G481 Mark Scheme January Question Expected Answers Marks Additional Guidance 1 (a) Correct lines from:

Transcription

1 G48 Mark Scheme January 00 G48 Mechanics (a) Correct lines from: B Note: marks for all correct joule (J) to N m mark for two correct watt (W) to J s - 0 marks for none or one correct newton (N) to kg m s - (b) (i) weight in the range 00 to 00 (N) (ii) area in the range 0.0 to 0.08 (m ) (iii) pressure = (b)(i)/b(ii) Allow: sf answer Total 5

2 G48 Mark Scheme January 00 (a) W mg Allow: Use of 9.8 (m s - ) weight (N) or 4. 7 (N) or 5 (N) Allow: Bald 5 (N); but not.50 0 = 5(N) (b) (i) Net / resultant force (on B) is less / (net) force (on B) is less than its weight / there is tension (in the string) / there is a vertical / upward / opposing force (on B) (ii) s ut at and u = t t =.60 (s) (iii) v / v v =.75 (m s - ) / v =.74 ( m s - ) (iv) change in velocity = (= 3.97 m s - ) 3.97 acceleration = acceleration = 3 (m s - ) Note: Must have reference to force Allow: marks for.75/.09 if answer from (iii) is used Allow: sf answer Allow: marks if.80 m is used; time =.7 (s) Possible ecf Allow:.7 or.8 (m s - ) Ignore sign for change in velocity Allow: 30 (m s - ) Special case: acceleration = 3. 3 or 3 (m s - ) scores mark Total 9

3 G48 Mark Scheme January 00 3 (a) mass = (= 40 kg) 40 Allow: 40 (reverse argument) 3.0 (b) (i) total mass = (= 480 kg) total weight = () / total weight = 450 (N) net force = / net force = 664 (N) tension = tension =.7() 0 4 (N) A0 Note: Omitting one of the masses can score maximum of 3 Omitting two masses can score maximum of Examples: 3 marks if mass of cable is omitted tension = = (N) marks if mass of cable and people are omitted tension = = (N) Note: 4 marks for tension = ( m( g a) ) 480 ( ) (ii) ( b)( i) stress = / stress = stress = 4.5(3) 0 7 (Pa) Possible ecf from (i) Note: A tension of (N) gives an answer of 4.4(7) 0 7 (Pa) Total 7 3

4 G48 Mark Scheme January 00 4 (a) The mass (of the electron) increases M Not: mass changes / electron becomes heavier as its speed approaches c / speed of light / m s - (b) (i) A line with correct arrow in the y direction has length of 4 to 6 small squares A line with correct arrow in the x direction has length of 4 to 6 small squares Note: If correct arrows are not shown, then maximum mark is (ii) component = ( 8.0 cos3 )6. 86 (m s - ) or 6.9 (m s - ) Allow: 6.85 as BOD (c) (i) Correct vector triangle drawn.4 (kn).50 (kn) Note: Expect at least one label on the sketch, eg:.4,.5, 90 0 The orientation of the triangle is not important The directions of all three arrows are required 90 0 (resultant force) = (ii) resultant force =.6 (kn).6() (kn) Allow: sf answer of.6 (kn) Allow a scale drawing; marks if answer is within 0. kn and mark if 0. kn Alternative for the marks:.50cos(55) or.4cos(35) resultant force =.50cos(55) +.4cos(35) resultant force =.6 (kn) Possible ecf (Constant velocity implies) zero net force / zero acceleration Total 0 Not: resultant force = drag since the first assumes this 4

5 G48 Mark Scheme January 00 5 (a) Energy cannot be created or destroyed; it can only Allow: Energy cannot be created / destroyed / lost be transferred/transformed into other forms or The (total) energy of a system remains constant or (total) initial energy = (total) final energy (AW) (b) Any suitable example of something strained (eg: stretched elastic band) (c) (i) E p= mgh and E k = mv (Allow h for h) (ii) mgh mv v gh v gh (d) (i) m = V 3 7 m =.0 0 ( ) mass of water = (kg) (ii) 8 loss in potential energy = % of GPE = (=.77 0 ) power = power =.9(63) 0 9 (W) ( GW) (iii) Any correct suitable suggestion; eg: the energy supply is not constant/ cannot capture all the rain water / large area (for collection) 3 A0 A0 Not: E k = mgh Allow any subject for the density equation Allow mark for (J) Allow marks for.77 0 (J) Note: ( 6.5 GW) scores marks Note: Do not allow reference to inefficiency / cost Total 5

6 G48 Mark Scheme January 00 6 (a) The graph shows length and not extension of the spring / spring has original length (of.0 cm) (AW) Allow: length cannot be zero (b) Straight line (graph) / linear graph / force extension / constant gradient (graph) (c).0 force constant = 0.04 force constant = 50 (N m - ) (d) work done = Fx or kx or area under graph Not force length Note: The mark is for any correct substitution Allow: mark for 0.5 (N m - ) 0 n error Allow mark for 5/ 0 - = 4.7 or 4/0 0 - = 40 or 3/8 0 - = 37.5 or /6 0 - = 33.3 or /4 0 - = 5 work done = or Possible ecf work done = 0.09 (J) (e) Find the gradient / slope (of the tangent / graph) Maximum speed at.0s / 3.0s / 5.0s / steepest part of graph / displacement = 0 Note: sf answer is allowed Allow: marks for steepest / maximum gradient Total 8 6

7 G48 Mark Scheme January 00 7 (a) (i) It has maximum / large / increased stress at this point Allow: it has same force but thinner/smaller area Not: Thin / small area (ii) The tape has (permanent) extension / deformation when the force / stress is removed (AW) (b) Measurement: Diameter Any two from: original / initial length (Not: final length) extension / initial and final lengths weight / mass X Note: Need reference to force or stress removed Allow:.. does not return to original size / shape / length when force / stress is removed The term diameter to be included and spelled correctly to gain the mark Equipment: Micrometer / vernier (calliper) (for the diameter of the wire) Any two from: Ruler / (metre) rule / tape measure (for measuring the original length / extension) Travelling microscope (for measuring extension) Scales / balance (for measuring the mass & mg equation is used or for measuring weight) / Newtonmeter (for the weight of hanging masses) / known weights used The term micrometer / vernier (calliper) to be included and spelled correctly to the gain mark. (ALLOW: Micrometer is used to measure area / radius / thickness as BOD) Allow: known masses & mg equation but not known masses Determining Young modulus: stress = force/(cross-sectional) area and strain = extension/original length Young modulus = stress/strain / Young modulus is equal to the gradient from stress-strain graph (in the linear region) Total 0 Allow: stress = F/A and strain = x/l Special case for determining Young modulus: EA Gradient from force-extension graph is L Young modulus = gradient L/A 7