2022 RI Yr 5 CT Sect B Soln

© Raffles Institution [Turn over Year 5 July Physics Common Test Solution 16 (a) units of g = m s−2 units of m = kg units of ρ = kg m−3 units of v = m s−1 units of d 2 = m2 ( )( ) ( )( ) ( ) 2 23 12 kg m s units of kg m m s m 1 dC − −− = = M1 A1 Some candidates incorrectly answered “0”. Take note that having no units is not the same as saying the unit of a constant is “0”. (b) (i) Vernier callipers B1 Many candidates incorrectly identified micrometer screw gauge. Take note that the resolution of a micrometer screw gauge is 0.01 mm. (ii) ( ) ( )( ) 22 2 2 3.648 9.818 1000 0.5064036 9.61.23 39.8 1000 d mgC vdπρ π  = = =   22 0.01 0.03 0.02 0.1 0.1220.5064036 3.64 9.81 1.23 39.8 9.6 0.0243 0.02 (to 1 s.f.) d d d d d C mg v d C mg v d C C C ρ ρ ∆ ∆∆∆ ∆ ∆=+++ + ∆  =+++ +    ∆= ∆= Cd = 0.51 ± 0.02 B1 M1 B1 A1 Some candidates used the max -min method which is also accepted. Candidates who did not obtain full credit either made careless calculation errors or forgot to change the units to SI units. Some candidates did not provide the 5sf value for the intermediate value of C d and used the 3 sf value instead for further calculations. (iii) 1 Use a n instrument with greater resolution such as micrometer screw gauge/travelling microscope. B1 2 Measure the length of two or more steel balls placed side by side and divide the length by number of steel balls. B1 Many candidates gave the answer of taking multiple readings of diameter of the same steel ball. This method does not reduce uncertainty in the measurement of diameter. It merely reduces the random error of measurements. The resolution of the instrument directly affects the uncertainty in the measurements taken. Some candidates gave the incorrect answer of using a larger steel ball to reduce uncertainty.

2 © Raffles Institution 17 (a) 1 1 15cos30 13.0 m s 15sin30 7.50 m s x y u u − − = ° = = ° = *1 mark to be awarded if student uses radian mode instead B1 B1 (b) Length of 14 crates = 1.5 × 14 = 21 m Distance between top of ramp and top of crates = 2.5 – 1.5 = 1.0 m To go over the 14 crates, the motorbike has to travel at least distance of 21 m after it has moved 1.0 m below the top of the ramp. ( ) ( ) ( ) ( )( ) ( ) 2 2 2 1 2 11.0 15sin30 9.81 2 11.0 7.5 4 9.81 1.0 2 12 9.812 1.65243 s or 0.12338 s (N.A.) s ut at tt t = + − = °+ − −− ± − − −=   = − ( )15.0cos30 1.65243 21.4657 m xxs ut= = ° = Since sx is greater than 21 m, the motorbike will be able to go over the crates. M1 M1 A1 To go over the 14 crates without hitting them, the motorbike should be 1.5 m above ground at the end of the 14 th crate. Many students either used ground level or 2.5 m height above ground to determine instead, which is wrong. (c) By conservation of energy, KEinitial + GPEinitial = KEfinal + GPEfinal ( ) ( )( ) 2 2 1 11 15 0 9. 81 2.522 16.6 m s m mv m v − += + − = 12.99cos 16.55 38.3 θ θ = = ° Direction of velocity is 38.3° clockwise from the horizontal Or Taking upwards as positive, 22 22 1 2 7.50 2( 9.81)( 2.5) 10.26 m s y y v u as v v − = + = +− − =− 22 110.26 12.99 16.6 m sv −= += 10.26tan 12.99 38.3 θ θ = = ° Direction of velocity is 38.3° clockwise from the horizontal 10.26 m s−1 12.99 m s−1 θ v 16.55 m s−1 12.99 m s−1 θ

3 © Raffles Institution [Turn over Many students did not state the direction of velocity clearly. Stating that the velocity is 38.3° with the horizontal is not clear enough as it does not indicate whether it is above or below the horizontal. We do not accept using North, South, East, West to indicate direction in this question as the question did not indicate which direction is North. (d) With a greater angle and same horizontal component of velocity, the initial vertical velocity is higher. Hence, the motorbike will be able to stay in the air longer and the motorbike will travel further horizontally. Calculation of initial vertical component of velocity, 1 15cos30 cos35 sin35 15cos30 sin35cos35 9.1 m s y u uu − °= ° = ° °= °° = B1 B1 Explaining that initial velocity is higher is insufficient. Students must explain that longer time duration is due to higher initial vertical velocity. One mark is deducted for students who just wrote vertical component has increased as it can be interpreted as vertical component of displacement or velocity. The component that you are referring to must be clearly stated. The question stated that Captain America left the ramp with a velocity that has the same horizontal component as indicated in (a) . However, a number of students did not read the question carefully and stated that the initial horizontal component of velocity has changed, hence changing the conditions stated in the question. No marks are awarded even if parts of the explanation are correct as students have changed the question. 18 (a) 1. Weight 2. Tension A1 Both forces must be correct to score the mark. Some students put down normal contact force, but that is exerted by sphere A on sphere B and vice versa, and are therefore internal forces (since both A and B are part of the system). (b) Both the tension and the weight are acting in the vertical direction. (weight and tension are perpendicular to the velocity/momentum just before the impact) They do not affect the momentum in the horizontal direction. M1 A1 • The most common mistake is to assume that the upward tension is equal and opposite to the weight. This is incorrect for sphere A because it is swinging in a vertical circular path and hence the tension is larger than the eight so that T − W is equal to the required centripetal force. • Weight and tension are negligible compared to normal contact force is not acceptable as this assumption is not required in this scenario. Students should take note that “the external forces do not act in the same direction as the momentum” is not the same as the two being perpendicular which is what is required in the answer.

4 © Raffles Institution (c) Show understanding that area under graph is impulse and demonstrate a reasonable method on how to estimate the area. Counting squares (5 × 5) = 12 squares × 2.5 × 0.005 = 0.15 N s Accuracy of answer: 0.142 to 0.158 Ns. An answer that is way out of the expected value is not awarded any marks. M1 A1 Students should state clearly that they are calculating the area under the graph. Showing some calculations without any explanation will be panelised. (d) 10.050 4.0 0.150speed of A 1.00 m s0.050 −×−= = 10.150speed of B 3.00 m s0.050 −= = B1 B1 The most common mistake is to use the relative speed of approach = relative of separation which is true only for elastic collisions. If the question did not state the collision is elastic, please do not make such an assumption. The only exceptions are the collisions between elementary particles which will always be elastic. (e) (i) t = 0.0125 s A1 For graph readings, students should not round off the answer. They should read to the correct number of dp following the rules in practical. 0.013 is not acceptable. (ii) By the principle of conservation of momentum, 2mv = m (4.0) v = 2.0 m s−1 M1 A1 Some students were penalised for very sloppy workings which are essential to score the marks. Simply scribbling some arithmetic calculations without explaining the physics will not score any marks. 19 (a) 1. The net force acting on the window panel is zero. 2. The net moment/torque about any point acting on the window panel is zero. B1 B1 Students need to refer to the panel (or the system) in some way in t