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.