VJC 2022 Prelim P3 Guide

1 2022 VJC Prelim H2 P3 Suggested solution 1(a) At maximum height, Loss in KE of mass = gain in GPE of mass + gain in EPE in cord [CONCEPT] 2211 022mv mgh kx− = + 3 2 3 211(150 10 )(5.7 ) 0 (150 10 )(9.81)(1.12) (45)22 x−− − =  + 0.19x= m (b) Length of unstretched cord = 1.12 – 0.19 = 0.93 m (c) (G should end lower so that G + E = initial K at 1.12 m) (d) Since the tension acting on the ball is always towards the center and perpendicular to the displacement of the ball [APPLICATION], there is no work done by the tension. [CONCLUSION] [CONCEPT: WD = force x displacement in the direction of the force] 2(a) When the sound wave hits the wall, it will be reflected in the opposite direction and interfere with the original wave. [APPLICATION] Now we have 2 waves of the same amplitude, frequency and wavelength moving in opposite directions interfering with each other [APPLICATION], giving rise to a stationary wave. [CONCLUSION] [CONCEPT: conditions for formation of stationary wave] Energy/J displacement/m K G E 0.93 1.12

2 (b) (c) From above diagram,  = 10 cm Frequency f = v  = 360 0.10 = 3600 Hz d(i) After hitting the wall, the wave will continue to spread out as before, but in the opposite direction. From diagram, reflected wave travels 25 cm. r1 = 5 cm, amplitude A1 = 3.0 x 10-5 m r2 = 25 cm, A2 = ? Intensity I  A2  2 1 r [CONCEPT]  A  1 r 21 12 Ar Ar = A2 = A1 1 2 r r = 3.0 x 10-5 x 5 25 = 6.0 x 10-6 m Source Wall 15 cm  = 10 cm Source X 5 cm 10 cm

3 (ii) Since X was originally a node, it is a place of destructive interference.  resultant amplitude = (3.0 – 0.6) = 2.4 x 10-5 m 3(a) The electric field strength at a point is defined as the electric force exerted per unit positive charge placed at that point. (b)(i) (Before x = 10 cm, the electric field points to the right. After 10 cm, it points to the left.) Since the electric field changes direction between A and B, the charges have the same sign. OR At a point between A and B, the electric fields due to each charge cancel out to give a resultant field strength of zero. So the charges have the same sign. (ii) Any 1 point: • Charge A is not the only charge present. • The electric field strength is also influenced by charge B. • The electric field strength is due to two/both charges. • The electric field strength is the resultant of the two fields due to charges A and B. (iii) From the graph, E = 1.8  103 N C-1 where x = 6.0 cm. Electric force on proton, FE = qE = (1.60  10−19)(1.8  103) = 2.88  10−16 N Acceleration of proton, 16 27 2.88 10 1.67 10 EFa m − − ==  = 1.72  1011 m s−2 (iv) At x = 10 cm, field strength due to A = field strength due to B (in magnitude) ABEE= [CONCEPT] 22 0044 AB AB QQ rr = 2 2 22 (10.0 cm) 4(5.0 cm) AA BB Qr Qr = = = 4(a) Magnetic force FB = Bev = (4.0 x 10-3) x (1.60 x 10-19) x (0.60 x 10-3) = 3.84 x 10-25 N (b) [Note: Direction of conventional current is opposite to that of electron flow. Using FLHR, magnetic force is downwards, so electrons accumulate at the bottom.] (c)(i) See above diagram. Electrons accumulate here E

4 (ii) 1. • The electric force FE points upwards [APPLICATION]. • As the electric field gets stronger, FE also gets stronger [APPLICATION]. The net force Fnet = FE – FB gets weaker [CONCEPT: net force = vector addition]. So the rate of accumulation drops [CONCLUSION]. • Eventually FE = FB, there’s no more net force acting on the electron [APPLICATION], and the accumulation stops [CONCLUSION]. 2. Accumulation stops: eE = Bev [CONCEPT] E = Bv Potential Difference, V = Ed = Bvd [CONCEPT] = (4.0 x 10-3) x (0.60 x 10-3) x (1.5 x 10-2) = 3.6 x 10-8 V 5(a) Higher frequency/lower wavelength means higher energy photons OR E = hf [CONCEPT] violet photons are energetic enough to liberate electrons, while red are not OR Energy of violet photons is higher than work function, while red photons is not [APPLICATION] Increasing intensity means higher rate of photons [CONCEPT]. Higher rate of photons (incident on potassium leads to) more electrons produced for violet light [CONCLUSION]. Since red light photons are not energetic enough to liberate electrons, even at very high intensity it will still not emit any electrons from the metal surface [CONCLUSION]. (b) Work function, where f is the threshold frequencyoohf = 19 34 14 14 4.5 10 6.63 10 6.78 10 6.8 10 J of h  − − ==  = = So photons below threshold frequency of 6.8 x 1014 Hz will not release electrons from the metal surface. FE FB ‘Show’ question: Write to more sf than the show answer

5 (c)(i) Ek is 0 for f below the threshold frequency 6.8×1014 Hz (allow ±0.4×1014 Hz) because no electrons are emitted from the metal surface. Ek = hf – , so plotting Ek against f gives a graph of constant gradient/straight line is obtained (after threshold frequency). (ii) − −−= =  − 18 34 14 (5.0 0) 10gradient = 6.6 10 J s(14.4 6.8) 10h 6(a) Binding energy is defined as the amount of energy needed to split a nucleus into its individual nucleons. (b) (i) 𝐻𝑒 → 𝐻𝑒2 3 + 𝑛0 1 2 4 (ii) Q = Difference in total BE [CONCEPT] = 4(6.8465) – 3(2.2666) = 20.5862 MeV (iii) mass of neutron + mass of 𝐻𝑒2 3 – mass of 𝐻𝑒2 4 = 𝑄 931.494 mass of 𝐻𝑒2 4 – mass of 𝐻𝑒2 3 = mass of neutron - 𝑄 931.494 = 1.0097 u - 20.5862 931.494 = 0.9876 u (iv) Since energy is supplied to make the process happen, this is a mass excess reaction. So mhelium-4 < mhelium-3 + mneutron mhelium-4 - mhelium-3 < mneutron 7 (a) Newton’s law of gravitation states that the force of attraction between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

6 (b) (i) ( ) ( )( ) ( ) 22 3 11 24 263 1 350 10 6.67 10 6.0 10 6.4 10 350 10 8.78 EE E GM GMg r R Nkg − − == +  =  +  = It is along the line joining point P and the centre of the Earth and is pointing towards the Earth’s centre. (ii) ( )    2 2 2 6 32 [CONCEPT] 2 4 6.4 10 350 104 55108.78 GF m r gm m r T rTs g = =   +  = = = (c) (i) Gravitational potential at a point is defined as work done per unit mass by an external agent to move the unit mass from infinity to that point. (ii) Since gravitational force is attractive, the force that the external agent exerts (which points towards infinity) is opposite to the displacement of the unit mass. So the work done by external agent to move unit mass from infinity to that point is always negative [APPLICATION]. So the gravitational potential at that point is also negative [CONCLUSION]. [CONCEPT: WD by force = force x displacement in the direction of the force] (iii) Gravitational field strength, a vector, is given by the gradient of the potential graph. [CONCEPT] Near the surface of the Earth, the net gravitational strength is directed towards the Earth. Near the surface of the Moon, the net gravitational strength is directed towards the Moon [APPLICATION. Therefore, the gradient of the potential graph near the surface of the Earth and that near the surface of the Moon have opposite signs [CONCLUSION]. (iv) To reach the surface of the Earth, the spacecraft require a minimum kinetic energy that is just sufficient to overcome the gravitational potential energy difference betwee