct

Two angles correct

Labels (3)

2.2 Ff,smax = μsFN

Ff,smax = (0,6)(1,5 × 9,8)

Ff,smax = 8,82 N (3)

2.3 TA = Ff,smax = 8,82 N OR TA = Ff,s

max = 8,82 N

Fg = TA tan 30 Fg =TA

tan60

Fg = (8,82) tan 30 Fg =(8,82)

tan60

Fg = 5,09 N Fg = 5,09 N (4)

2.4 TB =TA

cos30 OR TB =

TA

sin30 OR TB = √TA

2 + FG2

TB =(8,82)

cos30 TB =

(8,82)

sin60 TB = √(5,09)2 + (8,82)2

TB = 10,18 N TB = 10,18 N TB = 10,18 N (3)

[13]

30o

60o

TB

Fg OR TC

TA

90o

30o

60o

TB

Fg OR TC

TA

90o

QUESTION 5 THAT’S ROUGH

Let right be positive.

5.1 Δ𝑥 = viΔt +1

2aΔt2

(+5) = (0) +1

2a(2,5)2

a = 1,6 m⋅s−2 (3)

5.2 For block B:

F⃗ net = ma⃗

F⃗ app,𝑥 + F⃗ T + F⃗ f = mBa⃗

(+40 cos 60) + F⃗ T + (−3,2) = (8)(+1,6)

F⃗ T = −4

FT = 4 N (5)

5.3 For mass A and B: OR For mass A:

F⃗ net = ma⃗ F⃗ net = ma⃗

F⃗ app,𝑥 + F⃗ fA + F⃗ fB = (mA + mB)a⃗ F⃗ T + F⃗ fA = mAa⃗

(+40 cos 60) + (−3,2) + (−3,2) = (mA + 8)(+1,6) (+4) + (−3,2) = mA(+1,6)

mA = 0,5 kg mA = 0,5 kg (4)

5.4 FN = Fg − Fapp,y Ff,k = μkFN

FN = (8)(9,8) − (40 sin 60) (3,2) = μk(43,76)

FN = 43,76 N μk = 0,07 (5)

5.5 F⃗ net = ma⃗

F⃗ g∥ + F⃗ f = ma⃗

mg sin 𝜃 − μmg cos 𝜃 = ma⃗ g sin 𝜃 − μg cos 𝜃 = a⃗

As the angle decreases, the parallel component of weight decreases and the friction increases ,

thus resulting in a greater acceleration. Therefore, block B will experience the lowest acceleration in

situation X . (5)

[22]

Vertical Projectile Motion in One Dimension (1D)

QUESTION 4 TAKING THE PLUNGE

4.1 vf = vi + aΔt

(0) = vi + (−9,8)(0,33)

vi = 3,234 m⋅s−1 (2)

4.2 The change in position. (2)

4.3 Δ𝑥 = viΔt +1

2aΔt2 OR 𝑣𝑓

2 = 𝑣𝑖2 + 2𝑎Δ𝑥

Δ𝑥 = (0) +1

2(−9,8)(0,33)2 (−3,234)2 = (0)2 + 2(−9,8)Δ𝑥

Δ𝑥 = −0,53 = 0,53 m down Δ𝑥 = −0,53 = 0,53 m down

OR 𝑣𝑓2 = 𝑣𝑖

2 + 2𝑎Δ𝑥

(−3,234)2 = (0)2 + 2(−9,8)Δ𝑥

Δ𝑥 = −0,53 = 0,53 m down (4)

4.4 Δ𝑥 = viΔt +1

2aΔt2 OR Δ𝑥 = viΔt +

1

2aΔt2

(−67) = (+3,234)Δt +1

2(−9,8)Δt2 (−67,53) = (0) +

1

2(−9,8)Δt2

Δt = 4.04 s Δt = 3,71 s

∴ Δttotal = 3,71 + 0,33 = 4,04 s (4)

4.5 vf2 = vi

2 + 2aΔ𝑥

vf2 = (+3,234)2 + 2(−9,8)(−67)

vi = 36,38 m⋅s−1

vi = 130,98 km⋅h−1 (4)

4.6 v = √2×m×g

ρ×A×C

v = √2×(75)×(9,8)

(1,2)×(0,72)×(1,1)

v = 39,33 m⋅s−1

No, the jumper will not reach terminal velocity , as the terminal velocity is greater than the predicted

final velocity, which would most likely be less anyway due to air friction . (5)

[21]

Momentum and Impulse

QUESTION 8 SKATING

Let left be positive.

8.1 The product of the net force and the contact time. (2)

8.2 J = 10 N⋅s right (2)

8.3 System: Amanda (nonisolated)

J = m(v⃗ f − v⃗ i)

(+10) = (65)(v⃗ f − (−4))

v⃗ f = −3,85 = 3,85 m⋅s−1 right (5)

8.4 System: Amy (nonisolated) OR System: Amanda and Amy (isolated)

J = m(v⃗ f − v⃗ i) Σpi = Σpf

(−10) = (52)((−3,85) − v⃗ i) mAmyv⃗ i,Amy + mAmandav⃗ i,Amanda = (mAmy + mAmanda)v⃗ f

v⃗ i = −3,66 = 3,66 m⋅s−1 right (52)v⃗ i,Amy + (65)(−4) = (52 + 65)(−3,85)

v⃗ i,Amy = −3,66 = 3,66 m⋅s−1 right (5)

[14]

Work, Energy and Power

QUESTION 6 BIG JUMP

6.1 In the absence of air resistance or any external forces, the mechanical energy of an object is

constant. (2)

6.2 (EP + EK)i = (EP + EK)f

(mgh + 0)i = (mgh +1

2mv2)

f

m(9,8)(6,16) = m(9,8)(1,5) +1

2mv2

v = 9,57 m⋅s−1 (5)

[7]

QUESTION 7 HARD WORK

7.1 The work done by a net force on an object is equal to the change in the kinetic energy of the object.

(2)

7.2 ΔEK =1

2mvf

2 −1

2mvi

2

ΔEK =1

2(800)(2)2 −

1

2(800)(7)2

ΔEK = −18 000 J (3)

7.3 Wf = FfΔx cos θ

Wf = (200)(40) cos 180

Wf = −8 000 J (3)

7.4 Wnet = ΔEK

Wapp + Wg + Wf = ΔEK

Wapp + mgh cos θ + Wf = ΔEK

Wapp + (800)(9,8)(3) cos 180 + (−8000) = (−18000)

Wapp = 13 520 J (4)

7.5 P =W

Δt

P =(13520)

(80)

P = 169 W (3)

[15]

Electricity and Magnetism

Electrostatics

QUESTION 9 KEEP AN ION THOSE ELECTRONS

9.1 A region of space in which an electric charge experiences a force. (2)

9.2 N =QA

e

N =(10×10−6)

(1,6×10−19)

N = 6,25 × 1013 (2)

9.3 E =kQA

r2

E =(9×109)(10×10−6)

(2×10−2)2

E = 2,25 × 108 N⋅C−1 left/toward A (5)

9.4

Direction

Shape

More field lines on B than A

(3)

9.5 The force between two charges is directly proportional to the product of the charges and inversely

proportional to the distance between the charges squared. (2)

9.6 F =kqAqB

r2

F =(9×109)(10×10−6)(20×10−6)

(3×10−2)2

F = 2 000 N left (5)

9.7 F =kqBqC

r2

(2000) =(9×109)(20×10−6)qC

(1×10−2)2

qC = 1,11 × 10−6 C (4)

[23]

[17]

A B

Electric Circuits

QUESTION 10 MY HEAD HERTZ FROM TRYING TO THINK OF A PUN HERE

10.1 PX = IXV∥

(100) = IX(20)

IX = 5 A (3)

10.2 PY =V∥

2

RY

(150) =(20)2

RY

RY = 2,67 Ω (4)

10.3 PY = IYV∥ emf = V − Ir (both equations)

(150) = IY(20) (24) = (20) − (5 + 7,5)r

IY = 7,5 A r = 0,32 Ω

OR

PY = IYV∥ PX =V∥

2

RX Rtotal =

RX×RY

RX+RY emf = I(R + r) (both equations)

(150) = IY(20) (150) =(20)2

RX Rtotal =

(2,67×4)

(2,67+4) (24) = (5 + 7,5)(R + r)

IY = 7,5 A RX = 4 Ω Rtotal = 1,6 Ω r = 0,32 Ω (6)

10.4 Voltmeter (1)

10.5 It has a very high resistance and so draws very little current , which is why the devices still function

as rated, i.e. there is no change in total current or voltage. (2)

[16]

Electrodynamics and AC

QUESTION 11 WITH GREAT POWER COMES GREAT ELECTRICITY BILL

11.1 Step down transformer (1)

11.2 𝑁𝑆

𝑁𝑃=

𝑉𝑆

𝑉𝑃

𝑁𝑆

(330)=

(12)

(220)

𝑁𝑆 = 18 (3)

11.3 Alternating current in the primary coil creates a changing magnetic field in the primary coil

The soft iron core ensures that the secondary coil experiences this changing magnetic flux

According to Faraday’s law of electromagnetic induction

this changing flux induced a changing emf in the secondary coil (AC) (4)

11.4

shape

full wave rectification

(−1) no labels

(2)

11.5 Electrical (potential) energy to mechanical energy (2)

11.6 The motor effect (1)

11.7 When the armature is in the vertical position (perpendicular to the magnetic field), the brushes make

contact with the split in the ring, switching off the current in this position

Inertia enables the armature to continue rotating

and when past this position, the brushed now make contact with the opposite sides of the armature,

switching the direction of the current through the armature (3)

11.8 Negative (2)

11.9 Fleming’s left hand motor rule

(1)

11.10 Any of the following with explanation :

𝒆𝒎𝒇

𝚫𝒕

Use stronger magnets, which will increase the magnetic field strength and thus the force on the

armature

Use curved magnets, which will ensure that the magnetic field strength and thus the force on the

armature, is at a maximum for any position of the armature

Use a higher voltage DC supply, which will increaser the current trough the armature and thus the

force on the armature

Use more turns on the armature (coil), which will increase the effective length exposed to the

magnetic field and thus the force on the armature (2)

11.11 Spinning the armature changes the angle between the magnetic field and the normal to the area of

the armature

Thus, the magnetic flux experienced by the armature changes

According to Faraday’s law of electromagnetic induction

This will induce an emf which causes the light bulb to shine (4)

[25]

Waves, Matter and Materials

Photoelectric Effect

QUESTION 12 THERE’S LIGHT AT THE END OF THIS TUNNEL

12.1 The photoelectric effect (1)

12.2 Any one of the following :

It deonstrates the particle nature of light

It establishes Quantum Theory (1)

12.3 Electroscope (1)

12.4 The minimum frequency of incident radiation at which electrons will be emitted from a particular

metal. (2)

12.5 W0 = hf0

(3,1 × 1,6 × 10−19) = (6,6 × 10−34)f0

f0 = 7,52 × 1014 Hz (4)

12.6 UV (1)

12.7 c = λf OR E =hc

λ

(3 × 108) = (340 × 10−9)f E =(6,6×10−34)(3×108)

(340×10−9)

f = 8,82 × 1014 Hz E = 5,82 × 10−19 J = 3,64 eV

𝑓 > 𝑓0 𝐸 > 𝑊0

Therefore, electrons will be emitted Therefore, electrons will be emitted (3)

12.8 E = W0 + EKmax

hc

λ= W0 + EK

max

(6,6×10−34)(3×108)

(340×10−9)= (3,1 × 1,6 × 10−19) + EK

max

EKmax = 8,64 × 10−20 J (4)

[17]