Physical-Sciences-Physics-Grade-12-NSC-MEMO-March-2025-KZN.pdf

) KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 12 ‘ PHYSICAL SCIENCES : COMMON ASSESSMENT TASK MARCH 2025 TEST ‘ 1 MEMO 3 ee ee ee ee ee ee ee ee MARKS: 100 These marking guidelines consist of 9 pages. Copyright reserved Please turn over

Physical Sciences NSC March 2025 Test QUESTION 1 y Fa Av’ (2) 1.2 Byy (2) 1.3 Avy (2) 1.4 Byvv (2) 1.5 Dvv (2) 1.6 Cvv (2) [12] QUESTION 2 2 Accept force diagram: N N F f F ; Ww Ww Accepted labels w_| Fo/ Fu/ weight/588 N/gravitational forcev FI’ f_| (kinetic) friction/Fi/fk N_| Fu/Normal/Frormal V7 Notes e Mark awarded for label and arrow. e Do not penalise for length of arrows since drawing is not to scale. e Any other additional force(s): Max % e If everything correct, but no arrows: Max/Maks ¥, 3 (4) 2:2 LEFT AS POSITIVE LEFT AS NEGATIVE Fnet = ma | vy Any one Fret = ma vy Anyone F+f=ma x aloe t ¥ F-70 v =(60)(1,5) v -F + 70 v= (60)(-1,5) v F=160NV F=160NV (4) Note: DO NOT award the fourth mark if the final anwer is F = -160 N Copyright reserved Please turn over

Physical Sciences NSC March 2025 Test 2.3 Criteria for graph Straight line with a positive gradient (accept if line is drawn from the horizontal-axis) Both axes labelled. a (3) Note: If the extrapolated straight line or the straight line passes through the origin, max: 2/3 vv acceleration (m-s) F Applied Force (N) 2.4 The boy continues to move forwardv (1) 2.5 Newton’s First law” A body will remain in its state of rest or motion at constant velocity unless a non-zero net force acts on it.” v (3) [15] Copyright reserved Please turn over

Physical Sciences QUESTION 3 21 3.2 3.3.1 3.3.2 3.4.1 NSC March 2025 Test An object which has been given an initial velocity and then moves under the influence of gravitational force only. vv Marking criteria: If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark 9,8 m-s? Vv downwardsY UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE vr = vi + aAt Y vi = vi + alt ¥ 0 = 20 + (-9,8)At v 0 = -20 + (9,8)At ¥ At = 2,04 s v At = 2,04 s v OPTION 1 OPTION 2 POSITIVE MARKING FROM Q 3.3.1 Ay=vAt+ ; adty v7? = vi? + 2aly v 0? = 20? + 2(- 9.8)Ay v 1 Ay = 20,41 mv ={ (20)(2,04) + 5 (-9,8)(2,04)? }™ = 20,41 mv OPTION 3 OPTION 4 POSITIVE MARKING FROM Q 3.3.1 ay = (Op )ate 20 +0 ay =((——) 2.04 Ay = 20,40 mV POSITIVE MARKING FROM Q 3.3. maximum height =ibh 3 =4(2,04)(20) v = 20,40 mv 1 ACCEPT OPTIONS THAT USE CONS' ERVATION OF MECH ENERGY UPWARDS AS POSITIVE vr = vi + alt ¥ = 20 + (-9,8)(5) v =-29 m:s1 ove= 29 m:st ¥ DOWNWARDS AS POSITIVE vi = vit alt ¥ = -20 + (9,8)(5) v =29ms'v Copyright reserved (2) (2) (3) (3) (3) Please turn over

Physical Sciences 3.4.2 NSC March 2025 Test OPTION 1 UPWARDS AS POSITIVE Ay =viAt+iaAtv = (205) ¥ + 3(-9,8)(5)? ¥ OPTION 1 UPWARDS AS NEGATIVE Ay =viAt+iaAtv = (-20)(5) ¥ + $(9,8)(5)? ¥ = -22,50 = 22,50 mv height = 22,50 m v OPTION 2 OPTION 2 UPWARD AS POSITIVE POSITIVE MARKING FROM Q 3.3.1 At = 5 - 2x2.04 = 0,92s Ay =vAt+laAtv = (-20)(0,92) v + 4(-9,8)(0,92) v DOWNWARD AS POSITIVE POSITIVE MARKING FROM Q 3.3.1 At = 5 - 2x2.04 = 0,92s Ay =vAt+iaAtv = (20)(0,92) v +3 (9,8)(0,92)? v = -22,55 = 22,55mv height = 22,55 m v OPTION 3 OPTION 3 UPWARDS AS POSITIVE POSITIVE MARKING FROM Q 3.3.1 Starting from the maximum height Ay = vjAt +iadty = (0)(2,96) v + 5(-9,8)(2,96)" ¥ =-42,93m Height of the building = 42,93 — 20,41 = 22,52 mv UPWARDS AS NEGATIVE POSITIVE MARKING FROM Q 3.3.1 Starting from the maximum height Ay = v,At + aay = (0)(2,96) v + 5(9,8)(2,96)" ¥ = 42,93 m Height of the building = 42,93 — 20,41 = 22,52 mv OPTION 4 POSITIVE MARKING FROM Q 3.4.1 Ay ee 4) (5)Y Ay = 22,50 mv POSITIVE MARKING FROM Q 3.4.1 vy = vie + aby % (-29)? v= 20? + 2(- 9,8)Ay ¥ Ay = 22,50 m v OPTION 6 POSITIVE MARKING FROM Q 3.3.1 & 3.4.1 0-29 Ay= = 4) (2,96)” Ay = 42,93 m Height of the building = 42,93 — 20,41 = 22,52 mv Copyright reserved (4) [17] Please turn over

Physical Sciences NSC March 2025 Test QUESTION 4 44 80 peor a6 22,22 ms! Vv 4.2 POSITIVE MARKING FROM QUESTION 4.1 RIGHT AS POSITIVE RIGHT AS NEGATIVE Ap = m(vr - vi) “ Ap = m(vs - vi) ¥ = 1 600 (6 — 22,22) v = 1600 (-6 — (-22,22)) v = - 25 952 kg ms? = 25 952 kg ms! = 25 952 kg m:s" to the left v = 25 952 kg m:s" to the left v 4.3 POSITIVE MARKING FROM QUESTION 4.2 25 952 kg m:s"! to the right Vv 4.4 Marking criteria If any of the underlined key words/phrases in the correct context is omitted deduct {.1mark In an isolated system the total linear momentum is conserved/remains constant. vv 4.5 OPTION 1 POSITIVE MARKING FROM QUESTION 4.1 RIGHT AS POSITIVE Copyright reserved Dpi= Dpr McVie + MbVib = (Mc + Me)VF v Any one (1600)(22,22) + 0 v = (1600 + mu)(6) “ “Mb = 4325,33 kg V RIGHT AS NEGATIVE dpi = dpr a MeVic + MbVib = (Mc + Mb)vF Any one (1600)(-22,22) + 0 Y = (1600 + mb)(-6) ¥ “+ Mb = 4325,33 kg V OPTION 2 RIGHT AS POSITIVE App = - Ape v Mb(VbF —Vbi) = - Me(VeF - Vei) ANY-OUS mo(6 - 0) “= -(-25952) v 2. Mb = 4325,33 kg V RIGHT AS NEGATIVE App = - Ape Mb(VbF —Vbi) = - Mc(Vef - Vei) mbo(-6 - 0) v = -25952 v “Mb = 4325,33 kg Vv ¥ Any one (2) (3) (1) (2) (4) [12] Please turn over

Physical Sciences NSC March 2025 Test QUESTION 5 5.1 5.2.1 5.2.2 5.3.1 5.3.2 5.4.1 5.4.2 5.5.1 5.5.2 5.5.3 A series of organic compounds that can be described by the same general formula. vv OR A series of organic compounds in which one member differs from the next with a CHz group. (2) Aldehydes Vv (1) Butanone or Butan-2-one VY (2) 3,4-dibromo-2,2-dimethylpentane Marking criteria e Correct stem: pentane v e Correct substituents: dibromo and dimethyl ~ e IUPAC name completely correct v (3) CnH2nO2Y” (1) C and D (must have both) vv (2) F v (Accept the name of the compound) (1) H HH O}H lid Il} | a a a C-—H i H HH i Marking criteria e Correct functional group “ e Correct number of carbon atoms on either side of functional group v e Whole structure correct v (3) Propylv ethanoateY (2) Sulphuric acid / H2SO4 ¥ (1) Copyright reserved Please turn over

Physical Sciences NSC March 2025 Test 5.6 Marking criteria * Calculate moles of CsH1002 « Using the mole ratio: n(C5H4902):n(C3H7OH) = 1:1 * Substitute 60 g-mot" into n=" e Substitute into the formula for percentage purity Final answer m m n(CsH10O2) = n(C3H7OH) = i= = 0,89 mol O88 60v m=53,4g N(Cs5H49O2) : n(C3H7OH) 14 4, 03:4 % Purity = 60 x 100% ¥ = 89% v QUESTION 6 6.1 Flammable/inflammable v 6.2 BY and DY (accept the names of compounds) 6.3 The boiling points of carboxylic acids are higher than the boiling points of alcohols (of comparable molecular mass). ever OR The longer the chain length the higher the boiling point OR The boling points of alcohols are lower than the boiling points of carboxylic acids (of comparable molecular mass). 6.4 Vapour pressure of compound B will be higher than that of compound A. vv OR Vapour pressure of compound A will be lower than that of compound B. 6.5 e Carboxylic acid in experiment 2 (propanoic acid) has a longer chain Copyright reserved length/greater surface area than the carboxylic acid in experiment 1 (methanoic acid).v e The longer the chain length/greater surface area, the stronger the intermolecular forces.” e More energy needed to overcome the intermolecular forces in compound C .v (5) [23] (1) (2) (2) (2) (3) [10] Please turn over

Physical Sciences NSC March 2025 Test QUESTION 7 7.1 Hydroxyl (group) v (1) 2 4,4-dimethylpentan-1-ol OR 4,4-dimethyl-1-pentanol Marking criteria e Correct stem (pentanol) v e Correct substituent: dimethyl v e IUPAC name completely correct v (3) ts Esters Vv (2) 74 Alkenes ¥ (1) 7.5 Halogenation / bromination v (1) 7.6 H ! 18. ia H—-C-H moe reilly | | | rng C—C—H + Bre——» Br—-Cc—C-C—C-—C—H | ARAL S Hak |e H—-C—H a ed | H H Marking criteria e Correct structure for compound Y v e Correct structure for compound Z ¥ Addition of Br2v (3) Copyright reserved TOTAL: 100