RI 2024 Y5 H2 Physics Timed Practice solution

2024 Y5 H2 Physics Timed Practice Solutions 1 (a) (i) 21 21 35.7 25.4 10.3 C ( ) 0.2 C 0.2% uncertainty in rise in temperature 100 % 1.9 %10.3 TT TT −= − = ° ∆−= ° = ×= M1 A1 (ii) I I I −− = − = − = ∆−∆∆∆ ∆∆=+ ++ + − ∆ =+ ++ + ∆= + + + + 21 11 21 21 () (3.80)(12.00)(60.0) (0.764)(35.7 25.4) 347.68 J kg K () () 0.06 0.08 0.1 0.001 0.2 3.80 12.00 60.0 0.764 10.3 0.06 0.08 0.1 0.001 0.2 3.80 12.00 60.0 0.764 10. Vtc mT T TTc Vtm c V t m TT c c c −−  × ∆=  11 347.683 15.59 20 J kg Kc M1 C1 M1 A1 Marker’s comments: Students are strongly encouraged to show clear substitution and workings. (iii) 11350 20 J kg Kc −−= ± B1 Marker’s comments: A common mistake is not reporting the actual value of c to the same d.p. as its uncertainty, e.g. 348 ± 20. (iv) The metal block can be heated over a longer period of time such that there is a bigger difference between T2 and T1, so that 21 21 () () TT TT ∆− − can be reduced. or Higher current/potential difference to produce a bigger difference between T2 and T1. B1 Marker’s comments: Answers which explained ways to reduce the fractional uncertainty of I, V, t or ∆T are accepted. Increasing mass to reduce uncertainty of c is not accepted because the increase in mass will lead to a smaller rise in temperature. (b) 2 2 2 3 12 2 Base units of kg m s Base units of (m )(kg m )(m s ) kg m s base units of as shown above F Av F ρ − −− − = = = = B1 B1 Marker’s comments: Students should pay attention to how the workings should be presented for questions that require them to show how units are derived. Students should also take note that -2 -2 kg m s 1 kg m s = and not zero.

2 © Raffles Institution 2 (a) ( )( ) 22 1 2 1 1 1 2 0 2 9.81 1.10 4.65 m s v u as v v − = + = + = B1 (b) (i) B1 There must be a negative sign for v1 Marker’s comments: The most common mistake is taking v∆ to be the resultant velocity instead of vR. (ii) ( ) ( ) ( ) ( )( ) 222 1 4.65 25.0 2 4.65 25.0 cos130 796.0706 28.2 m s v v − ∆= + − ° ∆= = sin130 sin sin130sin 25 0.6787628.2147 42.7 Rvv θ θ θ ° =∆ °= ×= = ° Angle from the horizontal = 90° − 42.7° = 47.3° M1 A1 A1 Marker’s comments: Some students correctly calculated the angle between the arrows without indicating the angle on their diagram but went on to incorrectly state that that angle was to the horizontal/vertical. (iii) Vertically: ( )25.0sin40 25.0sin40 9.81 3.276 s yyyv u at t t = + − °= °+ − = Horizontally: 25.0cos40 3.276 62.7 m xxs ut= = °× = M1 M1 A1 Marker’s comments: Many students unnecessarily broke the first step down into two parts and found the time taken to the maximum height and then the time taken back down to the ground before calculating the horizontal range. (iv) There is air resistance, which acts against the ball’s motion. The actual