NYJC EJC 2024 Prelim (Answers)

EUNOIA JUNIOR COLLEGE, NAN YANG JUNIOR COLLEGE JC2 Preliminary Examination 2024 General Certificate of Education Advanced Level Higher 3 H3 PHYSICS (9814) SUGGESTED SOLUTIONS Paper 1 Question Suggested Solution Marks 1(a) Loss in GPE = Gain in KE B1 1(b) Using expressions in (a) , In order for the normal force on the beads to be radially inward, the centripetal force required (shown above) must exceed the radial component of weight. Thus, B1 C1 M1 A0 9814/2024NYEJPrelims/Suggested Solution

Question Suggested Solution Marks 1(c) Let the normal force be N . By Newton’s 2 nd Law, This normal force (hoop on beads) is radially inwards (when the beads are moving fast). By Newton’s 3 rd Law, the beads exert on the hoop a normal force radially outwards of the same magnitude . The sum of the upward components of N from the two beads is . If this vertical force on the hoop is greater than the weight of the hoop, the hoop will be lifted off the ground. Thus, we require To find the maximum value of the denominator, Hence, at its maximum, , which implies . The maximum value of the ratio is then . Note: Initially, the beads move slowly, the required centripetal force ( ∝ v 2 ) is small. The radial component of weight ( = mg cos θ ) is more than enough to supply the centripetal force. Hence, the normal force is radially outwards, and mg cos θ − N is the centripetal force. As the beads speed up, the centripetal force required exceeds mg cos θ (which decreases with θ ). The normal force will then be radially inwards, such that mg cos θ + N is the centripetal force. Since the hoop exerts on the beads an inward normal force, by Newton’s 3 rd Law, the beads exert on the hoop and outward normal force, which has an upward component. If this upward force component is greater than the weight of the hoop, the hoop will be lifted off the ground. B1 C1 B1 B1 B1 B1 9814/2024NYEJPrelims/Suggested Solution

Question Suggested Solution Marks 2(a) 9814/2024NYEJPrelims/Suggested Solution

Question Suggested Solution Marks 2(b) 3(a)(i) So that there is a net pressure and thus a net force pointing to the right to provide the centripetal force . 3(a)(ii) 9814/2024NYEJPrelims/Suggested Solution

Question Suggested Solution Marks 3(b)(i) 3(b)(ii) 3(b)(iii) 4(a) The first law of thermodynamics states that the increase in internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system. 4(b)(i) From the First Law of Thermodynamics, For the isothermal process A to B, 9814/2024NYEJPrelims/Suggested Solution

Question Suggested Solution Marks 4(b)(ii) From the First Law of Thermodynamics, For the adiabatic process B to C, , similarly, 4(c)(i) Work done on the steam from A to B is = Similarly, the work done on the steam from C to D is = 9814/2024NYEJPrelims/Suggested Solution

4(c)(ii) For one complete Carnot cycle, the net work done on the steam is where ∴ the net work done by the steam is = For the adiabatic processes B to C and D to A, But, , and , 9814/2024NYEJPrelims/Suggested Solution

Question Suggested Solution Marks Hence, 5(a) Conservation of momentum: Conservation of energy: B1 B1 5(b)(i) Rearranging the above equations: (1) + (2) to eliminate λ ′, and rearranging: Solving the quadratic equation for v : M1 M1 A0 5(b)(ii) Since M1 A1 9814/2024NYEJPrelims/Suggested Solution

Question Suggested Solution Marks 5(b)(iii) Conservation of energy: Rearranging, M1 M1 A0 6(a)(i) weight must be larger than tension 9814/2024NYEJPrelims/Suggested Solution

Question Suggested Solution Marks 6(a)(ii) 6(b)(i) 6(b)(ii) 6(c) 9814/2024NYEJPrelims/Suggested Solution