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 + GPEiniti