2024 DHS Prelim H1 Phy P2 Ans

2024 DHS H1 Physics Prelim Paper 2 Suggested Solutions 1 (a) Rate of change of velocity B1 (b) (i) Using s = ut + ଵ ଶ gt2 2 2 2 2 2(3.59) 0.860 9.70795 9.70 m s sg t      C1 A1 (ii) 2 2 1 2 109.70795 359 860 0.25 0.3 m s (to 1sf) g s t g s t g               2(9.7 0.3) m sg    C1 M1 A1 (iii) In the presence of air resistance, the time taken to travel the same vertical distance is longer. Hence, the value of g calculated by the student Y will be lower than student X. B1 B1 (c) (i) The line of best fit minimises the distance between the line and the data points and hence, reduces the uncertainty due to random error B1 (ii) Systematic errors are suggested by the best -fit line not passing through the origin. B1 2 (a) Resultant force in any direction is zero. Resultant torque about any axis is zero. B1 B1 (b) (i) The single point at which the entire weight of the body can be considered to act. B1

(ii) Take moments about the hinge, by principle of moments, Clockwise moments = Anticlockwise moments cos40 5.0 120cos40 5.0 5(9.81) 1.8741 1.9 m T W d d d         C1 A1 (iii) By resolving the forces, the force by hinge on the beam R is 120cos 40 5(9.81) 42.875 N yR    120sin 40 77.135 N xR   2 2 2 242.875 77.135 88.250 88 N x yR R R      Let ߠ be the angle measured clockwise from the horizontal beam to R 1 42.875tan 77.135 29          C1 C1 A1 A1 3 (a) The velocity of the object in uniform circular motion has constant magnitude, but it is continuously changing direction , so its velocity is not constant. Since acceleration is the rate of change of velocity, and not speed, the acceleration of the object is non-zero. However, as the centripetal force is always perpendicular to its velocity, the work done by the centripetal force is zero , hence there is no change in kinetic energy of the object and therefore it travels at constant speed. OR The acceleration of the object has constant magnitude, but its direction is always perpendicular to the velocity and towards the centre of its circular path, so the speed will not change. B1 B1 B1 B1 (b) (i) The gravitational force exerted by the sun on the planet provides the centripetal force needed for the planets to undergo uniform circular motion. M1

2 2 2 mv GMm r r GMv r GMv r    M1 A0 (ii) From a v against r graph, the gradient of the graph is 4 7 10 (4.70 1.15) 10 (41.0 10.0) 10 1.14516 10 GM gradient         2 10 2 11 30 30 (1.14516 10 ) 6.67 10 1.966 10 kg 1.97 10 kg gradientM G         M1 C1 A1 (iii) 2 11 30 24 8 3 2 21 (6.67 10 )(1.966 10 )(0.642 10 ) (2.279 10 10 ) 1.62 10 N G GMmF r          C1 A1 (iv) Any one of the following reasons:  Accuracy of graph due to limitations in scale, hence the gradient is an estimate  Gravitational forces due to other planets were not taken into consideration  The planets do not orbit the sun in a perfect circle but rather in an elliptical orbit, hence the orbital radius is an estimate. B1 (c) (i) Planet m / 1024 kg r / 108 km v / 104 m s-1 a / 10-4 m s-2 Mercury 0.330 0.579 4.74 388 Venus 4.87 1.082 3.50 113 Earth 5.97 1.496 2.98 58.0 B1 B1

Mars 0.642 2.279 2.41 25.4 Note: a can be calculated using 2v r (ii) 4 8 3 7 1 3.50 10 1.082 10 10 3.23 10 rad s v r v r            C1 A1 (d) For a satellite to be geostationary, the orbital period must be the same as the time taken for the Earth to rotate 1 round about its own axis (T = 24 hours) and orbits the Earth at a fixed distance from the centre of the Earth. Since the gravitational acceleration provides the centripetal acceleration, 2 2 3 2 11 24 2 3 6.67 10 (5.97 10 ) 2 24 60 60 42230 km E E GMr r GMr                 Hence, satellite C is a geostationary satellite. B1 C1 M1 A1 4 (a) (i) A region of space where a (non-contact) force is felt. B1 (ii) B1: Correct direction of arrows B1: Correct field pattern

(b) (i) Out of the page B1 (ii) The magnetic force provides the centripetal force, 2 27 6 19 1.67 10 4.5 10 0.12 1.6 10 0.39141 0.39 m mvBqv r mvr Bq            C1 A1 (c) (i)  Field lines must be drawn with a ruler  Draw at least 5 field lines directed upwards with equal spacing between adjacent field lines. B1 (ii) For the proton to pass through undeflected, the electric force must be equal in magnitude but opposite in direction to the magnetic force on the proton such that the net force on the proton is zero. By Newton’s first law, when the net force is zero, the proton will continue its state of motion and travel straight through undeflected. Since the direction of electric field denotes the direction of force exerted on a positive charge (such as protons), in order to exert an upward electric force on the proton, the electric field lines are directed upwards. B1 B1 B1

(iii) 6 5 1 0.12(4.5 10 ) 5.4 10 N C Bqv qE E Bv        C1 A1 5 (a) (i) 214 4 210 83 2 81Bi He Tl  214 0 214 83 1 84Bi e Po  B1 B1 (ii) reactants 213.9987 209.9901 4.0015 0.0071 productsm m m u u u u        2 27 8 2 12 12 (0.0071)(1.66 10 )(3.0 10 ) 1.06074 10 1.06 10 J releasedE mc             C1 C1 A1 (iii) The gamma rays are produced from the decay of different daughter nuclei, hence the energy of the gamma ray for each decay is different B1 (b) (i) 6 23 15 15 Number of nuclei 2.0 10 6.03 10214 5.626 10 5.6 10 An N           OR 6 3 27 15 15 Number of nuclei 2.0 10 10 213.9987 1.66 10 5.6300 10 5.6 10 total Bi m m             M1 A0 M1 A0

(ii) 2.7 0 2.7 15 14 1 2 15.6 10 2 8.6 10 N N              C1 A1 (iii) When a cell is exposed to ionising radiation, it creates ions which breaks the bonds within DNA and cause mutations. This leads to creation of tumour cells or cell death. OR Ionising radiation can cause the formation of free radicals which are highly reactive and can form many harmful compounds such as hydrogen peroxide. This initiates harmful chemical reactions within the cells and as a result, the cells undergo a variety of structural changes which lead to altered functions of the cells. B1 B1 B1 B1 6 (a) (i) As the hammer is released and is in freefall, the gravitational potential energy of the hammer is being converted into kinetic energy of the hammer. Upon striking the pile, the hammer does work on the pile. Since the collision is perfectly inelastic, some of the initial kinetic energy possessed by the hammer is lost while the remaining is converted into kinetic energy of the pile-hammer system. As the pile-hammer system is driven into the riverbed, kinetic energy of the pile-hammer system is dissipated by the force exerted by the riverbed as heat and sound energy until it comes to a stop. B1 B1 B1 (ii) 31.0 10 9.81 5.0 49050 49000 J GPE mgh       C1 A1

(iii) 2 3 2 3 1 loss in GPE gain in KE 149050 2 149050 (1.0 10 )2 2(49050) 1.0 10 9.9045 9.9 m s mv v v          C1 A1 (b) (i) The Principle of Conservation of linear momentum states that the total momentum of a system of interacting bodies is constant pro