2023 VJC Prelim P1 AS

1 VICTORIA JUNIOR COLLEGE SUGGESTED SOLUTIONS TO 2023 PHYSICS H1 PRELIM EXAM PAPER 1 Q1 C Q7 A Q13 C Q19 D Q25 B Q2 C Q8 D Q14 D Q20 B Q26 A Q3 B Q9 C Q15 B Q21 B Q27 A Q4 A Q10 C Q16 A Q22 C Q28 B Q5 D Q11 D Q17 C Q23 D Q29 C Q6 D Q12 B Q18 B Q24 C Q30 D Q1 Check for homogeneity of equations. An inhomogeneous equation is an incorrect equation. Let base units be represented by BU. A: LHS: BU = kg m2 s-2 RHS: BU = (kg m s-1)(m s-1) = kg m2 s-2 LHS = RHS  homogeneous equation (wrong answer) B: LHS : BU = (kg m2 s-2)2 = kg2 m4 s-4 RHS: BU of p2c2 = (kg m s-1)2 (m s-1)2 = kg2 m4 s-4 BU of m2c4 = kg2 (m s-1)4 = kg2 m4 s-4 All terms on both sides of the equal sign have the same units  equation is homogeneous (wrong answer) C: LHS: BU = kg m2 s-2 RHS: BU =(kg m s-1) 2(m s-1) = kg m3 s-3  inhomogeneous equation (correct answer) D: LHS: BU = kg m2 s-2 RHS: BU =   2  -1 2 -2kg m s kg m skg homogeneous equation (wrong answer) Ans: C Q2 2 26 AD B A D CB C B A D C         20.1 0.01 0.052 7.7 10 7.7%2.0 3.45 4.10 B xB       Ans: C Q3 The second student’s measurements are closer to one another and are therefore more precise compared to the measurements of the first student. Ans: B

2 Q4 Area under v-t graph is the displacement. For graph P, the area is always positive, meaning that there is positive displacement. For graph Q, the area during the negative velocities and the area during positive velocities are equal. Hence the total displacement is zero. Displacements for graphs P and Q are thus different. The magnitudes of the total area under the graphs P and Q are different. Thus the total distances travelled for P and Q is different. Ans: A Q5 2 2 2 2 2 for vertical motion 3 2( 9.81)( 2.4) taking upward as posi tive direction 56.09 y y y y yv u a s v        2 -2 m s Horizontal velocity, 5.0x xv u  -1 ms Speed of toy pilot,   2 2 2 1 56.09 5.0 9.0msy xv v v      Ans: D Q6 Here the vertical forces are gravitational and magnetic forces. The horizontal forces are the electric force and air resistance. The electrically charged oil drop is moving horizontally at constant velocity. The oil drop is in dynamic equilibrium and the resultant force is zero in all directions. Vertically, the gravitational force is acting downwards. Therefore, the magnetic force must be acting upwards to ensure equilibrium. The electric force cannot act vertically in this case because the electric field is horizontal, and air resistance should be acting in the direction opposite to the motion, which is horizontal in this case. Ans: D Q7 By N2L and taking the loads-pulley system as a whole, 8.0g – 2.0g – f = (8.0 + 2.0)a 6.0 x 9.81 – 3.0 = 10.0a  a = 5.6 m s-2 Ans: A

3 Q8 From Newton’s 3rd Law, force of P on Q (action force) is equal and opposi