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PHYSICAL SCIENCES PHYSICS PAPER 1 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS JUNE 2016

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Physical Sciences
P1
Grade 12
June 2016 Memorandum
Nsc Past Papers And Memos

MEMORANDUM

QUESTION 1

 1.1 B                        1.2 C                 1.3 D               1.4 C                           1.5A
1.6 D                         1.7 C                 1.8 C              1.9 A                            1.10 C

                                                                                                                           [20]

QUESTION 2

 2.1  9,8 m·s-2 ✓ , downwards✓ (2)
 2.2 Yes✓
Only gravitational force is acting on the ball ✓ OR
The ball is moving under the influence of its weight✓

 

(2)
 2.3  
Upwards +OPTION 1
Vf2 = Vi2 + 2gΔy ✓
= (-8)2 ✓ + 2(-9,8)(-1,8) ✓
  Vf = 9,96 m·s-1 ✓		 Downwards + OPTION 2
Vf2 = Vi2 + 2gΔy ✓
= (8)2 ✓ + 2(9,8)(1,8) ✓
Vf = 9,96 m·s-1 ✓
OPTION 3
2.3		 OPTION4
EM (Top) = EM (Floor)
(mgh+½mv2)Top = (mgh+½mv2)Floor ✓
m(9,8)(1,8) + ½ m(8)2✓= 0 + ½ mv2 ✓
              17,64 + ½ (64) = ½ v2
∴ v = 9,96 m·s-1✓

 

 

 

 

 

(4)
2.4  POSITIVE MARKING FROM 2.3
OPTION 1
Downwards +
Vf = Vi + gΔt ✓
9,96 = 8 + 9,8Δt ✓
Δt = 0,2 s ✓
Upwards +
Vf = Vi + gΔt ✓
-9,96 = -8 + (-9,8)Δt ✓
Δt = 0,2 s ✓		  OPTION 2
Downwards +
Δy = viΔt + ½ gΔt2 ✓
1,8 = 8Δt + ½ (9,8) Δt2 
∴ Δt = 0,2 s ✓
Upwards +
Δy = viΔt + ½ gΔt2 ✓
-1,8 = -8Δt + ½ (-9,8) Δt2
∴ Δt = 0,2 s ✓
 OPTION 3
Downwards +
Δy = 𝑉𝑓+ 𝑉𝑖 Δt ✓
            2 
1,8 = 9,96+82 Δt
                2
∴ Δt = 0,2 s ✓		  OPTION 4
Upwards +
Δy = 𝑉𝑓+ 𝑉𝑖 Δt ✓
            2
-1,8 = −9,96+(−8) Δt
                    2
∴ Δt = 0,2 s ✓

 

 

 

 

 

 

 

 

 

 

 

(3)

 

 2.5


 POSITIVE MARKING FROM 2.3
OPTION 1
80% of 9,96 = 7,97 m·s-1
Upwards +
Vf2 = Vi2 + 2gΔy ✓
02 = (7,97)2✓ + 2(-9,8) Δy ✓
∴ Δy = 3,24 m ✓
No, the ball will not reach the ceiling. ✓
Downwards +
Vf2 = Vi2 + 2gΔy ✓
02 = (-7,97)2✓+ 2(-9,8) Δy ✓
Δy = -3,24 m
∴ Δy = 3,24 m ✓
No, the ball will not reach the ceiling✓

OPTION 2
80% of 9,96 = 7,97 m·s-1
Upwards +
2.1
No, the ball will not reach the ceiling
Downwards +
2.2
No, the ball will not reach the ceiling✓

OPTION 3
80% of 9,96 = 7,97 m·s-1
Upwards +
3

No, the ball will not reach the ceiling ✓

OPTION 4
80% of 9,96 = 7,97 m·s-1
Downwards +
4
No, the ball will not reach the ceiling ✓

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(5)

 

2.6  2.6   
 

 Upwards as positive

Marking Guidelines  Marks 
Initial v at -8 m.s-1  ✓
Positive marking from 2.3 & 2.4
Time and velocity shown with which ball hits the floor/
Velocity (-9,96 m.s-1 ) and time (0,2 s) with which ball hits the floor.		  ✓
Positive marking from 2.3 & 2.4
Time and velocity shown with which ball leaves the floor/
Velocity (7,97 m.s-1) and time (0,21 s) with which ball leaves the floor./		  ✓
Two parallel lines ending at v = 0 m.s-1 on t-axis.  ✓

Upwards as positive

Marking Guidelines  Marks 
Initial v at 8 m.s-1/  
Positive marking from 2.3 & 2.4
Time and velocity shown with which ball hits the floor/
Velocity (9,96 m.s-1) and time (0,2 s) with which ball hits the floor.		  ✓
Positive marking from 2.3 & 2.4
Time and velocity shown with which ball leaves the floor/
Velocity (-7,97 m.s-1) and time (0,21 s) with which ball leaves the floor.		  ✓
Two parallel lines ending at v = 0 m.s-1 on t-axis.  ✓

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(4) 

                                                                                                                                           [20]

QUESTION 3

3.1  A force for which the net work done in any closed path is dependent on the path the object travelled ✓✓ (2) 
 3.2  Fk =μkN = 0,22 × 86 × 9,8 Cos 25º✓ = 168,04 N ✓ (2)
 3.3  The net work done on the object is equal ✓ to the change in the object’s kinetic energy
                                    OR
The work done on an object by a resultant/net force is equal ✓ to the change in the object’s kinetic energy

 

(2)
 3.4

 Accepted labels
3.4

w  Fg/Fw/Weight/Gravity/Gravitational force/FgI and Fg⊥
 N  Normal
 ƒ  Ff/frictional force
 F  Applied force/ FA/Fapplied

 

 

 

 

 

 

 

(4)
 3.5  
OPTION 1 Downwards as +
V2f =v2i + 2aΔx
V2f - v2i = 2(1,54)(0,9) ✓
             = 2,772
Wnet = Ek✓
Wman + Wgrav+Wf =½m(v2f -v2i)
Wman + (86×9,8) sin25º (0,9)cos0º + 168,04(0,9) Cos180º ✓=½ (86)(2,772) ✓
Wman = -50,13 J ✓

OPTION 2 Downwards as +
Fnet = ma
-Fman +Fgsin𝜃+ f =ma
-Fman +86 × 9,8sin25º – 168,04=86(1,54) ✓
Fman = -356,18 +168 04 +132,44
Fman =-55,7N

Wman = Fman × cos𝜃 ✓
Wman =(55,7) ✓ (0,9) cos 180º ✓
Wman = -50,13 J ✓

OPTION 3 Upwards as +
Fnet= ma
Fman - Fgsin+ f =ma
Fman - 86 × 9,8sin25 + 168,04 = 86(-1,54) ✓
Fman = 356,18 -168 04 -132,44
Fman =55,7 N

Wman = Fman × cos𝜃✓
Wman =(55,7) ✓ (0,9) cos 180º✓
Wman = -50,13 J ✓

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(5) 

                                                                                                                                                                                   [15]

QUESTION 4

 4.1 The total Mechanical Energy is conserved/ remains constant ✓ in an isolated system / in absence of external forces/ non-conservative forces ✓ 
OR
The sum of gravitational potential energy and kinetic energy of an object remains constant ✓  in an isolated system✓ 

 

(2) 
 4.2
OPTION 1
Em at A = Em at B
(mgh+½mv2)A = (mgh+½mv2)B ✓   
[(5)(9,8)(5)+½(5)(0)2]✓ =[(5)(9,8)(1)+½(5)v2]
∴ v = 8,85 m·s-1✓ 		 OPTION 2
When B is taken as ground surface:
Em at A = Em at B
(mgh + ½ mv2)A = (mgh + ½ mv2)B ✓
[(5)(9,8)(4)+½(5)(0)2]✓=[(5)(9,8)(0)+½(5)v2]
∴ v = 8,85 m·s-1✓
(4)
 4.3 Weight / force of gravity✓  
Normal force✓ 		 (2)
 4.4 C to✓  (1)
 4.5 Equal to✓  (1)

 

QUESTION 5

 5.1.1  The apparent change in the detected frequency (pitch ) or (wavelength) ✓ as a result of the relative motion between a source and an observer (listener) ✓  (2) 
 5.1.2  Away (from submarine)✓
The detected /observed frequency is lower than the actual frequency✓		 (2) 
 5.1.3  
 OPTION 1
FL = 𝑣 ± 𝑉𝑙 Fs✓
        𝑣 ± 𝑣𝑆 
0,985 Fs✓ = 1470 × Fs✓
                    1470+ 𝑣𝑆 ✓
𝑣𝑆 = 22,39 m·s-1✓		  OPTION 2
FL = (0,985)Fs
437 = (0,985) Fs
Fs = 443,655 Hz
FL = 𝑣 ± 𝑉𝑙 Fs✓
        𝑣 ± 𝑣𝑆
437✓ = 1470 ✓× 443,655✓
         1470+ 𝑣𝑆
 𝑣𝑆= 22,39 m·s-1✓
(5)
 5.2  Red shift implies that light emitted by stars shows a shift towards the lower frequencies of the spectrum✓✓ (2)
 5.3  Blue shift✓ (1)

                                                                                                                                                                                                                                                                    [12]

QUESTION 6

6.1   6.1 (2) 
6.2  
OPTION 1  OPTION 2

Fnet = maFnet = ma
At/by  A:4 kg block
In x-direction.
T – ff = mAa
T – μN = mAa✓
T – μmAg =mAa ( equation 1)

At/by A:4 kg block
In y direction (up)
N – Fg = 0
N = Fg = mg

At/by B:8 kg block
In the x direction
–T + Fg = mBa✓
–T + mBg = mBa (equation 2)

Solving the system of equations
T – μmAg – T + mBg = mAa + mBa✓
Removing T and isolating a:
–μmAg + mBg = (mA + mB)a✓
–(0,6)(4)(9,8) +(8)(9,8)✓
= (4+8)a✓
54,88 = 12a
a = 4,57 m·s-2 ✓

6.2 

Adding Eq 1 & Eq 2
4a +23,52 + 8a – 78,4 = 0✓
12a = 54,88
∴ a = 4,57 m·s-2✓
(7)
6.3  
OPTION 1
T – ff = mAa
T -μmAg = mAa
T -23,52 ✓= 4(4,57) ✓
T = 41,8 N ✓		  OPTION 2
–T + mBg = mBa
–T+ (8)(9,8) ✓ = (8)(4,57) ✓
T = 41,84 N✓
(3)
6.4  
OPTION 1
FF = μN✓
N = mg
FF =μmg=(0,6)(4)✓ (9,8) ✓=23,52N✓		 OPTION 2
T – FF = mAa ✓
41,8 - FF✓ = 4(4,57) ✓
FF = 23,52 N✓
(4)

                                                                                                                                        [16]

QUESTION 7

7.1

 7.1

Choose RIGHT as positive
(mmb vimb + mcvic) = (mmb vfmb + mcvfc) ✓
(3050)(15) ✓ + (1650)(-25) ✓ = vf (3050 + 1650) ✓
vf = 0,96 m.s-1 to the right✓

 

 

 

 

 

(5)
7.2   Σ𝐾𝑖 = ½ mMB v2iMB + ½ mC v2iC ✓
= ½(3 050)(15)2 + ½(1 650)(-25)2✓ = 85 8750 J✓
Σ𝐾𝑓 = ½mMB v2fMB + ½ mC v2fC ✓
= ½(3 050)(0,96)2✓+ ½(1 650)(0,96)2✓ = 2 165,76 J ✓
Since Σ𝐾𝑖 ≠ Σ𝐾𝑓 ✓
∴ the collision was inelastic✓

 

 

 

(9)
7.3  Crumple zones in a car ensure that the car comes to rest over a longer period of time✓, (Δt), during an accident / stopping time is increased.
From FNET ∝ 1/Δt , it follows that FNET decreases with an increase in Δt.
∴ The magnitude of FNET determines the extent of passenger’s injuries. ✓

 

 

(3)

                                                                                                                                                                                                                                       [17]
QUESTION 8  

 8.1 Every object in the universe attracts every other object with a force ✓that is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers. ✓ (2) 
 8.2  Data: ME (Mass of the earth) = 5,98 × 1024 kg
           MS (mass of the spaceship) = 3 500 kg
r = 8,53 × 106 m
G = 6,67 × 10-11 N·m2·kg-2
F = 𝐺 𝑀𝐸 ∙𝑀𝑆✓ = (6,67 × 10−11)(5,98 × 1024 )(3500)✓ = 19 186,55 N✓
              𝑟2                         (8,53 × 106 )2✓

 

 

 

(4)
 8.3  Equal to✓ (1)
 8.4  Newton’s third law of motion✓
When object A exerts a force on object B, object B simultaneously exerts an equal force✓ on object A in an opposite direction✓.		 (3)
 8.5 F = 𝐺 𝑀𝐸 ∙𝑀𝑆✓
           𝑟2
4 × 19 186,55 ✓ = (6,67 × 10−11)(5,98 × 1024 )(3500)✓
                                           𝑟2
𝑟2 = (6,67 × 10−11)(5,98 × 1024 )(3 500)
                                76 746,2
𝑟2 = 1,82 × 1013 m2
𝑟 = 4 265 000,54 m✓
= 4,27 × 106 m

 

 

 

 

 

(4)

                                                                                                                                                                                                                                                                                                                                                 [14]

QUESTION 9

9.1  Impulse = Area under graph ✓ Any ONE
= ½ bh + ½ bh + lb
= ½( 2)(10) ✓ + ½ (2)(-10) ✓ + 6(-10) ✓
= - 60 N.s in opposite direction✓

 

(5) 
9.2  Greater than✓✓ (2)
 9.3 East positive
Impulse = mvf – mvi✓ -60 = 10vf – (10)(5,5) ✓ Vf = 0,5 m·s-1✓

 

(4)

                                                                                                                [11]

QUESTION 10

 10.1

10.1 

CRITERIA TO MARK THE PATTERN
Shape of the pattern✓
Direction( arows from positive to negative)✓
Angle of contact / field lines not touching each other✓

 

 

 

 

 

(3) 
 10.2

EA = 𝐾 𝑄𝐴✓
           𝑟2
= (9 × 109)(4 × 10−9)✓
           (20 × 10−3)2✓
= 90 000 N·C-1, to the right

EC = 𝐾 𝑄𝐶 = (9 × 109)(6 × 10−9) ✓ = 86 400 N·C-1, to the right
           𝑟2             (22×10−3)2
∴ ENET = EA + EC = 90 000 + 86 400 (for adding EA and EC)
= 176 400 NC-1
= 1,76 × 105 N·C-1✓, to the right✓

 

 

 

 

 

 

 

(7)
 10.3 F= 𝐾 𝑄𝐴𝑄𝐶✓
         𝑟2
= (9 × 109)(4 ×10−9)✓(6 ×10−9)✓
                  (30×10−3)2✓
= 7,2 ×10-3 N, (attraction)✓

 

 

 

(5)

                                                                                                               [15]

 Total : 150

Last modified on Tuesday, 21 September 2021 09:35
Published in Grade 12 2016 June Exams, Past Papers

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