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 = 40.0125 W Option A: IV = 18 W Option B: IV = 39.6 W Option C: IV = 14.4 W Option D: IV = 67.2 W Option B is the most feasible answer which is Less than the maximum possible value of 40.0125 W and close to it. There could be other losses e.g. not all translational KE of the air transferred to the turbine. The 60% efficiency of the generator accounts for only what has been transferred to the generator and then the efficiency of it in being converted into electrical energy. Answer: B 7 Essential Question(s):  What is an ‘orbit’?  What is meant by ‘angular displacement’? Solution: 1 year = Period  angular displacement = 2π rad ½ year  angular displacement = π rad Answer: C 8 Essential Question(s):  What are the definitions of the quantities listed in A, B, C and D? Solution: By definition of gravitational field strength. Answer: B 9 Essential Question(s):  What affects the motion / the path travelled by an object?  Net force Solution: Draw the free body diagram of the spacecraft for paths A, B, C and D. Without the force from firing the rocket, the only force is gravitational pull of Earth which is always attractive in nature. Path A: Possible. Spacecraft speeds up moving towards Earth in a straight line. Path D: Possible. Spacecraft slows down moving away from Earth in a straight line.

4 Path B: Possible. Gravitational pull exerted by Earth always acting perpendicularly to the spacecraf t’s instantaneous velocity, hence changes only the direction of its velocity and not the magnitude of velocity. Spacecraft move in uniform circular motion. Path C: Not possible without additional force from firing rockets. The gravitational force along path C is not acting parallel to the velocity in path C and will therefore tend to cause a change in the direc tion of the velocity. An additional force equal in magnitude and opposite in direction to the component of the gravitational force perpendicular to the velocity is needed for the spacecraft to move in the straight line C. Answer: C 10 Essential Question(s):  How to apply the kinetic theory of gases to derive an expression for the force/pressure exerted by the molecules with a wall of the container?  What are the basic assumptions for an ideal gas? Solution: By Newton’s 2nd law of motion, Average total force exerted on the shaded wall exerted by all N molecules <F> = Average total rate of change of momentum of all the N molecules = N. ∆௣ ∆௧ = N ௠∆௩ ∆௧ Ideal gas assumption  all gas molecules collide elastically with the wall of the container  no change in KE before and after collision, so ∆v = 2v. ∆t is the time interval between successive collisions of per molecule with the shaded wall, ∆t = ଶ௣ ௩ <F> = N ௠(ଶ௩) ቀమ೛ ೡ ቁ = ே௠௩మ ௣ Answer: A 11 Essential Question(s):  What is the relationship between Internal Energy and Temperature for an ideal gas? Solution: For an ideal gas, no intermolecular forces so p.e. between all molecules equal to 0. Internal energy = Total k.e. of the molecules only = 3/2 NkT for a monoatomic ideal gas Since 3/2 Nk remains constant, Internal energy ∝ T, where T is in kelvins 0°C = 273.15 K  internal energy = U 273 °C = 546.15 K  internal energy = 546.15 / 273.15 x U = 1.00045 U Answer: B 12 Essential Question(s):  What are the definitions of specific heat capacity (c), and, specific latent heat of fusion (l)?  How to relate c and l to information on rate of heat supplied, mass, temperature and time?

5 Solution: For specific heat capacity (c), look at the graph where T is changing. Q = mc∆θ P ∆t = mc∆θ c = ௉ ௠ ∆௧ ∆ఏ where P and m are same for both solids, and ∆௧ ∆ఏ is the reciprocal of the gradient of the graph where T is changing. c = ௉ ௠ ଵ ௚௥௔ௗ௜௘௡௧ Since gradient of graph of P is greater, c of P is smaller than Q  either option C or D. For specific latent heat (l), look at the graph where T is constant. Q = ml P ∆t = ml l = ௉ ௠ ∆ݐ where P and m are same for both solids, and ∆t is the duration for each solid to melt completely. Since ∆t is longer for solid P, l of P is greater than Q  Answer is C not D. Answer: C 13 Essential Question(s):  What kind of motion is the molecule undergoing?  x-t graph is sinusoidal, hence SHM.  How to calculate maximum KE for SHM? Solution: x-t graph is sinusoidal, hence SHM. Maximum KE = ½ m vmax2 = ½ m (ω xo)2 = ଵ ଶ ቀ ெ೘ ேಲ ቁ ቀ ଶగ ்ݔ௢ቁ ଶ = ଵ ଶ ቀ ଶ.ଽ௫ଵ଴షమ ଺.଴ଶ௫ଵ଴మయቁ ቀ ଶగ ଷ௫ଵ଴షయݔ 2ݔ10ି଺ቁ ଶ = 4.23 x 10-31 J Answer: B 14 Essential Question(s):  How to determine phase different between 2 waves from the graph? Which graph feature to use?  How to determine ratio of intensities from the graph? Which graph feature to use? Solution: Phase difference between X and Y: ∆߶ 360° = ∆ݐ ܶ ∆߶ 360° = 1 3 ∆߶= 120° Ratio of intensities: Intensity ∝ Amplitude2 ܫ௑ ܫ௒ = ൬ܣ௑ ܣ௒