RI 2020 Promo H2 Physics Solutions

© Raffles Institution 2020 Year 5 Promotion Examination H2 Physics Solutions 1 D Eh f= ( ) ( ) 2 21 11 kg m s munit of Junit of k g m sunit of ss Eh f − − −−= = = = 2 A The magnitude of the resultant vector of option A is 2.83 times the magnitude of one vector. The magnitude of the resultant vector of the other options are 2 times the magnitude of one vector. 3 C 2 22 1distance travelled by ball P 9.81 1.5 11.0 362 distance travelled by ball Q 25 11.036 13.9 64 11using , 13.964 (1.5) ( 9.81)(1.5 )22 16.66 s ut at u u = ××= = −= =+ = +− = 17 17 m s −= 4 C Change in momentum, ∆p = Area under the graph from t = 1 to 5 s = (1)(4) + (0.5)(1)(4) + (0.5)(2)(−2) = 4 N s ∆p = m∆v 4 = (4.0)(∆v) ⇒ ∆v = 1 m s−1 ∴vf = 1.8 + 1 = 2.8 m s−1 5 B A floating object displaces its own weight = mg in fluid. Since g is a constant , an object displaces its own mass in fluid as well. A: Fully submerged object displaces less than its own weight in fluid that’s why it sinks C: Object can have no net forces on it but may not be in rotational equilibrium. D: Upthrust is ρgV regardless of depth in an incompressible fluid. 6 B Since there is no friction (smooth surface), applied force F is the resultant force which is constant. dvF ma m dt= = P Fv= Since F is constant and v increases at a constant rate, P will also increase at a constant rate with respect to time. 7 B The circular motion is in the horizontal plane, hence the horizontal component of the normal force (normal force × sinα) provides for the centripetal force, whereas its vertical component (normal force × cosα) is equal in magnitude to the weight (because the marble ball does not accelerate in the vertical direction). Hence the magnitude of the normal force must be larger than that of the weight.

2 © Raffles Institution 8 B 2 1 2 1 1 2 1 1 12 21 1 22 1Hence, and K K K mv GMm r r GMmmv r ErE r Er = = ∝= 9 A ( ) ( ) 22 10 2.5 2 planet G Mm GMmW mg RR = = = × 10 A ( ) ( ) ( ) 22 2 2 2 22 2 22 11 22 11 22 2 1 20 2 1 41 cm o o mv kx m x x mx km xx x. x. ω ωω = −= = = = = Shortest distance moved = 2.0 – 1.41 = 0.59 cm 11 C When damping is decreased, the frequency at which the system responds with the highest amplitude should increase i.e. the peak of the curve should shift to the right. A: In fact, all points on the frequency response curve are plotted when the amplitude of the oscillations have stabilised such that the rate of energy gained is equal to the rate of energy lost in the system. B and D : When damping increases (or decreases), t he whole curve becomes flatter (sharper) and every point on the curve is lower (higher) than the original curve. 12 C 2 2 &4 P kArπ= =II For the sound to be as loud as before, amplitude and hence intensity will have to be as it was originally. Since power increased, intensity increases as well. At 3 m from the source: ( ) ( ) ( ) 1 21 12 12 22 2& 43 43 43 P PP π ππ = = = >II I The person will have to move away from the source to experience the same intensity (and hence amplitude) as before. ( ) ( ) ( ) 21 211 22 For 2 2 3 4.243 m 4 43 PP r rππ = = ⇒= = II

3 © Raffles Institution 13 A 14 D Path difference = 1.6 − 1.3 = 0.3 m = 3 wavelengths The waves meet at P in phase and constructive interference occurs. Amplitude = 2A + A = 3A 15 D There must be an antinode at each open end. There can be any number of nodes with the tube. 16 (a) The acceleration of the mass in the y direction is constant and there is a uniform velocity in the x direction. (no acceleration in x-direction) B1 B1 (b) (i) u sin 60° or 0.866 u B1 (ii) −g sin θ (accept g sin θ) B1 (iii) since 9.81sinya θ=− and at the maximum height, 0yv = using 0 sin60 ( 9.81sin )(0.25) sin60 2.4525sin 2.8sin (shown) yyyv u at u u u θ θ θ = + = °+ − °= = M1 (iv) cos60 2.8sin cos60 xuu θ = ° = ° 0.24 2.8sin (cos60 )(0.25 2) 20.1 xs ut θ θ = = °× = ° M1 M1 A1 Using u = 2.83 sin θ, θ = 19.8° (c) marking points: lower height smaller distance along AB and symmetrical B1 B1 trajectory of mass 60° u 0.24 m A B C D y x Sketch the position of the string at the next instant in time.

4 © Raffles Institution 17 (a) (i) An arrow originating from the object, pointing towards point O. B1 (ii) The frictional force acts as the centripetal force. B1 (b) (i) Using points ( 6.00, 5.90) and ( 9.00, 8.90) on the line: (coordinates read to half smallest square) 8.90 5.90Gradient 1.009.00 6.00 −= =− B1 B1 (ii) 2 c 2 c 1 F mr F mr ω ω =  =   or 2 2 1 ar a r ω ω = = × cgradient = F am= Hence, the gradient is numerically equal to the maximum centr ipetal acceleration. M1 A1 (c) c gradient gradient 1.00 0.40 0.400 N fFf m fm = ⇒= = × = × = M1 A1 18 (a) The gravitational field strength at a point in space is defined as the gravitational force experienced per unit mass at that point. B1 (b) (i) Gravitational field strength due to one star ( ) ( ) 11 30 2 22 12 12 1 6.67 10 5.0 10 1.0 10 1.0 10 N kg GM r − −− × ××= = × +× = × 41.67 10 Resultant gravitational field strength ( ) ( ) 2244 1 1.67 10 1.67 10 N kg −− −− = × +× = × 42.36 10 B1 C1 M1 A1 (ii) Gravitational potential 11 30 12 1 2 2 6.67 10 5.0 10 1.41 10 J kg − − =− × × ××=− × = −× 84.72 10 GM r M1 A1

5 © Raffles Institution (iii) Gravitational potential at M 11 30 12 8 -1 2 2 6.67 10 5.0 10 1.0 10 6.67 10 J kg GM r − =− × × ××=− × = −× By the principle of conservation of energy, gain in Ek = loss in Ep ( )( ) ( ) 224 81 2.0 10 4.72 6.67 102 mv m − × = ×− −− ×  1 m s−= × 42.81 10v B1 B1 A1 19 (a) Angular velocity is defined as the rate of change of angular displacement, Angular frequency is defined as the rate of change of phase angle of an oscillation. (angular velocity is a vector and angular frequency is a scalar, award 1 mark over all) *Marks are not awarded for students who just state the difference in the units. B1 B1 (b) (i) ( ) ( )( ) cos 2.0cos 2 2.0cos 2 0.50 2.0cos3.1 xr t ft t t ω π π = = = = M1 (ii) The camera is moving in simple harmonic motion. B1 (iii) 11 2.0 s0.50T f= = = A1 (iv) 22 max 1 From , speed is maximum when 0 2 0.50 2.0 6.28 m s o o v xx x vx ω ω π − = −= = = ×× = M1 A1 (v) At t = 0.25 s, ( ) ( )( ) ( ) 2 2 2 2.0cos 3.1 0.25 1.4288 2 0.50 1.4288 14.1 m s x ax ω π − = × = =− =− =− C1 M1 A1 20 (a) 1. The waves or sources must be coherent. 2. The waves must have similar (same, equal) amplitude. 3. The waves must be unpolarised or polarised in the same plane. B1 B1

6 © Raffles Institution (b) (i) 3 3 7 2.43.0 10 0.60 10 7.5 10 m 750 nm Dx a λ λ λ − − − = ××= × = ×= M1 (ii) 89 14 3.0 10 750 10 4.0 10 Hz vf f f λ − = ×= ×× = × M1 A1 (iii) 1. As violet light has shorter wavelength, the fringe separation becomes smaller. M1 A1 2. As the amount of light passing through the slits is smaller, the bright fringes become dimmer. or The amount of diffraction at each slit increase and creates a larger overlap of the two diffracted waves, resulting in an increase in the number of fringes. M1 A1 M1 A1 (c) 3 7 sin 1 10 sin 7.5 10500 22.02 tan22.02 2.4 0.971 m dn x x − − = × = × = ° °= = θλ θ θ M1 M1 A1

7 © Raffles Institution 21 (a) (i) ( ) ( )δ − = × = − = ×= 39 lg lg 4.07 10 2.390, lg 8.57 10 9.933 E lg Plo