DHS 2023 H2 Phy P1 Ans

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 1 Answers to Pr elim Exam H2 Physics Paper 1 1 2 3 4 5 6 7 8 9 10 A B B D A B C C D C 11 12 13 14 15 16 17 18 19 20 A C D B D D B A B B 21 22 23 24 25 26 27 28 29 30 D C A D C C A A B D 1 A energy = power x time = 3.0 x 109 x 2.0 x 10-12 = 6.0 x 10-3 J = 6.0 x 10-15 TJ 2 B Recall: ∆v = vf - vi The horizontal component of vi = vf since there’s no horizontal acceleration, so ∆v is vertical. ∆v = 12 sin 25o = 5.1 m s-1 3 B Before time instant B, the cars are moving towards each other. After time instant B, they are moving away from each other. 4 D Kinetic energy is conserved only if the collision is elastic. Hence, only momentum and total energy is conserved as there are no net external forces acting on the system when the two objects collide. 25o vf -vi = 12 m s-1 ∆v

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 2 5 A By considering the free body diagrams of the 5.0 kg and 3.0 kg masses: 55 33 gT a Tga −= −= .( 1 28 1 4 2 5 m s to 2 s.f.) ga ag − ∴= = = 6 B Taking moments about the pivot, W (0.900 ) (1.60 ) [ (4.00 )( ) ](0.900 ) [ ( )( ) ](1.60 ) (4.00)(0.900) (1.60) 4.00(0.900) 1.60 2.25 wood rubber wood rubber wood rubber rubber wood LW L L Ag L L Ag Lρρ ρρ ρ ρ = = = = = 3.0 kg 5.0 kg 3g 5g T T 2.90 L 2.10 L Wood Rubber WRubber WWood 2.00 L 0.50 L

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 3 7 C (0.50)(1000)(9.81) 4910 Pap hpg∆= = = 8 C The work done by the driving force F was used to increase the car’s kinetic energy as it gains speed from u to v. Hence the useful work done by the force F is equals to the increase in KE of the car. 9 D 2D kv= ,where k is a constant of proportionality At 12 m s-1, the driving force provided by the 2 engines is equal and opposite to the total drag. () () 3 3 36000 2 12 720000 12 P Dv k k = ×= = If only 1 engine is on, the new maximum speed v’ can be found by: 3 3 33 1 3 720000360000 ( ') (12) 360000' (12 )720000 12 9.5 m s 2 v v − = = = = 10 A At Q, 2mvN mg r+= As the bead is just in contact with the wire, N = 0. Thus, 2 0.20v rg g= = Using conservation of energy, 21 (0.40)2 1 (0.20 ) (0.40)2 0.50 m mgh mv mg mgh m g mg h = + = + =

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 4 11 A period of rotation = 365 days 7 -1 11 7 4 - 1 22 1.99 10 rad s(365 24 60 60) (1.50 10 )(1.99 10 ) 2.99 10 m s T vr ππω ω − − = = = ×××× = =× ×=× 12 C Gravitational field strength g= GM r2 (vector) Due t o different masses of the Earth and Moon, the field strength due to the Earth (points to the left) is larger than the field strength due to the Moon (points to the right). Hence the resultant field strength is not zero and points towards the Earth. A and B are wrong. Gravitational potential φ = − GM r (scalar) Due to di fferent masses of the Earth and Moon, the potential does not add to zero at point P. That would eliminate D. 13 D Heat gain by A = heat loss by water mA cA (1) = mw cw (4) Heat gain by B = heat loss by water mB cB (2) = mw cw (3) Hence, c A = 2.7 cB 14 B Internal energy of a system is the sum of all the kinetic energy and potential energy of all the atoms of the system.

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 5 15 D ( )( )( ) ( ) 1.5 8.31 273.15 25 20 3.71000 50222 Pa 50kPa R pV nRT nRTp V mRT MV = = = +=   = = 16 D ( ) 2 2 energy of oscillator amplitude 1.25 1fraction of initial energy remained in the first 4.0 s, 5.00 16 15fraction of initial energy lost in the first 4.0 s 1 16 f f ∝ = = =−= 17 B Initially, the angle between the polarising filters is 90 o. After rotating 30o, the angle between them is 60o. Using Malu’s law, 2ocos 60 1 4 o o = = II I I Since intensity is proportional to square of amplitude, '1 2o A A = = I I 18 A

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 6 Standing wave on a rope has nodes at both ends. 1.02Ln λ= = For wavelength 0.4 m, this corresponds to the 5th harmonic (n = 5) mode, thus A is correct. B is wrong as there are 6 nodes for the 5th harmonic. C is wrong as the fundamental wavelength is 2 m. D is wrong as the midpoint of the rope may be moving (e.g. for fundamental mode, the midpoint is the antinode. 19 B Diffraction is the most pronounced when the size of the opening is comparable to the size of the wavelength. Option B is correct because halving the frequency means doubling the wavelength to 1.0 m which is the size of the doorway. 20 B 2 0 0 and , therefore, 4 4 QQ VVE E rr rπε πε= = = 21 D Option C is incorrect since the current is the same in both sections. Since l ll ρρ= ⇒= ⇒∝RRR A AA 1 (ρ is constant for same material), resistance per unit length of the narrow section is twice that of wide section since the constant current flows through the narrow section. Since ∝VR for constant I, the resistor with the larger resistance has a larger p.d. across it. Hence, the narrow section (with larger resistance per unit length), has larger p.d. per unit length across it. Hence options A and B are wrong. 22 C

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 7 Effective resistance of the 3 parallel resistors, Ω =     + + = − 588 . 00 . 5 1 0 . 2 1 0 . 1 1 1 effR Potential difference across each resistor, V = IReff = (5.0)(0.588) = 2.94 V Current through 2.0 Ω resistor, I2 = V/R = 2.94/2.0 = 1.47 ≈ 1.5 A 23 A The bulb with the largest current flowing through it will brightest as power = I2R. Hence bulb P is the brightest. 24 D The direction of the force is given by Fleming’s left -hand rule. The direction of the force is 90o to the plane containing the magnetic field and the current. 25 C Before the rod lands on the slope, it is traveling in the direction of the Earth’s magnetic field. There is zero change in magnetic flux linking the rod. As the rod lands and rolls down the slope, the rate of change of magnetic flux increases uniformly. The magnetic flux changes constantly after it rolls off the slope. 26 C <P> = 1 2 I02R = P <P> = 1 2 [(I0)new]2(2R) = 4P → (I0)new = 2 I0, Irms = 1 2 (I0)new = I0 27 A

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 8 hp λ= , momentum of photon is independent of the intensity of light. 28 A As a result of wave- particle duality, we cannot simultaneously know both the position and momentum of a particle, such as a photon or electron, with perfect precision. 29 B By conservation of charge and mass number, X = 2. No. of hydrogen nuclei = 0.001 / (1.008 x 1.66 x 10 -27) = 5.98 x 1023 Energy released, in MeV, in 1 reaction = 1.6 x 10 12 / (5.98 x 1023 x 1.6 x 10-13) = 16.73 MeV Total BE He-4 – BELi-7 = energy released, 16.73 MeV (note BE of H-1 is 0, as it is only 1 p