CIE A Level Physics Solved Past Paper May/June 2019 P21
1a) Since velocity is a vector quantity, you must give your answer in terms of displacement as that is also a vector quantity rather than distance which is a scalar quantity.
bi) Your final answer needs to be corrected to 2 significant figures.
|v||3.3 × 10^2ms^?1||± 3%|
|P||9.9 × 10^4Pa||± 2%|
For absolute uncertainties, look carefully to the following working:
We need the absolute uncertainty of value k so for that we need to re-arrange the formula given in the question and make k the subject of the formula i.e k=(v^2*?)/P. For k, speed of a sound wave, v is taken twice in the formula so we’ll multiply v’s PERCENTAGE uncertainty with 2 and that gives us 6%. In the formula, ? and P are taken only once so we’ll take their PERCENTAGE uncertainties only ONCE. Third step is to take the sum of all the PERCENTAGE uncertainties we just considered giving us the total of 12%. This explains that PERCENTAGE uncertainty of k (1.42) is 12% however the question asks you to find the absolute uncertainty which would be (12/100)*1.42 and that is equal to 0.1704. Last but not the least, the examiner wants you to make absolute uncertainty corrected to appropriate significant figures. Since in the table above, uncertainties are of 1 significant figure, your absolute uncertainty needs to be corrected to 1 significant figure too so I will turn 0.1704 to 0.2. REMEMBER whenever you are asked to find absolute uncertainty, you will first get the % uncertainty of each quantity like P, v and ?, if some quantity is taken more than once then multiply its % uncertainty to however times its taken in the formula. then sum them up and divide by hundred followed by multiplication with whichever value you’re finding absolute uncertainty for.
FOR MULTIPLICATION AND DIVISION, WE ADD % UNCERTAINTIES FOLLOWED BY OTHER STEPS. FOR ADDITION AND SUBTRACTION, WE ADD ABSOLUTE UNCERTAINTIES.
2 bi) Until t=60 ms, block Y has zero momentum or is at rest position. From t=60 ms to t=100 ms, the momentum of block Y increases at a constant rate (straight line with positive gradient) and then on it moves with the same exact momentum until t=160 ms. There are 3 phases to block Y’s momentum. Block X approaches block Y with momentum 0.40 kgms^-1 while block Y is at rest, they both collide and remain in contact with each other for a short span of time and then separate. While they are in contact, block Y must gain momentum and that’s what can be seen on the graph from t=60 ms to t=100 ms and as soon as they separate, there is no force any longer to make block Y increase its momentum that’s why after t=100 ms, momentum of block Y remains constant. Thus time interval for their contact is 40 ms from t=60 ms to t=100 ms.
ii) 1. Change in momentum between t=80 ms and t=100 ms is constant means velocity is increasing at constant rate or acceleration is constant.
2. Change in momentum is zero means no difference between initial and final velocity and thus no acceleration between t=100 ms and t=120 ms.
iii) Force exerted=change in momentum/time taken for that change. Here you can take any two co-ordinates from t=60 ms to t=100 ms since the line is straight so gradient will be equal throughout that portion.
3b) When the spring was not loaded with 2.5N, it had gravitational energy stored because it was at height. When the spring was loaded with the block and released, it extends converting its gravitational energy into kinetic energy (that’s because it now is moving) and elastic potential energy (that’s because spring now has extended). In short g.p.e=k.e+e.p.e
4 aii) The least charge possible is 1.6×10^-19C and charge will always be a multiple of this value hence 8×10^-20C can never be the charge of oil drop.
b iii) The question already mentions that the oil drop is in equilibrium which means no resultant force acts on it. Oil drop’s weight acts downwards which means its electric force (due to presence of charged plates) must be acting upwards as to balance its weight. The plate on the top is positively charged which would only attract a negatively charged particle. This means oil drop must be negatively charged.
ci) Now that voltage has been decreased, this results in drop in electric force experienced by oil drop acting upwards which means weight is now acting more on the oil drop than electric force. Consequently, oil drop accelerates downwards.
ii) Closer field lines symbolize stronger electric field strength: Now that voltage has decrease, E.F.S also decreases and distance between electric field lines increases.
di) Force X is upthrust force which is a force that acts against weight of oil drop whether its moving or stationary.
ii) Force Y is air resistance which is a frictional force and acts only when the object is speeding up or moving.
5 ai) 1. On displacement-distance graph, x-axis tells distance travelled by the wave. So each oscillation covers a distance of ? (distance between 2 crests) and in total N oscillations were made so total distance travelled by the wave is N?.
distance from part i) is N? and speed of sound wave, v=?*f
c ii) At point X, there is loudness, L and distance travelled by both waves is equal which means phase difference is 0 degrees. At point Q (one closest to X) has a minimum point which means that S1 has travelled an extra distance and that extra distance is half its wavelength. L in the middle is a maximum point which means S1 is now covering an extra distance of one complete wavelength and that means it has a phase difference of 360 degrees. Now Q, one which is closest to Y, is a minimum point having S1 travelled an extra distance of 3/2 wavelength and phase difference of 540 degrees. At point Y, there is a maximum point which means S1 has travelled an extra distance of double its wavelength. Total distance travelled by S1 to point Y is 7.4 + (2*wavelength)= 9.1m