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 fr