GCE Physics A. Mark Scheme for June Unit G484: The Newtonian World. Advanced GCE. Oxford Cambridge and RSA Examinations

Transcription

1 GCE Physics A Unit G484: The Newtonian World Advanced GCE Mark Scheme for June 06 Oxford Cambridge and SA Examinations

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3 G484 Mark Scheme June 06 Annotations Annotation Meaning Benefit of doubt given BP Blank Page Contradiction Incorrect esponse Error carried forward Follow through Not answered question Benefit of doubt not given Power of 0 error Omission mark ounding error Error in number of significant figures Correct esponse Arithmetic error Wrong physics or equation

4 G484 Mark Scheme June 06 Annotation Meaning / alternative and acceptable answers for the same marking point () Separates marking points reject Answers which are not worthy of credit not Answers which are not worthy of credit IGNOE Statements which are irrelevant ALLOW Answers that can be accepted ( ) Words which are not essential to gain credit Underlined words must be present in answer to score a mark ecf Error carried forward AW Alternative wording OA Or reverse argument Subject-specific Marking Instructions All questions should be annotated with ticks where marks are allocated; One tick per mark. 4

5 G484 Mark Scheme June 06 CATEGOISATION OF MAKS The marking schemes categorise marks on the MACB scheme. B marks: M marks: C marks: A marks: These are awarded as independent marks, which do not depend on other marks. For a B-mark to be scored, the point to which it refers must be seen specifically in the candidate s answers. These are method marks upon which A-marks (accuracy marks) later depend. For an M-mark to be scored, the point to which it refers must be seen in the candidate s answers. If a candidate fails to score a particular M-mark, then none of the dependent A-marks can be scored. These are compensatory method marks which can be scored even if the points to which they refer are not written down by the candidate, providing subsequent working gives evidence that they must have known it. For example, if an equation carries a C-mark and the candidate does not write down the actual equation but does correct working which shows the candidate knew the equation, then the C-mark is given. These are accuracy or answer marks, which either depend on an M-mark, or allow a C-mark to be scored. Note about significant figures: If the data given in a question is to sf, then allow to or more significant figures. If an answer is given to fewer than sf, then penalise once only in the entire paper. Any exception to this rule will be mentioned in the Guidance. Penalise a rounding error in the second significant figure once only in the paper. 5

6 G484 Mark Scheme June 06 Question Answer Mark Guidance (a) (i) Gradient /It is the acceleration which is the same (for both) (AW) Note: acceleration must be spelled correctly for this mark Allow: Gradient /It is the acceleration and acceleration is free fall/g/9.8 () Collision is inelastic / kinetic energy is lost (on impact with the ground) Idea that area is height (above ground) / Height (at E) is less (than height of A) (AW) Not heights are not the same Allow: displacement or distance travelled by ball for height (b) (i) u 9.8().7 u 5.8 (m s ) (.) Not g = 0 Note answer to sf is 5.78 (m s - ) (iii) EITHE F t m( v u) and v.5 v h g (ms h 0. 6 (m) 0. ) F t v ( 5.78) O a = F/m = 6/0. (a =) (upwards positive) v = x 75 x 0 - =.5 (m s - ) C Allow ECF from b(i) Allow v = 4 x 5.78 (from graph for C mark) Note: answer to sf is.46 (ms ) Using u = leads to v =.4 scores / Using u = leads to v = 5 scores / Using equation of motion with a = 9.8() is WP scores 0/ NO ECF Allow graphical method using h v² Allow answer in range (m) Total 7 6

7 G484 Mark Scheme June 06 Question Answer Mark Guidance (a) A body will remain at rest or keep travelling at constant velocity unless acted upon by a resultant/net (external) force (AW) Allow speed in straight line for velocity Allow uniform motion (b) (i) They have equal magnitude/ same size They are the same type / nature Allow act for the same time Allow have same line of action Act in opposite directions Act on different bodies Not act in different directions (c) (i) dm Av dt ( 8.5kg s - ) Weight (of fireman) = 9g / W = 9 x 9.8() (= 90 N) Vertical component of water force = 8.5 x 5 sin 55 (= 69 N) Vertical component of contact force = = 00 N C M Allow use of 8. leading to 70 N Note answer to sf is 070 N 9g Note: a bald 00 is WP scores 0/ sin 55 Total 9 7

8 G484 Mark Scheme June 06 Question Answer Marks Guidance (a) (i) C and F G (iii) 5π/4 (=.5 π ) or.9 (rad) (b) (i) Correct shape graph (by eye) Through the points (-5,0) (0,50) and (5,0) Note : Max KE = 80 0 = 50 (mj) ½ (0.45)v max ² = 50 x 0 - v max = 0.47 (m s - ) Allow ECF if max value on y axis from b(i) is used. If max KE = 80 mj then v max = = 0.60 (m s - ) (iii) A vmax = T (5.00 T = 0.47 T = 0.67 (s) ) C Total 8 Allow C mark for correct frequency =.5 (Hz) ECF from b Using v max = 0.60 leads to T = 0.5 (s) and using v max = leads to T = 0.5 (s) 8

9 G484 Mark Scheme June 06 Question Answer Marks Guidance 4 (a) (i) g M G M M g g h h g s 0.9 h (N kg kg 6 ) [any subject] 6 C If square is omitted from.4 x 0 6 score is 0/. Allow mark for M = 6.4 x 0 7 (Mars radius km not converted to m) Allow: h = so g h = ¼ g s GM Allow use of g h h Allow ECF from a(i) b (i) T² ³ with T = period and = orbital radius Allow separation / distance between bodies Do not allow bald radius for D TD P TP (c) D D Speed will increase (km) [any subject] C M0 C mark is for correct substitution GMT Allow use of with possible ECF 4 from a(i) [Note M=6.4 x 0 7 leads to. x 0 km] Because a decrease in orbital radius results in a decrease in period (by Kepler s law) / Correct reference to centripetal force = gravitational force or v =Gm/ Allow GPE decreases so KE increases Total 7 9

10 G484 Mark Scheme June 06 Question Answer Marks Guidance 5 (a) (i) GMM Ignore sign F 4 M F T (b) Centripetal forces on both star are same magnitude / F = F / answer to a equated to similar expression for S Correct working starting from correct a forces M Allow F M T 4 ² M 4 ² M Eg T T (c) M M and (m) (m) A0 C Allow marks if =.6 x 0 (m) And =. x 0 (m) (d) v v. 0 T m s 7 C Possible ECF Mark is for substitution Max mark if T is not converted to seconds ( leads to speed =.9 x 0 ) 0

11 G484 Mark Scheme June 06 Question Answer Marks Guidance (e) Mv 4 M GMM C Allow ECF from (c) and (d) only if method is T correct Allow this C mark if M 4.80 has been cancelled M C M.00 (kg) Total Special case Use of T will lead to.7 x 0 (kg) this scores mark. Do not allow any ECF if this method is used.

12 G484 Mark Scheme June 06 Question Answer Marks Guidance 6 (a) (Gravitational) potential energy is converted to kinetic energy Not GPE to KE and thermal which is then converted to thermal energy/heat Statement that KE to thermal takes place on impact (b) GPE converted in one inversion = 0.05 x 9.8 x. (= 0.94) GPE converted in 50 inversions = 0.94 x 50 = 4.7 (J) (Use of Q =mcδθ to give) 4.7 = 0.05 x c x 4.5 c = 0 (J kg - K - ) C C Allow follow through from their total GPE converted Note answer to sf = (J kg - K - ) (c) No heat is absorbed by the tube/ lost (by conduction) through the tube/all heat goes to pellets All the lead falls through the same height or length of tube/ Lead does not bounce on impact Ignore heat lost to surroundings/air (d) Temperature change is the same M (Since mass is doubled) (max) GPE/KE/total energy is doubled AND Q is doubled Allow mgh = mcδθ and m is same or m cancels Alternative answer Allow marks for any sensible practical suggestions why T is not the same eg double mass means more lead which will not fall full length of tube. Total 0

13 G484 Mark Scheme June 06 Question Answer Marks Guidance 7 (a) An ideal gas has zero/negligible (electrical) PE / All internal energy is (translational) KE (translational) KE absolute/ thermodynamic /kelvin temperature Allow internal energy absolute/ thermodynamic /kelvin temperature (b) (i) Number of moles of helium = 80/0.004 (= x 0 4 ) 4 nt V 5 p. 0 0 V 490 (m ) C C Note: absolute/thermodynamic/kelvin must be used and spelled correctly for second mark Allow use of pv=nkt Use of T in ºC is WP giving max out of Allow follow through(ft) from an error in n number of moles remaining pv T C Use of T in ºC is WP 0/ = 8.68 x 0 Number of moles escaping = x x 0 =. x 0 4 Total 7

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