TJC 2022 Prelim P1 Guide

2022 H2 Prelim Paper 1 Solutions: 1 2 3 4 5 6 7 8 9 10 B D A C D B C C B A 11 12 13 14 15 16 17 18 19 20 D C A C D A C C A C 21 22 23 24 25 26 27 28 29 30 B D B A D A D B C B 1 B 2 d n FC vA=  2n dC v A F = Unit of v = m s-1 Unit of  = kg m-3 Unit of A = m2 Unit of F = kg m s-2  1  kg mn-1s-n = kg m s-2  n = 2 2 D Radius of the object = 6 6 0.12 1.7 106.4 10  = 0.032 m = 3.2 cm 3 A Velocity of motorcyclist relative to car, VR = VM – VC 4 C Separation between P and Q is maximum when P is just hitting the ground. Time for P to reach the ground is given by: s = 2 2 1 gtut+ = 2 2 1 gt 80 = 5t2 t = 4 s Position of Q when P is hitting the ground: s = 2)1(2 1)1( −+− tgtu = 2)1(2 1 −tg = 5(4-1)2 = 45 m Hence, max. separation between P and Q = 80 – 45 = 35 m

5 D As the object is in static equilibrium, the resultant force of the three external forces acting on the object is equal to zero. i.e. WFT  ++ = 0 and WF  + = T  − vertical: F cos  = W = mg horizontal: F sin  = T  T = mg tan  At the instant the thread is suddenly cut, the force T  is removed. Resultant force acting on the object = WF  + = T  − Hence, acceleration of the object is direction II and magnitude of acceleration = g tan  6 B Total momentum of wagon remain unchanged because the vertical force by rain does not affect the horizontal momentum of the wagon. The horizontal speed of wagon decreases because the amount of water in the wagon increases which increases the total mass of the system and so horizontal speed will decrease. The kinetic energy of the system decreases using the equation ke = 2 2 p m since m increases while p remain constant. 7 C The 3 forces, the 2 tensions in the string and the weight of the rod need to intersect at a point for it to be in equilibrium. 8 C Driving power = driving force x velocity = 500 x 40 = 20 MW Power lost due to air resistance = air resistance x velocity = 200 x 40 = 8 MW Rate at which kinetic energy is increasing = 20 – 8 = 12 MW 9 B As the bob undergoes circular motion, there must be a resultant force acting on the bob in radial direction = centripetal force = T – W cos  In the tangential direction, resultant force acting on the bob = W sin  Hence, the resultant force should be in direction Q. 10 A At the bottom of the hill, the net force = N - mg = centripetal force = ma Hence, N = mg + ma > 0. Therefore, the car can never lose contact at the bottom of the hill. At the top of the hill, the net force = mg – N = ma Hence, N = mg – ma Therefore, the car will lose contact (feel weightless) when N < 0