TJC 2023 H2 Prelim Paper 2 Solutions

2023 H2 Prelim P2 Solutions 1. (a) ( ) ( ) ( ) ( ) 22 22 3 -2 -2 -2 11Using 22 2 0.6002 354 10 9.5758 m s 2 210.001 9.5758 0.600 354 0.07 m s 9.58 0.07 m s − = + → = ==  =   =+  =+ = = s ut gt h gt hg t g g h t g h t g g g C1 C1 A1 (b) (i) Steel is a hard magnetic material. The delay is due to the time taken for the electromagnet and steel ball to lose their magnetism. A1 (ii) Systematic error. The constant delay causes the timing measured to be consistently too long. A1 C1 (iii) Place light gate just below steel ball. Recorded time is time delay. OR Graphical method: Measure the timings for different heights of fall and plot a graph of h against t2. Details: The constant delay is the square root of the x-intercept. OR Experimental method: Use of light gates / high speed camera / etc. Details: Positioning of light gates / view video in slow motion / etc. M1 C1 2 (a) head-on : collision takes place along the line joining the centres of the colliding bodies. elastic collision: both the momentum and the kinetic energy are conserved. C1 C1 (b) speed of separation = speed of approach (1)--- 0 211 112 vvuor uvv =+ −=− A1 (c) By Conservation of momentum, 1 1 2 1 1 2 1 2 2 1 1 2 1 1 1 1 1 1 12 12 22 13 13 2 11 13 13 11 13 ---(2) (1) (2) from (1) - Thus ratio = 0.846 mu mv mv u v v u v v u v v u u u u v u =+  − = + =  = = = − =− = C1 C1 A1

(d) ( ) 22 2 11 1 2 1 1 2 1 111 0.2813 1 2required fraction 1 2 m u v v umu − = = −   = − = C1 A1 (e) This is a head-on elastic collision between two equal masses. After collision, the incident neutron will stop while the target neutron moves off with the velocity of the incident neutron. (The ratio of the final to the initial speed of the incident neutron would be zero.) The fraction of the kinetic energy of the incident neutron that is transferred to the target neutron would be 1. C1 A1 3(a) The work done by a force on an object parti cle is defined as the product of the magnitude of the force and the component of the displacement in the direction of the force. B1 3(b)(i) Change in KE = Net work done ( )( ) ( ) 2 max 2 110 22 0.400 2.5 14 0.18 m mu F x x x − =− = = M1 A1 (ii) B1 – shape B1 – x and work done values B1 B1 x / m work done by F / J 0 1.25 0.18

(iii) Since F proportional to x, and v is constant, power also proportional to x B1 4 (a) The direction of the object changes continuously and hence there is a change in velocity of the object. By Newton’s second law, there must be a force acting on the object. Since the speed remains unchanged, this force cannot have a component tangential to the circle / does no work . This implies that the force is perpendicular to the velocity/displacement of object acting towards centre of circle. M1 C1 (b) i For the block to just make it through the top of the loop, the weight is just providing for the centripetal force. mg = mvtop2 / R (0.50)(9.81) = (0.50)(vtop2) / 1.5 vtop = 3.84 m s-2 [1, also award if student leave as KE = 3.68 J] C1 A1 ii By Conservation of energy, KEi + EPEi + GPEi – work done by friction = KEf + EPEf + GPEf 0 + ½ ke2 + 0 – Fd = ½ mvtop2 + 0 + mg(2R) ½ (78.4)(e2) – (1.47)(2.5) = ½ (0.50)(3.8362) + 0.50(9.81)(2 x 1.5) Solving, e = 0.750 m C1 A1 (c) vmin remains unchanged. At C, to stay in contact, centripetal acceleration of the block = acceleration of free fall, 2 min min v g Hence v rgr == , independent of the mass of the block. A1 C1 power / W x / m 0

5 a free: oscillation without any loss of energy / no external forces e.g. resistive forces forced: body is made to vibrate by an (external) periodic driving force such that continuous energy is input B1 B1 b From the graph, maximum displacement = 23.00 cm minimum displacement = 11.00 cm amplitude = 0.5 (23.00 – 11.00) = 6.00 cm = 0.0600 m A1 c From graph, T = 2.000 s [M1] 122 3.142.000 radsT  −= = = A1 d Energy of Oscillation = Maximum KE = ( )( ) ( ) 2222 0 11 10 3.14 0.06 0.17722m x J == M1 A1 e For loss of contact, a > g At this setting, amin = 9.81 = 3.142 (xmin) xmin = 0.995m M1 A1

6 a (ii)1 . 5 4 2.4 10 112.2 10 WE Q = = =  V M1 2 EQ It t R== 42.2 10 3800 11  = =QRt E 67.6 10 s= M1 A1 3 fraction = 2 22 1800 1800 2000 TT TT I R R I R I R R R==+ + + = 0.474 Note: fraction must be written in decimal. -1 for answer not written in decimal M1 A1 b (i) 8.0 2.0 1.4558.0 0.50 2.5 XYV =  =++ V At null deflection, VXJ = e.m.f. of cell B. 0.90 1.455 1.311.0E =  = V M1 M1 (ii)1 At null deflection, XJ RSVV = 0.75 1.455 1.0911.0 XJV =  = V 1.091 0.16796.5= = =RS VI R A 1.31 1.091 1.300.1679 −−= +  = = =EVE V Ir r I Ω allowed range of E: 1.20 to 1.40 depending on substitution M1 M1 A1 With resistor connected in parallel, load resistance of cell B would decrease. This would cause terminal p.d. of cell B to decrease. Since no change in potential difference per unit length along wire XY, Balance length XJ would decrease. B1 A1

7 (a) (i) force = qE in the direction of the field M1 A1 (ii) no force (since current/velocity is parallel to B-field) B1 (b) (i) B field pointing into the page/plane of the paper A1 (ii) resultant force is zero/the 2 forces cancel out (good to show FE acts downward and FB acts upwards) Eq = Bqv  v = E/B B1 B1 (c) (i) - Horizontal component of velocity experience zero force, so constant horizontal motion to the right. -Vertical component of velocity results in a magnetic force into page which provides centripetal force for circular motion into the page -Since radius r = mv/Bq, as B decreases, radius r increases B1 B1 B1 (ii) Helical path drawn with increasing radius and longer pitch (Note: The period of each circle T = 2m/Bq, as B decreases T increases, so horizontal pitch sx = vxT increases) B1 pitch

8 (a)(i) Activity A of a radioactive substance is defined as the number of disintegrations per unit time. B1 (a)(ii) use of λ= (ln2)/t½ C1 shows that years have been converted to seconds λ = 2.5x10-10 s-1 A1 (a)(iii) N = A/ = 0.63x1012/2.5x10-10 per g (must use value given in the table) = 0.63x1012/2.5x10-10 x 0.16 = 4.0 × 1020 1 mark awarded if use N = (0.16/238) x 6.02 x 1023 = 4.0 x 1020 C1 C1 A0 (b) (i) Energy per unit time = activity x energy per decay C1 Energy per sec = (0.63x1012 x 0.16) x 5600 × 103 x 1.60x10-19 C1 = 0.090 J A1 (b) (ii) e = 7.5 × 10-4/0.090 x 100% C1 = 0.83% A1 (c) Since Power output is proportional to Activity, P = Po e-t 0.60/0.75 = e-t t = 8.9 × 108 s C1 A1 (d) use E= mcΔθ 0.68x10-3x12x3600 = 6.7x10-3x410xΔθ Δθ = 11 ºC C1 A1 (e) -A slow neutron may cause nuclear fission of Pu-238 and trigger a chain reaction, which will cause a high amount of energy to be released in a short time. This is undesirable for th