EJC Physics 2023 J2 H1 MYE PAPER 1 MS

©EJC 2021 9749/J1H2MYE/2021 EUNOIA JUNIOR COLLEGE JC2 Mid-Year Examination 2023 8867 PHYSICS MARK SCHEME Paper 1 – Multiple Choice Questions 1 Answer: D Volume of car tyre ≈ (0.3)3 = 0.027 ≈ 0.03 m3 2 Answer: B N = kg m s−2 V = J C−1 = (kg m2 s−2) (A s)−1 N m V−1 = A s Option A: m s−2 Option B: A s Option C: A Option D: Unit of V I = kg m2 s−3 A−1 3 Answer: D Precise: The values are close to one another. Not accurate: The mean is not close to the true value. 4 Answer: D In between bounces, the ball experiences constant free-falling acceleration. The time of contact is taken to be similarly negligible. The duration of free-fall in between bounces decreases as the ball loses energy. The acceleration while in contact with ground decreases due to decreased normal contact force with the ground.

2 ©EJC 2023 8867/J2H1MYE/2023 5 Answer: C 2 1 2 2 20 = 10002 0.20 20 5002 0.40 3000 1000 500 1500 1500 7520 100 75 50 225 s const const total s s s t t = × = = × = −−= = = = ++= 6 Answer: D v = u + at Time to reach highest point, 0 = 5.2 + (–1.62)t t = 3.2 s Time to return to starting point = 2 x 3.2 = 6.4 s v/ m s−1 t/ s 0 20 0.20 1 0.40 1 1 20 0.2 100 s t∆= = 2 20 0.4 50 s t∆= =

3 ©EJC 2023 8867/J2H1MYE/2023 7 Answer: A For constant force hence constant acceleration, v= u + at p = mv = mat (for u = 0) ∴ p is proportional to t OR F = dp/dt = gradient of graph Gradient has to be constant 8 Answer: D 12 – 4 cos 37o – 3 cos 53o = (12 / 9.81) a a = 5.7 m s −2 9 Answer: C By Covservation of Linear Momentum, (5000)(2.00) + (5000)(–1.00) = (5000 + 5000) v v = 0.50 m s−1 KE lost = Total KE initial – Total KE final = ½ (5000) (2.00)2 + ½ (5000)(–1.00)2 – ½ (5000 + 5000) (0.50)2 = 11 250 J 10 Answer: D 3 W kx= ⇒ 3 Wk x= 'W kx= ⇒ '3 WWx x= x’ = 3x 11 Answer: C Step 1: Draw vector triangle Step 2: Apply sine rule 12 Answer: C Apply Principle of Moments including moment caused by weight of ruler.

4 ©EJC 2023 8867/J2H1MYE/2023 13 Answer: A By Principle of Conservation of Energy, Loss in GPE = Work done against resistive forces 600 9.81 300 1500 76.4 m 80 76.4 3.5 m mg h F s h h h ∆=∆ × ×∆ = × ∆= = −= 14 Answer: C Since car is travelling at constant speed, net force and acceleration is zero. FDr = FR ∝ v2 Since P = FDr v P ∝ v3 3 22 11 3 2 40 2320 184 Pv Pv P =  = × = 15 Answer: A Elastic Potential Energy is the area bound by the graph and the extension axis. Be careful of the axis !

5 ©EJC 2023 8867/J2H1MYE/2023 16 Answer: B Kinetic energy is scalar but linear momentum is vector. 17 Answer: D 2 GMmF r= Be careful of the words “above the surface….” 18 Answer: C ( ) ( ) To find total charge transferred dur