2024 RI Prelim H2 Phy Paper 3 Answers

Raffles Institution Year 5-6 Physics Department 1 2024 H2 Physics Preliminary Examination Solution Paper 3 1 (a) The internal energy of an ideal gas is just the sum of the microscopic kinetic energy, due to the random motion of its molecules since the microscopic potential energy of an ideal gas is zero. Its internal energy depends only on its state (pressure, volume, temperature) and amount of gas. (b) (i) 1. U Q W = + Since volume is constant, 0W = . ( ) ( )( ) 21 21 2 5 5 33 22 3 2 3 2.0 10 1.5 10 1.0 102 1500.0 1500 J (shown) QU p V pV V p p − = =− =− =   −  == 2. Since 21 1 21 1 23 3 6.2 10 J2 2 6.2 10 299.52 K3 1.38 10 kE kT T − − − = =   ==   From nRpT V= , since volume is constant, 1 1 pnRp T TVT =  =  ( ) 1 1 5 5 0.5 10 299.52 1.0 10 149.76 150 K pTT p = =  ==

Raffles Institution Year 5-6 Physics Department 2 OR 1 1 3 2 2 3 k k E kT ET k = = Since volume is constant, 12 12 22 21 11 2 3 k pp TT p p ETT p p k = ==  2 21 1 2 1 21 5 23 5 22 33 2 13 2 6.2 10 1.5 10 13 1.38 10 1.0 10 149.76 150 K 150 C kk k p E ETT p k k Ep kp − − − = −  =−     =−      = = =  OR 1 kU NE= where N is the number of gas molecules 1 k UN E= ( ) 21 21 21 33 22 3 2 U U U NkT NkT Nk T T  = − =− =− ( ) ( ) ( )( ) ( )( )( ) 21 1 3 12 21 5 2 23 2 3 2 3 2 3 6.2 10 15004 9 1.0 10 2.0 10 1.38 10 149.76 150 K 150 C k k UTT Nk U U E k EU pV k − −− −= = =  =    = = = 

Raffles Institution Year 5-6 Physics Department 3 (ii) The first law of thermodynamics states that the increase in internal energy U is equal to the sum of the work done on the system W and the heat supplied to the system Q, i.e. U Q W = + . Hence Q U W= − . When a unit mass of the gas , heated under constant volume or under constant pressure, experiences a unit rise in temperature, U will be the same since UT  . When the gas is heated at constant volume, there is no work done. W is zero and VQU= . When the gas is heated at constant pressure, work is done by the gas as it expands. W is negative and pQU . The specific heat capacity of a gas is the heat supplied to a unit mass of the gas to cause a unit rise in its temperature i.e. ( ) Qc mT=  . Since pVQQ the specific heat capacity at constant pressure is higher than that at constant volume.

Raffles Institution Year 5-6 Physics Department 4 2 (a) Since k and m are constant, ax− . This implies that the block’s acceleration is proportional to its displacement from the equilibrium position. The negative sign implies that the direction of its acceleration is always opposite to its displacement, pointing towards the equilibrium position. This satisfies the definition for simple harmonic motion. (b) (i) From the graph, 0.60 sT = , 2 0 2.0 10 mx −= ( ) 00 0 2 1 2 2 2.0 100.60 0.20944 0.209 m s vx xT    − − = = = == (ii) -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 v / t / s -0.209 0.209

Raffles Institution Year 5-6 Physics Department 5 (iii) ( ) ( ) 2 2 2 2 2 2 2 00 1 1 1 2 2 2 KE mv m x x m x x = =  − = − ( ) 2 2 2 2 2 2 2 00 1 1 1 2 2 2 P T KE E E m x m x x m x  = − = − − = ( ) 2 2 2 2 2 0 22 0 2 0 11 22 2 2.0 10 0.014142 m 1.4142 cm 22 PKEE m x m x x xx xx  − = =− = = = = = At equilibrium, 16.0 cmL= . When PKEE= , 16.0 1.4142 17.4142 17.4 cmL= + = = (block is below equilibrium) OR 16.0 1.4142 14.5858 14.6 cmL= − = = (block is above equilibrium) (c) 2 2 2 eff eff eff F F mg k e ke mg kk == == = The same mass results in half the extension in each spring at equilibrium compared to a single spring in Fig. 2.1. Hence the effective force constant is twice the force constant of one spring ( 2effkk = ). From kax m=− , angular frequency 2 k Tm  == . Hence period 2 mT k= . When the force constant is twice that in (a), the period decreases to 1 2 the period in (a). block mg F F

Raffles Institution Year 5-6 Physics Department 6 3 (a) amplitudes: Particles in a progressive wave have the same amplitude. Particles in a stationary wave have amplitudes ranging from zero at the nodes to maximum amplitude at the antinodes. phases: Particles within a wavelength in a progressive wave have different phases. Particles in a stationary wave between adjacent nodes oscillate with the same phase and particles between adjacent segments oscillate  rad out-of-phase. Particles at the node do not oscillate. (b) (i) Wavelength is the distance between two adjacent particles that oscillate in phase. 2 1.20 m = (2 d.p.) 0.600 m = (3 s.f.) (ii) There are 6 intervals between 0.00 m and 0.30 m (1/2 wavelength) indicating that the separation between particles when they are at equilibrium is 0.05 m. The particle with equilibrium position at 0.15 m is at the amplitude position of 0.19 m at 0t . 0 0.19 0.15 0.04 mx =−= (2 d.p.) (c) (i) (ii) 1. Pressure at the nodes is highest or lowest compared to the initial atmospheric pressure. *Correct high and low pressure positions. 2. For stationary waves, energy is not transferred and the waveform does not progress forward. Positions of nodes and antinodes remain unchanged. Only the displacement of the particles between the nodes changes. *Graph of smaller amplitude at 1 8 Tt + compared to at 1t . 0.0 0.2 0.4 0.6 0.8 1.0 1.2 N N N N N A A A A x / m 0.0 0.2 0.4 0.6 0.8 1.0 1.2x / m Y Z p

Raffles Institution Year 5-6 Physics Department 7 (d) (i) Stationary wave is formed in the air column in the tube when an anitnode is at the mouth of the tube and a node is at the water surface. When first loud note is heard, 1 1 4Lc += ( ) 1 1 4 1 0.600 0.1444 0.150 0.144 0.006 m (3 d.p.) cL =− =− =− = (ii) When the next loud note is heard, 3 2 4Lc += ( ) 2 3 4 3 0.600 0.0064 0.444 m (3 d.p.) Lc =− =− =

Raffles Institution Year 5-6 Physics Department 8 4 (a) (i) The electric field strength due to each particle is directed towards the left . As the resultant electric field strength at any point is the vector addition of the individual electric field strengths of A and B, it is always towards the left. Hence there will be no point where the electric field strength is zero. (ii) The electric potential due to A is negative while that due to B is positive. The total electric potential at any point is the scalar addition of the individual electric potentials of A and B. Hence, there will be a point in between the charges where electric potential is zero. (b) (i) By the principle of conservation of energy, considering energy changes of particle from the point it enters the electric field to the point it hits point P, increase in kinetic energy = decrease in electric potential energy ( ) ( ) ( ) ( ) ( ) ( ) 22 22 27 2255 19 11 22 6.6 10 6.5 10 4.1 10 3.2 10 5247 5250 V fi fi m v v q V mV v vq − − − =    = −  =  −   == Since the negatively charged particle accelerates towards plate Y which is at 0 V, plate X must be at a lower potential with respect to plate Y. 5250 VV =− (ii) ( )( ) ( )( ) 19 27 2 12 12 2 3.2 10 5250 6.6 10 3.6 10 7.0707 10 7.07 10 ms EF ma Vq mad qVa md − −− − =  = =  =  =  =  (iii) Take direction to the right and downwards as positive. consider horizontal motion, ( ) ( ) ( ) 2 2 5 12 2 78 1sin 2 10.5 3.6 10 4.1 10 sin32 7.07 10 2 1.0842 10 s or 4.6963 10 s (NA) s u t at tt t  − −− =− +   =−   +  =  −  consider vertical motion, ( ) ( )( ) 57 cos 4.1 10 cos32 1.0842 10 0.037698 0.0377 m d u t  − = =    ==

Raffles Institution Year 5-6 Physics Department 9 5 (a