EJC Physics 2021 J2 H1 MYE P2 MS

EUNOIA JUNIOR COLLEGE JC2 MID YEAR EXAMINATION 2021 General Certificate of Education Advanced Level Higher 1 PHYSICS MARK SCHEME Section A 8867/02 July 2021 Question Solution Marks 1 (a) Rate of change of velocity (with time) B1 (b) Constant acceleration OR Motion in a straight line B1 (c) (i) =xxs ut ( )=25 50cos20 o t = 0.53ts C1 A1 (ii) = +y yyv u at = + = -1 50sin20 (9.81)(0.53) 22.3 m s o y y v v 22.3tan 50cos20 y o x v vθ = = 25 oθ = C1 C1 A1 (iii) = + 212yy ys ut at = + 21(50sin20 )(0.53) (9.81)(0.53)2 o ys =ys 10.4 m = + 22 xysss = + 2225 10.4s = 27 ms C1 C1 A1

(iv) Accept positive or negative velocities. B1 B1 velocity time 0 A B

Question Solution Marks 2 (a) Total momentum of a system remains constant provided no external resultant force acts on the system. B1 (b) (i) ∆𝑃𝑃 = 𝑃𝑃𝑓𝑓 − 𝑃𝑃𝑖𝑖 = 25 × 103 − 50 × 103 = −25 × 103 kg m s−1 A1 (ii) Initial momentum of Y is 30 × 103 kg m s-1 Momentum of Y increases by 25 × 103 kg m s-1 B1 B1 (iii) 𝑢𝑢𝑋𝑋 = 25 m s−1, 𝑢𝑢𝑌𝑌 = 10 m s−1, 𝑣𝑣𝑋𝑋 = 12.5 m s−1 , 𝑣𝑣𝑌𝑌 = 18.3 m s−1 total kinetic energy before collision = 775 kJ total kinetic energy after collision = 660 kJ Since the total kinetic energy before collision (775 kJ) is not equal to the total kinetic energy after collision (660 kJ), the collision in not elastic. OR relative speed of approach = 15 m s−1 relative speed of separation = 5.83 m s−1 Since relative speed of approach (15 m s−1) is not equal to relative speed of separation (5.83 m s−1), the collision is not elastic. M1 A1 (M1) (A1) (iv) ( ) 22511 (1240) 22 2 3055608 1022 KE mv . .= = = × J C1 momentum / 103 kg m s−1 time / s 0 0.4 20 40 60 0.0 0.8 1.2 1.6 X X

2K K E mv Emv ∆ ∆∆= + 5 20 0823055608 10 1240 22 2 KE . .. ∆ = +× ∆EK = 0.3 × 105 J EK = (310 ± 30) kJ C1 A1 Examiner’s Comments

Question Solution Marks 3 (a) F = ma Driving force – mgsin9.0o = ma Driving force – (2500)(9.81)sin9.0o = (2500)(2.0) Driving force = 8837 N Power = Fv Power = (8837)(8.5) Power = 7.5 × 104 W C1 C1 A1 (b) loss in KE = gain in GPE + work done against friction 2211 ( )( )22mu mv mgh friction s −= + ( ) 2211(2500)(12) (2500)(8.0) (2500)(9.81) sin9.0 (500)( )22 ss −= + s = 23 m C1 A1 Examiner’s Comments

Question Solution Marks 4 (a) gravitational force provides centripetal force GMm / r2 = mrω2 ω = 2π/T Showing algebra clearly to reach answer B1 B1 (b) Any two of the followings: appears to remain above the same point on the Earth has a period of 24 hours. equatorial orbit/orbits (directly) above the equator from west to east B1 B1 (c) (24 × 3600)2 = 4 π r3 ÷ (6.67 × 10-11 × 6.0 × 1024) r = 4.2 × 107 m ω = 2π ÷ (24 × 60 × 60) = 7.27 × 10−5 rad s−1 v = r ω = (4.2 × 107)(7.27 × 10−5) = 3.1 × 103 m s−1 M1 M1 A1 (d) gravitational force is just sufficient to provide the centripetal force 'weight'/sensation of weight/contact force/reaction force is difference between FG and FC which is zero B1 B1 Examiner’s Comments

Question Solution Marks 5 (a) (i) t = sx/v = (40 × 10-3) ÷ (1.5 × 107) = 2.7 × 10-9 s A1 (ii) 1. change in velocity, ∆v = change in y-component = vfsin20o – 0 = 1.6 × 107 (sin 20°) = 5.5 × 106 ms-1 C1 A1 2. By Newton’s 2nd Law, the rate of change of momentum of the body must be in the same direction as the resultant force acting on the body. Hence the direction of the change in velocity must be in the same direction as the electric force which is vertically upwards. B1 B1 (iii) F = qE = ma where a= (Vy-u)/t and E = V/d (1.6 × 10-19)V/(20 × 10-3) = 9.11 × 10-31 × (5.5 × 106)/(2.7 × 10-9) V = 230 V M1 A1 (iv) For no deflection: Net force on the electron is zero Electric force = magnetic force qE = Bqv 230 / (20 × 10-3) = B (15.0 × 106) B = 7.7 × 10-4 T M1 A1 (b) To find the radius of the semi-circle: Magnetic force provides the centripetal force for the electron to complete the circular path. Bqv = mv2/r r = mv/Bq = 9.11 × 10-31 × 1.5 × 107 / [(7.7 ×10-4 )(1.6 × 10-19)] = 0.11 m Circumference of circle = 2π r = 2 × π × 0.11 = 0.691 m Total distance travelled in 1 cycle = 2 × (40 × 10-3)+(0.691) = 0.771 m Time = 0.771 / (1.5 × 107) = 5.14 × 10-8 s M1 C1 A1 Examiner’s Comments

Question Solution Marks 6 (a) assuming uniform deceleration, ( ) 22 23 222 2 185 1 0 2 0 2 2 80 0 6 17 m s v as v u a u s − =  × − = = − = − + deceleration = 17 m s-1 C1 A1 (b) (i) ( )( ) 2 max 1 4 4 9 81 30 34 m s a v r v ar gr . − = = = = = M1 A1 (ii) From Fig. 6.1 to increase the value of contact force by increasing the value of μ at high temperature. From Fig. 6.2, when tyres are heated, it becomes softer and can increase the contact area with the surface by covering up the air gap B1 B1 (c) the wing is shaped to deflect air upwards. By N2L, air experiences a rate of change of momentum upwards and hence experiences an upward force By N3L, the wing experiences a force equal in magnitude and opposite in direction of that experienced by air. the force presses down on the car and increases the apparent weight (no marks awarded if only mention about force acting down on the car) M1 M1 A0 (d) (i) car is in rotational equilibirum. by principle of moments, about centre of gravity, sum of clockwise moments = sum of anti-clockwise moments RR FFxxNN Dh= + car is in vertical translational equilbirum, vector sum of forces along vertical is 0 RFN NW+= Sub NF : ( ) ( ) RR FF RR R F RR F F F R RF x x Dh x W N x Dh x x Wx Dh Wx D N N x N x N N h = + − + = = = + + + + B1 B1 A0

(ii) F R RF F F RF F F RF RFF RF R RF Wx Dh xx Wx Dh xx Wx Dh xx x x Wx Dh x N x xD WN NW W xx W h W = + − + += += −= − − − = + + + + + A1 (iii) as magnitude of D increases as the car accelerates magnitude of NR increases and magnitude of NF decreases since by vertical translational equilibirum RFN NW+= a greater share of the weight is supported by the rear wheel as driving force increases B1 B1 Examiner’s Comments

Section B Question Solution Marks 7 (a) (i) Magnetic flux density is defined as the force per unit length per unit current acting on a conductor carrying a current, placed perpendicularly to the magnetic field. B1 B1 (ii) kg s-2 A-1 A1 (iii) Clearly labelled diagram: Diagram – Presence of Pivots, Coil and Current Magnetic Force must be acting downwards Rider/Mass acting downwards Application of principle of moment: Showing the full equation. By measuring the current I, the mass m of the rider, the distances L, x and y, the magnetic flux density B can be determined. B1 B1 B1 B1 B1 (b) When switch is opened: PD across the 4.0 Ω resistor = 12 / (0.5 + 4) x 4 = 10.667 V Power across the 4.0 Ω resistor = V2/R = 28.45 W When switch is closed: 4//3 = 1.714 Ω PD across the 4.0 Ω resistor = 12/ (0.5 + 1.714) x 1.714 = 9.2899 V (Since 4 Ω and 3 Ω resistors are parallel, they have the same PD) Power across the 4.0 Ω resistor = V2/R = 21.58 W Decrease in power = 28.45 – 21.58 = 6.87 W = 6.9