HCI SPECIMEN PAPER C Higher Paper Answers

1 Hwa Chong Institution Examinations Secondary 3 Integrated Programme CANDIDATE NAME CLASS REGISTER NUMBER HIGHER PHYSICS SPECIMEN PAPER C 1 hour Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class and register number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Write your answers in the spaces provided on the Question Paper. Section B Answer only 2 questions out of the 3 questions. Write your answers in the spaces provided on the Question Paper. Electronic calculators may be used. You may lose marks if you do not show your working or if you do not use appropriate units. The number of marks is given in brackets [ ] at the end of each question or part question. Take g to be exactly 10 m s-2 or 10 N kg-1 unless otherwise stated. This document consists of 15 printed pages and 1 blank page.

2 Section A Answer all the questions in the spaces provided. 1 A simple pendulum of length L consists of a small mass attached to the end of a light string. The time T taken for the mass to swing through one cycle is given by T = 2π√L g where g is the acceleration due to gravity. The student modifies a simple pendulum of length L so that, after release, it swings for a quarter of a cycle before the string strikes a hor izontal thin edge. For the next half cycle, the pendulum swings with a shorter length x. The string then leaves the horizontal thin edge to swing with its original length L. The length L of the string is kept constant during the ex periment. The vertical position of the horizontal thin edge is varied to change x.

3 With the data obtained from the investigation, the student plotted the graph of time period T against √x . The plotted graph is shown below. (a) On the graph, draw the best-fit line for the data. Best fit straight line drawn [1] (b) Determine the value of x when T is 1.40 s. When T = 1.40 s, √x = 0.42 m-1/2 [1] x = 0.18 m [1] x = [2] (c) Explain why the time period for one complete oscillation of the pendulum is given by T= π √g (√L+ √x) As length of pendulum is still L m for half of the oscillation, the time for this half of oscillation is π√ L g . [1] For another half oscillation, the length of the pendulum is x m, hence the time for this half of oscillation is π√ x g. [1] 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 0.40 0.50 0.60 0.70 0.80 0.90 √𝐱/m1/2 T/s

4 Hence time for one complete oscillation is T= π √g (√L+ √x) [2] (d) Determine g from the graph. Determine gradient from graph [2] (1.78 – 1.50)/(0.80-0.52) = 1.0 π √g = 1.0 g = 9.9 m s-2 [1] g = [3] (e) Determine L. Choose a point on the graph for substitute into equation [1] Correct computation of L with unit [1] When √x is 0.80 m-1/2, T = 1.78 s, Hence 1.78 = π √9,9 (√L + √0.80) L = 0.79 m L = [2]

5 2 Gravitational potential is the potential energy per unit mass at a point in a gravitational field. V = mass energy potential The below shows the variation of the gravitational potential V due to Earth with distance R from the centre of the Earth. The radius of the Earth is 6400 km. (a) Use the graph to determine (i) the gravitational potential at the surface of the Earth. Gravitational potential = -6.3 MJ kg-1 [1] [1] (ii) the increase in the potential energy of a 1200 kg satellite when it is raised from the surface of the Earth into a circular orbit of two times the radius of the Earth. difference in gravitational potential = V2R – VR = - 3.2 – (-6.3) [1] = 3.1 MJ kg-1 [1] Increase in potential energy = 3.1 x 1200 = 3700 MJ [1] Increase in potential energy = [3] -7 -6 -5 -4 -3 -2 -1 0 0 5 10 15 20 25 30 35 V/MJ kg-1 R/Mm

6 (b) An object of mass m placed within the gravitational field of the Earth of mass M experiences a force F given by the equation F = G m M R2 where G = 6.7 x 10-11 N m2 kg-2 and R is the distance between object and the center of the Earth. (i) Derive the expression of gravitational potential, V, in terms of G, M and R mg = F = GmM/R2 [1] Work done by moving mass m through a distance R, mg*R = GmM/R Since gravitational potential V = work done per unit mass, V = gR V = GM/R [1] [2] (ii) Calculate the gravitational field strength at a distance two times the radius of the Earth. The mass of the Earth is 5.97 x 1024 kg. g = GM/R2 = (6.7 x 10-11)( 5.97 x 1024)/(2 x 6400 x 103)2 [1] = 2.4 N kg-1 [1] Gravitational field strength = [2] (iii) Explain how your result for part (b) (ii) is consistent with the fact that the gravitational field strength on the surface of the Earth is about 10 N kg-1. Since distance is twice as large as radius of Earth, gravitational field strength is four times smaller. [1] Hence gravitational field strength on the surface of the earth is 4 x 2.4 = 9.6 N kg-1. [1] [2]

7 Section B Answer two out of the three questions in this section. 3 On 14 October 2012, Felix Baumgartner did a sky dive from a height of 36 km, in the upper atmosphere. He exceeded the speed of sound for parts of his dive and landed safety by parachute after freefalling for about 6 minutes. The figure below shows the phases of his daring act. Picture from https://lovetime.fr/wp-content/uploads/2012/10/36855.jpg Phase 1 Balloon launches, carrying Felix in the capsule. Phase 2 Balloon reaches the edge of space in less than 3 hours. Phase 3 At 36 km altitude, Felix jumps. Phase 4 Felix breaks the speed of sound. Phase 5 Free fall duration for about 6 minutes, reaching terminal velocity in this phase. Phase 6 Felix deploys parachute at about 1500 m. Phase 7&8 Felix lands approximately 10 minutes after parachute was deployed at constant velocity.

8 (a) The table below list some data at sea level and at 36 km altitude. Sea Level 36 km altitude Temperature/⁰C 15 -25 Atmospheric pressure/Pa 101 325 318 Air density/kg m-3 1.225 0.00447 The volume of the balloon at 36 km in altitude was 8.22 x 105 m3. What should be the volume of the balloon at sea level? PV/T = constant (318)(8.22 x 105)/(-25 + 273) = (101325)(V)/(15 + 273) [2] V = 3.0 x 103 m3 [1] OR Mass of air at 36 km altitude = (8.22 x 105)(0.00447) kg [1] Volume of air at sea level = (8.22 x 105)(0.00447)/ 1.225 = 3.00 x 105 m3 [1] Volume of balloon = [2] (b) Explain how jumping from 36 km allowed Baumgartner to reach a terminal velocity that is greater than the speed of sound. At greater altitude, there is less air, hence less air resistance encountered. [1] He experienced greater downward resultant force for a longer period of time, accelerate for a longer period of time. [1] [2]

9 (d) Maximum velocity of 1342 km h-1 was achieved after 42 s of his jump. Calculate the mean resultant force which was acting on him during the 42 s if total mass of Baumgaetner and his gears was 120 kg. 1342 km h-1 = 372.8 m s-1 [1] F = ma = 120 x (372.8 – 0)/42 [1] = 1.1 x 103 N Mean resultant force = [3] (c) Sketch a graph to show how acceleration varies with time when he first jumped off the capsule till he landed on the ground. [3] a/ m s-2 t/ s Decreasing acceleration – [1] Zero acceleration for two durations of fall – [1] Decreasing deccleration – [1] [-1] for not labelling axes with unit