RI 2024 H2 Physics Promo solutions ABC

2024 Promotion Examinations H2 Physics Solutions 1 B Actual mass of a single coin is 7.62 g. The total mass of eight coins is about 60 g which is 0.6 N 0.06 N is too small as it means that the mass of a single coin is less than 1 gram. 6 N is about the weight of a full 500 ml water bottle, which is too high. 60 N is about the weight of a 6 kg mass, which is definitely too high. 2 C Acceleration 282 6 m s1 −−= = Distance ( ) ( ) ( )11 12 1 8 2 1 8 10 3 33 m22 2s =× ++ × ++ × = 3 A 22 2v u as= − where a is positive. 2 2 2 2 2 v u as uas a = − = −   The graph of v against s is a “square-root” graph, i.e., a parabola. The gradient at the x- intercept should be infinite and the gradient at the y-intercept should be non-zero. Option C is wrong as the gradient at x-intercept is not infinite and the gradient at y-intercept is zero. 4 D Change in momentum occurs in the horizontal direction (no change in the vertical direction) Impulse = change in momentum = (p cosθ ) – (– p cosθ ) = 2p cosθ 5 B The box is at rest on the seabed and there is no resultant force. W = T + N + S If the box is lifted, N = 0 and WTS≤+ . Hence, option A is wrong as it means that the box can be lifted. The box has to be at rest on the seabed before the attempt was made to lift it. Option C is wrong as it implies that the box is moving into the seabed (the total downward force is greater than the total upward force). Upthrust acts on the box and is not included in the equation. Option D is wrong as S has to be in the equation. 6 D Angle of inclination of escalator 1 30sin 3060θ − = = °  To move the escalator at constant speed v without passengers, sin30 Power needed 2. 0 kW drive drive F fW PFv = +° = = With 15 passengers on the escalator, to move at the same constant speed v, ' sin30 15(60)(9.81)(sin30 ) ' 4414.5 Power needed ' ( ')( ) ' ( 4414.5)(0.6) ' 2000 2648.7 4648.7 W 4.6 kW drive drive drive drive drive F fW FF PF v PF P = + °+ ° = + = = + = += = Fdrive W θ

2 © Raffles Institution 7 C 58 8 6 8 GPE per unit time (1.5 10 )(9.81)(120) 1.7658 10 W P 1.7658 10 W Since P 100 MW, 100 10Efficiency = 100% 56.6%1.7658 10 in out = ×= × = × = × ×=× 8 B The hour hand takes 12 hours to make a revolution. 5120.25 3.6 10 m s12 60 60vr πω −−= = ×= × ×× 9 A The highest potential between P and Q is less than zero but greater than −60 MJ kg−1. To project the 1.0 kg mass from P to Q, it will require energy E < 60 MJ to reach this point of highest potential. 10 D ( ) ( ) ( )( ) 2 00 2 0 00 2 0 22 0 2 22 2 1 2 1 2 12 2 2 2 2 0.25 0.050 0.64 0.030 J 30 mJ K mv mx v x mx TT mx T ωω ππ ω π π = = =   = =     = = = =   11 C This is a free oscillation as there are no driving forces acting on it. 12 D 6 42 2.9 10intensity 0.60 10 4 (3.0) 5.5 W P P π − − ×= =× = 13 B 2 1 00 2 2 10 Intensity (cos20 ) 0.88 Intensity (cos30 ) 0.66 o o = = = = I II I II 14 C Q and R are in phase as they are between the same two adjacent nodes. For option A, all particles separated by a node are in antiphase. For option B, R and Q are equal distance from the antinode and have the same amplitude. For option D, s ince P is the antinode, its amplitude and hence max imum speed are the largest. 15 A 7 sin 1 sin 42500(1000) 6.7 10 m2 o dn θλ λ − =  = = ×

3 © Raffles Institution [Turn over 16 (a) (i) ( )( ) ( ) ( ) 2 2 2 Nmunits of N m mm NPa units of m E E −= = = = B1 Marker’s comments: - The working for this question was very poorly done with students equating quantities to units, or wrongly using square brackets to indicate dimensions e.g. F = [N]. - Many students went on auto pilot and derived the base units of E and equated it to Pa without explaining that the units of E is equivalent to that of force divided by area which is therefore Pa. (ii) ( )( ) ( ) ( ) 23 3 11 11 10.0 0.500 0.5 10 0.21 104 1.2126 10 1.21 10 Pa FLE Ae π − − = = × × = ×=× M1 A1 Marker’s comments: - Common mistakes were: o Forgetting to convert mm to m o Using the diameter instead of the radius in the formula 2rπ o Using the wrong formula for cross-sectional area o Not being able to round of the answer to three significant figures o Rounding up instead of rounding off the final answer. (iii) ( ) ( ) 11 11 11 2 0.1 0.001 0.1 0.01210.0 0.500 0.5 0.21 0.4596 1.2126 10 0.6 10 Pa (1 sf) 1.2 0.6 10 Pa EFLAe E F LAe FL d e FL d e E E ∆ ∆ ∆∆∆= +++ ∆∆ ∆ ∆ = ++ +  = ++ +  ∆= × × = × =±× M1 M1 A1 Marker’s comments: - An alternative method would be to use the maximum minimum method. o However, a large number of students failed to recognise that to get maximum E, maximum values for F and L as well as minimum values for A (and hence d or r) and e have to be used. The opposite is true to find minimum E. o Also, many students forgot to divide the difference between maximum and minimum by 2. - Common mistakes following the determination of E E ∆ were o Forgetting to multiply this value by E to get E∆ o Not rounding off E∆ to 1 s.f. o Not rounding off E to the correct number placement.

4 © Raffles Institution (b) d contributes most significantly. Use a micrometer screwgauge to measure the diameter. M1 A1 Marker’s comments: - Many students correctly identified diameter as the quantity with the largest percentage uncertainty. - Many students suggested to use a wire with a larger diameter. This is wrong as it changes the whole experiment. - There were suggestions to measure the diameter of the wire several times in different places and take the average of the readings. This does not affect the uncertainty in E. - Students who suggested to use a “more precise instrument” should provide the name of the instrument. - Vernier Calipers (VC) is not accepted unless it is specifically stated that a digital VC with a higher precision to at least 0.01 mm is to be used. Otherwise, a standard VC’s precision is 0.1 mm which is the same as the uncertainty in the diameter. - Putting several wires together to measure the diameter will not work . The wires are assumed to be round and will not be able to stay in a straight line that allows you to measure all the diameters in one straight line. This introduces more uncertainty and random error into the measurement. - Winding up the wires to do the same is a slightly better method but also introduces a significant amount of random error into the measurement. - Using the “instrument used to measure e” to measure d is not acceptable as the method and instrument used to measure extension cannot be used to measure diameter. 17 (a) Consider motion in the x-direction. 4.0 cos60 4.0 cos60 ut t u = °× = ° Consider motion in the y-direction. By 21 2s ut at= + , ( ) 2 1 4.0 1 4.01.0 sin60 9.81cos 60 2 cos 60 7.28 m s u uu u −  = °−   °°  = M1 M1 A1 Marker’s comments: Most students are able to write down the corrections describing the horizontal and vertical motion. However, a significant number of students encounter difficulties solving the two equations simultaneously. There were also a number of students who mixed up t and t2. Some students set vy to 0 to find t, without realizing that the value of t found is for the highest point of the trajectory, not when the ball enters the net. (b) 4.0 4.0 1.10 scos60 7.28cos60t u= = =°° A1 Marker’s comments: This part is generally well done. However, students who quoted a value for t obtained in (a) using wrong physics are not given any credit here. ECF is given only if they used a wrong value from (a) and apply it to a correct equation in (b).

5 © Raffles Institution [Turn over (c) By 22