2021 SH1 Promo P1 (sol)

Solutions to NJC 2021 Promo Paper 1 Qn Ans Solution 1 D tera is 1012 2 C high precision → readings are close to each other → low random error low accuracy → readings are far from actual value → high systematic error 3 D Consider z as resultant. Vector x and y must be arranged as follows to obtain z. 4 D Ball at rest with respect to the elevator initially → ball and elevator at same velocity u displacement of ball is 2 + 𝑠 = 𝑢𝑡 + 1 2 (9.81)𝑡2 …… (1) displacement of elevator 𝑠 = 𝑢𝑡 + 1 2 (5.8)𝑡2 …… (2) (1) – (2): 2 = 1 2 (9.81 − 5.8)𝑡2 ⇒ 𝑡 = 0.999 s 5 B “X advances by 100 m relative to Y during overtaking” means that X is 100 m behind Y before the overtaking. So, the shaded area = 100 m 100 = 1 2 (𝑇 + 𝑇 + 2 × 5)(35 − 30) ⇒ 𝑇 = 15 s z y –x 2 m when ball is released when ball reach floor displacement s of elevator acceleration of elevator ae = 5.8 m s–2 acceleration of ball = 9.81 m s–2

6 B Initial vertical velocity = 0, so vertical displacement = 2 = 1 2 𝑔𝑡2 ⇒ 𝑡 = √4/𝑔 Horizontal displacement = 12 = 𝑣𝑡 ⇒ 𝑣 = 12 √4/𝑔 = 18.8 m s–1 7 A Area under F– t graph gives impulse (= change in momentum) on the body area = 𝑚𝑣𝑓 − 𝑚𝑣𝑖 1 2 × 80 × (100 × 10−3) = 2𝑣𝑓 − 2 × 15 𝑣𝑓 = 17 m s−1 8 C k.e. of elastic is equal before and after collision k.e. before collision = 2 × 1 2 𝑚𝑣2 = 𝑚𝑣2 9 B k.e. of car 𝐾 = 1 2 𝑚𝑣1 2 Conservation of momentum gives 0 = 4𝑚𝑣2 − 𝑚𝑣1 ⇒ 𝑣1 = 4𝑣2 k.e. of launcher = 1 2 × 4𝑚 × 𝑣2 2 = 1 2 × 4𝑚 × ( 1 4 𝑣1) 2 = 4 16 ( 1 2 𝑚𝑣1 2) = 1 4 𝐾 10 B force on gun = force on bullet = Δ𝑚 Δ𝑡 (𝑣𝑓 − 𝑣𝑖) = 750×0.0056 60 (840 − 0) = 58.8 N 11 C The forces form a right–angled triangle. So, 𝐹2 + 3002 = 6002 ⇒ 𝐹 = 520 N velocity of car v1 velocity of launcher v2 600 N 300 N F 600 N 300 N F 30o 60o

12 A Consider the forces acting on the system (hinge + shelf). At equilibrium, • the lines of action of the forces must pass through the same point. • forces form a closed triangle → force on system by wall as shown Use Newton’s 3rd law to deduce direction of force on wall exerted by system. 13 D Origin of upthrust due to pressure difference To float, upthrust = weight 14 D Shape of k.e. same for 4 options Velocity of ball increases as it falls, so distance travelled for equal interval of time increases. Therefore, p.e. decreases with time at increasing rate. There is drag force, so p.e. → k.e. + work done against drag. Final k.e. < initial p.e. 15 B Mass of water passing through turbine per unit time = 1000 × 6 = 6000 kg s−1 g.p.e loss per unit time = 6000 × 9.81 × 80 J s−1 energy converted to electrical energy per unit time = 0.6 × 6000 × 9.81 × 80 = 2.8 MJ s−1 = 2.8 MW 16 B Component of weight down the slope = 𝑊 sin 10° Driving force 𝐹 = 𝑊 sin 10° + 𝑊/5 Output power of energy = 𝐹𝑣 = (1100𝑔 sin 10° + 1100𝑔/5) × 15 = 60500 W 17 D Centripetal acceleration = 𝑟𝜔2 = 𝑟 ( 2𝜋 𝑇 ) 2 = 1 × ( 2𝜋 1 ) 2 = 4𝜋2 18 C At minimum speed at the to