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