Download Archimedes Principle - Physics - Solved Past Exam and more Exams Physics in PDF only on Docsity! 1 PC1141 Physics I – AY03/04 Exam Solutions 1. Let be the area of a bar of soap facing top/bottom. The weight of the bar of soap is × × the depth of soap immersed under water = 1000kg m × × 1.5 × 10 m. Let the thickness of the oil layer when the surface of bar soap is equal level to the oil surface be . Using Archimedes’ Principle, × × 600kg m + !2.0 × 10 m− $ × × 1000kg m = 1.25cm 2. i) Total force acting on the platform in the downward direction is 65g + 10g = 75g . ' is caused by the person standing on the platform. So '( must be greater than 75g in order to lift up the platform. ' = '( ⇒ 2' = 75g. ∴ ' = 367.9N ii) In finding the reaction force, consider the free body diagram on the man, . − ' −/0 = /1. As the man is pulling up the rope, the same tension is also pulling down the man other than his own weight. Thus, . = ' +/!0 + 1$ = 2186N. 3. i) = sin 30° × 0.24 = 0.1m . Let 56 be the centripetal force, and 57 be the force which balances the weight. 56 = 57 tan 30° = 489 , ⇒ 9 = ;< =>?@° AB = 5.32rad s(. ii) The tension in the cord, C56 + 57 = 2.27N 4. There is no torque acting on the ball either in or opposite their rotating direction, so the angular momentum will be conserved. Let 9D be the final angular velocity. 0.65 × 4.0 × 0.15 × 2 = 0.65 × 2 × 0.05 × 9D, ⇒ 9D = 36.0rad s(. But the kinetic energy is not conserved, FG = GD − GH = 2!0.5 × 0.65 × 0.05 × 36 $ − 2!0.5 × 0.65 × 0.15 × 4 $ = 1.872J. 5. The man is the source as well as the observer. As he is moving towards the reflection wall, the source is moving towards the observer and at the same time, the observer is moving towards the source. In this case, the apparent frequency should be J′ = J LMLN LLO = 503.5Hz. The beat frequency is then 3.5Hz. 6. i) Let ( be the length stretched by spring of spring constant R( = 50N m(. ( + = 0.2m, 50 ( = 150 ⇒ ( = 3 , 4 = 0.2m. ∴ = 0.05m, ( = 0.15m. ii) When displaced by F , the force, 5 = 41 = !R( + R $F , and so S TU = 9 = VWMVX A . ∴ ' = 2Y 9 = 0.2s iii) The total potential energy, 0.5!R( + R $!0.054$ . Let the velocity at the centre be Z. So we have 0.54Z = 0.5!R( + R $!0.054$ . ∴ Z = 1.58m s(. Conservation of momentum for horizontal component, 0.2 × 1.58 = !0.2 + 0.1$ZD, ∴ ZD = 1.053m s(. Kinetic energy converted into max potential energy. Solutions provided by: NUS Physics Society (Q1, Q2i, Q3, Q4, Q6-Q8), Tan Meng Ho (Q2ii, Q5) Edited, 2012 © NUS Physics Society → 0.5mV 2 = 0.5(k1 + k2)x 2 0.5(0.3)(1.053)2 = 0.5(200)x2 x= 0.04m (max amplitude) 7. • Given Period,T = 100minutes ⇒ GMm r2 = mV 2 r = mrω2 4π2 T 2GM = 1 r3 r = √ T 2GM 4π2 = 7.138× 106m from the center of earth. Height above the surface of the earth = 7.138 ×106m− 6.37× 106m = 0.768× 106m Speed of the spacecraft = V ⇒ V 2 = GM r V = √ GM r = √ 6.67×10−11×5.98×1024 7.138×106 = 7475m/s • Let Ecentricity = e The Semi-major axis = 2a+a 2 =1.5a Let C = the distanc between the center point of the major axis to the center of the earth. ⇒C = 1.5a-a = 0.5a →e · 1.5a = 0.5a e = 1 3 • Angular momentum conserved Let Vi = velocity at the point nearer to earth, Vf = velocity at the point farther from earth. ⇒ mVir = mVf2r Vf = 0.5Vi The total energy should be conserved as well ⇒ 1 2 mV 2 i − GMm r = 1 2 mV 2 f − GMm 2r 1 2 mV 2 i − GMm r = 1 2 m(0.5Vi) 2 − GMm 2r 3 8 V 2 i = GMm 2r Vi = √ GM8 2r·3 = √ 6.67×10−11×5.98×1024×8 2×3×7.138×106 = 8631m/s 8. • For calculation of Inertial moment of Cylinder with respect to the axis passing through its center, please refer to textbook. • Total mass of yoyo = 0.08kg 3
