QUESTION 2 2.1 The rate of change of velocity. ✔✔ Accept Change in velocity per unit time (2)
NOTE: If any one of the underlined key words in the correct context is omitted deduct 1 mark.
2.2
OPTION 1 Δy = viΔt + ½ aΔt2✔ 0,5 = (0)(3) + ½ (a)(32) ✔ a = 0,11 m·s-2✔ (3)
OPTION 2 Δy = (vi + vf) Δt ✔ 2 0.5 = (0 + vf) (3) 2 vf = 0,333 m∙s-1 OR vf = vi + aΔt 0,33 = 0 + a(3)✔ a = 0,11 m∙s-2✔ 1 mark for either of the two (3)
OPTION 3 vf = vi + aΔt = 0 + 3a vf- = 3a…………..(i) OR vf2 = vi2 + 2aΔy = 02 + 2a(0,5) vf = √a ………….(ii) 9a2 = a ✓ ∴a = 0,11 m·s-2✓ 1 mark for either of the two (3
2.3
POSITIVE MARKING FROM 2.2 OPTION 1 For the 3 kg mass: Fnet = ma ✔ (mg – T)/(mg + T) = ma (3)(9,8) – T = (3)(0,11) ✔ T = 29,07 N ✔
OPTION 2 For the 3 kg mass: vf = 0,333 m∙s-1 OR POSITIVE MARKING FROM 2.2 vf = vi + aΔt = 0 + (0,11)(3) = 0,33 m∙s-1
Wnet = ∆EK✔ Ww + WT = ½ mvf2 - ½ mvi2 mg∆xcosθ + T∆xcosθ = ½ mvf2 - ½ mvi2 (3)(9,8)(0,5)cos0º + T(0,5)cos180º = ½ (3)(0,332 – 02)✔ 14,7 – 0,5T = -0,16 T = 29,72 N ✔ (3)
OPTION 3 Wnc = ΔEk + ΔEp ✔ TΔxcos180º = ½ mvf2 – ½ mvi2 + mghf - mghi T(0,5)cos180º = ½ (3)(0)2 – ½ (3)(0,33)2 + (3)(9,8)(0) – (3)(9,8)(0,5) ✔ T = 29,72 N✔ (3)
2.4 (4)
Accepted labels
w
Fg / Fw / weight / mg / gravitational force
✔
w
Friction/ Ff / fk / 27 N/ wrywing / Fw
✔
N
Normal (force) / FnormaL / FN /Freaction
✔
T
FT/tension
✔
Notes
Mark awarded for label and arrow
Do not penalise for length of arrows since drawing is not to scale.
Any other additional force(s) Max 3/4
If force(s) do not make contact with body : Max: 3/4
(4)
2.5
POSITIVE MARKING FROM 2.2 AND 2.3
For P Fnet = ma ✔ T- f = ma 29,07 - 27 = m(0,11) ✔ m = 18,82 kg ✔ (Range: 18,60 – 18,82)
OR
For P Fnet = ma ✔ T- f = ma 29,72 - 27 = m(0,11) ✔ m = 24,73 kg ✔ (3)
[15]
QUESTION 3 3.1
Motion under the influence of gravity/weight/gravitational force only. ✔✔ OR
Motion in which the only force considered is gravitational. ✔✔ (2)
NOTE: If ONLY is omitted minus 1 mark.
3.2
OPTION 1 UPWARDS AS POSITIVE Δy = viΔt + ½ aΔt2✔ = (0)(1) + ½ (-9,8)(12) ✔ = - 4,9 m Height = 2Δy = (2)(4,9) = 9,8 m✔
DOWNWARDS AS POSITIVE Δy = viΔt + ½ aΔt2✔ = (0)(1) + ½ (9,8)(12) ✔ = 4,9 m
OPTION 2 Δy = viΔt + ½ aΔt2 -9,8 = 0 + ½ (-9,8)Δt2 Δt = 1,41 s OR vf = vi +aΔt = 0 + (-9,8)(1,41) ✔ = -13,82 m·s-1 Magnitude = 13,82 m·s-1 ✔ 1 mark for either of the two
3.4
OPTION 1 POSITIVE MARKING FROM 3.3 UPWARDS AS POSITIVE vf2 = vi2 + 2aΔy ✔ 0 = vi2 + (2)(-9,8)(4,9) ✔ vi = 9,8 m∙s-1 Fnet∆t = m∆v OR Fnet∆t = m (vf – vi) 1 mark for any Fnet(0,2) ✔ = 0,4[9,8 –(-13,86)] ✔ Fnet = 47,32 N✔
UPWARDS AS POSITIVE vf2 = vi2 + 2aΔy✔ 0 = vi2 + (2)(9,8)(-4,9) ✔ vi = -9,8 m∙s-1 If calculation of 9,8 m∙s-1 is not shown and it is substituted correctly award 1 mark Fnet∆t = m∆v OR Fnet∆t = m (vf – vi) 1mark for any Fnet(0,2) ✔= 0,4[-9,8 - (13,86)]✔ Fnet = -47,32 N
Fnet = 47,32 N✔
OPTION 2 POSITIVE MARKING FROM 3.3 Emech top = E(mech ground) (Ep +Ek)top = (Ep +Ek)bottom (mgh + ½ mv2)top = (mgh + ½ mv2)bottom (0,4) (9,8) (4,9) + 0 = ½ (0,4)vf2 ✔ vi= 9,8 m∙s-1 If calculation of 9,8 m∙s-1 is not shown and it is substituted correctly award 1 mark Fnet∆t = m∆v Fnet∆t = m(vf – vi) Fnet(0,2) ✔ = 0,4[9,8 - (-13,86)]✔ Fnet = 47,32 N✔ (6)
OPTION 3 POSITIVE MARKING FROM 3.3 vf = vi + aΔt ✔ 9,8 = -13,86 ✔+ a(0,2) ✔ a = 118,3 m·s-2 Fnet = ma ✔ = (0,4)(118,3) ✔ = 47,32 N ✔ (6)
[14]
QUESTION 4 4.1 (4)
E(mech top) = E(mech bottom) (Ep + Ek)top = (Ep + Ek)bottom (mgh + ½ mv2)top = (mgh + ½ mv2)bottom (1,5)(9,8)(2) + 0 ✔= 0 + ½ (1,5)v2✔ v = 6,26 m∙s-1✔
4.2 In a/an closed/isolated✓ system, the total✓ linear momentum is conserved. (2) 4.3 (4)
POSITIVE MARKING FROM 5.2 OPTION 2 Wnet = ∆K ✔ Wnc = ∆K + ∆U WT + Wg = ∆K Wnc = (½ mvf2 - ½ mvi2 )+ (mghf - mghi) 9405 ✔= (½ (75)vf2 -0) ✔+ (75)(9,8)(12 -0)✔ vf = 3,95 m⋅s-1✔ T – mg = ma T – 75(9,8) = 75(0,65)✔ T = 783,75 N WT = 783,75 (12) cos0º✔ = 9405 J 9405 – (8820) = ½ (75)(vf2 – 0)✔ vf = 3,95 m∙s-1 (3,949 m∙s-1)✔
[13]
QUESTION 6 6.1 It is the (apparent) change in frequency (or pitch) 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 the relative motion between a source and an observer ✔ (listener). (2)
NOTE: If any one of the underlined key words in the correct context is omitted deduct 1 mark.
6.2 OPTION 1 (7) OPTION 2 (7)
6.3 Greater than ✔(1) [10] QUESTION 7 7.1
The two forces must be equal in magnitude ✔ but in opposite directions ✔ OR
The force experienced by Q due to P, must be equal in magnitude ✔but opposite in direction to the force experienced by Q due to V.✔ (2)
7.2
The magnitude of the electrostatic force exerted by one point charge on another point charge is directly proportional to the product of the (magnitudes of the) charges✔ and inversely proportional to the square of the distance (r) between them. ✔ OR
The force of attraction or repulsion between two point charges is directly proportional to the product of the charges ✔ and inversely proportional to the square of the distance between them. ✔ (2)
NOTE: If mass is mentioned instead of charges. 0/2
NOTE: If any one of the underlined key words in the correct context is omitted deduct 1 mark.
6,708(1 - x) = 7,937x x = 0,458 m x is 0,46 m (away from P)
OPTION 2
E = kQp r2 Enet = kQp - kQv r2 r2 (Enet = Ep - Ev = 0) 0 = (9 × 109)(5 × 10−6 ) - (9 ×109)(7 ×10−6) x 2 (1− x)2 (9 × 109)(5 × 10−6 ) = (9 ×109)(7 ×10−6) (equating two equations) x 2 (1− x)2
∴x = 0,46m (away from P) (5) [9]
QUESTION 8 8.1 (2)
Criteria for sketch
Marks
Lines are directed away from the charge
✔
Lines are radial, start on sphere and do not cross.
✔
8.2 (7)
Q = ne✔ = (8 x 1013)(-1,6 x10-19) ✔ or (8 x 1013)(1,6 x 10-19) = -12,8 x 10-6 Net charge on the sphere = (+ 6 x 10-6 ) + (-12,8 x 10-6) ✔ Qnet = - 6,8 x 10-6 C E = kQ ✔ r2 E = (9 × 109) (6,8 × 10-6) ✔ (0,5)2 = 2,45 x 105 N⋅C-1✔ towards sphere ✔
[9]
QUESTION 9 9.1.1 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. OR The ratio of potential difference across a conductor to the current in the conductor is constant, provided the temperature remains constant. (2)
NOTE: If any one of the underlined key words in the correct context is omitted deduct 1 mark.
9.1.2 (3)
V1 = IR ✔ = (0,6)(4) ✔ = 2,4 V ✔
9.1.3 (2)
POSITIVE MARKING FROM 9.1.2 I6Ω = V R = 2,4 ✔ 6 = 0,4 A✔
OR
6 (I) = 0,6 ✔ 10 I = 1 A I6Ω = 0,4 A✔
OR V4Ω = V6Ω I4ΩR1 = I6ΩR2 (0,6)(4) = I6Ω(6) ✔ I6Ω = 0,4 A ✔
9.1.4 (2)
POSITIVE MARKING FROM 9.1.3 V2 = IR = (0,4 + 0,6)(5,8) ✔ = 5,8 V ✔
9.1.5 (3)
POSITIVE MARKING FROM 9.1.4 AND 9.1.2 OPTION 1 Vext = (5,8 + 2,4) ✔ = 8,2 V V int = Ir = (1) (0,8) ✔ = 0,8 V Emf = 0,8 + 8,2 = 9 V✔
OPTION 2
9.1.6 (3)
POSITIVE MARKING FROM 9.1.5 AND 9.1.3
W = V I ∆t ✔ = (0,8)(1) (15) ✔ = 12 J ✔
W = I2R ∆t ✔ = (1)2 (0,8)(15) ✔ = 12 J✔
W = V2Δt ✔ R = 0,82(15 ) ✔ 0,8 = 12 J ✔
9.2.1 (2)
R = V I = 2,8 0,7 = 4 Ω ✔
9.2.2 Increases ✔
Total resistance decreases, ✔ current/power increases✔, motor turns faster (3) [20]
QUESTION 10 10.1 10.1.1 Split ring / commutator ✔ (1) 10.1.2 Anticlockwise ✔✔(2) 10.1.3 Electrical energy ✔to mechanical(kinetic) energy ✔ (2) 10.2 10.2.1 DC generator: split ring/commutator and AC generator has slip rings✔ OR AC generator: slip ring and DC generator has split rings✔ (1) 10.2.2 (3)
POSITIVE MARKING FROM 10.2.2 OPTION 1 Irms = Vrms = 226,27 ✔ = 6,46 A ✔ (4) R 35
[13]
QUESTION 11 11.1 Work function (of a metal) is the minimum energy needed to eject an electron from the metal/surface (2)
NOTE: If any one of the underlined key words in the correct context is omitted deduct 1 mark.
11.2 (Maximum) kinetic energy of the ejected electrons✔ (1) 11.3 Wavelength/Frequency (of light) ✔ (1) 11.4 Silver✔
According to Photoelectric equation, hf = Wo + ½ mv2 (For a given constant frequency), as the work function increases the kinetic energy decreases. ✔ Silver has the smallest kinetic energy✔ and hence the highest work function. (3)
1.1.5
hf = Wo + ½mv2max ✔ hc = Wo + Ek(max) λ (6,63×10-34)(3×108) = Wo + -18 9,58 ×10-8 2×10-8 9,945 x10-18 = Wo + 9,58 x 10-18 Wo = 3,65 x 10-19 J✔
11.6 REMAINS THE SAME✔
Increasing intensity increases number of photons(per unit time) but frequency stays constant✔ the energy of the photon is the same ✔therefore the kinetic energy does not change. (3) [14]
(2) 
(7) 
(7) 
(2) 