DHS 2023 JC1 Physics Revision (Set B Solutions)

1 @ DHS 2023 Dunman High School Year 5 Holiday Revision Set B Answers 1 (a) Use v2 = u2 + 2gs At highest point, v = 0 0 = (800 sin )2 − 2(9.81)s or gs 2 sin640000 2= [A1] Minimum height = s + 300 g2 sin640000 2= + 300 [A1] (b) Time of flight t is found using v = u + gt − 800 sin  = 800 sin  − gt gt sin1600= [A1] Use x = ut = (800 cos )       g sin1600 [M1] g ))(cos)(sin2()800( 2 = g )2(sin640000 = [A0] (c)(i) units of R = units of E 0.2  −0.2 t z m = (kg m2 s−2) 0.2 (kg m−3) −0.2 s z = m sz−0.4 [M1] z = 0.4 [A1] (c)(ii) R = E 0.2  −0.2 t 0.4 E = R5  t − 2     ++    = t t R R E E 25   [M1]     ++    = 006.0 001.022.1 1.0 80 25 E = 0.6 × 1014 J [A1] E = (1.1 ± 0.6) × 1014 J [A1] 2 (a) Velocity is defined as the rate of change of displacement with respect to time. (b) (i) 𝑠 = (12 × 2) + (1 2 × 12 × 2) − (1 2 × 18 × 3) − (18 × 4) = −63 m (ii) Curve from 2 to 7 s, highest point at 4 s, smooth line at 2 and 7 s.

2 @ DHS 2023 Straight line with positive gradient from 0 to 2 s, straight line with negative gradient from 7 to 11 s . Correct displacements at 2, 4, 7, 11 s. (iii) Horizontal lines for 0 to 2 s, 7 to 11 s, a = 0 m s−2. Horizontal line from 2 to 7 s, a = −6 m s−2. 3 (a) If body A exerts a force on body B, then body B exerts a force of the same type that is equal in magnitude and opposite in direction on body A. [1] (b) Force by rocket on water = vt m t p waterwater      =  = mass of water ejected per unit time  v [1] By Newton’s 3rd Law, the thrust (force) T, of rocket due to ejected water is equal to the force by rocket on ejected water. [1] Taking upwards as positive, net force on the rocket F is F = T – W where W is the weight of rocket [1] = r2v  v – mg Hence F = r2v2 – mg −63 s / m t / s 7 11 4 24 9 36 2 2 7 11 a / m s−2 t / s 4 −6

3 @ DHS 2023 (c) (i) Initial thrust will be the same as the gas pressure remains the same. [1] Alternative Solution Initial thrust will be greater as the total pressure at the nozzle is greater due to a larger volume (height) of water above it. Further elaboration Pressure at nozzle = Initial air pressure + water pressure = Initial air pressure + g(V/A) Where g is gravitational field strength, V is volume of water and A is cross sectional area of bottle, assuming cylindrical bottle. BUT the additional water pressure is about 1000  10  0.1 = 1000 Pa (not very significant compared to 160000 Pa) assuming A of about 20 cm3 (ii) Magnitude of initial acceleration will be smaller as mass of rocket and its contents is greater. [1] Acceleration = gm vr − 22  = gV vr −  22 = gV vr − 22  Since velocity of ejected water v remains the same (increase is insignificant), the initial acceleration experienced is smaller. (iii) Maximum height reached is smaller because the acceleration experienced by the toy is smaller. [1] 4 (a) (i) Consider if the block (with penny) start to slide first, 𝐹1 = 0.40 (𝑚1 + 𝑚2)𝑔 = (𝑚1 + 𝑚2)𝑟𝜔2 (𝑚1 & 𝑚2 are mass of block and penny respectively) 𝜔 = (0.40 × 9.81 0.12) 0.5 = 5.72 rad s−1 M1 A1 (ii) Consider if the penny starts to slide first, 𝐹2 = 0.52 (𝑚2)𝑔 = (𝑚2)𝑟𝜔2 𝜔 = (0.52 × 9.81 0.12) 0.5 = 6.52 rad s−1 M1 A1 (iii) Max allowable 𝜔 = 5.72 rad s−1 M1 Number of revolution per minute = 𝜔 2𝜋 × 60 = 54.6 min-1 A1 (b) Friction provides the centripetal force for mud to stay in circular motion. 𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝑚𝑟𝑤2 = 𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒. The required centripetal force increases as r increases M1

4 @ DHS 2023 so flies off at edge first if the required centripetal force is greater than frictional force. A1 5 (a) The gravitational potential at a point is defined as the work done per unit mass by an external agent in bringing a small test mass from infinity to that point, without producing any acceleration of the test mass. B1 (b) (i) ∅𝑝 = 4∅𝑑𝑢𝑒 𝑡𝑜 1 𝑚𝑎𝑠𝑠 𝑎𝑡 𝑃 M1 = −4 𝐺𝑚 √𝑅2 + ℎ2 A1 (ii) Loss in Ek (from P to infinity) = Gain in EP (from P to infinity) 1 2 𝑀𝑣2 − 0 = 𝑀 [0 − (−4 𝐺𝑚 √𝑅2 + ℎ2)] M1 𝑣 = √8𝐺𝑚 √𝑅2 + ℎ24 A1 6 (a) A straight line passing through the origin ⇒ acceleration is directly proportional to displacement. B1 Negative gradient ⇒ direction of acceleration is opposite to the direction of displacement B1 The features of the graph show that 𝑎 ∝ −𝑥 (defining equation for SHM) B1 (b) (i) Energy stored in spring = ½ 𝑘𝐴2 = ½ (100)(0.200)2 = 2.00 J A1 (ii) Loss in EPE = Gain in KE 2.00 = ½ (𝑚1 + 𝑚2)𝑣2 2.00 = ½ (9.00 + 7.00)𝑣2 M1 v = 0.500 m s-1 A1 OR 𝜔 = (k/m)1/2 = (100/(9.00+7.00))1/2 = 2.50 rad s-1 M1 𝑣 = 𝑟0𝜔 = (0.200)(2.50) = 0.500 m s-1 A1 (iii) Loss in KE = Gain in EPE 1 2 𝑚1𝑣2 = 1 2 𝑘𝐴′2 1 2 (9.00)(0.500)2 = 1 2 (100)(𝐴′)2 M1 A’ = 0.150 m A1 OR 𝜔 = √ 𝑘 𝑚 = √100 9.00 = 3.33 rads-1 M1 𝑣 = 𝐴′𝜔 0.500 = A’(3.33) A’ = 0.150 m s-1 A1

5 @ DHS 2023 7 (a) 𝑣0 = 𝑥0 = ( 2𝜋 𝑇 ) 𝑥0 [M1] = ( 2𝜋 𝜋/5) (0.080) = 0.80 m s−1 [A1] (b) sine graph [B1] y-axis labeled correctly [B1] (c) 𝑣 = √𝑥0 2 − 𝑥2 𝑥 = √𝑥0 2 − 𝑣2 2 [M1] = √(0.080)2 − (0.42)2 102 = 0.068 𝑚 = 6.8 cm Displacement = −6.8 cm [A1] (d) At equilibrium point, 𝑚𝑔 = 𝑘𝑒 𝑒 = 𝑚𝑔 𝑘 = 0.200×10 20 = 0.10 = 10 cm [M1] At highest point, EPE = 1 2𝑘𝑒2 = 1 2(20)(0.10−0.080)2 = 0.0040 = 4.0 mJ [A1] (e) Lowest Point Equilibrium Point Highest Point Gravitational Potential Energy / mJ 0 160 320 Elastic Potential Energy / mJ 324 100 4 Kinetic Energy / mJ 0 64 0 v / m s−1 t / s 0.80 −0.80 /5 2/5

6 @ DHS 2023 8 (a) (i) Total charge of particles = (e)(n)(A)(L) B1 (ii) Time taken for all particles to pass through shaded area = L / v B1 (iii) Current = charge flow per unit time = [(e)(n)(A)(L)] / (L / v) = n A v e M1 B1 A0 (b) Faulty: C B1 Nature: lamp is shorted OR Lamp has no resistance. B1 (c) R = 15 Ω A1 (d) V = IR 𝑅 = 6.0 0.20 = 30 Ω A1 (e) Filament is cold when measuring with ohm-meter B1 and the resistance of the filament lamp rises as temperature increases B1 OR The increased temperature led to increased resistance B1 Temperature of 2 cases highlighted to have increased OR Lattice vibrations increased with temperature increase. B1 (f) (i) B1 1. S is open to measure E; 2. S is closed to measure V. B1 B1 (ii) 1. 22 11.6 20.0 6.73 VR P== = 2. Rearranging the equation V E r=− I , we obtain 1 15.01 6.73 1 1.97 11.6 Er VR ErR V =−    = − = − =        M1 A1 M1 C1 A1 End S