CJC 2025 A level H2 Physics Answers

H2 Physics 9749 – 2025 A Level Exam Paper 1 1 Essential Question(s):  How can magnetic flux density be expressed in terms of other physical quantities? Approach: Recall the formulas that contain Magnetic flux density (B) 1. B = Φ/A 2. F = BIL 3. emf = -dΦ/dt = - d(BA)/dt Unit of B = T Suggested Solution 1: (Proving if options are correct) Option A: Using B = Φ/A Unit of B = unit of Φ / unit of A T = Wb m−². Option B: Using emf = -dΦ/dt = -d(BA)/dt Unit of emf = unit of BA / unit of time V = (T m2)/t T = V s m−2 Option C: Using F = BIL, Unit of B = N A−1 m−1 = N m A−1 m−2 = J A−1 m−2 Option D: Using F = BIL F = B(Q/t)L Unit of B = [Ft]/ [QL] = N m−1 s C−1. ✗ Answer: C Suggested Solution 2: (Check if the options are correct) Option A: Wb m2 = (T m2) m2  unit of Wb = T m2 Option B: V s = Wb s-1 s  unit of V = Wb s-1 = T m2 s-1 s Option C: Use Solution 1 Option D: Use Solution 1 2 Essential Question(s):  How do you calculate percentage uncertainty when quantities are multiplied and divided?  What is the formula for combining uncertainties in multiplication and division?  How do you handle powers in uncertainty calculations? Solution: a = bc²/d, Δa/a = Δb/b + 2(Δc/c) + Δd/d Δa/a = 3/24 + 2(2/13) + 1/18 Δa/a = 0.125 + 0.308 + 0.056 Δa/a = 0.489 (Δa/a) x 100% = 49% Answer: D 3 Essential Question(s):  What are the key equations for projectile To derive range R

motion?  How do you relate maximum height and range in projectile motion?  What is the relationship between launch angle and trajectory parameters? Solution: For projectile motion: Range R = (2u2sinθcosθ)/g -- (3) Maximum height H = (u²sin²θ)/(2g) – (4) Divide eqn (4) by (3), H/R = (tan θ) / 4 tan θ = 4H/R Given: H = x and R = 6x tan θ = 4/6 = 2/3 Answer: B Let t be the time to return to ground level. Then t/2 is time taken to reach maximum height. Horizontal direction: T ake right as positive direction. Use “sx = uxt” R = (ucosθ)t -- (1) Vertical direction: T ake upwards as positive direction. Use “vy = uy + ayt” At max. height, vy = 0 0 = usinθ + (-g)(t/2) usinθ = g(t/2) t = 2usinθ/g -- (2) Sub (2) into (1): R = (ucosθ)(2usinθ/g) R = (2u2sinθcosθ)/g = (u²sin2θ)/g To derive max height H Vertical direction: Use “vy2 = uy2 + 2aysy“ 0 = (usinθ)2 + 2(-g)H (usinθ)2 = 2gH H = (usinθ)2/2g 4 Essential Question(s):  What determines acceleration?  net force  What are the forces acting on the ice cube?  Draw free body diagram to visualize. Buoyant force acts upwards and Weight of ice cube acts downwards.  How do you calculate buoyant force when an object is fully submerged?  Which force, buoyant force or weight of ice cube, is larger? What then is the net force?  Since ice cube was floating initially, it was initially in vertical equilibrium. After it was pushed downwards, its weight is unchanged but because it displaced more fluid the buoyant force increased. So buoyant force exceeds weight of ice cube & the net force acts upwards and equal to (buoyant force – weight of ice cube).  How to relate acceleration and net force?  Newton’s 2nd law. Solution: When fully submerged: Buoyant force = ρwater × Vcube × g = 1000 × (0.02)³ × 9.81 = 0. 0785 N  This formula is not in Syllabus for 2026 Exam onwards. Weight of cube = ρice × Vcube × g = 920 × (0.02)³ × 9.81 = 0.0722 N Net upward force = 0.785 - 0.722 = 0.0063 N Mass of cube = 0.92 × (2.0)³ = 7.4 g = 0.074 kg Initial acceleration = net force / mass = 0.0063 / 0.0074 = 0.85 m/s² Answer: B 5 Essential Question(s):  What features of a velocity-time graph relates to resultant force?  gradient gives acceleration, and, resultant force = mass x acceleration.  What does it mean when velocity is constant on a v-t graph?  How do you interpret the forces acting on a falling object with air resistance?

Solution: From Newton's second law, Fnet = ma, where a is the gradient of the v-t graph. The velocity is constant (acceleration = 0) at t = 30 s (terminal velocity when parachute is not opened yet) and t = 60 s (terminal velocity when parachute is fully opened). This means the net force on parachutist is zero (weight of parachutist equals air resistance) at both of these timings. Answer: C 6 Essential Question(s):  How do you calculate torque about a pivot?  What happens to torque when both force magnitude and distance change? Solution: Let distance between any two labelled points be d. Take clockwise as positive direction. Initially: F1 = F2 = F in magnitude Torque about pivot = F1d – F2(3d) = Fd – F(3d) = -2Fd = -T After changes: F1 = F2 = F/2 in magnitude F2 moves to R. New torque about the same pivot = F1d – F2(2d) = (F/2)d – (F/2)(2d) = -Fd/2 = -T/4 Answer: B 7 Essential Question(s):  How does Hooke's law apply to wires under tension?  What is the relationship between extension and applied force for elastic materials?  How does the configuration of the wire affect its effective force constant? Solution: Method 1: Initially: extension d produced by force W, so force constant k = W/d for the whole wire. When folded in half: Each half has force constant 2k (length shortened by half, same material). Two halves in parallel: Effective force constant = 2k + 2k = 4k New extension produced by the same force W: x = W/(4k) = W/(4W/d) = d/4 Method 2: Initially: T1 = W kd = W where k is the force constant for the wire -----(1) When folded in half: 2T2 = W 2(kx2) = W same k considering same continuous length of wire -----(2) (1) = (2): x2 = d/2 This is the total extension of the wire to produce tension equal to T2. Since wire is folded in half, vertical distance moved = ½ x d/2 = d/4

Answer: B 8 Essential Question(s):  What conservation laws apply for elastic collisions?  conservation of total momentum throughout, and, conservation of total kinetic energy before & after the collision.  What is the relationship between initial and final velocities in elastic collisions?  Solving the 2 conservation laws simultaneously, we can derive “relative speed of approach = relative speed of separation”. Solution: v = vQ - vP represents the relative speed of separation. In elastic collisions, the relative speed of approach equals the relative speed of separation. Since Q is initially at rest: relative speed of approach = u - 0 = u Therefore: v = u , v is directly proportional to u. The graph should be a straight line through the origin with slope = 1. Answer: A 9 Essential Question(s):  What is the relationship between power, work done, and time?  How do you calculate work done against gravity?  What is the energy transformation taking place? Solution: Power out P = (work done against gravitational force) / time = (weight × vertical displacement) / time = (W × h) / t Rearranging: h = Pt / W OR Assume no change in kinetic energy, All energy supplied by motor transforms into gain in GPE. Pt = mgh Pt = Wh since W = mg h = Pt / W Answer: B 10 Essential Question(s):  How do you calculate the speed of a point on a rotating object? Solution: 4 12 2 0.30 4.8 10 m s66 60 v r v r T                         OR Circumference of circle = 2πr = 2π × 0.30 = 1.885 m Actual time for one revolution = 66 minutes = 66 x 60 = 3960 s Average speed = distance/time = 1.885/3960 = 4.8 × 10⁻⁴ m s-1

Answer: A 11 Essential Question(s):  How does gravitational field strength vary with distance from Earth?  Why is g ≈ 9.81 m s-2 only valid at or near Earth's surface? Solution: The expression W = 9.81 × m assumes gravitational field strength g is constant at 9.81 N kg-1. However, g = GM/r², so g decreases with