Physical Sciences Grade 12 P2 Eastern Cape Mock 2019 Errata

Downloaded from hlayiso.com Assessment and Examination Directorate Steve Vukile Tshwete Complex, Private Bag X0032, Zwelistsha, 5605 REPUBLIC OF SOUTH AFRICA, Website: www.ecdoe.gov.za E-mail: Nomvuyo.Mbeleki@edu.ecprov.gov.za Ref. no. 13/P Tel.: (040) 602 7028/082 391 1342 Enquiries: Ms N. Mbeleki Fax: 040 602 7295 TO: DISTRICTS HEADS OF EXAMINATIONS PRINCIPALS OF SCHOOLS IN THE FET BAND FROM: MS N. MBELEKI CES: INSTRUMENT DEVELOPMENT AND MODERATION SECTION SUBJECT: ERRATA – PHYSICAL SCIENCES P2 GRADE 12 SEPTEMBER 2019 DATE: 12 SEPTEMBER 2019 The Physical Sciences P2 Grade 12 September was written on Monday, 09 September 2019. We were made aware of certain amendments and omissions that were discovered during the marking process. In order to address this and to ensure that learners are not disadvantaged, the following standardised approach to marking must be adopted across the Province. The following guidelines with regard to marking was prepared in conjunction with the examiner and moderator. ERRATA THE FOLLOWING IS THE CORRECT ANSWERS. QUESTION 7 Identified problem In QUESTION 7.3 an incorrect value (0,02) was used on the memorandum instead of the correct value (0,2). Implications ✓ This affected 7.3, 7.4.1 and 7.4.2. ✓ Learners who would have done it correctly may be marked incorrect by other markers. Recommendations The solution provided must followed for marking the above-mentioned questions.

Downloaded from Corrected solution for hlayiso.com 7.3, 7.4.1 and 7.4.2: 7.3. T sin𝜃 = FE 480 sin 25o = FE  FE = 202,8567656 N 𝑘𝑄1 𝑄2 FE =  𝑟2 9 𝑥 109 𝑥 𝑄 2 202,8567656 =  0,22 Q = 3,00 x 10-5 C  (5) 7.4.1 POSITIVE MARKING FROM 7.3 OPTION 1 OPTION 2 BOTH CHARGES POSITIVE BOTH CHARGES NEGATIVE 𝑘𝑄 𝑘𝑄 E = 𝑟2  E = 𝑟2  9 𝑥 109 𝑥 3,00 𝑥 10−5 9 𝑥 109 𝑥 3,00 𝑥 10−5 E1 =  E1 =  0,152 0,152 E1 = 12 000 000 N.C-1 right E1 = 12 000 000 N.C-1 left 9 𝑥 109 𝑥 3,00 𝑥 10−5 9 𝑥 109 𝑥 3,00 𝑥 10−5 E2 =  E2 =  0,052 0,052 E2 = 108 000 000 N.C-1 left E2 = 108 000 000 N.C-1 right Enet =E1 + E2 Enet =E1 + E2 Enet = 108 000 000 – 12 000 000  Enet = 108 000 000 – 12 000 000 Enet = 96 000 000 N.C-1 left.   Enet = 96 000 000 N.C-1 right.  (5) 7.4.2 POSITIVE MARKING FROM 7.3 OPTION 1 OPTION 2 𝐹 𝑘𝑄 𝑄 Enet = 𝑛𝑒𝑡  F = 𝑟12 2  𝑞 𝐹 9 𝑥109 𝑥 3,00 𝑥10−5 𝑥1,6𝑥10−19 96 000 000 = 1,6 𝑥𝑛𝑒𝑡  F1 = 10−19 (50 𝑥 10−3 )2 Fnet = 1,536 x 10-11 N F1 = 1,728 x 10-11 N right both Fnet = ma  9 𝑥109 𝑥 3,00 𝑥10−5 𝑥1,6𝑥10−19 F2 = 1,536 x 10-11 = 9,11 x 10-31 a  (150 𝑥10−3 )2 a = 1,69 x 1019 m.s-2  F2 = 1,92 x 10-12 N left Fnet = F1 + F2 Fnet = 1,728 x 10-11 – 1,92 x 10-12 Fnet = 1,536 x 10-11 N right Fnet = ma  1,536 x 10-11 = 9,11 x 10-31 a  a = 1,69 x 1019 m.s-2  (5)

Downloaded from hlayiso.com QUESTION 8 Identified problem Internal resistance was omitted in the circuit diagram of the Afrikaans version. Implication Afrikaans candidates will not be able to answer 8.3.1 with a total of 5 marks. Recommendation Afrikaans learners must be marked out of 145 marks and be converted to 150 marks. We request that this must be brought to the attention of all educators marking these papers and sincerely apologise for the inconvenience. Yours in education. 12 September 2019 MS N. MBELEKI DATE