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