MEMORANDUM

QUESTION  1
1.1 A √√ (2)
1.2 D √√ (2)
1.3 C √√ (2)
1.4 A √√ (2)
1.5 B √√ (2)
1.6 C √√ (2)
1.7 D √√ (2)
1.8 A √√ (2)
1.9 D √√ (2)
1.10 D √√ (2)                   [20]

QUESTION 2
2.1
2.1.1 F1 = Normal force  √ (1)
2.1.2 F2 = Weight √ (1)
2.2
2.2.1 Force of the block on the EARTH  √(2)
2.2.2

• F₄ = mg √
  = 5 x 9,8 √
  = 49 N √  (3)

2.3.1 Newton’s First Law √ - A body will remain at rest or continue moving with constant velocity unless a net force acts on it. √√ (3)
2.3.2 F_net = 0 N √ (1)
[11]

QUESTION 3
3.1
3.1.1 To ensure that tension is uniform. √√ (2)
3.1.2

• A net force produces acceleration of an object in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass of the object. √√  (2)
  OR
• The net force acting on a body is equal to the rate of change of momentum in the same direction as the net force.

3.1.3

• F_H = F cosθ √
  = 40 cos 35° √
  = 32,77 N
  
  F_net = ma √
  
  80 kg mass:         50 kg mass:
  F_net = ma              F_net = ma
  T – 6 = 80 a √       32,77 – T √ = 50 a √
  a = 0,21 m.s^-2 √   (7)

3.1.4 DECREASES √ Horisontal component (of 40 N force) decreases √√ (3)
3.2
3.2.1 Mass (of the trolley) √ (1)
3.2.2 a α F √√ (if m constant and net force decreases, the acceleration decreases) OR (as net force increases acceleration increases) (2)
3.3 Experiment B √
Gradient of A > gradient of B  √
1/m_A > 1/m_B √√
Therefore m_A < m_B (4)
[21]

QUESTION  4
4.1 In an isolated system the total linear momentum remains constant. √√  (2 or 0)  (2)
4.2

• Σ pi = Σ pf
  m1v1i + m2v2i = m1v1f + m2v2f √
  16 000 + (2 000)(10,61) √= (800)(12) + 2 000 v2f √
  v2f = 13,81 m.s^-1 √      (4)

4.3

• Ektotal before  = ½ mvA2 +½mvB2
  = ½ x 800 x 202 √ + 112 500
  = 272 500 J
• Ektotal after = ½(800)122 √+ ½ (2 000)(13,81)2 √
  = 248 316,1 J √
• Inelastic: Ek before > Ek after √ OR Ek before ≠ Ek after  (5)

4.4

• F_net Δt = Δp √
  F_net = (9 600 – 16 000 √)/0,15 √
  = - 42 666,67 N
  = 42 666,67 N, left/links √  (4)

[15]

QUESTION 5
5.1 Energy a body has because of its position above the ground √√  (2)
5.2

• Ek = ½ mv² √
  = ½ x 0,5 x 19,8² √
  = 98,01 J √    (3)

5.3

Related Items

• E_p lost= - E_k gained
  = - 98,01 √√
  98,01 = - mg(yf –yi)
  98,01 √= - 0,5 x 9,8(yf – 0) √
  yf = - 20 m
• E_mech bottom = E_mech top
  (mgh + 1/2mv²)_bottom = (mgh + 1/2mv²)_top √
  0 + ½ (0,5 x 25²) √ = ½ x 9,8 x h_P √
  hP = 31,89 m
  h = 31,89 + 20 = 51,89 m √ (8)

5.4

• P =W/Δt √
  = 37 350/20 √
  = 1 867,5 W
  P = 1867,5 / 746 √ = 2,50 hp √ (4)

[17]

QUESTION 6
6.1 Stress √  (1)
6.2
6.2.1

• Strain = Δl/L √
  = (405 – 400) /400 √
  = 0,0125 √     (3)

6.2.2

• Strain = F/A √
  A = π r² √
  = π x (5/1000)² √
  = 7,85 x 10 ^-5 m²
• Strain = 30 000/7,85 x 10 ^-5 √
  = 3,82 X 10 ⁸ Pa √ (5)

6.3
6.3.1 Modulus of elasticity √ (1)
6.3.2 Hooke’s law √ (1)
6.3.3 Within the limits of elasticity√, stress is directly proportional to strain √ (2)
6.4
6.4.1 B √ (1)
6.4.2 A √ (1)
6.5
6.5.1 Elasticity √ (1)
6.5.2 Elastic band, bow, trampoline, spring diving board (Any TWO) √√ (2)
6.5.3 Restoring force √ (1)
[19]

QUESTION 7
7.1
7.1.1 Viscosity is the property of a fluid to oppose relative motion between two adjacent layers. √√ (2)
7.1.2 R √ Time of flow in R is highest . √√  (3)
7.1.3 smaller than √
Viscosity increases with increase in temperature. √√  (3)
7.1.4

• p = ρgh √
  = 1000 x 9.8 x 10 √
  = 98 00 Pa √  (3)

7.2
7.2.1 In a continuous liquid at equilibrium, the pressure applied at a point is transmitted equally to the other parts of the liquid. √√ (2)
7.2.2

• F₁/A₁= F₂/A₂ √
  F₁/(0,045) √= 20 000/0,28 √
  F₁ = 3 214,29 N √  (4)

[17]

QUESTION 8 
8.1
8.1.1 Communication  √ (1)
8.1.2 Images of bones √/Security machines at airports etcetera.(1)
8.2 Gamma rays √ (1)
8.3

• c = fλ √
  3 x 10⁸ = 5 x 1017 λ √
  λ = 6 x 10^-9 m √ (3)

8.4

• E = hf √
  = h c/ λ
  = 6.63 x10^-34 x 3 x10⁸/ 502x10^-9 √
  = 3.96 x 10^-19 J √ (3)

[9]

QUESTION 9
9.1
9.1.1 Reflection √ (1)
9.1.2 Angle of incidence equal to angle of reflection √
Angle of incidence, angle of reflection and normal are in the same plane √ (2)
9.1.3 b is the angle of reflection √ b = 60º √ (2)
9.2 9.2.1 Breaking up of white light into its constituent colours √√ (2)
9.2.2 Red √ Highest wavelength √ The higher the wavelength the smaller the refraction√ (3)
9.3
9.3.1 Refraction √ (1)
9.3.2 Critical angle √ (1)
9.3.3 Total internal reflection √(1)
9.3.4 Communications √and medicine √ (2)
9.4
9.4.1 (5)
9.4.2 Upright √/ larger than the object/Virtual (Any one) (1)
[21]

TOTAL: 150