CJC 2023 A level H2 Physics Answers

1 H2 Physics 9749 – 2023 A Level Exam Paper 1 1 Essential Question(s):  What is the difference between systematic error & random error? Solution: Options A, B and D contributes to deviation of reading from the true reading by the same amount in t he same direction. Answer: C 2 Essential Question(s):  How to relate displacement (goal) to the v-t graph (given info)? Solution: Change in displacement = Area under v-t graph Answer: D 3 Essential Question(s):  How is force related to energy?  via work done  What is the energy transformation here? Solution: Loss in KE of the arrow = Negative work done by F on arrow E – 0 = F.∆s E = F. vavt where vav is average velocity of the arrow moving through the target Since E is the ‘arrival’ value, to relate velocity to E, we need to express vav in terms of the ‘arrival’ speed. Assuming arrow is brought to rest with a approximately uniform acceleration, E = F 2 u v     t E = F 0 2 u     t E = ½ Ft u ------(1) E = ½ m u2 Since m is constant for all arrows used, E ∝ u2 E ∝ u Let u = k E -----(2) Sub (2) into (1): E = ½ Ft (k E ) Ft ∝ E Answer: A

2 4 Essential Question(s):  What equations apply for an elastic head-on collision between 2 bodies? Solution: Take right as positive direction. Relative speed of approach = Relative speed of separation. Note that u1, u2, v1 and v2 are ‘speeds’ meaning only the magnitude. u1 – (-u2) = v2 – v1 u1 + u2 = v2 – v1 ---------(1) By Conservation of Momentum, m1u1 + m2(-u2) = m1v1 + m2v2 ---------(2) Since no information on whether the two masses are equal, equation (2) cannot be further simplified. Answer: C 5 Essential Question(s):  What is the meaning of ‘centre of gravity’?  point on a body through which its entire weight can be modelled as acting through. Solution: By symmetry, c.g. of the lower half of the sheet alone (CL) is at 2.0 cm vertically above P (let mass be M); c.g. of the upper half of the sheet alone (CU) is at 6.0 cm vertically above P (let mass be 2M). To find the c.g. of the entire sheet (G), consider balancing the sheet HORIZONTALY about G so that t he sheet is horizontal when balanced. Side view: Take moments about G, Sum of anticlockwise moments = Sum of clockwise moments (2Mg)(6.0 cm – y) = (Mg)(y – 2.0 cm) 12 – 2y = y – 2 3y = 14 y = 4.6667 = 4.7 cm Answer: B 6 Essential Question(s):  What is the energy transformation? 2Mg Mg CU CL G P 2.0 cm 4.0 cm 4.0 cm y = ? 6.0 cm

3  What is efficiency? How to relate to the output V and I?  use the definition of efficiency in terms of power rather than energy. Solution: Useful power output 0.60 ---(1)Useful power input  Total power input = Rate of KE loss by air to turbine blades = ½ x (mass flow rate of air) x (vi2 – vf2) = ½ x (9.7 kg s-1)(4.02 – 1.52) = 66.6875 W From (1), Useful power output = 0.60 x 66.6875 W =