Uniform Acceleration/ SUVAT Equations

Transcription

1 Uniform Acceleration/ SUVAT Equations Scheme 1 Level International A Level Subject Physics Exam Board Edexcel Topic Mechanics Sub Topic Booklet Uniform Acceleration / SUVAT Equations Scheme 1 Time Allowed: 64 minutes Score: /53 Percentage: /100 Grade Boundaries: A* A B C D E U >85% 777.5% 70% 62.5% 57.5% 45% <45%

2 1 B 1 2 C 1 3 C 1 4 D 1

3 Question Answer 5(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 1 3 5(a)(ii) 5(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 5(b)(ii) 5(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 5(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 5 12

4 Question Answer 6 Use of s = ut + ½at 2 with u = 0 Calculated displacement of (m) Or 0.44 (m) seen Use of scaling factor (0.215/0.50) for the candidate s value of displacement seen or implied from position on diagram (This can be awarded if the candidate has used an incorrect displacement) Correct positions of the two times at 0.02 m and 0.19 m from the 0 line 4 Total for question 6 4

5 Question 7(a) Answer Use of area under the graph Use of counting squares method to find the area under the curve Or approximates the area to at least one triangle and one rectangle Height = 5.2 to 5.5 (m) 3 7(b)(i) 7(b)(ii) 7(c) Area under the graph = (27 cm squares 0.2 s 1 m s 1 ) Height = 5.4 m Use of gradient of the graph Or use of a = (v-u)/t (using any pair of points on graph) Using a tangent up to t = 0.2 s Or a pair of values up to t = 0.2 s Maximum acceleration = 7.6 to 8.4 m s 2 (5.0 0) m s 1 Gradient of tangent = (0.64 0) s Maximum acceleration = 7.8 m s 2 Their maximum acceleration value marked on vertical axis (ecf from bi) Maximum acceleration at t = 0 Curve with decreasing gradient drawn from t = 0 to t = 1.4s Acceleration of 0 or close to 0 from 1.4s to 1.58 s (The region of 0 acceleration may begin at any time from 1.2 to 1.4 s) Acceleration/ m s 2 Time/ s Repeat experiment (using the same method) 3 4 Check these values are similar to initial values Or compare 2 results Total for question 7 12

6 Question 8(a)(i) 8(a)(ii) 8(b)(i) 8(b)(ii) Save My Exams! The Home of Revision Answer Use of v 2 = u 2 + 2as a = 2.9 (m s 2 ) a = a = 2.88 m s 2 Use of F = ma to find a or F Maximum a = 3.2 m s 2 Or Force in (a)(i) F = 580 N(or 600 N) (3.2 m s 2 is the maximum acceleration because) the box must have the same acceleration as the lorry a = 630N/200 kg a = 3.15 m s 2 W parallel = Wsinθ W perpendicular = Wcosθ (Accept mg, 200g or 1962 for W) F = Wsinθ Or F= W parallel Or R = Wcosθ Or R= W perpendicular Substitute F = 0.32R into candidate s equation for F or R Use of sinθ/cosθ = tanθ θ = Total for question 8 11

7 Question 9(a) Answer Correct trajectory e.g. 1 9(b)(i) 9(b)(ii) Use of trig function appropriate to calculate the horizontal component of velocity Or 2.25 (m s 1 ) seen Use of v =s/t time = 0.67 (s) u h = 4.5 m s 1 cos60 = 2.25 m s 1 t = t = 0.67 s Use of trig function appropriate to calculate the vertical component of velocity Or 3.9 (m s 1 ) seen Use of suitable equation(s) of motion to find the vertical displacement from the release point after 0.67 s Displacement from release point = m (ecf for t from (b)(i)) (show that value of 0.7 s gives displacement = 0.32 m 0.33 m) Statement to explain why the ball will miss/overshoot the ring e.g. the ball passes below the net Or the ball will not have reached the height of the ring yet Or 0.41 < 0.7 Or ball undershoots ring (Explanation must be consistent with the calculated value of displacement) 3 4 9(b)(iii) u v = 4.5 m s 1 sin60 = 3.9 m s 1 s = (3.9 m s s) + ( ½ 9.81 m s 2 (0.67 s) 2 ) s = 0.41 m The ball would be travelling with a decreasing (horizontal) speed Or there would be a (horizontal) deceleration The (calculated) time would increase 2 Total for question 9 10