Physical Sciences Grade 11 Limpopo March Test 2023 QP & Memo

Downloaded from hlayiso.com 7 CIE PROVINCIAL GOVERNMENT REPUSLUC OF SOUTH AFRICA DEPARTMENT OF EDUCATION MOGALAKWENA DISTRICT PHYSICAL SCIENCES NATIONAL SENIOR CERTIFICATE TERM 1 PRE-CONTROL TEST 07 MARCH 2023 GRADE 11 MARKS : 100 DURATION: 2 hours STARTING TIME: 08HO00 This question paper consists of 13 pages including this one Page 1 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test INSTRUCTIONS 1. This question paper consists of 13 pages including the cover page 2. Answer all the questions in the answer book 3. You are advised to use the attached DATA SHEETS. 4. Round off your final answer to a minimum of TWO decimal places 5. Show all your calculations including formulae where applicable. 6. Candidates may use non-programmable calculators. 7. Write neatly and legibly. Page 2 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A-D) next to the question number (1.1 - 1.10) in the ANSWER BOOK 1.1 A boy applies a force of 400 N to push a crate as shown below. The frictional force acting on the crate is 300 N. 400 N i—> — 300 N The magnitude of the force exerted by the crate on the boy is... A ON B 50N c 300 N D 400 N (2) 1.2 The following graph represents the relationship between acceleration and net force F for two objects, X and Y, moving on the same frictionless surface. & | Object X Object Y Which one of the following statements is TRUE? Page 3 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test A X has a greater mass than Y. B X has a smaller acceleration than Y for equal forces. Cc X experiences a greater force than Y for equal accelerations. D The gradient of graph X represents the inverse of the mass of (2) 1.3 1.4 1.5 1.6 X. Which ONE of the following physical quantities is a measure of the inertia of a body? A Mass B Energy Cc Velocity D Acceleration (2) Acar moving forward collides with a tree and comes to rest very quickly. The passenger who was not wearing a seat belt continued moving forward and collided with a car’s windscreen. Which law explains this scenario? A Newton’s first law of motion B Newton’s second law of motion Cc Newton's third law of motion D Newton's law on universal gravitation (2) A laptop rests on a table. According to Newton's 3rd law, what is the reaction force to the weight of the laptop? A The upward force of the table on the laptop. B The upward force of the laptop on the earth. Cc The downward force of the Earth on the laptop D The normal force on the laptop. (2) A lady pushes against a heavy crate on a horizontal surface. One action-reaction force pair applicable to this situation is indicated in the diagram below. Page 4 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test Ferate on lady F lady on crate ty Which one of the following statements is correct for this situation? A The crate accelerates because the magnitude of the force of the lady on the crate is greater than the magnitude of the force of the crate on the lady. B The crate remains stationary because the force of the lady on the crate is equal in magnitude to the force of the crate on the lady. Cc The crate remains stationary because the force of the lady on the crate is equal in magnitude to the force of friction on the crate. D The crate moves at constant velocity because the force of the (2) lady on the crate is equal in magnitude to the force of the crate on the lady. 1.7. The gravitational force between two uniform spheres will change to one half of the original force if the ... A Mass of each sphere is halved. B Distance between them is halved. Cc Distance between them is doubled. D Mass of one of the spheres is halved. (2) Page 5 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test 1.8 Two small objects with masses M and m exert a gravitational force of F on each other when separated by a distance d. What gravitational force would be exerted by the objects with masses 3M and 2m at a separation distance of 6d? A -36F B 6F Cc OF D F (2) 6 1.9 Three point charges of magnitudes -3 uC, +3 uC and +2 UC are placed at the three corners of a right angled triangle as indicated in the diagram. -3 uC e 42 uC e e t3uC Which one of the following best represents the direction of the resultant force exerted on the +2 uC charge by the other two charges? D. Sa (2) Page 6 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test 1.10 Which one of the following sketch graphs best represents the relationship between the electric field strength E and the distance d from a given charge Q? > > >» | > (2) [20] QUESTION 2 Two forces, of magnitude 50 N and 80 N, act at a point on a Cartesian plane in the directions shown in the sketch below. x 50|N IW 30° INS 80N 2.1 Give the correct term for the following description: A single vector (1) having the same effect as two or more vectors together 2.2 Calculate the: 2.2.1 Magnitude of the vertical component of the 50 N force (2) 2.2.2 Magnitude of the horizontal resultant (net) force (3) 2.2.3 Magnitude of the resultant (net) force (2) 2.2.4 Direction of the resultant (net) force (2) Page 7 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test 2.3. An athlete at point A runs 100 m north, then 70 m east, then 100 m south and then 70 m south. 2.3.1 Draw a neat labelled vector diagram in the Cartesian place (2) acting from the origin using the tail-to-head method 2.3.2 Explain why the vector diagram from Question 2.3.1 is referred (1) to as a closed vector diagram [13] QUESTION 3 Grade 11 learners investigate the relationship between the net force and acceleration by pulling a trolley across a surface which is slightly inclined to compensate for friction. The trolley is connected to different masses by a string of negligible mass. The string passes over a frictionless pulley. Refer to the diagram below. f oe trolley mass Ticker-tape attached to the trolley passes through the ticker-timer. The acceleration of the trolley is determined by analysing the ticker-tape. The results of the net force produced by the different masses and the acceleration of the trolley were recorded in the table below. Net Force (N) Acceleration (m.s) 0,3 0,41 0,6 0,81 0,9 1,22 1,2 1,63 3.1 Write down a hypothesis for this experiment (2) 3.2 — Identify the: 3.2.1 Independent variable (1) 3.2.2 Dependent variable (1) Page 8 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test 3.2.3 Law being investigated (1) 3.3. Use the graph paper on the ANSWER SHEET and draw a graph of (4) the acceleration versus net force. 3.4 Calculate the gradient of the graph (3) 3.5 Use the gradient of the graph calculated in Question 3.4 to determine (2) the mass of the trolley. 3.6 Use the information gained from your calculation and investigation to (2) explain why overloading the car is dangerous. [16] QUESTION 4 A moving sled pulled by dogs begins to climb a slope which is inclined at 20° to the horizontal. The total mass of the sled and the two passengers is 150 kg. The coefficient of kinetic friction for the sled is 0,145. The dogs exert a combined force of 600 N on the sled. an i 4.1 Drawa labelled free-body diagram (4) 4.2 Calculate the kinetic frictional force acting on the sled (3) 4.3. Calculate the acceleration of the sled. (6) 4.4 — If the direction of the acceleration is down the slope, what can you (1) say about the motion of the sled? 4.5 Consider a car engine which hangs from a chain. A mechanic pulls the engine sideways by applying a force of 100 N to the right. The engine is held in such a position that the chain makes an angle of 30° to the vertical Page 9 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test Calculate the tension T in the chain. (3) [17] QUESTION 5 © 5.1 5.2 5.3 5.4 State Newton’s law of universal gravitation in words. (3) Distinguish between the mass and the weight of an object (2) Mars has a mass of 6.4 x 1023 kg and its moon Phobos has amass of 9,6 = (5) x 10% kg. If the magnitude of the gravitational force between the two bodies is 4,6 x 107° N, calculate how far apart Mars and Phobos are. Calculate the weight of an astronaut whose mass is 72kg on the (3) Phobos if the radius of Phobos is 7,4 x 106 m. [13] Page 10 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test QUESTION 6 Two small conducting spheres which carry charges of +8 nC and +5 nC are placed a distance of 40 mm apart as shown in the diagram below. +8 nC ia +5 nC . « 2 40 mm 6.1 State Coulomb’s Law in words (2) 6.2 Calculate the magnitude of the electrostatic force that one charge (4) exerts on the other. 6.3 Define an electric field. (2) 6.4 Draw the electric field that would surround these two small conducting (2) spheres. Show the conducting spheres as small circles. 6.5 Along the line joining the two small spheres, there is a point P, which (8) is 10 mm from the +8 nC charge. Show that the magnitude of the resultant or net electric field strength at P is 6,7 x 105 N.C7. 6.6 Anelectron is placed at point P. Calculate the force experienced by (3) the electron due to the electric field. [21] GRAND TOTAL: 100 MARKS Page 11 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleasted From Stanmore~akhysics.com Term 1 Control Test DATA FOR PHYSICAL SCIENCES GRADE 11 PAPER 1 (PHYSICS) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Sas! ven cle Inarde Re 6,38 x 10° m Nason ven ‘io Aarde Me 5,98 x 10° kg TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING Vv, =V, +aAt AX = vjAt+4aat? V2 =v2 +2aax ax=[ "ae FORCE/KRAG Fee =Ma w=mg Gm,m Fo smanm F=- 7 2 b= ( A laks ) = fi My = N ELECTROSTATICS/ELEKTROSTATIKA F F= Kai, (k=9,0x 10°N-m?-C*) | E “5 r g-kO (k= 9,0 x 10°Nm?-c) | n=2 r € Page 12 of 13 Mogalakwena District

Downloaded from hlayiso.com Dowaleacted From Stanmore~leysics.com Term 1 Control Test ANSWER SHEET HAND IN THIS ANSWER SHEET TOGETHER WITH THE ANSWER BOOK. NAME: CLASS: Question 3.3 Graph of acceleration versus net EEE a ro Toor (m.s%) Hi oem BA) Hee oH HH ASSSRGGen Gene a a Lt t oe BETEEETEOEETUGSEETE . ro I Fret (N) Page 13 of 13 Mogalakwena District

Downloaded from hlayiso.com m Stanmorephysics.com gece LIMPOPO PROVINCIAL GOVERNMENT REPUSLUC OF SOUTH AFRICA CEPARTMENT OF EDUCATION MOGALAKWENA DISTRICT PHYSICAL SCIENCES NATIONAL SENIOR CERTIFICATE TERM 1 CONTROL TEST 07 MARCH 2023 GRADE 11 MEMORANDUM MARKS : 100 DURATION: 2 hours STARTING TIME: 08H00 This question paper consists of 8 pages including this one Page 1 of 8 Mogalakwena District

Downloaded from hlayiso.com Poswaloactied From Stanmoreyhsysics.com Term 1 Control Test MEMO QUESTION 1 1.4 Dvv 1.2 Dvv 1.3 AVY 1.4 AVY 1.5 BYY 1.6 BYY 1.7 D¥vv 1.8 Dvv 1.9 Dvv 1.10 BYY QUESTION 2 2.1 Resultant (net) vector ¥ 2.2 2.21 F =Fsin@ =50sin 30° v OR 50cos 60° =25NV 2.2.2 F,=Fcos 8 =50cos 30° OR 50sin60° =43.30 N to the left ¥ Substitution marks awarded within the question even if calculations for F, and Fy are wrong Horizontal resultant Frx=(-43.30)+(80) ¥ =36.70 N to the right ¥ Mogalakwena District [20] (1) (2) (3) Page 2 of 8

Downloaded from hlayiso.com Poswaloactied From Stanmoreyhsysics.com Term 1 Control Test MEMO 2.2.3 2.2.4 2.3 2.3.1 2.3.2 Positive marking from Questions 2.2.1 and 2.2.2 (2) Vertical resultant Fry= 25 N upwards Resultant Force FRo= Frx + Fry” FR’ = (36.70)*+ (25)° ¥ Fr=44.41N¥ Positive marking from Question 2.2.1 and 2.2.2 (2) tane=*Z v 25 0=55.74° v OR inQ= 20 sinO= a 8=55.74° OR Also accept direction as: cos@=—28 @ = 90 — 55.74° aa @ = 34.26° 0@=55.74° . (2) 70m ¥v Correct diagram , 100m 100m 70m The first vector starts at the origin and the last vector ends at (1) the origin. (The resultant would have a magnitude of zero since itis drawn from the tail of the first vector to the head of the final vector) ¥ [13] Page 3 of 8 Mogalakwena District

Downloaded from hlayiso.com Poswaloactied From Stanmoreyhsysics.com Term 1 Control Test MEMO QUESTION 3 3.1 3.2 3.3 3.4 Marking criteria e The dependent and independent variables are stated correctly. e State the relationship between dependent and independent variable The (net) force is directly proportional to acceleration if the mass of the trolley is kept constant VW 3.2.1 Net force V 3.2.2 Acceleration ¥ 3.2.3 Newton's second law of motion ¥ The relationship between acceleration and net force ra Pe 00 PP oy s P or N Soo roy Acceleration (m.s?) o nS os Nu ° Fret (N) Marking criteria for the graph Axes with correct/ appropriate scale v 3 or more coordinates correctly plotted vv Drawing a line of best fit v 1,63-0 v Gradient= 72-07 = 1,36 ¥ (2) (1) (1) (1) (4) (3) Page 4 of 8 Mogalakwena District

Downloaded from hlayiso.com Poswaloactied From Stanmoreyhsysics.com Term 1 Control Test MEMO OR — 1,63-1,22 _ Gradient 7205 ~ 1,36 OR 1,22—0,81 _ Gradient= Fo-96 — 1,36 3.5 POSITIVE MARKING FROM QUESTION 3.4. a 1 Gradient= m a FNet 1 1,36=-—v m m=0.74kg ¥ 3.6 For aconstant applied force, an increase in mass decreases the negative acceleration. This cause the stopping time to increase and therefore the vehicle has a greater stopping distance. VW QUESTION 4 4.1 Normal force N Applied force F = 600 N Friction f: Weight w Accept the following symbols Nv | Fy/ Normal force Ff¥ | Kinetic friction force/ f,/ Fl F, Fg¥ | W /mg FY | Fa/Force Applied/600N NB: Do not penalise for the length of the arrows (2) (2) [16] (4) Page 5 of 8 Mogalakwena District

Downloaded from hlayiso.com Poswaloactied From Stanmoreyhsysics.com Term 1 Control Test MEMO 4.2 {= uN (3) = umgcosé ¥ = 0,145 (150)(9,8) cos 20° ¥ = 200,30 N ¥ 4.3. Wy,=mgsiné (6) =(150)(9,8) sin 20° =502,77 N down the slope ¥ Fret=ma (Fapp)+(-Wi)+CH=ma v (600)+(-502,77)+(-200,30)”=150a ¥ -103.07=150a a=-0,69 m.s?¥ a=0,69 m.s* down the slope ¥ 4.4 It means that the sled will slow down. ¥ (1) 45 (3) 30° T w 90° ~=60° F=100N «ano 100 sin 30 => v 100 ~ sin 30° T=200NV¥ [17] Page 6 of 8 Mogalakwena District

Downloaded from hlayiso.com Poswaloactied From Stanmoreyhsysics.com Term 1 Control Test MEMO QUESTION 5 5.1 Every object in the universe attracts every other object in the universe (3) with a force that is directly proportional to the product of their masses vand inversely proportional to the square of distanceY between their centres, ¥ 5.2. The mass of an object is the amount of matter found in an object. ¥ (2) The weight of an object is the force with which the centre of a planet attracts an object. ~ 5.3 F= Gmime v (5) r 46x10 oy = (6:87 10"1 106.4% 1079)(9,6x10')w r2 r=9,44x10°m ¥ 5.4 — Smimg 3 Fe Stim (3) _ (6,67x107"')(9,6x10'°)(72)¥ 2 (7,4*10°) =8,42x10 NV [13] QUESTION 6 6.1 The force between two charges is directly proportional to the product (2) of the charges and inversely proportional to the distance between the charges squared¥¥ 6.2 =p Q1Q2 4 Fok ov (4) -9 -9 8x10 5x10™ |v “en 9 SNe" (40x10) v =2,25 x104N ¥ 6.3 Aregion in space in which an electric charge V will experience a (2) force ¥ Page 7 of 8 Mogalakwena District

Downloaded from hlayiso.com Poswaloactied From Stanmoreyhsysics.com Term 1 Control Test MEMO 6.4 (2) Y direction v shape 6.5 Eenc = kS Esne = kS (8) I r — (9%10°)(8x10%v) _ (9*10°)(5%10) v re i, er i, (10x10 v)* (30x10)? v = 720000 N.C? ¥ =50000N.C? ¥ Enet = 720 000-50 000 Y (towards the right — positive direction) 670 000 NC? ¥ 6.6 F = QE = (1,6x10"!°) x (670 000) ¥ (3) =1,07x 10" NV towards the +8nC charge ¥ [21] GRAND TOTAL: 100 MARKS Page 8 of 8 Mogalakwena District