Accelerations of Objects - Physics - Solved Past Exam, Exams of Physics

Download Accelerations of Objects - Physics - Solved Past Exam and more Exams Physics in PDF only on Docsity! PC1141 Exam Solution - AY06/07 Page 1 of 10 Question 1 Part A Defining T as tension in string, m1 = 4.00 kg and m2 = 6.00 kg as masses of objects 1 and 2, a1 and a2 as accelerations of objects 1 and 2. T = m1a1 (1a) m2g − T = m2a2 (1b) a1 = a2 (1c) From above, m2g = m2a+m1a a = m2g m1 +m2 ≈ 5.88 m s−2 T = m1m2 m1 +m2 g ≈ 23.5 N Part B m2g − 2T = m2a2 (2a) m1a1 = T (2b) a1 = 2a2 (2c) From above, m2g = m2a2 + 4m1a2 a2 = 1 2 m2g m2 + 4m1 ≈ 2.67 m s−2 a1 ≈ 5.35 m s−2 T ≈ 21.4 N Question 2 Part A Defining U as upthrust, M as total mass of balloon and helium, ρa = 1.20 kg m −3 as density of air, ρHe = 0.178 kg m −3 as density of helium, V = 0.0045 m3 as volume of balloon, m = 0.0025 kg as mass of deflated balloon. Fnet = U −Mg = ρaV g − (m+ ρHeV )g ∴ a = ρaV −m− ρHeV m+ ρHeV g PC1141 Exam Solution - AY06/07 Page 2 of 10 From kinematics, s = 1 2 at2 t = √ 2s a Putting everything together gives t ≈ 0.694 s . Part B Speed of balloon immediately before string is taut: v = at ≈ 4.323 m s−1 By Conservation of Momentum and defining ms = 0.015 kg as mass of stone, Mv = (M +ms)vf vf = M M +ms v ≈ 0.780 m s−1 Question 3 Defining G as the gravitational constant, M = 5.98×1024 kg as the mass of the Earth, m as the mass of the meteoroid, r1 = 2.8×108 m as the distance between the meteoroid and the Earth initially and r2 = 8.5×106 m as the distance between the meteoroid and the Earth at the closest approach, v1 = 1.1 × 103 m s−1 as the initial speed of the meteoroid and v2 as the speed of the meteoroid at closest approach. Part A By Conservation of Energy, 1 2 mv21 − GMm r1 = 1 2 mv22 − GMm r2 v2 = √ 2 [ v21 2 −GM ( 1 r2 − 1 r1 )] v2 ≈ 9.60× 103 m s−1 Part B As the total energy of the system, i.e. 12mv 2−GMmr , is less than zero, the system is bound — the meteoroid does not have sufficient energy to escape Earth’s gravitational field. Hence it will return to Earth’s vicinity. Question 4 Defining k as the wave number in rad m−1, ω as the angular frequency of the wave, v as the propagation speed of the wave, T as the tension in the cord, µ1 = 0.10 kg m −1 and µ2 = 0.20 kg m −1 as the linear densities of the cord at different sections, f as the frequency of the wave, λ1 and λ2 as the wavelengths of the wave at the first and second section of the cord respectively. PC1141 Exam Solution - AY06/07 Page 5 of 10 Find roots by: U(x) = 0 6x2 − x3 = 0 x2(6− x) = 0 ∴ x = 0 or x = 6 Particle with energies below 32 within the region −2 < x < 4 will oscillate within the regions. Particles with energies greater than 32 or particles outside the region −2 < x < 4 will eventually move in the positive x direction forever. x = 0 is a stable equilibrium point; x = 4 is an unstable equilibrium point. Part B Section 1 Defining vbottom to be speed of object at the bottom and by Conservation of Energy, 1 2 mv20 +mgh = 1 2 mv2bottom v20 + 2gh = v 2 bottom vbottom ≈ 6.943 m s−1 Time taken for object to stop when moving from bottom to maximum displacement along Plane B, vbottom = atB tB = vbottom a = vbottom g sin(15°) ≈ 2.737 s Time taken for object to stop when moving from bottom to maximum displacement along Plane A, vbottom = atA tA = vbottom a = vbottom g sin(25°) ≈ 1.676 s Period is then T = 2(tA + tB) ≈ 8.83 s Section 2 Maximum Height Using Conservation of Energy, we consider the energy of the particle when it reaches the bottom, where we define v = 3.00 m s−1 and h = 2.00 m: E = 1 2 mv2 +mgh+ fs = 1 2 mv2 +mgh+ µkmg cos(25°) h sin(25°) PC1141 Exam Solution - AY06/07 Page 6 of 10 We equate the above energy to the energy of the particle when it is at its maximum height on Plane B: 1 2 mv2 +mgh+ µkmg cos(25°) h sin(25°) = mgh′ + f ′s′ = mgh′ + µkmg cos(15°) h′ sin(15°) h′ = 1.99 m Permanent Stop Considering whether static friction will stop subsequent motion on Plane B, mg sin(15°) > µsmg cos(15°) Hence static friction will not stop subsequent motion on Plane B. Considering whether static friction will stop subsequent motion on Plane A, mg sin(25°) > µsmg cos(25°) Hence static friction will not stop subsequent motion on Plane A. This suggests that object will continue to move back down upon reaching its maximum, albeit decaying, height on either planes. It will only come to a permanent stop at O — the bottom — when it has exhausted all its energy to friction. Question 7 Part A Section 1 Defining z to be position of center of mass from the bottom, (3m+m)z = 3m · 3b+m · b z = 5 2 b ∴ ∆z = 3b− z = b 2 Section 2 Defining Inet to be the moment of inertia about the centre of mass of the system (the two hoops combined), Inet = I3m + Im = [ 3m · (3b)2 + 3m · ( b 2 )2] + [ m · b2 +m · ( 5b 2 − b )2] Inet = 31mb 2 PC1141 Exam Solution - AY06/07 Page 7 of 10 Part B A 3 b b 2 θ Centre of Mass of System 4mg Analysing torque about Point A and using the parallel axis theorem, τ = Iθ̈ −(4mg) · ( 3b+ b 2 ) · sin(θ) = [ Inet + 4m ( 3b+ b 2 )2] θ̈ −14mgb sin(θ) = 80mb2θ̈ Using small angle approximation (sin(θ) ≈ θ), θ̈ = − 7g 40b θ =⇒ ω2 = 7g 40b 2π T = √ 7g 40b T = 4π √ 10b 7g Part C When the centres of the hoops lie in a vertical line, the point where the two hoops meet is in contact with the table. We need to find the KE of the hoop system. We shall do this by looking at the instant where the hoop system seems to revolve around the contact point with the table. The KE will be given by the moment of inertia about that point, and the angular velocity about that point.