NJC 2024 H1 Physics Prelim P1 Ans

[Turn over 1 A pico nano micro 10–12 10–9 10–6 2 D The product PS has the same units as Q and R. 3 A 4 B Accurate when the average is close to the true value. Precise when the spread of readings is small. 5 A Constant acceleration: ݒଶ =ݑଶ − 2ݏܽ  ଵ ଶݒ݉ଶ = ଵ ଶ݉ݑଶ − (ܽ݉)ݏ Since (ܽ݉) is constant, graph is a straight line with a negative gradient. 6 B Constant acceleration = ݃Initial speed ݑ= 0 Let the time interval between successive drops be ݐ . 9.0 = ଵ ଶ݃(3ݐ)ଶ  ݐ= ට ଶ ௚ Second drop from the roof: ݏଶ = ଵ ଶ݃ݐଶ = 1.0 m Third drop from the roof: ݏଷ = ଵ ଶ݃(2ݐ)ଶ = 4.0 m Distance =ݏଷ −ݏଶ = 3.0 m 7 D ݔ= 500 = (ݑݏ݋ܿ ݏ݋ܿ ߠ )× 2ݐ  ݑݏ݋ܿ ݏ݋ܿ ߠ = ଶହ଴ ௧ ----------------- (1) ݕ= 210 = ଵ ଶ (ݑ݊݅ݏ ݊݅ݏ ߠ +0)ݐ  ݑ݊݅ݏ ݊݅ݏ ߠ = ସଶ଴ ௧ ----------------- (2) (2)  (1) gives: ݊ܽݐ݊ܽݐ ߠ = 420 250  ߠ= 59° 8 A Consider both masses and the rod as one system: (݉ଵ +݉ଶ)݃݊݅ݏ ݊݅ݏ ߠ =( ݉ଵ +݉ଶ)ܽ  ܽ= ݃݊݅ݏ ݊݅ݏ ߠ Consider forces acting on m1 (or m2) ݉ଵ݃݊݅ݏ ݊݅ݏ ߠ −ܶ= ݉ଵܽ= ݉ଵ݃݊݅ݏ ݊݅ݏ ߠ ܶ= 0 m a m1 m2 rod

2 9 A The reason why a person is unable to move a heavy box when he pushes it is because he cannot overcome the frictional force exerted by the floor on the box. 10 A A B 70 N Consider both A and B as one system: 70 = (20 + 6.0)ܽ  ܽ= 2.69 m s-2 Consider forces acting on B: ݂ᇱ = 6.0ܽ= 16 N  frictional force on A =݂ Alternatively, consider forces acting on A: 70 −݂= 20ܽ  ݂= 16 N 11 A Moment = Force x perpendicular distance = 4 x 2 = 8 Nm. Moment for B = C = D = (4݊݅ݏ݊݅ݏ 45 ) × 2 = 5.7 Nm 12 C Resultant force = zero. Resultant torque = 6x clockwise x = half the length of the one side. Hence for equilibrium, we need an anticlockwise Torque of 6x Nm. 13 C Impulse = area under force – time graph = ଵ ଶ (3 + 6)(20) = 90 N s Impulse = change in momentum: (20)ݒ− 0 = 90  ݒ= 4.5 m s-1 B Upward force = 6 N Weight = 6 N

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4 14 B Horizontally, momentum is conserved: (2.0)(0.50) = (2.5)v  v = 0.40 m s-1 Change in momentum of trolley = (2.0)(0.40) (2.0)(0.50) = 0.20 Ns 15 C From the relative speed relation: ݑ− 0 = 0.5ݑ−ݒ  ݒ= 0.5ݑ−ݑ= −0.5ݑ From momentum conservation: ݑ݉= ݉(−0.5ݑ)+ ݉ଶ(0.5ݑ)  ݉ଶ = 3݉ 16 D At maximum speed, ܨ௘௡௚௜௡௘ =ܦ= ݇ݒଶ: ܲ= ܨ௘௡௚௜௡௘ݒ= ݇ݒଷ With two engines: 72,000 =݇(12)ଷ With one engine: 36,000 =݇ݒଷ Dividing gives: ݒ3 (12)3 = 1 2 ݒ= 9.52 m s-1 17 C Tension becomes zero and the only force acting on the mass is its weight. Hence, it undergoes projectile motion (i.e, constant vertical acceleration = g and non - zero constant velocity in the horizontal direction). For projectile motion, path is parabolic. 18 B Centripetal acceleration, ܽ= ௩మ ௥ From energy conservation, gain in k.e. = loss in g.p.e. ଵ ଶ݉ݒଶ − 0 =݃݉(25 − 25ݏ݋ܿݏ݋ܿ (48.2) ) ݒଶ = 163.6  ܽ= ௩మ ௥ = ଵ଺ଷ.଺ ଶହ = 6.54 m s-2 m u m2 m m2 0.5u v 25cos48.2

5 [Turn over 19 B Horizontally: ܮ݊݅ݏ ݊݅ݏ ߠ = ௠௩మ ௥ i.e. centripetal force = ܮ݊݅ݏ ݊݅ݏ ߠ or = ௠௩మ ௥ Ratio of centripetal force to the weight = ௠௩మ ௥ ÷݃݉= ௩మ ௥௚ = ଺ହ଴మ ଼଴,଴଴଴×ଽ.଼ଵ = 0.538 20 A Weight = Gravitational force exerted by Earth on satellite: ܹ= ீெ௠ ௥మ = ீெಶೌೝ೟೓௠ೞೌ೟೐೗೗೔೟೐ ோమ At a distance of ݎ= 5ܴ+ ܴ= 6ܴ from the centre of the Earth, gravitational force exerted by Earth on satellite = ீெ௠ ௥మ = ீெಶೌೝ೟೓௠ೞೌ೟೐೗೗೔೟೐ (଺ோ)మ = ଵ ଷ଺ ቀீெಶೌೝ೟೓௠ೞೌ೟೐೗೗೔೟೐ ோమ ቁ = ௐ ଷ଺ 21 C Since satellite is in circular orbit, from Newton’s 2nd Law: ܨ௡௘௧ =݉௦௔௧௘௟௟௜௧௘ܽ  ீெ௠ ோమ = ௠௩మ ோ  ݉ݒଶ = ீெ௠ ோ Kinetic energy of satellite = ଵ ଶ݉ݒଶ = ଵ ଶ ቀீெ௠ ோ ቁ 22 B Since satellite is in circular orbit, from Newton’s 2nd Law: ܨ௡௘௧ =݉௦௔௧௘௟௟௜௧௘ܽ  ீெ௠ೞೌ೟೐೗೗೔೟೐ ோమ =݉௦௔௧௘௟௟௜௧௘ܴ߱ଶ mass of satellite cancels off.  ீெ ோయ =߱ଶ = ቀଶగ ் ቁ ଶ For period T to be 24 hrs, there can only be 1 value of R. 23 A Length of wire in one turn of the coil = 2ݎߨ= 2ߨቀ஽ ଶቁ =ܦߨ Total length of wire used in N turns ܮ= ܰܦߨ L W r = 80,000 m a

6 ܴ= ܮߩ ܣ= ߩ(ܰܦߨ) ߨቀ݀ 2ቁ ଶ = 4ߩ(ܰܦ) ݀ଶ 24 D 66,000 =ܴ+ ܴ 2 +ܴ 3 ܴ= 36 k 25 D Energy dissipated =ܸܳ  potential difference ܸ= ா௡௘௥௚௬ ஽௜௦௦௜௣௔௧௘ௗ ொ p.d. across r = ଵ଴.଴ ଶ଴.଴ = 0.500 V p.d. across R = ହ଴.଴ ଶ଴.଴ = 2.50 V emf = 2.50 + 0.50 = 3.00 V 26 B When light intensity decreases, resistance of the LDR increases. From potential divider rule, p.d. across LDR = Q increases while p.d. across fixed resistor = P decreases. 27 C Let the e.m.f. of the battery be E. In diagram 1, p.d. across each of the 4 bulbs is ܸ= ா ଶ. In diagram 2, p.d. across each of the 4 bulbs is still ܸ= ா ଶ. Points A and B are at the same potential, hence connecting A to B will not result in any current flowing between A and B. A B

7 [Turn over 28 A Only forces on WZ and XY produces turning effect. Maximum torque = NBILsin90 x WX = (20)(0.80)I(0.17) x (0.11) = 1.35 ܫ= ଵ.ଷହ (ଶ଴)(଴.଼଴)(଴.ଵ଻)(଴.ଵଵ) = 4.51 A 29 A From Newton’s 2nd law: Resultant force = mass x centripetal acceleration ݒݍܤ= ߱ݒ݉  ߱= ஻௤ ௠ Since ߱= ଶగ ்  ଶగ ் = ஻௤ ௠  ܶ= ଶగ௠ ஻௤ 30 D 90232ܶℎ→ 224ߙ+ 2 − 10ߚ+ ܺܣܼ ܣ= 232 − 8 = 224 and ܼ= 90 − 4 + 2 = 88