MEMORANDUM

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

QUESTION 2
2.1 When object A exerts a force on object B, √ object B simultaneously exerts an oppositely directed force of the same magnitude on object A. √ (2)
2.2
 (4)
2.3 Data:

• M_B = 25 kg
  f_B = 15 N
  a = 1,5 m.s^-2
  F_A on B = ?
  F_B on A = ?
  F_net = ma
  F_{A on B} + f_B + F_g = ma          √ Any one
  F_{A on B} + fB + mgSin35°= ma
  F_{A on B} + (- 15) + (- 25x9,8x0,574) √ = 25x1,5 √
  F_{A on B} – 155,63 = 37,50 FA on B = 193,13 N √
  F_{B on A} = - F_{A on B}
  F_{B on A} = - 193,13 N
  F_{B on A} = 193,13 N √ down the slope  √          (6)

2.4

• F_net = ma
  F_Nceba + F_B on A + f_A + Fg = ma √          Any one
  F_Nceba + FB on A + f_A + mgSin35°= ma
  F_Nceba + (- 193,13) + (- 4,5) + (- 10x9,8x0,574) √ = 10x1,5 √
  F_Nceba – 253,88 = 15
  F_Nceba = 268,88 N √     (4)

[16]

QUESTION 3
3.1 When a net/resultant force acts on an object of mass m, the object accelerates in the direction of the net force. √ The accelereation is directly proportional to the net force and inversely proportional to the mass. √  (2)
3.2

• F_net = ma
  Fg + FN + FY = ma √ Any one
  Fg + FN + Fa Sin40° = ma
  
  Fg + (-623,3) + (-500Sin40°) = mx0 √
  Fg - 623,3 – 321,39 = 0
  Fg = 944,69 N
  mg = 944,69 √
  mx9,8 = 944,69 √
  m = 96,40 kg √ (5)

3.3 For 96,4 kg box 

• F_net = ma
  Fa Cos40° + f + T = ma
  500 Cos40° + (-115) +(- T) √ = 96,4a √
  383,02 – 115 – T = 96,4a
  268,02 - T = 96,4a
  T = 268,02 – 96,4a (1)

For 15 kg box

• Fnet = ma √ Any one
  mg + T = ma
  -15x9,8 + T√ = 15a √
  -147 + T = 15a (Substitute)
  
  -147 + (268,02 – 96,4a) = 15a
  121,02 = 111,40a
  a = 1,09 m.s^-2 √     (7)

[14]

QUESTION 4
4.1 It (projectile) is an object upon which the only force acting is the force of gravity. √√
OR/OF
Projectile is an object that experiences only gravitational force. √√ (2)
4.2

(3) 
4.3 The maximum height reached by ball A (Δymax) = 1,28 + 5 = 6,28 m
Maximum height reached by ball B is given by: 

(5)
4.4

(4)
4.5 (Upwards positive)
For ball A
vf = vi + g∆t √
vf = (5) + (-9,8 ). ∆t (1) √
For ball B
vf = vi + g∆t
vf = (8) + (-9,8 ). ∆t (2) √

Equating (1) and (2)/ Stel (1) gelyk aan (2):
Since ball A will be moving downwards then:
-(5 – 9,8.∆t) = 8 – 9,8.∆t √
19,6.∆t = 13
∆t = 0,66 s
∆t = 0,011min √        (5)
[19]

QUESTION 5
5.1 A system on which the net external forces acting is zero. √√  (2)
5.2 Let east be positive

Related Items

• p = mv √
  = 1 140 x 30 √
  = 34 200 kg.m.s^-1, (east) √  (3)

5.3

• Σpi = Σpf √
  mvviv + mcvic = mvvfv +mcvfc
  1 650 _viv+ 1 140 x 30 √ = 1 650 x - 12 + 1 140x 10 √
  1 650 _viv = 11 400 – 34 200 – 19 800 viv = - 25,82
  viv = 25,82 m.s^-1 √, west √     (5)

5.4 Principle of conservation of linear momentum √
The total linear momentum of an isolated system remains constant (preserved). √√  (3)
5.5

• ∆p = m(vf - vi) √
  ∆p = 1 650 (-12 – (-25,82)) √
  ∆p = 22 803 kg.m.s-1 √ (3)

[16]

QUESTION 6
6.1 The net/resultant force acting on an object is equal to the rate of change in momentum of the object. √√ (2)
6.2

• Total E_{k Before} = E_{k tree} + E_{k car}
  = ½m v²_tree + ½m
  = ½(m)(0²) √ + ½(1 125)(25)² √
  = 351 562,5 J √ (4)

6.3 Impulse of the tree:
Let original direction be positive

• F_net∆t (car) = m(vf - vi) √
  = 1 125(-6 – 25) √
  = - 34 875 N
  F_net∆t (car) = 34 875 N, backwards/terugwaarts
  F_net∆t (tree) = 34 875 N √, original direction of the car √ (4)

6.4 

• F_net (car) = m(vf - vi ) √
                          Δt
  F_net (car) = 1 125(-6 - 25) √
                             3,5
  F_net (car) = - 9 964 N
  F_net (car) = 9 964 N, backwards  √ (3)

6.5 When the airbags deploy during collision, the contact time of the driver with the airbag increases. √ According to the equation Fnet∆t = ∆p, when contact time increases the net force experienced by the driver decreases. √ Then, the extent of injury is reduced. (2)
6.6 Decrease √
The final velocity will now be zero. √ Then, according to the equation:
∆p = m(vf – vi), it follows that ∆p will decrease. √ If ∆p decreases then Fnet decreases as they are directly proportional to each other.  (4)
[19]

QUESTION 7
7.1

• F = Gm¹m² √
             r²
  F = (6,67 x10^-11)( 3,4 x10³ )( 5,98 x10²⁴ ) √
                               (2,32 x10⁶ )² √
  = 251 959,05
  = 2,52 x 10⁵ N √ (4)

7.2 Newton’s Law of Universal Gravitation √
Any two objects or particles in the universe attract each other with a force which is directly proportional to the product of their masses √ and inversely proportional to the square of the distance between their centres. √  (3)
7.3(3)
[10]

QUESTION 8
8.1
8.1.1 The net work done on an object is equal √ to the change in the kinetic energy of the object. √
OR 
The amount of work done by a net force √ on an object is equal to the change in the object’s kinetic energy. √ (2)
8.1.2

(4)
8.1.3 (4)
8.2 (5)
[15]

QUESTION 9
9.1
9.1.1 It is the apparent change in frequency (or pitch/wavelength) of the sound detected by a listener because the sound source and the listener have different velocities relative to the medium of sound propagation. √√
OR
An apparent change in observed/detected frequency (pitch/ wavelength) as a result of relative motion between a source and an observer. √√ (2)
9.1.2 (4)
9.1.3 When the listener moves away from the source the sound waves that reach him have longer wavelength √ and a sound with a lower pitch is registered because of a lower frequency. √ (2)
9.1.4 It would be the same √√ (2)
9.1.5

• FL = 0,85fS
  = 0,85 x 190 √
  = 161,5 Hz √  (2)

9.2 To measure the velocity of blood flowing through blood vessels. √
To scan a fetus/foetus. √ (2)
[14]

QUESTION 10
10.1 The magnitude of the electrostatic force exerted by one point charge (Q) on another point charge (Q) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them. √√ (2)
10.2 East is positive
(5)
[7]

TOTAL: 150