2024 JPJC Prelim H2 Physics P3 Solutions_Printed

2024/JPJC/Prelim/9749/03 Answers to 2024 JC2 H2 Preliminary Examinations Paper 3 Suggested Solutions: No. Solution Remarks 1(a) v = u + at = −2.5 + (9.81)(2.0) = 17.1 Speed is 17 m s−1 [1] substitution [1] answer 1(b) v / m s−1 0 1.0 2.0 t / s [1] helicopter: horizontal straight line (-ve value) [1] parcel: diagonal straight line (+ve gradient) (reverse sign accepted) [-1] if not labelled 1(c) Distance between helicopter and parcel = area of right-angled triangle = 1 2 (2.0)(17.1 2.5) 19.6 m+= OR From point of release, For helicopter, upwards s = ut = (2.5)(2.0) = 5.0 m For parcel, downward s = ut + 21 2 at = (−2.5)(2.0) + 21 2 (9.81)(2.0) = 14.6 m Hence their separation = 5.0 + 14.6 = 19.6 m [1] working [1] answer 2(a) The change in height is negligible compared with radius of the planet. Thus, the field strength 2 GMg R= remains relatively constant for any small change in R. The gravitational field lines are effectively parallel. [1] 2(b)(i) 2 GMY R = where G is the gravitational constant [1] P H –2.5 17

2 2024/JPJC/Prelim/9749/03 2(b)(ii) [1] for correct shape of curve from 3R− to R− ; ending at ( , 1.0 )RY− [1] for correct shape of curve from R to 3R ; starting at ( , 1.0 )RY− [1] for curves passing through ( 3 , 0.11 )RY− ; ( 2 , 0.25 )RY− ; (2 , 0.25 )RY− ; (3 , 0.11 )RY− (don’t need to mark for ±3R) 2(b)(iii) Initial total energy = Final total energy 221 1 1 2 2 2 GMm GMmmv mv Rx − = −  ( ) ( )( ) 2 23 11 23 6 6 1 1 1 4 1 1 14.7 10 6.67 10 6.4 104 3.4 10 6.070 10 m v GM Rx x x − =−   =   −   = Therefore distance travelled 66 6 6 6.070 10 3.4 10 2.670 10 m 2.7 10 m xR=− =  −  = = [1] for correct equation for conservation of energy [1] for correct substitution [1] for correct answer 3(a) The electric field strength in both sphere A (between x = 0 and x = 1.4 cm) and sphere B is zero (between x = 11.4 and x = 12.0 cm). [1] 3(b)(i) The resultant field strength is zero at a point between the spheres and this shows that electric fields are in opposite directions in the region between the two spheres. This shows that the polarity of the two charges are the same. Hence, since sphere A is positively charged, sphere B must also be positively charged. [1] expl [1] state

3 2024/JPJC/Prelim/9749/03 [Turn over 3(b)(ii) At x = 0.08 m, the electric field strength due to sphere A cancels out the electric field strength due to sphere B. EA = EB ( ) ( ) 2 o B 2 o A 04.0 408.0 4  QQ = 404.0 08.0 2 B A =    =Q Q [1] sub [1] ans 3(c)(i) The electric field strength is negative of the electric potential gradient, i.e. dx dVE −= [1] don’t accept dx dVE −= 3