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NATIONAL SENIOR CERTIFICATE EXAMINATION

MARKS: 150

TIME: 3 HOURS

This question paper consists of 18 pages including data sheets

PHYSICAL SCIENCES (P1)

JUNE 2015 EXAMINATION

QUESTION PAPER

GRADE11

Physical sciences /P1 CAPS/Grade 11 MPU DBE /June 2015

2

INSTRUCTIONS AND INFORMATION

1 Write your name in the appropriate space on the FOLIO PAPER.

2 This question paper consists of TEN questions. Answer ALL the questions.

3 Start EACH question on a NEW page in the FOLIO PAPER.

4 You may use a non-programmable calculator.

5 You may use appropriate mathematical instruments.

6 YOU ARE ADVISED TO USE THE ATTACHED DATA SHEETS.

7 Show ALL formulae and substitutions in ALL calculations.

8 Number the answers correctly according to the numbering system used in this question paper.

9 Round off your final numerical answers to a minimum of TWO decimal places.

10 Give brief motivations, discussions, et cetera where required.

11 Write neatly and legibly.


3

QUESTION 1

Four options are provided as possible answers to the following questions. Each question has only

ONE correct answer. Choose the answer and write the correct answer (A-D) next to the Question

number (1.1 – 1.10) in the FOLIO PAPER.

1.1 Which of the following forces always acts perpendicular to the surface on which the

object is placed?

A Tension force

B Normal force

C Frictional force

D Gravitational force (2)

1.2 Identical isolated bodies A and B are 200 km apart and experience an attractive

gravitational force F between each other. If the two bodies move away from each

other until a distance of 400km separates them, the magnitude of the new

gravitational force will be…

A ½F

B ¼F

C 4F

D 2F (2)


4

1.3 The magnitude of the resultant force in the diagram below is …

A 18N

B 8N

C 0N

D 13N (2)

1.4 Which one of the graphs below best represents the relationship between

acceleration and the mass of an object.

(2)

1.5 Which of the forces below is a contact force?

A Frictional force.

B Gravitational force.

C Electrostatic force.

D Magnetic force. (2)


5

1.6 A crate of mass 6kg is pulled up a frictionless incline at a constant velocity as shown below.

The magnitude of the tension force in the string is…

A 58,8 N

B 6 N

C 29,4 N

D 3 N (2)

1.7 The law that explains why passengers in a moving vehicle must wear seat belts.

A Newton’s First Law,

B Newton’s Second Law.

C Newton’s Third Law,

D Newton’s Universal Law of Gravitation. (2)

1.8 “An imaginary line that connects points that are in phase”. This statement describes a..

A Diffraction Pattern,

B Nodal line,

C Wave front,

D Wavelet (2)

1.9 Which one below describes an angle of incidence that gives rise to a 90° angle of

refraction

A Angle of reflection

B Total internal reflection

C Angle of refraction

D Critical Angle (2)

30°


6

1.10 A musician is playing a guitar in a large hall. Peter cannot see the musician around the corner but can still hear the music. The phenomenon that allows Peter to hear the musician is…

A Reflection.

B Refraction.

C Diffraction.

D Photoelectric Effect. (2)

TOTAL [2x10 = 20]


7

QUESTION 2 (Start on a new page)

The diagram below shows two forces X and Y of equal magnitude which act on an object

as shown below.

2.1 Define the term resultant Vector (2)

2.2 Calculate the…

2.2.1 horizontal components of X and Y (2)

2.2.2 vertical components of X and Y (2)

2.3 Use answers in questions 2.2.1 and 2.2.2 to draw a single vector diagram including

the resultant force.

(3)

2.4 Calculate the resultant of X and Y. (3)

[12]

Y 200N 200N X

20° 30°


8

QUESTION 3

Study the diagram below and the forces acting on the crate. The mass of the crate is 50 kg. The

forces acting on the crate are in equilibrium. The crate experiences an upward force of 300 N.

3.1 Explain what is meant by the term equilibrium as applied to forces. (2)

3.2 Calculate the weight of the block (3)

3.3 Determine the magnitude of the vertical component of F. (3)

3.4 The block experiences a frictional force of 250N. What is the magnitude of the

horizontal component of F.? (1)

3.5 Calculate the magnitude of F.

(3)

[12]


9

QUESTION 4

Two blocks of mass 2kg and 5kg are connected by a light inextensible string. The system is

pulled up a rough plane inclined at 30° to the horizontal by a force of 60N parallel to the

plane as shown in the diagram below.

The magnitude of the kinetic frictional force between the surface and the 5kg block is 15 N.

The co-efficient of kinetic friction between the 2kg and the surface is 0, 30.

4.1 State Newton’s Third Law of motion.

(2)

4.2 Draw a free body diagram showing ALL the forces acting on the 2kg block.

(5)

4.3 Calculate the magnitude of the:

4.3.1 Frictional force between 2kg block and the surface. (4)

4.3.2 Tension in the string connecting the two blocks. (6)

[17]


10

QUESTION 5

A light inelastic string connects two identical blocks A and B each of mass 5kg as shown below. The two blocks are dragged across a rough surface by a 60N force on block B at 300 to the horizontal. The tension in the string linking A and B is 16N. The blocks accelerate to the right at 2m.s-2

5.1 State Newton’s Second Law of motion in words. (2)

5.2 Draw a free body diagram indicating ALL the force acting on block B. (5)

5.3 Calculate the frictional force experienced by each block. (6)

5.4 If the angle at which the 60N force applied is increased what effect will this have on the frictional force on the block B. Write down INCREASES, DEREASES or REMAIN THE SAME? (1)

5.5 Explain your answer in QUESTION 5.4. (2)

[14]

5 kg

5 kg 16 N

60 N

300

A B


11

QUESTION 6

Grade 11 learners perform an investigation to investigate the relationship between force and acceleration. They obtained the results below.

Force Acceleration (m.s-2) 1 0,25 2 0.50 3 0.75 4 1,00

6.1 Formulate an investigative question for the investigation. (2)

6.2 Write down the.

6.2.1 Independent variable. (1)

6.2.2 Controlled variable. (1)

6.3 Formulate a possible hypothesis for this investigation. (2)

6.4 Apart from the trolley state any TWO apparatus needed to carry out this experiment. (2)

6.5 Draw a graph using the recorded results on the graph paper provided. (5)

6.6 Deduce a conclusion from the graph. (2)

6.7 Using the graph calculate the mass of the object. (4)

[19]

QUESTION 7

7.1 State Newton’s Law of Universal Gravitation in words. (2)

7.2 Calculate the magnitude of the acceleration due to gravity on planet Venus given

that mass of Venus = 4,87 x 1024 kg, Radius = 6.05 x 103km

(4)


12

7.3 Consider an object of mass m placed on the Earth’s surface. Given that the radius of

the Earth is 6, 4 x 106m, G = 6.67 x 10-11Nm2kg-2 and g = 9, 8m.s-2.

7.3.1 Calculate the mass of the Earth. (3)

7.3.2 If the mass of the object is halved, what effect will this have on the force

between them? Write only INCREASES, DECREASES OR REMAIN THE

SAME.

(1)

[10]

QUESTION 8

8.1 A ray of light moves from air into a transparent rectangular glass block as shown in

the diagram. Take the refractive index of air to be 1.

8.1.1 Write down the value of this angle of incidence. (1)

8.1.2 Write down the value of the angle of refraction. (2)

8.1.3 Calculate the refractive index of the block. (4)

8.1.4 Calculate the speed of light through the glass block. (3)

8.1.5 A light ray enters a block (with the same angle of incidence as in QUESTION 8.1) with a higher refractive index than the one calculated in QUESTION 8.1.3. How does the angle of refraction of the block compare to the one calculated in question 8.1.2? Answer only Larger, Smaller or remain the same. (2)


13

8.2 The critical angle of a particular type of glass with respect to air is 42°. A ray of light

is incident on the block of this glass as shown in the sketch below.

8.2.1 Using a ray diagram, show what happens to the light ray as it enters the

glass block.

(4)

[16]

QUESTION 9

9.1 Define refraction. (2)

9.2 Study the diagram below

9.2.1 In the diagram above label A, B, C, and . (5)

9.2.2 Define the term optical density. (2)


14

9.3 Study the table below and answer the questions that follow.

Table 1

Substance Refractive Index Air 1,00

Perspex 1,49

Diamond 2,42

Water 1,33

Glass 1,52

9.3.1 Arrange the substances in the table in ascending order of optical density. (3)

9.3.2 Use the refractive index of glass to calculate the speed of light in glass. (4)

9.3.3 For the same angle of incidence which one would refract light more between Perspex and water?

(1)

9.3.4 In which one between glass and diamond does light travel faster? (1)

[18]


15

QUESTION 10 Munashe a grade 11 learner conducted an experiment to investigate the relationship between slit width and its diffraction. Monochromatic light of wavelength 600nm is first passed through slit A which is at a fixed distance from the screen and the angle of the first minimum is recorded. The experiment is repeated using the same light but slit A is replaced with slit B.

The following results are recorded.

Slit Angle of 1st Minimum A 5° B 10°

10.1 Define the term diffraction. (2)

10.2 For the above investigation write down the following,

10.2.1 Independent variable. (1)

10.2.2 Dependent variable (1)

10.2.3 Controlled variables. (2)

10.3 Using the results, which one between slit A and slit B is larger. (1)

10.4 Explain your answer in 10.4. (1)

10.5 State Hugens’ Principle. (2)

10.6 Briefly explain the formation of dark bands and bright bands on a screen during diffraction.

(2)

[12] GRAND TOTAL: 150 MARKS


16

DATA FOR PHYSICAL SCIENCES GRADE 11

PAPER 1 (PHYSICS)

TABLE 1: PHYSICAL CONSTANTS

NAME SYMBOL VALUE

Acceleration due to gravity g 9,8 m·s–2

Speed of light in a vacuum c 3,0 u 108 m·s–1

Gravitational constant G 6,67 u 10–11

N·m2·kg–2

Coulomb’s constant k 9,0 u 109 N·m2·C –2

Charge on electron e� �1,6 u 10–19

C

Electron mass me 9,11 u 10–31

kg

Permittivity of free space H0 8,85 u 10–12

F·m–1


17

TABLE 2: FORMULAE

MOTION

vf = vi + a't 'x = vi't +

��

12

a't2

vf2 = vi

2 + 2a'x 'x =

��

vf � vi

2§�©�¨�

·�¹�¸�'t

FORCE

Fnet = ma U = mv

F = 221

rmGm F't ='U = mv – mu

Ps =

��

fs (max)

N Pk =

��

fkN

WEIGHT AND MECHANICAL ENERGY

Fg = mg U = Ep = mgh

K = Ek =

��

12

mv2

WAVES, LIGHT AND SOUND

v =

��

f O T =

f1

n1sinө1=n1sinө2


18

ELECTROSTATICS

F =

��

kQ1Q2

r2 (k = 9,0 u 109 N·m2·C –2) E =

qF

E =

��

kQr 2 (k = 9,0 u 10

9 N·m2·C –2) E =

��

Vd

V = QW

W = QEs

U =

��

kQ1Q2

r (k = 9,0 u 10

9 N·m2·C –2) C =

VQ

C =

��

H0Ad

ELECTROMAGNETISM

H =

��

�N ')' t

) = BA

��

Vs

Vp =

��

Ns

Np F = qvB

CURRENT ELECTRICITY

I =

��

Q' t

R = IV

emf = I(R + r) R = r1 + r2 + r3 + …

...1111321

�� rrrR

NATIONAL SENIOR CERTIFICATE EXAMINATION

MARKS: 150

This memorandum consists of 9 pages

PHYSICAL SCIENCES (P1)

JUNE 2015 EXAMINATION

MEMORANDUM

GRADE11

Physical sciences /P1 CAPS/Grade 11-Memorandum MPU DBE /June 2015

Question 1 1.1. B (2) 1.2. B (2) 1.3. C (2) 1.4. C (2) 1.5. A (2) 1.6. C (2) 1.7. A (2) 1.8. C (2) 1.9. D (2) 1.10. C (2) [20] Question 2 2.1. A vector that has the same effect as two or more vector quantities.

33 (2) 2.2. 2.2.1 &F = 200 Cos 200

= 187,94 N3

YF = 200 Cos 300

= 173,21 N 3 (2) 2.2.2 &y = 200 Sin200

= 68,40 N 3

Yy = 200 Sin300

= 100 N 3 (2) 2.3 Xres = 14,73 N due East

Yres = 168,40 N due North (3)

168,40 N3

14,73 N3

Fnet 3


2.4

= (168,40)2 + (14,73)2 3

Fnet = √ 3

= 169,04N 3 (3) [12] Question 3 3.1 The net force / the sum of the forces acting on the object is equal to

zero. 33 (2) 3.2 Fg = mg 3

= 50(9,8) 3

= 490,0N 3 (3) 3.3 Fnet y = 0 3

Fy + 300 = 490 3

Fy = 190N 3 (3) 3.4 Fx = 250 N 3 (1) 3.5 F2 = Fx

2 + Fy2

= 4(190)2 + (250)23

F = √ 3

= 314,01N 3 (3) [12] Question 4 4.1 If an object A exerts a force on object B, object B also exerts an equal

but opposite force on object A. 33 (2) 4.2 OR

(5)

z z

N3

60 N3

w⏊ w//

f3 T3 f3

60 N3 N3

T3

w3

3


4.3 4.3.1 N = Fg

= mgCosϴ

= 2(9.8)Cos3003

= 16,974 N

Ff = µkN3

= 0,3(16,974) 3

= 5,09 N3 (4) 4.3.2 Fnet = ma 3 (on 5kg block)

T + f + Fg// = ma T – 15 – 5 (9,8)Sin300 = 5a3 T - 39,5 = 5a - - - - - - - 1

Fnet = ma (on the 2kg) F + T + f + mgSinθ = ma3 60 – T – 5,02 – 2(9,8)sin300 = 2a3 –T + 45,18 = 2a - - - - - 2

M + N: T- 39,5 + (- T+45,18) = 7a

5,68 = 7a a = 0,81 m·s-13 From 1 : T – 39,5 = 5a T - 39,5 = 5(0,81)

T = 43,55 N3 (6) [17]


Question 5 5.1 It states that when a net force is exerted on an object it causes the

object to accelerate in the direction of the net force, the magnitude of this acceleration is directly proportional to the net force and inversely proportional to the mass of the object. 33 (2)

5.2

OR

(5) 5.3 For block A

Fnet = ma 3

T + fk = ma

16 + (-fk) = 5(2) 3

Fk1 = 16 - 10

Fk1 = 6N 3

For block B

Fnet = ma 3

F// + fk + T = ma (6)

y

F// f

T

N

w

F⏊

3

y

F

f T

N

w 3

3 3

3

3

3

3

3

3


60 Cos 300 - fk - 16 = 5(2) 3

51,9615 – fk -16 = 10

fk = 25,96N 3

5.4 Decreases. 3 (1) 5.5 When the angle increases F⏊ (FSin θ) increases and N = Fg - FSinθ

decreases.3 From f = µkN = f D N 3 hence friction decreases. (2) [14] Question 6 6.1 What is the relationship between the force applied on an object and

its acceleration? 33 (2) 6.2 6.2.1 Force 3 (1) 6.2.2 Mass 3 (1) 6.3 Force is directly proportional to acceleration

OR Force is inversely proportional to acceleration OR As force increases/decreases as acceleration increases/decreases.33 (2)

6.4 Ticker timer 3

Runway 3 Mass pieces 3 (Any two) (2)

6.5

(5)

Marking Guidelines Correct Heading3 Correct Labels3 Correct Plotting of all points33 Correct Shape of graph3


6.6 Force is directly proportional to the acceleration of the object. 33 (2) 6.7

Gradient =

3

=

3 (Any other points will do).

=

= 0,25

Gradient =

0,253 =

Mass =

= 4 kg 3 (4)

[19] Question 7 7.1 It states that every object in the universe will attract every object with

a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. 33 (2)

7.2 g =

3

= 3

3

= 8,746

= 8,75 m·s-1 3 (4) 7.3 7.3.1

g = 3

9.8 = 3

M = 6.02 x 1024kg 3 (3) 7.3.2 Decreases 3 (1) [10]


Question 8 8.1 8.1.1 590 3 (1) 8.1.2 900 - 550 3 = 350 3 (2) 8.1.3 n1Sinθ1 = n2Sinθ23

n1Sin590 = n2Sin3503

n2 =

n2 = 1,493 (4) 8.1.4 n =

3

1,49 =

3

V = 2,01 X 108 m·s-13 (3) 8.1.5 Smaller than 33 (2) 8.2 8.2.1

(4)

[16] Question 9 9.1 It is the bending of light as it passes from one medium to another

medium with a different optical density. 33 (2) 9.2 9.2.1 A – Incident ray 3

B – Normal 3

C – Refracted ray 3

θ1 – Angle of incidence 3

θ 2 – Angle of refraction 3 (5)

3

3

3

3


9.2.2 Optical density is a measure of how fast light travels through

a substance compared to through a vacuum. 33 (2) 9.3 9.3.1 air,water, perspex,glass, diamond 333 (3) 9.3.2 n =

3

1,52 3 =

3

= 1,97 x 108ms-1 3 (4) 9.3.3 Perspex 3 (1) 9.3.4 glass 3 (1) [18] Question 10 10.1 Diffraction is the bending of waves as they pass through a small

opening/an aperture. 33 (2) 10.2 10.2.1 Slit width. 3 (1) 10.2.2 Degree of diffraction/angle of diffraction. 3 (1) 10.2.3 Monochromatic light and distance between the slit and the

screen. 33 (2) 10.3 Slit A 3 (1) 10.4 The larger the slit width the smaller the degree of diffraction. 3

OR As slit width increases degree of diffraction decreases. 3 (1)

10.5 Every point on a wave front acts as a source of secondary wavelets

which spread out in all directions with the same speed as that of the original wave. 33 (2)

10.6 Wavelets that meet out of phase result in destructive interference

leading to the formation of nodal lines resulting in a dark band on the screen. 3 Wavelets that meet in phase result in constructive interference leading to the formation of anti-nodal lines resulting in a bright band on the screen. 3 (2)

[12] TOTAL = 150

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