CJC 2017 A level H2 Physics Answers

1 H2 Physics 9749 – 2017 A Level Exam Paper 1 1 Essential Question(s):  What are the 7 base units?  Just memorise! (Base units vs Derived units) Answer: A 2 Essential Question(s):  Question asks for “uncertainty” – In which form? Absolute, Fractional or Percentage?  Percentage uncertainty  How to compute the percentage uncertainty of a derived quantity (i.e. xy2 / z ) from the individual percentage uncertainties (i.e. x, y and z)?  Apply Product-Quotient Rule of combining uncertainties. ݁݃ܽݐ݊݁ܿݎ݁ܲ ݕݐ݊݅ܽݐݎ݁ܿ݊ݑ ݂݋ݔ ݕଶ ݖ= ൬∆ݔ ݔ+ ૛ ∆ݕ ݕ+ ∆ݖ ݖ൰ݔ100% = 6% For options A to D, substitute the values in the 3 columns into the expression above. See which gives 6%. Note that the effect of ∆௬ ௬ is doubled. Option A: 1 + 2(1) + 4 = 7% Option B: 2 + 2(1) + 2 = 6% Option C: 3 + 2(2) + 2 = 9% Option D: 4 + 2(1) + 1 = 7% Answer: B 3 Essential Question(s):  What remain constant/the same over the period of 12 s?  Which are the variables that are involved in the question?  Distance, S  Time, t  Initial Speed, u = 0  Which kinematic equations links up all these variables? s = ut + ½ at2 According to the examiner’s report, student performed well for this question. Approach 1: Acceleration is constant as can be seen from the gradient of this single line graph. Therefore, valid to use the kinematics equation. s = ut + ½ at2 where a = gradient and u = 0. 2 2 0 6 0 6 2 2 0 12 0 12 6 1 12 4 s at s at        Therefore, s6-12 = s0-12 - s0-6 Since it is ratio, we can subtract the parts : s6-12 = 3 6 12 0 6 3 1 s s   

2 Approach 2: Area under a speed time graph is the distance travelled. Red area is 3 times the blue area. Answer : C 4 Essential Question(s):  How will the object’s velocity change as it falls? (it continues to increase until terminal velocity is reached)  What are the forces acting on the object? (weight downwards and air resistance upwards)  Under what conditions will object reach max velocity? (when air resistance equals and opposite to weight) To find vmax, Air resistance = Weight 0.60 x vmax = 3 x 9.81 v = 49 m s-1 acc when v = 12 m s-1, R = 0.6 x 12 = 7.2 N To find acc, equate resultant force = mass x acc mg – R = ma 3 x 9.81 – 7.2 = 3a a = 7.4 m s-2 Answer: D 5 Essential Question(s):  Ball collides inelastically – What happens to the ball’s KE? (decreases)  What happens to the magnitude of momentum after collision? (reduces)  How to determine change in momentum? (pf – pi; Note that these are vector quantities) Max Change in momentum (if elastic) = pf – pi = p – (-p) = 2.0 p (cannot be option D since collision is inelastic and pf less than p) If object comes to rest after collision and pi = 0 Then change in momentum = pf – pi = 0 – (-p) = 1.0 p (cannot be option B since object rebounds upwards so it has some momentum after collision) Change in momentum should fall between 1.0 to 2.0 p for this scenario, therefore option C is a possible option). Option A means that ball continues to move in the same direction as before, after collision which is not the case.

3 Answer: C 6 Essential Question(s):  Which are the objects in interaction? (earth with brick; floor with brick) W, weight of brick = force the earth exerts on the brick Newton’s 3rd law pair = force the brick exerts on the earth S, support force from floor on brick Newton’s 3rd law pair = force from brick on the floor Answer: A 7 Essential Questions:  What kind of motion will the golf ball undergo when it is hit?  What are the characteristic of this motion? Since the golf ball is undergoing a projectile motion, acceleration (or rate of change of vertical component of velocity) is always constant (i.e. downward acceleration due to gravity) and horizontal velocity will always stay constant as there are no horizontal forces. However, since the velocity (& speed) of golf ball changes, the total kinetic energy will change. Answer: D 8 Essential Questions:  What does the total energy of the satellite comprise of? Total energy of a satellite = GPE of satellite + KE of satellite in the orbit = − ଵ ଶ ீெ௠ ோ Therefore, change of total energy = final total energy at the smaller orbital radius – initial total energy at the higher orbital radius. = - 2 x 109 J Negative sign implies that energy is lost. Answer: C 9 Essential Question(s):  What kind of motion is this?  Which physical quantities remain the same for both position of P and Q?  Angular Velocity  Which quantity changes between point P and point Q?  Radius  Which is the relevant equation to use for this question? Angular velocity at P and Q is the same, because when P rotate 1 round, Q also rotate 1 round. 2 1 2 8 Q Q P P Q a m r a r a r a     Answer: D

4 10 Essential Question(s):  What is the formula for gravitational potential?  How to apply it using some mathematical manipulation.  =  E GM r -6.257 x 107 =  E GM r  -6.257 x 107 rE = GM So  new = + 50000 E GM r = 7- 6.257 x 10 + 50000 E E r r =     7 6 6 - 6.257 x 10 6.371 x 10 6.371 x 10 + 50000 = - 6.208 x 107 J kg-1 Answer: A 11 Essential Question(s):  What equation links the temperature in the Kelvin scale to the degrees Celsius scale?  How to apply it using some mathematical manipulation. T/K = θ/C + 273 Hence by linearising the equation, the gradient is 1 and the y-intercept is 273. Answer: D 12 Essential Question(s):  What is the question requirement for Thermal Physics? (Is it just Internal Energy = Sum of all KE and PE of atoms/molecules/particles of the system?)  If not then is the first law of thermodynamics required? Since the answers in the table provides the headings as internal energy/heat supplied and work done, therefore it must be agreed that the First Law of Thermodynamics must be used. As there is an increase in temperature, internal energy must increase. As the container is insulated, no heat is supplied. And when stirring happens, work is done on the system hence work done increases. Answer: B 13 Essential Questions:  What are the different degrees of damping?  How to represent the different degrees of damping through graphs?  Which degree of damping is best for a car suspension system? Option A: Light damping oscillations. An “oscillation” is seen from the graph alternating between the positive and negative displacement throughout time. Option B: Heavy damping. It takes relatively a longer time to reach the equilibrium position as compared to option D. Option C” Also a light damping oscillation, but is more damped than option A.

5 Option D: Critical damping. Reaches equilibrium position is relatively the shortest time. A critical damping is required for a car suspension system. Whenever the car goes over a hump, it is expected that the car returns to the equilibrium position as soon as possible. Answer: D 14 Essential Questions:  How do we know whether a wave profile is a representation of the whole wave or the oscillation at one point in space?  How do I calculate the phase difference when the time axis is shown?  ∆ߠ= ∆௧ ் × 2ߨ From the graph, the dotted graph lags the solid graph by one-quarter of the wave. Therefore,  ∆ߠ= భ ర் ் × 2ߨ= గ ଶ The graph can also be interpreted as the solid graph lags the dotted graph by three-quarter of the wave.  ∆ߠ= య ర் ் × 2ߨ= ଷగ ଶ Other learning points:  Phase difference is independent on the amplitude of oscillation  It is not meaningful in this case to ask if there is a గ ଶ rad as one of the options, which is the correct answer? Both answers are essentially the sam