NJC 2022 Prelim P2 Guide

[Turn over National Junior College 2022 Preliminary Examination 9749 H2 Physics Paper 2 1 (a) (i) –8.75 = (15.0 sin 60o) t + ½ (–9.81) t2 M1 8.75 = (–15.0 sin 60o) t + ½ (9.81) t2 (M1) solving t = 3.21 s A0 (ii) range of projectile = (15.0 cos 60o) (3.21) = 24.075 m M1 distance travelled by ship = 0.450 × 3.21 = 1.4445 m M1 D = 24.075 + 1.4445 = 25.5 m A1 (b) horizontal component of air resistance causes horizontal deceleration M1 so range of projectile shorter and D will be shorter A1 (c) (i) straight line with negative gradient with final speed > original speed M1 label initial speed on y-axis and time of flight on x-axis A1 (ii) correct trend for graph (ii) M1 velocity = 0 at time earlier than (i) A1 Vy / m s–1 t / s0 13.0 –18.5 3.21 (i) (ii)

2 2 (a) (i) force proportional / equal to rate of change of momentum (Newton II) M1 force on (block) A equal and opposite to force on (block) B (Newton III) M1 (Do not award this mark if the general definition of third law is given.) equal magnitude of forces and collision time for A and B hence change in momentum same magnitude A1 opposite forces hence change in momentum opposite A1 (This mark is awarded only if “opposite forces” is explicitly linked to direction of change in momentum.) (ii) if both blocks at rest, magnitude of change in momentum of A = ( –)1.2M AND change in momentum of B = 0.25M M1 not equal hence not possible A1 (b) (i) either 3M × 0.40 + M × (–0.25) = 3M × 0.20 + Mv (conservation of momentum) or 3M × (0.20 – 0.40) = – M × (v + 0.25) M1 v = 0.35 m s–1 A1 (ii) either show total kinetic energy after < total kinetic energy before collision or show relative speed of approach ≠ relative speed of separation M1 inelastic A1

[Turn over 3 3 (a) (i) point which weight of ship M1 appears / seems to act A1 (ii) water pressure on lower surface greater than air pressure on upper surface of ship B1 (b) upthrust = weight 1030 V × g = 2.20 × 108 × g C1 volume = 2.14 × 105 m3 A1 (c) (i) vertically downward arrow for W at G, vertically upward arrow for U at B and length of W and U approximately equal B1 (ii) weight (or W) and upthrust (or U) forms a couple M1 which gives anticlockwise torque (or moment) A1 to right / correct the roll / tilt of the ship (d) labelled weight through higher c.g. and labelled upthrust that produce overturning torque B1 weight and upthrust forms a couple that give clockwise torque (or moment) to overturn ship B1 G B W U B G W U

4 4 (a) (i) effective resistance across BN = ( 1 2𝑅 + 1 𝑅+𝑅) −1 = 𝑅 effective resistance across AM = ( 1 2𝑅 + 1 𝑅+𝑅𝐵𝑁 ) −1 = ( 1 2𝑅 + 1 𝑅+𝑅) −1 M1 = 𝑅 A0 or effective resistance across AM = ( 1 2𝑅 + 1 𝑅+( 1 2𝑅+ 1 𝑅+𝑅) −1) −1 (M1) (ii) use potential divider principle, 𝑉𝐴𝑀 = 𝑅 𝑅+𝑅 × 𝑉 = 1 2 𝑉 M1 𝑉𝐴 − 𝑉𝑀 = 𝑉𝐴 − 0 = 𝑉𝐴𝑀, so 𝑉𝐴 = 1 2 𝑉 A1 (iii) 𝑉𝐷 = 1 2 𝑉𝐶 = 1 2 × 1 2 𝑉𝐵 = 1 4 × 1 2 𝑉𝐴 = 1 8 × 1 2 𝑉 C1 𝑉𝐷 = 1 16 𝑉 A1 (b) either 𝑅𝑃 = 6 6×10−3 (= 1000 Ω) C1 𝑅𝑄 = 6 3×10−3 (= 2000 Ω) C1 𝑅𝑀𝑁 = ( 1 1000 + 1 2000) −1 = 667 Ω A1 or p.d. across MN = 6 V (C1) total current = 6 + 3 = 9 mA (C1) 𝑅 = 6 9×10−3 = 667 Ω (A1)

[Turn over 5 5 (a) (i) region of space which a force is felt/experienced B1 (ii) explanation that relates field strength to the closeness of lines B1 (b) force on the magnet is downward B1 force on current upwards by Newton’s third law M1 magnetic field directed towards P, P is South A1 (c) (i) arrow from below foil to above the foil B1 (ii) centripetal force of charged particle in circular motion provided by magnetic force 𝑚𝑣2 𝑟 = 𝐵𝑞𝑣 ⇒ 𝑚𝑣 = 𝐵𝑞𝑟 ⇒ momentum ∝ radius C1 ratio = 5.7 / 7.4 = 0.77 A1

6 6 (a) (i) ++→ HePoRn 4 2 216 84 220 86 total proton and nucleon numbers equal on both sides B1 equation include photon B1 (ii) 1.0 MeV = 1 × 106 × 1.6 × 10–19 = 1.6 × 10–13 J A1 (iii) 1. energy released = 6.29 MeV + 0.55 MeV (= 6.84 MeV = 1.094 × 10-12 J) C1 E = mc2 m = 1.094 × 10-12 / (3 × 108)2 = 1.2 × 10-29 kg A1 2. 𝐸 = ℎ𝑓 = ℎ𝑐 𝜆 ⇒ 𝜆 = ℎ𝑐 𝐸 wavelength = (6.63 × 10-34 × 3 × 108) / (0.55 × 1.6 × 10–13) M1 = 2.3 × 10-12 m A1 (b) (i) N0 is initial number of C-14 atoms, Nt is number of C-14 atoms now and N12 be number of C-12 atoms. So 𝑁𝑡 = 𝑁0 exp (− ln 2 𝑡1/2 𝑡) C1 𝑁𝑡 𝑁12 = 𝑁0 𝑁12 exp (− ln 2 𝑡1/2 𝑡) 1 8.6×1010 = 1 3.3×1010 exp (− ln 2 5700 𝑡) C1 t = 7900 years A1 (ii) 1. mass of a Carbon-14 ion different from Carbon-12 ion B1 same magnetic force on each atom because they have same speed and charge B1 so, radius of curvature of path of each type of ion in magnetic field is different 2. method enables determination of amount of Carbon-14 exactly B1 instead of measuring the radioactivity which is probabilistic in nature B1

[Turn over 7 7 (a) A photon is a quantum (or packet/ discrete bundle) of electromagnetic radiation (energy) B1 Each quantum has energy equal to hf where h is Planck constant and f is the frequency of the electromagnetic radiation B1 (b) (i) 1. 𝐸𝑖 = 𝐸𝑠 + 1 2 𝑚𝑣2 2. 𝑝𝑖 = 𝑝𝑠 cos 𝜃 + 𝑚𝑣 cos 𝜙 (ii) electron gain kinetic energy after collision, so scattered photon has lower energy than incident photon (refer to (b)(i)1.) B1 energy of photon inversely proportional to wavelength M1 so, wavelength of scattered photon larger than incident photon A1 (c) calculation for Δλ M1 i / 10-12 m s /10-12 m  /  Δλ / 10–12 m 191.92 193.27 57 1.35 153.30 154.65 57 1.35 965.04 966.84 75 1.8 valid conclusion comparing 𝜆𝑖, 𝜃 and Δλ for the three sets of data e.g. A1 – compare first two sets of data, Δλ independent of incident photon wavelength for same scattering angle – compare third set of data with the first two, Δλ dependent on incident photon wavelength for different scattering angles (d) cos 70° = 0.342, cos 75° = 0.259, cos 80° = 0.174 C1 either Δ(cos 𝜃) = 0.342 − 0.259 = 0.08 or Δ(cos 𝜃) = 0.259 − 0.174 = 0.09 or Δ(cos 𝜃) = 1 2 (0.342 − 0.174) = 0.08 C1 cos 𝜃 = 0.26 ± 0.08 (or ±0.09) A1 (e) (i) line of best fit assessed by even distribution of points either side of line along full length B1 (ii) [Method 1] Δ𝜆 = k(1 − cos 𝜃) = −k cos 𝜃 + k, so best fit straight line of Δλ against cos 𝜃 gives M1 gradient equal –k and y-intercept equal k A1 [Method 2] obtain values of Δ𝜆 and cos 𝜃 from a point on the best fit straight line and substitute into Δ𝜆 = k(1 − cos 𝜃) to calculate for k. B1

8 (iii) read-offs accurate to half a small square in both the x and y directions, gradient calculated correctly using Δ𝑦 Δ𝑥 and sign of gradient matches graph C1 𝑘 = −gradient and with unit m A1 OR 𝑘 = y-intercept and with unit m (M1) precision of y-intercept to half the smallest square (A1) OR Substitute Δ𝜆 and cos 𝜃 from a point on the line into Δ𝜆 = k(1 − cos 𝜃) (C1) k calculated correctly and with unit m (A1) (𝑘 ≈ 2.5 × 10−12 m) (f) binding energy of electron less than / about ( 10 30000 × 100% =) 0.03% of energy of photon M1 photon energy >> binding energy so assumption of free electrons is valid / justified A1 (0.625, 1.0 × 10-12) (–0.75, 4.4 × 10-12)