JPJC 2026 Energy and Fields Assignment Solutions

2026/JPJC/PHYSICS/9478 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS ENERGY AND FIELDS ASSIGNMENT SOLUTIONS Qn. Solution Mark Allocation 1(a) The tension in the supporting cord does no work. The direction of motion of the pendulum is always perpendicular to the cord and therefore perpendicular to the tension. [1] tension [1] reason 1(b) The air resistance does negative work on the pendulum at all times. The air resistance is always acting in a direction opposite to that of the motion, with the exception of the highest points of the swing where the work done by air resistance is zero. [1] air resistance [1] reason + no WD at highest point 1(c) The gravitational force always acts downward. Thus, the work done by gravitational force is positive on the downward swing and negative on the upward swing. However the work done by gravitational force is zero at the lowest point of the swing. [1] +ve WD during downward swing & -ve WD during upwards swing [1] no WD at lowest point of swing 2(a) Let x be the compression of the spring. 2 2 (6.0 2.0) 10 4.0 10 m x − − = −  = Energy stored in the spring 2 2 2 2 11 (25)(4.0 10 )22 2.0 10 J kx − − = =  = Assume that all the elastic potential energy stored in the spring is transferred to the toy car as kinetic energy and no energy is lost due to friction and air resistance. By conservation of energy, loss in KE = gain in GPE 22.0 10 mgh−= (m = mass of car, h = the height attained by car) 2 2 2.0 10 (0.200)(9.81) 0.010194 1.0194 10 m h − − = =  Distance travelled along the slope [1] correct answer of k [1] assumption [1] correct answer of h

2026/JPJC/PHYSICS/9478 2 Qn. Solution Mark Allocation 2 2 sin30 1.0194 10 sin30 2.04 10 m h − − = = =   [1] correct final answer 2(b) In order for the toy car to move at constant speed up the slope, the net force on the toy car must be zero. A force F, parallel to the downward slope, equal to magnitude sin30 0.200(9.81)(0.5) 0.981 Nmg ==  must be acting on the toy car parallel to the slope, to keep it moving at constant speed. Power required to keep the toy car moving at constant speed, 0.981 4.0 3.9 W P Fv= = = [1] for explanation [1] for correct expression and substitution [1] for correct answer 3(a) There is a point between the Earth and the Moon where the gravitational force from both bodies acting on an object are equal in magnitude but opposite in direction. At this point, the gravitational forces cancel out , so the gravitational field strength is zero. [1] [1] 3(b) When the gravitational field strength is zero, let r is the distance from the centre of the Earth. EMgg= ( ) EM 22 E60 GM GM r Rr = − EM E 22 24 60 7.4 10 6.0 10 0.111 R r M rM − = =  = [1] for correct expression and substitution weight, mg 30° direction of motion mg sin(30°)

2026/JPJC/PHYSICS/9478 3 Qn. Solution Mark Allocation E E 60 0.111 54 R r r rR −= = [1] for correct answer 4 Resultant horizontal electric field strength at the centre of the circle is zero. The horizontal components of the electric field strengths E1 and E2 cancel out. 12 12 22 00 2 1 cos65 cos15 cos65 cos15 44 cos65 cos15 0.438 0.44 EE qq rr q q   = = = =      [1] for correct explanation [1] for correct expression and substitution [1] for correct answer