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 o