2021 H2 Phy Prelim Paper 3 Solution (forJCs)

2021 JC2 H2 Preliminary Examination Paper 3 Suggested Solutions 1 Thus, mass of the astronaut = (65  2) kg 1 1 1 2 (a) (i) 1 1 (ii) 1 1 1

2 (b) 3 (a) The spring does not obey Hooke’s Law. The straight line in the graph does not pass through the origin OR The tension force in the spring is not proportional to the extension. * Answer must derive evidence from the graph. 1 1 (b) Remove the masses and the spring should return to its original length if the spring is not permanently deformed. 1 (c) Work done is obtained from the area under the graph multiplied by g . * Student must attempt to explain the approach taken. Work done = ½ [(70 + 145)  103]  120  103  9.81 = 0.127 J (0.13 J) 1 1 1 4 (a) By the principle of conservation of energy, KEXi + KE Yi + GPE Xi + GPE Yi = KE Xf + KE Yf + GPE Xf + GPE Yf + WD against friction 0 + 0 + 0 + 0 = KEXf + KEYf + (4.0)(9.81)(3.0 sin300) + (5.0)(9.81)(- 3.0) + (10.0) (3.0) KEXf + KEYf = 58 J (or 58.3 J) * Correct calculation of change in GPE of X and change in GPE of Y [M1] * Correct substitution in COE equation [M1] 2 1 (b) Since X and Y are connected by an inextensible cord, both bodies attain the same final speed. vX = vY = 3.6 ms1 1 Maximum height is lower and range is smaller [1] (highest point should therefore be to the left of the original path) Path is asymmetrical [1]

3 5 (a) (i) The centripetal force is provided by the gravitational force acting on stars A and B. By Newton’s Third Law, the gravitational force acting on stars A and B are of equal magnitude and opposite direction. Hence, the centripetal force acting on both stars has the same magnitude. 1 1 (ii) 1 1 (b) (i) Since the centripetal force acting on star A and B are of equal magnitude. MA 2 d = MB 2 (2.8  108  d) 4d = 2.8  108 d = 7.0  107 km 1 1 1 (ii) Gravitational force provides for the centripetal force acting on star A. Hence, MB = 2.0  1029 kg 1 1 1 6 (a) (i) Upthrust and Weight 1 (ii) At equilibrium, the magnitude of upthrust and weight are equal. When the tube is pushed downwards, the magnitude of upthrust increases. Since the weight of the tube remains constant, the resultant force is upwards 1 1 (b) Since Aρg M is a constant, acceleration of the tube is proportional to its displacement The negative sign indicates that acceleration and displacement are in the opposite direction 1 1 (c) (i) Period of the oscillation = 2.2 s Angular frequency = 2π/T = 2π/2.2 1

4 = 2.9 rad s–1 (ii) ω2 =Aρ g M 2.92 =(4.5 × 10−4 ) ρ (9.81) (0.17) ρ = 323.86 = 320 kg m–3 1 1 (iii) v0 = ωx0 v0 = (2.9)(0.03) = 0.087 m s-1 1 1 7 (a) (i) From definition,  1 (ii) From definition of work, 1 (b) From VQ = Fd , we have Electric field strength (force per unit charge) = (potential gradient since the potential changes linearly with dist