DHS 2023 JC1 Physics Revision (Set A Solutions)

1 1 DHS @2023 Dunman High School Year 5 Holiday Home Revision Set A Answers 1 (a) m Tv l= 0095.0 )50.1(20= [M1] = 56 m s-1 [A1] (b)      ++= m m T T v v l l 2 1 = ( )02.001.005.02 1 ++ = 0.04 [C1] Δv = 0.04 x 56 = 2.2 [M1] v = (56 ± 2) m s-1 [A1] 2 (a) s = ut 150 = (45 cos 50)t [M1] t = 5.19 s [A0] (b) s = 2 2 1 atut + = (45 sin 50 × 5.19)+ 2)19.5)(81.9(2 1 − [M1] = 47 m [A1] (c) v2 = u2 + 2as 0 = u2 + 2 (−9.81)(47) [M1] u = 30 m s-1 [A1] (d) v = u + gt 0 = 30 + (−9.81)t [M1] t = 3.1 s [A1] Time to throw the apple after launch of arrow is 5.19 – 3.1 = 2.1 s. [A1] 3 (a) The principle of conservation of momentum states that the total momentum of a system of bodies is constant [B1], provided no external resultant force acts on it [B1]. (b) Elastic potential energy stored by spring = 21 2 kx = 21(1330)(0.15)2 = 14.96 J [M1] Elastic potential energy stored by spring = Change in kinetic energy of A 2114.96 (1.5) 02 v=− v = 4.47 m s−1 [A1] (c) Let the common final velocity be v2. By conservation of momentum,

2 2 DHS @2023 m(4.47) = (2.5m) v2 (m = 1.5 kg) [M1] v2 = 1.79 m s−1 [A1] (d) By conservation of energy, ki pi f kf pfE E W E E+ + = + 2o1(2.5 )(1.79) 0 ( 8.5)( ) 0 (2.5 )(9.81)( )(sin3 0)2 m s m s + + − = + [M1] s = 0.223 m [A1] 4 (a) Resultant force on it must be zero in any direction. [B1] Resultant torque on it must be zero about any axis of rotation. [B1] (b) (i) (ii) Let x be the distance of c.g of beam from the hinge. By principle of moments, Taking moments about hinge, Sum of clockwise moments = sum of anticlockwise moments 0 0 0 0 0 0120sin40 (5sin70 ) 120cos40 (5cos70 ) (5 9.81 )( 5sin70 ) (20 9.81 )( sin70 )x+ =  +  [M1] x = 1.57 m [A1] (iii) Vertical summation of forces: 0(120)sin40 5 20YF g g+ = + FY = 168.1 N Horizontal summation of forces: [C1] 0(120)cos40XF = FX = 91.93 N 2222 )1.168()93.91( +=+= Yx FFF [M1] weight of beam force by wall on beam 40o weight of load tension, T β All forces correctly labelled. [B1] Correct directions of forces. [B1]

3 3 DHS @2023 = 192 N 168.1 91.93tan = β = 610 [A1] 4 (c) 5 (a) R cos θ = mg [C1] R cos 20.00 = 1450 (9.81) [M1] R = 1.51 × 104 N [A0] (b) 90 km h-1 = 25 m s−1 [C1] r mvR 2 sin = [C1] 2 4 1450(25)(1.51 10 )sin(20 )o r= [M1] r = 175 m [A1] 6 2r Gmg = 28 )200.0( )00.8(Gg = =1.33 × 10-8 m s−2 pointing to the left 215 )300.0( )0.15(Gg = = 1.11 × 10-8 m s−2 pointing to the right [C1] (i) (ii) (iii) 1. (iii) 2.

4 4 DHS @2023 gnet = (1.33 −1.11) × 10-8 m s−2 = 2.22 × 10−9 m s−2 [A1] to the left [A1] 7 (a) (i) For P: ‘U shaped’ graph: potential energy + kinetic energy = 15 mJ. [A1] (ii) For T: straight line, parallel to x-axis, at 15 mJ. [A1] (b) (i) Maximum kinetic energy = 2 1 m 2A2 15 × 10−3 = 2 1 (0.150) (2f)2 (5 × 10−2)2 [M1] f = 1.4 Hz [A1] (ii) Using a = (−) 2x = (21.4)2(0.02) [M1] = 1.55 or 1.6 m s−2 [A1] (c) (i) It is the (continuous/ gradual) loss of energy (from the system) or decrease in amplitude [B1] due to (additional) force (acting on the mass) always opposing motion. [B1] (ii) New kinetic energy = 0.8 × 15 = 12 mJ From (b) (i) kinetic energy ∝ A2 2 58.015 12     === A KE KE i f A = 4.5 cm [A1] OR Loss in kinetic energy = 0.2 × 15 = 3 mJ So draw a horizontal line with kinetic energy = 3 mJ [C1]. The point of intersection with existing kinetic energy graph gives the amplitude on the x- axis = 4.5 cm. [A1] 8 (a) (i) ( )0.03 60 60 108 CQt= =  =I B1 (ii) ( )2.5 108 270 JW VQ= = = OR ( )PV 0.030 2.5 0.075 W= = =I 270 0.075 W60 60P ==  B1 (iii) 2.5 830.03 VR = = = I B1 [M1]

5 5 DHS @2023 (b) (i) VR max max 5.0 0.2025= = = I 22 4 2 R A dd     = = =   l l l ( )d R 73 min max 4 4 15 4.8 10 6.77 10 m0.20 −− = =  =    l B1 C1 A1 (ii) 1. R A= l ( ) ( ) 77 22 3 1 4 4.8 104.8 10 0.35 10 2 R esista nce 5. per me 0 t re m R A d   −− − − = = =   = =  l 2. Resistance of filament, R = V2 / P = 122 / 30 = 4.8 Ω Length of wire, l = 4.8 / 5.0 = 0.96 m B1 A1 B1 B1