PHYSICS Units 3 & 4 Written examination (TSSM s 009 trial exam updated for the current study design) SOLUTIONS TSSM 017 Page 1 of 1
SECTION A - Multiple Choice (1 mark each) Question 1 Answer: D ( ) ( ) N C -1 Question Answer: B Using the right hand grip rule, thumb in direction of current, fingers gripping wire out of the page at Point P. Question 3 Answer: C Question 4 Answer: C Half the radius of Earth and a quarter of the mass using hence 9.8 N kg -1 will equate to the same as Earth TSSM 017 Page of
Question 5 Answer: A Using right hand slap rule thumb velocity direction, fingers out of the page, force right. Question 6 Answer: B A south end is created at the end closest to the magnet to oppose the change, hence the field is up through the magnet. Using RH rule current would be anticlockwise as viewed from above. Question 7 Answer: D Question 8 Answer: D m s - Question 9 Answer: C Acceleration of the system ( ) m s - TSSM 017 Page 3 of 3
Tension in Rope 1 ( )( ) N Question 10 Answer: B Length contraction applies for the moving muons, so they will have less time to decay and therefore more arrive at the end of the detection zone. Question 11 Answer: B Question 1 Answer: C Radiowaves have the longest wavelength Question 13 Answer: A Air from less dense to more dense slows down. As it slows down its wavelength increases but frequency remains same. TSSM 017 Page 4 of 4
Question 14 Answer: D All points are not in phase, there is an energy transfer, the distance between nodes in half a wavelength. D is correct Question 15 Answer: B Closed Pipe Open Pipe m Hz Question 16 Answer: C Less constructive and destructive interference as you move across the screen. Question 17 Answer: A 0.38 ev is the only energy that cannot be derived from a drop from a given excited state to a lower energy state TSSM 017 Page 5 of 5
Question 18 Answer: B hc Use E A represents a jump from ground to 1 st excited state (1.63 ev). C represents a jump from ground to 3 rd excited state (.85 ev). D represents a photon with enough energy to ionize the atom (3.8 ev) Question 19 Answer: C Heisenberg uncertainty principle states it is impossible to know exact location and momentum of a particle Question 0 Answer: B Systematic errors in experimental observations usually come from the measuring instruments TSSM 017 Page 6 of 6
SECTION B Short Answer Question 1 (8 marks) a. N C -1 As charge is negative and the force is down the field will be up b. N As electron is also negative its force will also be down c. N = 9.1 m s - Question (6 marks) a. Using area under Graph 1 to determine increase in kinetic energy (that is, loss of potential energy). b. KE KE area mass surface surface surface surface v 767 m s 400km KE mv KE KE 3 GM R 4 T 5 0.5 4 10 0.5 0.370 0.593 650 6.5810 1.510 8 1.9110 J -1 GMT R 3 4 6.6710 3 R 6 R.4610 m 11 7 7 10 4 4.110 TSSM 017 Page 7 of 7
Question 3 (5 marks) a. Using the RH slap rule for current, force on Cable A F BIL F B IL 0.1 B.5 0.1 B 0.48T Direction is out of the page b. Attracted Using the RH grip rule, the magnetic field out of the page at Cable A implies that the current through B must also be directed up. Using the RH slap rule for this current and the field due to Cable A (into the page) results in a force on B directed left. Thus cables are attracted to one another. Question 4 (6 marks) a. Current flows from C to B, within a field directed South. RH slap rule results in a force directed east. F F F BIL 0. 0.031.5 0.009 N b. The loop is rotating, but the current, length and magnetic field remain constant for side CD, so the force will not change. Therefore D c. Commutator. Only this device would be able to change the direction of the DC source so that the force on sides CD and AB reverses every 180 o and ensures continuous rotation. TSSM 017 Page 8 of 8
Question 5 (4 marks) a. F F F NET NET NET ma 85 4 340 N b. The net force on the stuntman is 340 N up. So the overall result of Tension and Weight must give 340 N T W F NET 340 T 850 T 1190 N Question 6 (7 marks) a. As per diagram. Note tension force should be greater than weight as the net force must be up (towards the centre of the circular path) Tension Weight b. For uniform circular motion, the centripetal force (net force) must be towards the centre of the circle. In this case, this means up. c. m s -1 At the bottom N TSSM 017 Page 9 of 9
Question 7 (7 marks) a. m s -1 b. Horizontal m s -1 c. m s -1 Question 8 (7 marks) a. Taking east as positive p mv b. p 1500 p 500 kgm s East p Ft p F t 500 F 0.4 F 93750 N East 3 1 1 TSSM 017 Page 10 of 10
c. The collision is not isolated, because external forces (that is, the wall) transfer momentum to the Earth. Question 9 (7 marks) a. v u as v u s a 7 s 9.8 s.5m b. Use conservation of energy principles E E E E TOP TOP TOP 0.5 mgh 5 9.8.5 1.5 J 1.5J 1.5 mgh 0.5mv 1.5 5 9.8 0.5 v 0.5 5 v 6.6m s 1 c. Use conservation of energy principles with elastic potential energy included (a compression of x = 0.3 m E 0. 1.5J 1.5 mgh 0.5kx 1.5-5 9.8 0. k 0.5 0.3 k 504N m -1 TSSM 017 Page 11 of 11
Question 10 (7 marks) a. Lo L b. 50 L 1.9 L 13 m T T o T 41.9 T 7.6hrs c. E E E k k k 1 m c o 8 1.9 1150003 10 1.1510 1 J Question 11 (5 marks) a. C b. With the switch disconnected, the flux through the secondary coil is zero. When the switch is connected, the primary coil creates a magnetic field in the core directed to the left. This left field means that the flux is now directed to the right in the secondary coil. The secondary coil induces a current and an associated opposing magnetic field to the left. This current must therefore be from A to B across the load. Note that once the switch has been connected and there is no more change to the system, the current will not continue in the secondary coil (which is why AC is required for ordinary transformers). TSSM 017 Page 1 of 1
Question 1 (9 marks) a. Use the voltage drop in the cables. V = IR V I = R 50 I = 5 I = 10 A b. : Current at the generator is 10000/50 = 40 A. Current in the lines is 10 A. Thus the transformer is a step-up in the ratio of 1:4 c. 1 4 n V V n 100 300 primstepdown secstepdown 50 4 50 950V 950 37.5V 4 d. V p p V RMS V V p p p p 50 707V TSSM 017 Page 13 of 13
Question 13 (5 marks) a. (Wb) 9.6 x 10-4 0.1 0. 0.3 0.4 Time (s) b. V Question 14 (1) a. Within this experiment, by using a Perspex block and light source, by using known angles of incidence the associated angle of refraction will be found. This will prove that Snell s Law, and for the Perspex will equal 1.5 b. Independent variable = angle of incidence Dependent variable = angle of refraction Controlled variable = material type = Perspex c. TSSM 017 Page 14 of 14
1. Sin θ r 1 0.8 0.6 0.4 0. 0 0. 0.3 0.4 0.5 0.6 0.7 Gradient equals refractive index d. Sin θ i e. Systematic error light not being 100% clear and making it hard to get true indication of angle. Random error errors in measurement of angle of incidence and angle of refraction by eye Question 15 (4 marks) a. A path difference of.5 m represents 4, based on the wavelength of the source. So, 4 dark bands (nodes) would be crossed, corresponding to path differences of: 0.5, 1.5,.5 & 3.5 b. When the path difference between Slit 1 and Slit is a half integer multiple of the wavelength (e.g. 1.5), the sources will arrive at the screen out of phase and complete destructive interference will result, leading to a dark band. TSSM 017 Page 15 of 15
Question 16 (6 marks) a. Both electrons and x-rays will diffract in similar ways, as long as the de Broglie wavelength of the electron is comparable to that of the x-rays. For the electrons: h h. Assuming both are directed towards a similar sized gap or obstacle, the p mv ratio: w will dictate the extent of the diffraction. b. KE Vq KE 600 1.6 10 17 KE 9.6 10 J 19 p mke p 9.110 p 1.3 10 3 31 kgm s 9.6 10 1 17 h p 6.6310 1.3 10 5.0 10 34 3 11 m c. J Question 17 (4 marks) a. KE Vq KE 1.7 1.6 10 v v 7.7 10 5 1 19 1.7 1.6 10 31 9.110 m s 19 TSSM 017 Page 16 of 16
b. Photoelectric effect states: Vq hf hf f f f 0 0 0 0 hf Vq h 6.6310 1.90 10 34 14 14 6 10 1.7 1.6 10 34 6.6310 Hz 19 TSSM 017 Page 17 of 17