# Physical Sciences Grade 12 Paper 1 EC June 2014 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the EC June 2014, 6 pages. Read online or get the free PDF download, question paper linked when available. - Page: https://www.hlayiso.com/library/physical-science-p1-2014-memo/13cfa6e3-7c62-4d34-97c1-dd37c83b59a6 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2014 - Exam period: EC June - Paper: 1 - Curriculum: CAPS - Pages: 6 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 Paper 1 EC June 2014 Memo](https://www.hlayiso.com/library/physical-science-p1-2014-memo/13cfa6e3-7c62-4d34-97c1-dd37c83b59a6) ## About this document Physical Sciences · Grade 12 · EC June · 2014. Memorandum, 6 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com Physical Sciences 1 June2014 Common Test NSC – Memorandum PHYSICAL SCIENCES P1 MEMORANDUM COMMON TEST JUNE 2014 NATIONAL SENIOR CERTIFICATE GRADE 12 N.B. This memorandum consists of 6 pages including this page. Copyright Reserved Please Turn Over ### Page 2 Downloaded from hlayiso.com Physical Sciences 2 June2014 Common Test NSC – Memorandum SECTION A QUESTION 1 1.1 C  1.2 D  1.3 C  1.4 A  1.5 A  1.6 B  1.7 A  (7 x 2) [10] QUESTION 2 2.1 FN FN – normal force Fapp Fg – weight Ff Ff – friction Fapp – applied force Fg (FN must be less than Fg) (4) 2.2 Fnet = ma 29 cos36o + (-Ff) = 20 x 0  Ff = 23,46 N (3) 2.3 Increase  (1) 2.4 FN increases  Co-efficient of kinetic friction remains constant  From the equation fk = μ k N Friction directly proportional to normal force  (3) [11] Copyright Reserved Please Turn Over ### Page 3 Downloaded from hlayiso.com Physical Sciences 3 June2014 Common Test NSC – Memorandum QUESTION 3 3.1  0,57 s  (2) 3.2 vf = vi + aΔt 0 = vi + (-9,8)(0,57)  = 5,586 m.s-1 Δy = viΔt + ½aΔt2 = (5,586)(0,57) + ½(-9,8)(0,57)2 = 1,59201 m y = 4 x y1 = 4 x 1,59201 = 6,37 m (8) 3.3 Inelastic – The ball does not bounce to the original height.  (2) 3.4 Δy (m) h 4 h 1 1,76 2,9 Δt (s) Checklist Marks Criteria for graph y and y1 correctly shown  1,76 and 2,9 correctly shown  Correct shape from 0 to 1,76 s  Correct shape from 1,76 to 2,9 s  Axes correctly labelled  (5) [17] Copyright Reserved Please Turn Over ### Page 4 Downloaded from hlayiso.com Physical Sciences 4 June2014 Common Test NSC – Memorandum QUESTION 4 4.1 The total linear momentum of an isolated system remains constant in magnitude and direction.  (2) 4.2 Total p before = Total p after mvi + mvi = mvf + mvf (75 x 3) + (2 x –5)  = (77) vf vf = 2,79 m.s-1 to the right  (4) [6] QUESTION 5 5.1 No.  The crate experiences a frictional force/non-conservative force.  The work done by the frictional force/non-conservative force changes the mechanical energy of the crate. (3) 5.2 A force for which the work done in moving an object between two points is independent of the part taken. (2) 5.3 E A B (5) D C 5.4 Fg// = mgsinθ = 120 x 9,8 x  = 176,4 N  (3) 5.5 O J force acts at right angles to the direction of motion.  (Cos 90° = 0) (2) 5.6 The net/total work done on an object is equal to the object’s change in kinetic energy.  (2) 5.7.1 W Ff = Ff∆xcosθ  = 50 x 10 x cos180°  = - 500 J (3) 5.7.2 W Fg// = Fg//∆xcosθ = 176,4 x 10 x cos 0° = 1764 J (2) Copyright Reserved Please Turn Over ### Page 5 Downloaded from hlayiso.com Physical Sciences 5 June2014 Common Test NSC – Memorandum 5.8 Wnet = ∆K  WA + WB + WC + WD + WE = ∆K (0) + (- 500) + 1764 + 0 + WE = 0 WE = - 1264 J (4) OR Wnc = ΔK + ΔU WB + W E = 0 + mghf - mghi - 500 + WE  = 0 - (120)(9,8)(1,5)  WE = - 1264 J 5.9 WE = F∆xcosθ - 1264 = F∆x cos180° F = 126,4 N P = Fv = 126,4 x 1,25 = 158 W (6) [32] QUESTION 6 v  vL 6.1 fL = fs  v  vS 580 = 340 + 0 x fs  340 – 37,5 fs = 516,03 Hz  (4) 6.2 The apparent change in the frequency of a wave when there is relative motion between the source of the wave and an observer.  (2) 6.3 No An increase in the speed of the source producing the sound waves results in a greater number of complete waves reaching the listener per second.  The apparent frequency increases and becomes greater than 580 Hz, the detector cannot detect frequencies greater than 580 Hz.  (3) 6.4 Equal to.  There is no relative motion between the source and the listener. (2) 6.5 6.5.1 change in pitch  (1) 6.5.2 change in colour (1) Copyright Reserved Please Turn Over ### Page 6 Downloaded from hlayiso.com Physical Sciences 6 June2014 Common Test NSC – Memorandum 6.6 The Doppler flow meter uses the Doppler effect to measure the speed of blood flowing through a person’s blood vessels.  Abnormal blood speeds could be an indicator of a health risk.  (3) [16] QUESTION 7 7.1 The badge is referring to a blue shift caused by the Doppler Effect.  For the badge to appear blue the car must be travelling at an extremely high speed.  (2) 7.2 The claim made by the badge is not possible.  The car will have to travel at a speed comparable to the speed of light.  This is too large for the car. 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