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1 Physics - Basics Scheme Level Subject Exam Board Topic Sub Topic Booklet International A Level Physics Edexcel Physics-Basics Scheme Time Allowed: 72 minutes Score: /60 Percentage: /100 Grade Boundaries: A* A B C D E U >85% 777.5% 70% 62.5% 57.5% 45% <45%

2 1 D 1 2 B s A: m can be used for both displacement (vector) and distance (scalar) C: m s 1 can be used for both velocity (vector) and speed (scalar) D: m s 2 can only be used for acceleration (vector) 1 3 C decreases increases A:viscosity change incorrect and flow rate change correct B:viscosity change incorrect and flow rate change incorrect D:viscosity change correct and flow rate change incorrect 1 4 C 1 5 D 1 6 D 1 7 C 1 8 B 1

3 9(a) (QWC work must be clear and organised in a logical manner using technical terminology where appropriate) 9(b) Measure the initial length (of the spring) Or record position of a fixed point Or record the position of the bottom of the spring (with no masses on the spring) Add mass/weight and record the new length/position Repeat for a range of masses/weights Reference to a precaution taken to ensure measurements were accurate e.g. use of set square, method to reduce parallax, hang spring close to rule, do not exceed proportional/elastic limit Plot appropriate graph of extension/length and force/mass Calculate the gradient (of linear region) Appropriate method to find k from their graph 4 3 9(c) (Max 1 if no graph is suggested i.e. use k = F/Δx and average k ) k would not be constant for the spring Or the graph would not be a straight line Or the idea that Hooke s law would not be obeyed Or F = k (Δ)x does not apply 1 Total for 9 8

4 10(a)(i) (i) Diag 2 (resultant) force is W/mg Or the acceleration is g Diagram 1 the (resultant) force is the component of the weight (along the plane) Or see a reference to (F =) mgsinθ Or see a reference to (a =) gsinθ (This can be inferred from a diagram) A comparison between either mgsinθ or gsinθ and mg or g leading to a smaller acceleration in diagram (a)(ii) 10(b)(i) (Max 2 for answer in terms of energy: Initial GPE in diagram 1 is less than in diagram 2, so KE at bottom is less, so max/final velocity is less Objects move the same distance, so time for diagram 1 is longer) Use of or see gsin35 Or gcos55 Acceleration = 5.6 m s 2 mgsin35 = ma a = 9.81 N kg 1 sin35 a = 5.63m s 2 Straight line or curve of time initially increasing with distance from the origin Correct shape curve 2 2 Distance travelled 10(b)(ii) 10(c) Time Time taken = 1/2 (t) Or t/ 2 Or t Or 0.71t Or t/1.4 1 Similar results indicate reliability/repeatability Or variation in pulse means results (on another day) might be different/unreliable 10(d) The time was to nearest second so measurements were not precise Rule video camera Or light gates (connected to a) data logger/computer/timer Or electromagnet, trap door(s), timer 2 2 Total for question 10 12

5 11(a) Construction of correct vector triangle or parallelogram (from which a measurement for the resultant could be made) 3N and 5N correctly scaled Correct directions for 3N, 5N, and a single resultant Weight = 6 N (accept 5.9 N to 6.1 N) 4 Examples of diagrams 3 N 5 N 5 N 3 N 5 N 3 N 11(b) To reduce parallax It ensured the string was directly above its image in the mirror (and hence the paper) Or so that the mark made on the paper is directly beneath the string Or so that the eye is directly above the string/mark Or so that the position marked on the paper of the string is in line with the string and eye 2 Total for question 11 6

6 12(a) 12(b) Straddle: centre of gravity to lie within rectangle 1 (QWC work must be clear and organised in a logical manner using technical terminology where appropriate) Either The idea that less work is done by the athlete using the FF (to reach that height) For the same height (of body) An athlete using the FF has a lower centre of gravity Or for the straddle jump the C of G goes over the bar and for the FF the C of G is below/at the bar Or the same work will be done by the athlete The idea that this will make the height of the athlete using the FF higher (so more likely to clear the bar) For the same centre of gravity 3 Total for question 12 4

7 13(a) 13(b)(i) There is a component of weight parallel to the ramp (opposing motion) Or work is done against weight component Opposing/resultant force decelerates vehicle (Max 1: kinetic energy (of the car) is transferred to gravitational potential energy) Use of W = mg Use of trigonometry to determine the component of the weight/g parallel to the ramp Component of weight parallel to ramp = (N) (b)(ii) W p = kg 9.81 N kg 1 sin10 W p = N Use of ΔW= F Δs Or Use of ΔW= mgδh (with use of the appropriate trig to determine the vertical height) ΔW = J (ecf for weight from (b)(i)) 2 13(b)(iii) (Using show that value ΔW = J) ΔW = N 180 m ΔW = J Use of E k = ½ mv 2 with E k = ΔW v = 25 m s 1 (ecf for ΔW from (b)(ii)) 2 (Using show that value v = 25.4 m s 1 ) E k = J = ½ ( kg) v 2 v = 24.8 m s 1 13(c)(i) Shorter (ramp) distance needed Or the ramp can be at a lower/no/downwards incline 1 13(c)(ii) Work is done to stretch the strips Or energy is transferred from the car to the strips The idea that this energy cannot be returned to the vehicle (beyond the elastic limit) 2 Total for question 13 12

8 14(a)(i) 14(a)(ii) Use of gradient Velocity = (m s 1 ) (accept ) Velocity = = (m s 1 ) 2 Displacement starts and ends at 0 Straight, diagonal line of increasing displacement from s = 0 Maximum displacement(s) of 0.2 m between times of 0.5 s and 1.25 s Dip in displacement near the middle of graph 4 14(a)(iii) 0 (m s 1 ), zero 14(b) Reduces uncertainties Or measurements more precise/accurate Max 2 No reaction time Can be paused/playback/rewound Can take a reading every frame Or more readings (in a given time) Allows values to be checked You can zoom in 3 Total for 14 10