RI 2026 H3 Physics Prelim (Ans)

1 2026 H3 Prelims - Solutions Section A 1 (a) According to Newton’s 2nd Law ( )d mv dvFm dt dt== (since mass is constant) Since the expression is in terms of position dv dv dx dv dvF m m m v mvdt dx dt dx dx   = =  =  =       Integrating both sides ( ) ( ) 1 0 22 2 1 22 2( 1) nn d v u n n F dx mv dv vuF n d m n Fdvu m − = − = −   −=+  (b) By the principle of conservation of momentum if i f f f mv MVmv mv MV v m−= +  = By the conservation of energy 2 2 21 1 1 2 2 2 i f fmv mv MV E= + + Solving for simultaneous equations ( ) 2 22 2 2 2 2 0 if if f i f mv MVmv m MV E m M m M V mMv V mE −= + +  + − + =  discriminant ≥ 0 ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 (min) 2 4 2 0 4 8 0 81 2 8 i i i K mMv M m M mE m M v mM m M E mM m M Emv mM mMEE M  − +  − +  + + (c) When nN=

2 2 22 2 1 1 2( 1) 22 1 ( 1)2 N N Fdmv m u m mu N Fd  −=+   = + − ( )21 ( 1)2 mMmu N Fd E M++ − = ( )( ) ( ) ( ) 2 14 2 13 2613 27 27 ( 1) 2 (62) 8.0 10 5.0 10 2.0 10 1.5 102.0 10 26.68 10 1.67 10 kg N Fd Em Eu M − − − − − − −−= −   −  =  − = 2 (a) Consider A+B, Resolving perpendicular to the incline (50 ) cosN M g =+ Resolving along the incline (50 ) sinf M g =+ At limiting friction 1 (50 ) sin (50 ) cos tan tan 0.3 16.7 fN M g M g       − = + = + = = = (b) Considering A only, Resolving perpendicular to the incline 50 cos30ABN N g= +  Resolving along the incline 50 sin30 0.3 cos30 ( 50) 0.3 cos30Ag T g M Mg= +  + +  ------ (1) Can just substitute values directly into this equation to obtain answer for m. No need to re-arrange into the expression shown. NA cos30Amg  NB cos30Bmg  cos30 ( ) BAg m m + sin30Amg  TA

3 Considering B only, Resolving perpendicular to the incline cos30BN Mg= Resolving along the incline sin30 0.3 cos30 BMg Mg T+ = ------ (2) For the rough cylinder, 0.2 1.8745B B BAT T e T T e = == ------ (3) Subst (1) and (2) into (3) 50 sin30 0.3 cos30 ( 50) 0.3 cos30 1.8745( sin30 0 .3 cos30 ) 50 sin30 0.3(50 cos30 ) 0.6 cos30 (1.8745)( sin3 0 0.3 cos30 ) 6.18 kg g g M Mg Mg Mg g g Mg M g g M =−  + −  +  −  − = +  = 3 (a) when lift is moving at constant velocity: 1 2 LTT g== when lift is accelerating vertically: 2 2 LT ga= − Explanation 1 for direction of a Since 21TT , ( )g a g− . Hence a is positive. This implies that a is in the same direction as g and a is directed vertically downwards. Lift is accelerating downwards. cos30Bmg  sin30Bmg  NB cos30Bmg  TB

4 1 2 1 2 2 1 2 2 1 2 2 2 1 L gT T L ga T ga Tg Ta g g T Tag T   = − −= =−    =−   Explanation 2 for direction of a Since 21TT , 2 1 2 1TT  . Hence a is positive. This implies that a is in the same direction as g and a is directed vertically downwards. Lift is accelerating downwards. (b) (i) At 0t = , the point of suspension of the pendulum at the arm of the retort stand and the bob are at rest relative to each other. As the lift begins to accelerate, t he point of suspension accelerates downwards at kg while the bob accelerates downwards at g due to free fall. Since the point of suspension accelerates downwards more than the bob, the point moves towards the bob causing the string to go slack instantaneously and the tension in the string is zero. Hence the only force acting on the bob is its weight. The bob’s relative acceleration, ( )g kg− , is vertically upwards, and its initial velocity is zero. Since the relative acceleration of the bob is vertical ly upwards and its initial velocity is zero, the string remains slack and the bob moves in a straight line vertically upwards from its initial position until the bob is above its suspension point and at a distance L from its suspension point where the string becomes taut again. (ii) Take upwards as positive. In the vertical direction, relative acceleration of bob with respect to the lift is ( ) ( )1rela g kg k g=− − − = − String becomes taut again when the bob is above its suspension point and distance L from the suspension point.

5 Since bob moves vertically upwards after release, its horizontal displacement is zero. Let y be the vertical displacement of the bob from the point where it is released till the point where the string becomes taut again. Using 21 2s ut at=+ , ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 01 2 20 1 2 0 1 1 2 1 2 2 0 1 12 2 0 1 12 2 0 1 1 2 12 cos 0 1 2 12 1 1 22 4 112 21 12 121 1 2 2 21 14 rely ut a t L k gt L k gt Lt kg Lt kg Lt kg Lt kg       =+ = + − − = −  =− −      = + −       −         = + −       −      =−   −  Shown. 4 (a) As all the nodes along Y have the same potential, there is no current in the horizontal resistors along Y. Therefore, the circuit simplifies to (the classic infinite resistors ladder problem). r r r r r r r r Z X r r r r r r r r Y X Z R equivalent to C D r r 2r

6 Since the network is infinitely long, the addition of the extra segment makes no difference to the overall resistance. ( ) ( ) ( ) ( ) ( ) ( ) 1 22 22 2 2 11 2 22 2 2 2 2 2 2 4 4 2 4 0 2 2 4 4 2 1 5 (reject negative ) 3.24 i.e. XZ rR rR rRrR rR R r R r r R rR rR R r rR R rR r r r r R rR r RR r − + + + = += + + = + + + = + − − = − −  − − − = =+ = =  (b) ( )( ) 2 3.24 1.2366 2 3.24 1.2366 1.2366 2 2 1.0 1.2366 0.61832 0.618 A vertical vertical vertical vertical rrRr rr rVV rr rr == + = + = = = I I OR Using Current Divider Rule, ( )3.24 1.02 3.24 0.61832 0.618 A vertical r rr = + = = I 5 (a) E = mv 212 Multiplying both sides by m yields mE = m v2212 = p212 Therefore pE= m22

7 (b) (i) Since both ends must be nodes, the length L must be equal to 2n  , where n is a positive integer (n = 1, 2, 3….). Therefore n 2L= n (ii) The momentum p of a particle is given by h h hnp L n Ln n (2 / ) 2= = = where h is the Planck constant and n is the wavelength for the nth state. (iii) 2 2 22 n n 2 ()2 22 hn p hnLE mm 8 mL= = = (iv) The energy difference E = En – En–1 ( )hnhn 8mL 8mL 2222 22 1−=− ( ) 2222 22 21h n nhnhf 8mL 8mL −+=− ( ) 2 2 1h 2nhf 8mL−= ( ) 2 1h 2nf 8mL−=

8 (v) Similar to the hydrogen atom, the energy levels are quantized. However, whereas the hydrogen’s energy levels become closer at higher orders n 2 13.6 eVE n =− , the energy levels here become further apart ( ) 22 n 28 hnE mL   = . There are two fundamental differences between the electron in this question and that in the hydrogen atom: 1. The energy levels of the electron in this question are purely kinetic, whereas the energy levels in the hydrogen atom comprise the total energy (kinetic + electric potential energy). 2. The potential well that traps this electron is one −dimensional, whereas the hydrogen atom is three−dimensional. For last part, either point 1. or point 2. gets 1 mark. 6 (a) At t = 0, 0, 32 0, 33 32 33 33 9 9 (1) = = AA NN Since 1/2, 33 1/2, 32 33 32 32 33 1.8 ln2 ln21.8 1.8 (2)   = = = tt Solving (1) and (2), 33 0.20=NN E h2/ (8mL2) h2/ (2mL2) 9h2/ (8mL2) 2h2/ (mL2) n = 1 n = 2 n = 3 n = 4

9 (b) At t = 0, 33 total 0.10=AA As t → , 33 total 1.0=AA [Explanation for mark allocation: B1 - Correct shape, B1 - Values of fractions correctly indicated on vertical axis] (c) 0, 32 0, 339=AA At time t, 32 33 32 33 32 33 32 33 32 33 () 9 9 9     −− − − −− = = = tt t t t AA ee A e A e e When 90% of decays is due to 33 1