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 top, centripetal force provided by weight only. 𝑚𝑔 = 𝑚𝑣2 𝑟 ⇒ 𝑣 = √𝑔𝑟 = √9.81 × 0.1 = 0.99 m s−1 force on system by wall force on shelf by cable W W/5 driving force F

19 A Object in non–uniform circular motion. conservation of energy gives 1 2 𝑚𝑣2 = 𝑚𝑔 × 0.2 ⇒ 𝑣2 = 0.4𝑔 Resultant of N and W provides centripetal force at the bottom of the bowl. 𝑁 − 𝑚𝑔 = 𝑚𝑣2 𝑟 ⇒ 𝑁 = 0.10 × 0.4𝑔 0.2 + 0.10 × 𝑔 = 3.0 N 20 D 𝑎 = −𝜔2𝑥 Gradient = −𝜔2 = [16−(−16)]×10−3 (−4−4)×10−3 = −4 s−2 ⇒ 𝜔 = 2 s−1 Since 𝜔 = 2𝜋 𝑇 , period 𝑇 = 2𝜋 2 = 𝜋 21 B Displacement of simple harmonic motion, 𝑥 = 𝑎 sin(𝜔𝑡) When 𝑡 = 𝑇 8, 𝑥 = 𝑎 sin ( 2𝜋 𝑇 × 𝑇 8) = 𝑎 sin 𝜋 4 = 𝑎 √2 v W normal reaction force N 20 cm a / mm s–2 x / mm 16 8 0 –8 –16 –4 –2 2 4

22 C Obtain the velocity and acceleration against time graphs When velocity or acceleration is zero, no direction → A and B are wrong Velocity and acceleration of D are both negative, so in the same direction 23 A B → Wrong. Amplitude decreases. C and D → Wrong. No oscillation when damping is critical. 24 D Particle P will move to a larger vertical displacement at the next instant. 25 D Amplitude of original wave = 4 units intensity ∝ amplitude2 ⇒ 0.5𝐼 𝐼 = ( 𝐴 4) 2 ⇒ new amplitude 𝐴 = 2.8 units Frequency remains constant 26 D intensity I of sound at distance x is given by 𝐼 = 𝑃 4𝜋𝑥2, where P is power of source for the sound to be as loud, intensity should be equal 𝐼 = 𝑃 4𝜋(8)2 = 𝑃/2 4𝜋𝑥2 ⇒ 𝑥 = 5.7 m

27 C 𝐼 = 𝐼0 cos2 45° = 𝐼0/2 28 B Resultant amplitude is 2Y. Phase difference = 𝜋 (see pic below) 29 B Figure shows 5 node-to-node segments, so 5 × 𝜆 2 = 1.2 ⇒ 𝜆 = 0.48 m 𝑣 = 𝑓𝜆 = 75 × 0.48 = 36 m s–1 Stationary wave with 2 loops, 2× 𝜆′ 2 = 1.2 ⇒ 𝜆′ = 1.2 m 𝑓′ = 𝑣 𝜆 = 36 1.2 = 30 Hz 30 D P Resultant wave