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