HCI SPECIMEN PAPER A Higher Paper Answers

1 Name: __________________________________ ( ) Class: _______ HWA CHONG INSTITUTION SPECIMEN PAPER A HIGHER PHYSICS Level: Secondary Three Duration: 1 hour INSTRUCTIONS TO CANDIDATES Do not open this booklet until you are told to do so. Write your name, index number and class on the top of this page. Answer the questions in Section A and B in the spaces provided. All workings must be shown. In Section B, answer only 2 questions out of the 3 questions. INFORMATION FOR CANDIDATES In Sections A and B the intended marks for each question or part of a question are given in brackets [ ]. Any working should be done in the space provided. When necessary, take acceleration due to gravity, g to be 10 m s-2. This question paper consists of 12 printed pages, including this page.

2 Section A [20 marks] Answer all the questions in this section. 1 An elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. An experiment to investigate coll isions has been set up as shown in figure A below. Two motion sensors are set up at each end of the runway. Trolley A is given a sharp push so that it runs along the runway and collides with trolley B. Figure B shows the velocity time graph of the motion o f the two trolleys. Both trolleys are of mass 500 g. (a) Estimate the distance moved by trolley A before collision. [2] Distance moved = area under the graph from t = 0.40 s to t = 1.60 s[1] = ½ (1.20 + 0.80)(0.35) = 0.35 m [1] (b) Explain why the velocity of one of the graphs is negative. [1] Trolley B travelled towards motion sensor 2 after the collision. [1] Figure A Figure B

3 (c) What is the average force acting on trolley A during the collision? [2] Change in momentum, ∆ρ = F x ∆t 0.500 (0.05 – 0.35) = F x (0.30) [1] F = - 0.50 N [1] (d) A student wrote the following observation in his lab report. Momentum before collision = 0.500 x 0.35 = 0.178 kg m s-1 Momentum after collision = 0.500 x 0.30 = 0.150 kg m s-1 Since momentum before collision is not equal to the momentum after collision, hence momentum is not conserved. Comment on the student’s observation. [2] Momentum is always conserved in collision. [1] Momentum after collision was incorrectly determined as momentum of trolley A was not considered. [1] (e) Explain with evidence if the collision between the trolleys is elastic or inelastic. [3] KE before collision = ½ mvA2 = ½ (0.500)(0.352) = 3.1 x 10-2 J [1] KE after collision = ½ mvA2 + ½ mvB2 = ½ (0.500)(0.05)2 + ½ (0.500)(-0.30)2 = 2.3 x 10-2 J [1] Since KE after collision after collision is not equal to KE before collision, hence collision is inelastic. [1]

4 2 Sea level rises provide one of the most immediate and direct consequences of global warming. A major contribution to sea level rise is thermal expansion of sea water. About 70% of the Earth’s surface is covered with water and the average depth is about 3800 m. The coefficient of volume expansion β, is defined as the fractional change in volume per unit change in temperature in Kelvin. ∆ V = βV0∆T where ∆V is the change in original volume V0 caused by a temperature change ∆T. Water has a rather unu sual expansivity, it contracts when heated from 0°C to 4°C before expanding. Since expansion of sea water is non -linear, the coefficient of volume expansion for water is not constant. The graph below shows the coefficient of volume expansion in SI unit for water at various temperatures. (a) On the graph above, draw the best-fit line for the data points. [1] Smooth curve [1] (b) State if the graph suggests that temperature T and the coefficient of volume expansion β are related by an expression of the form T = a β + c where a and c are constants. [1] No. The graph is non-linear. [1]

5 (c) Deduce the unit for volume expansivity. [1] ∆ V = β V0 ∆T β = ∆ V/( V0 ∆T) Hence unit of β is K-1 or °C-1 . [1] (d) Estimate the total volume of water on Earth. [3] Surface of Earth = 4πr2 = 4π(6.37 x106)2 [1] estimating radius of Earth in Mm Volume of water = surface area x depth = 70% (4π(6.37 x106)2 )(3.8 x 103) [1] = 1.3 x 1018 m3 [1] (e) If the temperature of the sea water is currently at 20°C, (i) calculate the approximate increase in volume of sea water if the average temperature of the sea water rose by just 1.0 °C. [2] ∆ V = βV0∆T = 20 x 10-5 x (1.3 x 1018) [1] reading β correctly from graph at 20°C = 2.6 x 1014 m3 [1] (ii) Predict the rise in sea level which would result from the increase in temperature of just 1.0 °C. [1] Increase in depth = ∆ V/surface area = 2.6 x 1014/(0.70 x 4π(6.37 x106)2 ) = 0.73 m [1] (f) Explain a possible impact on Singapore due to rise in sea level. [1] Less useable area as low lying coastal regions will be submerged. [1]

6 Section B [20 marks] Answer two out of three questions in this section. 3 Animals have different eye structures to enable them to see and be able to survive in their natural habitat. Several eye structures are shown in the diagram below. Rays of light from two points of a distance object are incident into the eyes. (a) A snail is only able to tell which direction light is coming from but cannot form an image, which eye structure does the snail have? [1] P [1] (b) Describe the vision of the animal having eye structure Q. [1] Sharp real image of the distance object formed. [1] P Q R S

7 The figure below shows an image formed by a single spherical surface (similar to eye structure R) separating media of refractive index n1 and n2. If the radius of curvature of the surface is r, then the focal length f, the distance from the centre of curvature C to the focus F is given by: f= n1r (n2-n1) (c) By applying Snell’s law to the refracted ray and given that OF is approximately the same as f+r, show that f = n1r (n2-n1) [4] n1sin a = n2sin b [1] need to indicate clearly where is angle a and b are sin a = x/r [1] basic trigo relation [sin b]/f = [sin (a-b)]/r = [sin (180 – a)]/(f+r) = (x/r)/(f+r) [1] sine rule Hence, n1sin a = n2sin b n1(x/r) = n2 (f(x/f+r)/r) proof [1] n1 = n2 (f/(f+r)) n1 (f+r) = n2f n1f + n1r = n2f n1r = f(n2-n1) f = n1r/(n2-n1) C F f r n2 n1 a x b a-b a O

8 (d) Explain the difference in the focal length of the eye when light rays enter the eye from air and water. [2] Focal length of eye when light rays enters the eye from water is greater than when light ray enters the eye from air [1] Because the refractive index of water is greater than refractive index of air [1] (e) The material in the eye structure R has a refractive index 1.40 and a focal length of 10 cm in air. Determine the focal length in water when the eye is completely in seawater of refractive index of 1.35. [2] f = n1r (n2-n1) 10 = 1.00 r/(1.40-1.00) [1] r = 4.0 cm f = 1.35(4.0)/(1.40 – 1.35) = 110 cm [1]

9 4 Next to the 3000-year-old Drombeg circle in Ireland is a stone-lined pit known as a Fulacht Fiadh. A fulacht fiadh is the name given to an ancient cooking pit dating from the Bronze Age. Fulachta fiadh is generally located beside a source of water, a pit o r hole was constructed and filled with clean water. Archaeology thinks that during that time, water was first heated by placing heated