EJC 2024 GCE A Level H3 Physics 9814 Suggested Solutions

©EJC 2025 9814/H3 PHYSICS 2024 PHYSICS SUGGESTED SOLUTIONS 9814 October/November 2024 1 (a) net yy y F ma T mg ma T T ma mg T ma mg T ma mg ma ag s ut at hg = −= −= −= −= = = = + =  2 12 1 2 2 From FBD drawn, and applying (N2L), --- ( 1 ) 2 --- ( 2 ) 3 3 --- ( 3 ) Taking (2) +(3), 3 5 --- ( 4 ) Taking (4) +(1), 26 1 3 1Using 2 11 23 t ht g  = 2 1 1 6 (b) (i) Take rightwards as positive. Determine the velocity of 2m when 3m just comes to rest:

2 9814/H3 PHYSICS 2024 1 6( )( )3 2 3 v at ghv g hgv = = = When the string gets slack, 2m and m continue with velocity v (to the right) with a force of mg (to the left) causing the two masses to decelerate. Hence considering 2m and m 21 21 21 2 2 0 20 ( )( ) 33 () 2 33 2 96() 3 66 6 2 4.9 v at hg mg ttm gt t hg hg htt gg hh h ht gg g g = − = −− − = −= × = =+= = (ii) (c) The student is incorrect because the acceleration of the mass m is constant and does not have a proportional relationship with the displacement from equilibrium position.

3 9814/H3 PHYSICS 2024 2 (a) The minimum amount of water that can be added to obtain 0.0°C temperature will be a situation whereby all water added just freeze to ice at 0.0°C. We assume the vaccine can be kept in solid ice as long as it is at 0.0°C. ice ice ice water water water water fmc θ mc θ ml    = +   °°    = + By principle of conservation of energy, Heat gain Heat loss Heat loss by ice to reach 0.0 C by water to drop to 0.0 C by water to freeze ΔΔ (0.120) water water .m m. . × −− = − = = 3(2.09 10 )[0 ( 4 2)] {(4180)[23.0 0.0] + 224000} 0 00329 kg 3 29 g (b) The maximum amount of water that can be added to obtain 0.0°C temperature will be a situation whereby all ice melted and the water reach 0.0°C. ice ice ice ice f water water watermc θ ml m c θ    +=   °°    += × By principle of conservation of energy, Heat gain Heat gain Heat loss by ice to reach 0.0 C by ice to melt by water to drop to 0.0 C ΔΔ (0.120)(2.09 water water . m. m. −− + = − = = 310 )[0 ( 4 2)] (0.120)(224000) (4180)(23.0 0 0 ) 0 291 kg 291 g

4 9814/H3 PHYSICS 2024 3 (a) (i) We are considering the situation of just before the ball -ball collision (shortly after the big ball has rebounded off the ground) The big ball (5M) is moving upwards with speed V (it hits the ground at speed V and rebound elastically) The small ball (M) is moving downwards with speed V. In the lab frame: M, M, ZMF ZMF u V , u V MV M VvV MM mgh mV V gh v V gh =+= − −= =+ = ∴== 5 lab lab 2 Taking upwards as positive, Lab - frame velocities : (5 )( ) + ( )( ) 2 53 By conservation of energy : Loss in GPE = Gain in KE 1= 2 2 22 2 (shown)33 (ii) ball ,ZMF ball ,lab ZMF M, M, v vv u VVV u VV V = − = −= = −− = − 5 ZMF ZMF Before collision (in ZMF) : 21 (upwards)33 25 (downwards)33 Because collision is elastic and ZMF has zero total momentum, the velocities in the ZMF sim M, M, M, Z M F M, M, M, Z M F vV v v v VVV vV v v v VV = − ⇒ = + = −+= = ⇒ = + =+= 5 ZMF 5 l a b 5 Z M F ZMF lab ZMF ple reverse sign on collision : After collision (in ZMF) : 1 (downwards) 3 12 1 (upwards)333 5 (upwards) 3 52 33 V7 (upwards)3

5 9814/H3 PHYSICS 2024 (b) (i) M ,ZMF M ,lab M vV v VVV M V Mg = − = −+= =  = 5 5 2 Height to which ball 5 rebounds. Its upward speed after the ball - ball collision 1 3 12 1 333 Using Principle of conservation of energy, Loss in KE Gain in GPE 11(5 ) (5 )( )23 M M H ghVHh gg= = = 5 22 5 () (2 )11 1 18 18 9 (ii) M ,ZMF M ,lab M M M vV v VVV M V MgH H = =+= =  = = 2 Height to which ball rebounds. Its upwa rd speed after the ball - ball collision 5 3 527 333 Using Principle of conservation of energy, Loss in KE Gain in GPE 17( ) ( )( )( )23 ghV hgg= = 22 (2 )49 49 49 18 18 9

6 9814/H3 PHYSICS 2024 4(a) Ampère’s law states that the line integral of B.ds around any closed path (Ampèrian loop) is equal to µoI, where I is the total current enclosed by the closed path: 0d µ=∫ IB. s . 4(b)(i) Using right-hand drip, direction is clockwise 4(b)(ii) For each turn, the line integral od µ⋅=∫ IBs Since there are N turns, the (2 ) 2 o o o dN B ds B r N NB r µ πµ µ π ⋅= = = = ∫ ∫   I I I Bs 4(b)(iii) From part (ii), we derived that inside the toroid (a<r<b), the magnetic flux density is: B=μ0NI/2πr This shows that B is inversely proportional to r (i.e., B∝1/r) in the region a<r<b. Outside the toroid: • For r<a: The Amperian loop encloses no net current (Ienc = 0), so B=0. • For r>b: The Amperian loop encloses equal and opposite currents (the current coming out and going in cancel), so Ienc = 0 and B=0. At r=a and r=b, there might be a discontinuity, but ideally, B is zero outside. Therefore, the graph of B vs. r should look like: • B=0 for r<a, • B increases sharply at r=a to a maximum value Bmax at r=a, • Then B decreases as 1/r for a<r<b, • Then B drops to zero at r=b and remains zero for r>b.

7 9814/H3 PHYSICS 2024 4(c) The maximum B occurs at the inner radius r=a (smallest r): 0 max 2 NIB a µ π= The wire is insulated and wound tightly so that turns are just in contact at the inner circumference. This means that at the inner radius, the number of turns per unit length is maximized. Specifically, at r=a, the circumference is 2πa. The width of one turn (including insulation) is equal to the diameter of the insulated wire = 2.00 mm = 2.00 × 10 −3 m. Therefore, the number of turns that can fit along the inner circumference is: N = circumference at r=a/width per turn 00 0 max 2 22 NI I IaB a ad d µµ µ π ππ = = = 33 7 6.00 10 (2.0 10 ) 4 10I π −− − ××= × = 9.55 A

8 9814/H3 PHYSICS 2024 5(a) Self-inductance is the ratio of the electromotive force (e.m.f.) induced in an electrical circuit/component to the rate of change of current causing it. = dIVL dt 5(b) At t = 0s, I = 0 A so emf opposes the initial rate of change of current. At t = 0s, 1 3 0.50 119 A s4.20 10 dI dt − −= =× 6.0 0.050 H119() EL dI dt = = = Alternative Method: In an RL circuit, the current as a function of time is: I(t)=Imax(1−e−t/τ) where τ=L/Rtotal is the time constant, and Rtotal is the total resistance (internal resistance + coil resistance). In practice, we can use the time to reach about 63% of Imax to find τ. From the graph, the time constant τ, the time taken for the current to reach 0.63×Imax =0.63×0.50=0.315 A is 4.2 ms The maximum current Imax = EMF/Rtotal =0.50 A. So, 0.50 = 6.0/Rtotal  ⟹ Rtotal =12.0 Ω the inductance is: L=τ×Rtotal = (4.2 × 10−3)(12) = 0.050 H So, the inductance is 0.050 H, and the unit is henry (H). (c) From the graph, The maximum current Imax = EMF/Rtotal =0.50 A. So, 0.50 = 6.0/Rtotal  ⟹ Rtotal =12.0 Ω The internal resistance is 1 Ω, Rtotal = 1+R = 12.0⟹ R=11.0 Ω So, the coil resistance is 11.0 Ω. (d) Opening the switch causes the current in the coil to collapse extremely rapidly. According to Faraday’s law, this sudden drop in current induces a very large back emf across the switch gap. The high voltage in large enough to exceed the breakdown voltage of air causing ionization and a spark across the switch.

9 9814/H3 PHYSICS 2024 6(a)(i) 1 19.1cos87Z = = ° 6(a)(ii) Possible suggestions: • Uncertainty in the half -moon phase: The Moon’s surface is rough and cratered, making it extremely difficult to judge the exact instant the lunar terminator (day -night line) is perfectly straight to signal a true half-moon. • Extreme sensitivity near 90°: The true angle is 89.8∘. Because cos 𝜃𝜃 changes extremely rapidly near 90∘ , even a tiny 1∘ measurement error leads to a massive percentage error in the calculated distance ra