2022 GCE H3PH 9814 P1 SS [HCI]

2022 H3 A Level Suggested Solutions 1 (a) (i) The net torque generated on rod due to its weight is given by, 𝜏 = ∫ 𝑔 (𝑀 𝐿 ) 𝑑𝑙. 𝑥 = ∫ 𝑔 (𝑀 𝐿 ) 𝑑𝑙. (𝑙𝑠𝑖𝑛𝜃) = 𝑔𝑀𝑠𝑖𝑛𝜃 𝐿 (𝐿2 2 ) = 𝑀𝑔𝐿 𝑠𝑖𝑛𝜃 2 (Alternatively, since the rod is uniform, you may obtain the torque as follows: The torque due to the weight of the rod about A would be sin2 LMg   . As A= I , and 222 12 2 3 A ML L MLM= + =  I (The term 2 12 ML is the moment of inertia about centre of mass of the rod) We have, 2sin 23 3 sin 2 MgL ML g L    = = Comments: Several students did not consider the factor of two in their answers. (ii) Comparing with the given form, 1.5k = . Mg sin2 L 

(b) As the total energy is conserved, the gravitational PE is converted to the rotational kinetic energy of the rod, 2 21cos 0 02 2 3 3 cos L MLMg g L    + = +   = Comments: Using the conservation of energy is the most straightforward method. Some students erroneously used SUVAT equations, however, the angular acceleration was not constant. 2 (a) ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ = 𝜇0𝐼𝑒𝑛𝑐 Where ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ is the net magnetic flux along the loop, 𝜇0 is the magnetic permeability of free space, 𝐼𝑒𝑛𝑐 is the enclosed current and 𝑑ℓ is the small length on the circular path (loop). (b) (i) Since 𝐼1 = 𝐼2, ∫ 𝐽1 𝑑𝐴 = ∫ 𝐽2 𝑑𝐴 ∫ 𝐽1 (2𝜋𝑟 𝑑𝑟) 𝑟1 0 = ∫ 𝐽2 (2𝜋𝑟 𝑑𝑟) 𝑟3 𝑟2 𝜋𝑟1 2𝐽1 = 𝜋(𝑟3 2 − 𝑟2 2)𝐽2 𝐽2 = 𝑟1 2 𝑟3 2 − 𝑟2 2 𝐽1 (ii)1 Draw a circular Amperian loop of radius 𝑟1 centred on the axis of the wire, ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ = 𝜇0𝐼𝑒𝑛𝑐 = 𝜇0𝐼1,𝑒𝑛𝑐 = 𝜇0(𝜋𝑟1 2𝐽1) 𝐵(2𝜋𝑟1) = 𝜇0𝜋𝑟1 2𝐽1 𝐵 = 1 2 𝜇0𝐽1𝑟1 (ii)2 Draw a circular Amperian loop of radius 𝑟2 centred on the axis of the wire, ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ = 𝜇0𝐼𝑒𝑛𝑐 = 𝜇0𝐼1 = 𝜇0(𝜋𝑟1 2𝐽1) 𝐵(2𝜋𝑟2) = 𝜇0𝜋𝑟1 2𝐽1 𝐵 = 𝜇0𝐽1𝑟1 2 2𝑟2

(iii) The shapes of the graph are: • 0 to r1: linear • (Maximum at r1) • r1 to r2: curve • r2 to r3: curve • r3 to r4: zero Calculations for each region (no need to show): Repeating the working in 2(b)(ii)1 and 2 but considering a loop with a general radius 𝑟, we have: 𝐵 = { 1 2 𝜇0𝐽1𝑟, 0 < 𝑟 ≤ 𝑟1 𝜇0𝐽1𝑟1 2 2𝑟 , 𝑟1 ≤ 𝑟 ≤ 𝑟2 That is, 𝐵 ∝ 𝑟 for 0 < 𝑟 ≤ 𝑟1 and 𝐵 ∝ 1 𝑟 for 𝑟1 ≤ 𝑟 ≤ 𝑟2 For 𝑟2 ≤ 𝑟 ≤ 𝑟3: Draw a circular Amperian loop of radius 𝑟 centred on the axis of the wire, where 𝑟2 ≤ 𝑟 ≤ 𝑟3. (Note that the current in the metal braid flows in the opposite direction, so the net current is 𝐼1 − 𝐼2,𝑒𝑛𝑐.) ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ = 𝜇0𝐼𝑒𝑛𝑐 = 𝜇0(𝐼1 − 𝐼2,𝑒𝑛𝑐) = 𝜇0(𝜋𝑟1 2𝐽1 − 𝜋(𝑟2 − 𝑟2 2)𝐽2) 𝐵(2𝜋𝑟) = 𝜇0 (𝜋𝑟1 2𝐽1 − 𝜋(𝑟2 − 𝑟2 2) 𝑟1 2 𝑟3 2 − 𝑟2 2 𝐽1) 𝐵(2𝜋𝑟) = 𝜇0𝜋𝐽1𝑟1 2 (1 − 𝑟2 𝑟3 2 − 𝑟2 2 + 𝑟2 2 𝑟3 2 − 𝑟2 2) 𝐵(2𝜋𝑟) = 𝜇0𝜋𝐽1𝑟1 2 𝑟3 2 − 𝑟2 2 (𝑟3 2 − 𝑟2) 𝐵 = 𝜇0𝐽1𝑟1 2 2(𝑟3 2 − 𝑟2 2) (𝑟3 2 𝑟 − 𝑟) If you expand the bracket you will get two terms: r1 r2 r3 r4 conductor insulator plastic jacket metal braid B x 0 0 Bmax

• The first term is proportional to 1/𝑟, and initially this dominates. However it gets smaller as 𝑟 increases. • The second term is proportional to 𝑟. This gets larger as 𝑟 increases Therefore, you get a curve with decreasing gradient, that eventually becomes a line with negative gradient. (Basically, a hyperbola that approaches an asymptote of 𝐵 = −𝑘𝑟 for some constant 𝑘). A kind of curve that is decreasing would suffice. For 𝑟3 ≤ 𝑟 ≤ 𝑟4: Intuitively, this must be zero because 𝐼𝑒𝑛𝑐 = 𝐼1 − 𝐼2 = 0 (two equal currents flowing in opposite directions). However, if you want to check, you could perform this calculation: Draw a circular Amperian loop of radius 𝑟 centred on the axis of the wire, where 𝑟3 ≤ 𝑟 ≤ 𝑟4 ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ = 𝜇0𝐼𝑒𝑛𝑐 = 𝜇0(𝐼1 − 𝐼2) Since 𝐼1 = 𝐼2 = 𝐼, 𝐵(2𝜋𝑟) = 0 ∴ 𝐵 = 0 (c) The magnetic field outside a standard transmission cable, unlike that of a coaxial cable, is not zero when a current is flowing through it. A high frequency signal will mean that energy will be dissipated in metal components in the surroundings. Comments: Several students did not talk about the effects outside of the cable.

3 (a) (i) ( ) ( ) 22 cos sin sin ( ) 2 sin 2 sin 2sin cos sin2cos flight y y flight flight flight R u t v u gt u u g t ut g u u uRu g g g        = =+ + − = + − = = = =  (ii) For maximum range R, ( )sin 2 1 45= → =  (iii) 22sin2 150 sin(2 45 ) 2290 m 2.29 km9.81 uR g  = = = = (b) (i) ( )cos cos flight flight R u t Rt u   = = (ii) tan tan h hRR= → = (iii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 2 2 2 22 22 22 2 1 2 10 sin 2 1sin 2 1tan sin cos 2 cos tan tan 2 cos 2 cos tan tan 2sin cos2cos tan cos y flight y flight flight flight flight flight s s u t a t h u t g t h u t g t RRR u g uu RgRR u uR g uR g           = + + + = + + − =− +    =− +        =− + =+ =+ ( ) 2 1 cos2 tan sin2u g    = + +  Comments: Many could handle the rigorous algebra here.

(iv) ( ) 2 2 1 cos2 tan sin2 0,tan 0 sin2 uR g uR g      = + + →→ → (v) ( ) 2 1 cos2 tan sin2 0 2cos2 2( sin2 )tan 0 cos2 sin2 tan 1tan2 tan tan 2 2 2 42 uR g dR d              = + + = → + − = = = = −  =− =− Comments: Many did not realize that  was a constant.

4 (a) (i) In Fig. 4.1, at the equilibrium, 0 2 LMg k =−  l , where 0 2 L − l is the extension. In Fig. 4.2, 00 22 hand left right LLMg F k k    = − + −       ll When it is released, 0handF → , 00 22 left right LLMg k k    − + −       ll . The upwards force is greater than Mg. Hence, when the system is released, it moves upwards. (ii) Fig. 4.2 must eventually be in equilibrium. ( )2 Mg ke= , where e is the extension for the spring either on left or right side in Fig. 4.2. The load is halved on either side. Hence, 2 Mge k= The total length on either side is 10 2 LLe= + +l , where 2 L is the length of string, 0l is the unstretched length of spring and e is the extension. From Fig. 4.1, 00 22 L L MgMg k k = − → = + ll Hence, 1 0 0 10 22 32 2 L Mg MgLe kk MgL k = + = + + + =+ ll l Comments: Very challenging. Many were confused with extensions in Fig. 4.1 and Fig. 4.2.

(b) (i) On either side, ( ) ( )1 5 5 2 2 4 new new MgMg k e e k= → = The new total length on either side is 20 2 new LLe= + +l , where 2 L is the length of string, 0l is the unstretched length of spring and newe is the new extension. 20 1 0 0 0 0 0 0 0 0 0 2 17 5 14 4 17 3 5214 2 4 34 5 17 3 6 122114 4 14 2 14 28 new LLe Mg MgL kk Mg Mg Mg k k k Mg Mg kk Mgk = + + = + + +    + = + + +          − = + − → =       = l ll l l l l l l l Comments: Very challenging for many. (ii) 10 3 3 722 2 2 2 Mg Mg Mg MgL k k k k = + = + =  l From the condition in Fig. 4.1, 0 22 4 L Mg Mg Mg Mg k k k k Mg L k = + = + = →= l 1 7 7 7 2 2 4 8 Mg L LL k = = =

5 (a) (i) The rod is light, finding the moment of inertia of spheres about the axis of rotation, 2 2 2 22 L mLm==  I 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2 42 22 mL a mL L mL LL mL mL mL TT         = = = =   = = → = = I Comments: Remember to include omega  as the motion is modelled as SHM and relate the torque to the moment of inertia. (ii) 15.0 cm + 2.5 cm + 0.1 cm = 17.6 cm Comments: Incorrect answers used diameters, not radii. (iii) For small angle 1 , ( ) 3 4 11 1 1 4 11 1 2 4.1 102 45.6 10 rad2 1.80 0.1 0.01 4.1 1.80 0.1 0.01 0.1 0.0145.6 10 0.0001 rad4.1 1.80 4.1 1.80 0.0046 0.0001 rad Lss L sL sL      − − −  = → = = =   = + = +     = + =  + =       = Comments: Some used the full length of the rod, rather than the distance to the centre. (iv) The gravitational force provides the torque on the system, 1 12 22 2 2 2 211 122 22 gFL GMm Lr rr mL LGr MmL Mm T MT        == = = = = Comments: The couple provides the torque.

(b) (i) The deflection of the