RI 2023 Promo H2 Physics Solutions

Raffles Institution Year 5-6 Physics Department 1 2023 Year 5 H2 Physics Promotional Examination Solution Section A Qn Ans Solution 1 B 3 344 3 32 3 dVr Vd VV ππ = =  ∆∆ = Gradient of the graph is 3. 2 D milli – 310− micro – 610− nano – 910− 3 C bus bus 12 30 360 ms ut= =×= ( ) 2 car car 2 2 1 2 1360 12 3.2 0 30 2 0.83378 0.83 m s s u t at a a − = + ++ = + = = 4 A from velocity-time graph, 1 0.15 0.2 0.35 m s v − ∆= − − = By Newton’s third law, ( ) ( )( ) ( ) force on cart force on sensor impulse on cart area under force-time graph of sensor 1 0.08 0.02 402 0.35 1.2 3.4286 3.4 kg mv m m =− =− ∆= − − −= − = = 5 D Option A (incorrect): Upthrust on an object is equal to its weight only if the object is floating (i.e. in equilibrium). Option B (incorrect): Upthrust does not depend on the mass of the object. It depends on the mass and hence the volume of the liquid displaced by the submerged object. The volume of liquid displaced depends on the volume of the fully submerged object, which can be different even if the two objects have the same mass. Option C (incorrect): Upthrust is due to the difference in fluid pressure at the top and bottom surfaces of the object. Option D (correct): Upthrust is numerically equal to the weight of the liquid displaced by the submerged object. Hence upthrust depends on the density of the liquid and the volume of the liquid displaced. Since both spheres have the same dimensions, both displace the same volume of liquid when fully submerged and hence experience the same upthrust.

Raffles Institution Year 5-6 Physics Department 2 6 C Taking moments about O, ( ) ( )4 sin sin cos 3sin cos 1tan 3 18.435 18.4 mg L mg L Lθ θθ θθ θ θ = + = = = = ° 7 C Power is the rate of transfer of energy. Hence the gradient of the graph is power. Maximum power occurs between 4 st = to 6 st = , when the gradient is the largest. 40 10maximum power output to lift container 15 kW64 −= = − Since power outputefficiency power input= , power output 15power input (supplied) 23.077 23 k Wefficiency 0.65= = = = 8 B ( ) 11 41 2 2 1.50 10 365 24 60 60 29886 2.99 10 m s vr rT ω π π − = = × = ××× = = × 9 A The centripetal force on the ball is the resultant of the contact force and weight. The centripetal force points horizontally towards the centre of the horizontal circle. 10 C Work done by the gravitational force, () 31 2 2 g extFF fi WW m GM GM RR GM R φφ =− = −−  = − − −−   = P Q R O L L L mg 4mg tabletop N W

Raffles Institution Year 5-6 Physics Department 3 11 A By conservation of energy, surface infinity 21 02 escape EE GMm mvR = −+ = 3 22 4228 3 3 escape escape GRGMv GR v RRR ρπ πρ ρ  == = ⇒∝ 2 , 2 , 2 , 1 41 (11.2)54 2.5044 2.50 km s escape M MM escape E EE escape M v R v R v ρ ρ − =   =     = = 12 D Option A (incorrect): Maximum transfer of energy from the driving system to the oscillating system occurs only at resonance. Option B (incorrect): With a decrease in damping, less energy is lost from the oscillating system. The total energy of the oscillating system should increase. Option C (incorrect): With a decrease in damping, less energy is lost from the oscillating system. The amplitude at resonance should increase. Option D (correct): With a decrease in the degree of damping, the frequency at which resonance occurs increases. Since 1Tf= , the period of the driving force decreases. 13 B The x-t graph shows the displacement from the equilibrium position of one particle at different times as it oscillates in simple harmonic motion. Option A (incorrect): At P, the particle is at maximum displacement from the equilibrium position. Its acceleration is maximum. Option B (correct): Since P and Q are 1 4T apart, their phase difference ( )14 360 90T Tφ = × °= ° . Option C (incorrect): Horizontal axis is time. OS represents two periods. Option D (incorrect): At P, the particle is at maximum displacement and is momentarily at rest. Hence its kinetic energy at P is zero. At R, the particle is at the equilibrium position and its speed is maximum. Hence its kinetic energy at R is maximum. The velocity of the particle at various times can also be determined from the gradient of the x-t graph.

Raffles Institution Year 5-6 Physics Department 4 14 B Option A (incorrect): In a stationary wave, points R and S oscillate π rad out of phase with each other. The phase difference between them is π rad. Option B (correct): In a stationary wave, points Q and S oscillate in phase with each other with different amplitudes. Amplitude varies from zero at the nodes to a maximum at the antinodes. Option C (incorrect): Point P is a node which does not oscillate. At the node, the incident and reflected waves always meet π rad out of phase with each other such that the resultant displacement is always zero. Option D (incorrect): The stationary wave is formed by the superposition of the incident wave and the reflected wave. Hence the amplitude of the incident wave is half the amplitude of point Q. 15 D The third order maxima in the interference pattern are missing as they coincide with the first order minima of the single slit diffraction envelope. single slit first order minima: sin ----- (1)b θλ= double slit maxima: sin ----- (2)an θλ= (2): (1) 0.60 3 0.60 0.20 mm3 a nb b b = = = =

Raffles Institution Year 5-6 Physics Department 1 2023 Year 5 H2 Physics Promotional Examination Solution Section B 1 (a) xx x x 1 20.0 1.40 14.286 14.3 m s s ut su t − = = = = = *Final answer to 3 s.f. (b) y x tan32.0 u u°= yx 1 tan32.0 14.286 tan32.0 8.9269 8.93 m s uu − = ° = ° = = *Final answer to 3 s.f. (c) Consider vertical motion of the ball from starting point to point of maximum height. ( ) yy 0 8.9269 9.81 8.9269 0.90998 0.910 s9.81 v u at t t = + = +− = = = *Final answer to 3 s.f. (d) (i) vertical velocity when the ball hits the wall: ( )( ) 1 yy 8.9269 9.81 1.40 4.8071 4.81 m sv u at −= + = +− = − = − OR gradient, y 1 y 0 9.81 4.8071 4.81 m s1.40 0.90998 v v −− = − ⇒= − = −− *Straight line with negative gradient, positive and negative vy *Correct label values vy / t / s 0 1.40 0 8.93 – 4.81 0.910

Raffles Institution Year 5-6 Physics Department 2 (ii) Height of the wall is the area under the graph from 0 st = to 1.40 st = . OR Height of the wall is the difference in the absolute areas from 0 st = to 0.910 st = and 0.910 st = to 1.40 st = . (e) (i) With air resistance, there is now a resultant force in the horizontal direction that causes the horizontal component of the ball’s velocity to decrease continuously / gradually / at a decreasing rate. (ii) Preferred answer (comparing the rate at which vertical component of velocity changes with and without air resistance) : With air resistance, when t he ball is moving upwards towards its highest point, the vertical component of its velocity decreases to zero at a faster rate compared to the rate when there is no air resistance. When the ball is movi ng downwards from its highest point, the vertical component of its velocity increases at a slower rate compared to the rate when there is no air resistance. Al ternative answer (describing the effect of air resistance on the variation of vertical component of velocity): With air resistance, the acceleration in the vertical direction is no longer constant. When the ball is mo ving upwards, the vertical component of its velocity decreases at a decreasing rate. When the ball is moving downwards, the vertical component of its velocity increases at a decreasing rate. Comments (a) (b) (c) • A number of stu