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Transcript

Yishun Junior College 2009 H2 Physics Prelim P1 (Solutions) 2009 Yishun JC Preliminary Examination H2 Physics Paper 1 Suggested Solution 1. 1 A 2 B 3 D 4 B 5 A 6 B 7 A 8 A 9 A 10 C 11 B 12 D 13 A 14 B 15 B 16 A 17 B 18 D 19 B 20 C 21 B 22 A 23 C 24 B 25 A 26 D 27 B 28 B 29 D 30 A 31 A 32 C 33 B 34 D 35 B 36 B 37 D 38 C 39 C 40 B d VT 0.3520 V L VA d L VT 0.3 0.001 0.3 0.214 VA d L VT 23.0 0.980 1.5 VA VA 0.08 V VA 0.35 0.08 V 2. By adding a resistor in series with the driver cell, the potential difference across resistance wire will become smaller. Hence the balance length will increase and this reduces the percentage uncertainty of VT. 3. Speed of the sphere decreases with time. Velocity is positive as downwards is positive. Therefore change in velocity is negative. 4. Initial velocity of balloon is resultant of the leftward velocity of skateboarder and the upward velocity during the throw. u uy ux After the balloon leaves the hand of the skateboarder, the trajectory is influenced by the weight of the balloon and air resistance to the balloon’s motion. 5. VI = (mc)/t + h. If rate of heat loss to surrounding, h, is to be accounted for, c is supposed to be smaller. Hence, experimental value of c will be overestimated if heat loss is not accounted for. 6. From ideal gas equation, pV = nRT pV T. From graph, (pV)Y < (pV)Z < (pV)X. Page 1 of 5 Yishun Junior College 7. 2009 H2 Physics Prelim P1 (Solutions) To create rotational equilibrium, the sum of moments exerted by the weight of each material about the knife-edge must be equal. Another way is to use visual inspection to estimate the location of the centre of gravity (CG) for the composite rod. The knife-edge must be below the CG for rotational equilibrium. For C and D, the CG lies in the middle of the rod so knife-edge is placed correctly below the CG. For B, the CG is off-centre to the right since material B is concentrated to the right. For A, the CG is also off-centre to the right since material B is concentrated to the right, but the knife-edge is at the centre. Rotational equilibrium cannot be established. 8. Forces acting on the plank are in equilibrium. Hence, they must be able to form a closed triangle. Also, the three forces must pass through a concurrent point, which is Q in this case. S R T W T F Q F P W 9. The largest force exerted by Q on P is the same magnitude as the force exerted by P on Q (by Newton’s 3rd law of motion). Such force can be determined from the product of the mass of Q and the acceleration given (by Newton’s 2nd law of motion). For A: For B: For C: For D: 10. FQP = 3 ma FQP = 2 ma FQP = 1.5 ma FQP = 2 ma Mass is the property of a body which resists change in motion. Magnitude of speed is irrelevant for consideration. Hence in the order of decreasing mass, the list is: aircraft, car, block, alpha particle. Page 2 of 5 Yishun Junior College 11. 2009 H2 Physics Prelim P1 (Solutions) Horizontal forces acting on man: v Fon man by cord Fon man by ground Fon trolley by cord Fon trolley by ground Work done is positive if the applied force and displacement are in the same direction. 12. 1 2 kx 2 kx mgh h 2 2mg 13. At top : mg N mv 2 r In contact N 0 mv 2 mg v gr r mv 2 r N always 0 ; weightlessness is felt when N = 0 At bottom : N mg 14. 15. 16. Speed and magnitude of acceleration are constant but direction of velocity and acceleration are changing. Velocity and hence momentum are tangential to the path. Acceleration and hence force (rate of change of momentum) are perpendicular to the path (towards centre). 1 Trend depicting g 2 (Note: since r is measured from the surface of the Earth, g r does not start from a very large value). 1 2 GMm 2GM . mv v 2 r r Larger mass of Sun results in stronger gravitational field, hence making it harder for hydrogen molecules to escape. 17. Period of wave, T = 4 x 4.0 ms = 16 ms Hence, frequency, f = 1/T = 63 Hz 18. If the new vertical sensitivity is applied to the original waveform, amplitude of trace would have reduced by half on the c.r.o. Since the same trace is obtained, the new intensity corresponds to twice of the original wave amplitude. As intensity (amplitude)2, the new intensity should be four times greater. 19. Since the target’s motion is simple harmonic, it will be the slowest in regions 1 and 5 and fastest in region 3. Page 3 of 5 Yishun Junior College 2009 H2 Physics Prelim P1 (Solutions) 20. Partial vacuum implies less damping on the pendulum. Thus maximum amplitude of forced oscillation would be higher. Frequency at maximum amplitude increases slightly due to lesser damping. 21. Using 22. Using d sin n , for the same order (n = 1) and wavelength , d is inversely proportional to (for small angles). (OR diffraction effects become poorer with larger grating spacing) 23. E 24. W F .s eE s cos 60 o . Work done is positive because the external force (acting ax , the fringe separation x would be largest if both wavelength and slit-toD screen distance are large and slit-to-slit distance is small. 2 E r' E r' 2 4.0 m 2 E' r E 4 4o r Q against the electric force) is in the same direction as the vertical displacement. 25. RA 3t 2t 2 3 2 , RB , RC . 6t 2t 3t Hence, maximum resistance can be established across face P and its opposite face. 26. Resistance of thermistor decreases as temperature increases. 27. The next lower wavelength corresponds to a higher frequency of light which is only slightly higher than f2. This is because energy states of a hydrogen atom are not equally spaced. VS B P2 P1 f f1 28. f2 f3 For fixed p.d. across PQ, maximum power occurs when equivalent R is smallest. Net RA = 4r, RB = r/4, RC = r, RD = r 29. Wb = T m2 = N m –1 A –1 m2 = (N m) A –1 = J A –1 Page 4 of 5 Yishun Junior College 2009 H2 Physics Prelim P1 (Solutions) 30. To find force on Y: Using right hand grip rule, B field on Y due to current in X is out of the plane. Using Fleming’s left hand rule, with B acting out of plane and current flowing left, an upward force is created. Using N3L, force on X is equal in magnitude but opposite in direction. 31. Since there is no change in magnetic flux for the rod when it moves, the induced e.m.f. is zero. 32. d which is the negative of the gradient of flux linkage-time graph dt Since the gradient is a negative constant, induced ε is a positive constant. 33. Vo 1 V 2 Vo 2 . For full sinusoidal wave, Vrms 2 2 For half-wave rectified version given, V2 34. P Vp Vs V 1 Vo 2 Vo 2 . Thus Vrms o 0.35Vo . 4 2 8 8 V2 Vs 50 2.0 10 V R Np 2000 Vp 10 400 V Ns 50 35. KEmax = 3.5 10–19 – 3.1 10–19 = 0.4 10–19 J = 0.25 eV. Thus stopping potential is 0.25 V. 36. Shortest wavelength largest photon energy largest difference between excited state and ground state. Highest possible excited state is closest and lower than energy W for atom B. 37. Doping does not change the energy gap of an intrinsic semiconductor. By introducing dopant atoms, a new energy level is found within the energy gap. Majority charge carriers are easily created because smaller energy difference is found between either a) donor level and conduction band, or b) acceptor level and valence band. 38. An important aspect of lasing action is to create more members at the excited state than the ground state. Thus, it is more important for members at the excited state to stay relatively longer than those at the ground state. Atoms at the ground state will stay at this stable state until they are excited by the energy supplied from the pump. 39. is unique and constant for a particular radioactive source. 40. Using C Co e t , 100 X e ( 8 ) 0.347 ; then e 0.347 ( 6 ) X 200 s 1 1600 1600 ~ End of Paper 1 Solution~ Page 5 of 5