RI 2024 H2 Physics Y5 TP Solutions

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 horizontal range of the ball’s motion is smaller. M1 A1 Marker’s comments: Many students overexplained this question. Students should use the marks of the question to decide on the length of their answers. θ θ vR −v1 ∆v 130°

3 © Raffles Institution [Turn over 3 (a) The rate of change of momentum of a body is proportional to the resultant force acting on it and occurs in the direction of the force. B1 B1 Marker’s comments: Many students reversed the order of the first statement and wrote that the resultant force is proportional to the rate of change of momentum. This suggests the resultant force appears as a result of the change of momentum, which is opposite to reality. Many students also missed out the second part of the statement of the law. (b) (i) T − mg = ma T − 3700(9.81) = 3700(0.39) T = 3.77 × 104 N M1 A1 (ii) 1. ( ) 2 2 M1 () 1.3 9.0 330 AxmV tt t xA t rv v v ρρ ρ ρπ π ∆∆∆ = =∆∆ ∆ ∆= ∆ = = ×× × = Rate of increase of momentum of air ( )20 m vt ∆= −∆ 2 2 330 v= × By Newton’s 2nd law, this is the force exerted on the air by the rotor. By Newton’s 3rd law, this is the equal to the force exerted by the air on the rotor. 2 2 330 150000v⇒ × = 115.1 m sv −⇒= M1 B1 A1 Marker’s comments: When the question states explicitly “explain your working”, students should make sure they explain clearly how the expressions used in their calculations are derived or how certain quantities can be equated. In this question, the force that acts on the air is equal to the rate of change of momentum of the air ( )∆ −∆ 0m vt (Newton’s second law). However, many students equated it to the force acting on the helicopter without any explanation, hence losing the mark for Newton’s third law. (iii) The density of air decreases with increasing altitude, reducing the upward force that can be generated by the helicopter. B1 4 (a) ( ) ( ) ( )( ) ( ) ( ) ( ) () () () () 0 0.250 15.0sin30 0.600 sin 0.250 15.0sin30 sin 3. 13 (shown)0.600 A Ay B By A Ay B By B B mu mu mv mv v v +=+ = + − = =   θ θ B1 Marker’s comments: Some students wrote both terms on the RHS of the equation as positive and ended up with vB sin(theta) = − 3.13. This could have been easily accounted for by stating the magnitude is +3.13. But instead, many students simply erased away the negative sign which resulted in a mathematical error and losing the mark.

4 © Raffles Institution (b) Applying the Principle of Conservation of Linear Momentum in the x-direction: ( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) () () () () 0.250 30.0 0 0.250 15.0cos30 0.600 cos 0.250 30.0 0.250 15.0cos30 cos 7. 0870.600 A Ax B Bx A Ax B Bx B B mu mu mv mv v v θ θ +=+ += + − = =   3.13tan 7.09 23.8 θ θ = = ° M1 M1 (c) ( ) 1 sin 23.8 3.13 7.76 m s B B v v − = =  Or: ( ) 1 cos 23.8 7.08734 7.75 m s B B v v − = =  A1 (d) Correct directions B1 Correct labelling of vectors and angles (30.0° and 23.8°) B1 Marker’s comments: This question asks for a diagram that shows the relationship between the total momentum and the momenta of A and B. Hence, a triangle is necessary. (e) ( )( ) 21Initial 0.250 30 112.5 J2 kE = = ( )( ) ( )( ) 2211Final total 0.250 15 0.600 7.75 46.1 J22 kE = += Since the final total kinetic energy is less than the initial kinetic energy, the collision is not elastic. M1 M1 A1 Marker’s comments: To justify the coll ision is inelastic, many students used the relative speed of approach is not equal to relative speed of separation. This is incorrect, because the two relative speeds are equal only if it is a one-dimensional elastic collision. Recall that the relation was derived from the equation of conservation of KE: In 2-D collisions, the velocities of the particles in the x-direction do not fully account for the KE of the particles. In fact, for elastic collisions: The presence of the four terms that arise due to velocit ies in the y- direction means that the derivation in the lecture notes is no longer valid. Hence, the only way to answer this question is to compare the kinetic energy before and after collision. pA pB pT 30.0° 23.8°

5 © Raffles Institution [Turn over 5 (a) (i) A closed polygon (triangle) (with arrows clockwise) shows that resultant force is zero. M1 A1 Marker’s comments: This question requires students to explain with reference to the diagram sketched. Hence, students who mentioned that the forces pass through a common point (indicating no net torque) are not given credit. (ii)