FND Semester 1, 2021: SAMPLE SOLUTIONS Page 1 of 6 PHYS1002 SAMPLE SOLUTIONS 2021 POTATO GUN Q13. When the potato strikes the coin, they have kinetic energy. As they swing upwards, this kinetic energy is converted into potential energy. Therefore the potential energy Uf at the end of the swing equals the kinetic energy Ki at the start of the swing (taking h = 0 at the height of the coin). Uf = (mp +mc)gh = (20 g + 1 g)× 9.8× 0.03 = 0.021× 9.8× 0.3 = 0.0062 J Hence the kinetic energy just after the collision is Ki = 1 2 (mp +mc)v 2 = 0.0062 J ⇒ v2 = 0.0062 0.021 = 0.294 Hence v = √ 0.294 = 0.542m s−1 (3 marks) Q14. Since there are no external forces, momentum is conserved during the collision. pafter = (mp +mc)v = 0.021× 0.542 = 0.0114 kgms−1 By conservation of momentum, this equals the momentum before, which is (since before the collision the coin was stationary) pbefore = mpv ′ = 0.001v′ = pafter = 0.0114 kgms−1 Hence v′ = 0.0114 0.001 = 11.4m s−1 = 41 kmh−1 So the claim on the box is not true, at least for this shot. (2 marks) Q15. When released from the gun, the potato had KE Kp = 1 2 mpv ′2 = 0.5× 0.001× (11.4)2 = 0.065 J (1 mark) Q16. We know that the potential energy of a compressed spring is Us = 12kx 2. The spring was com- pressed, giving it potential energy, which was converted into the kinetic energy of the potato when the gun was fired. Hence we can equate K with Us to find k: 1 2 kx2 = K ⇒ k = 2K x2 = 2× 0.065 0.022 = 325Nm−1 (2 marks) Total: 8 marks FND Semester 1, 2021: SAMPLE SOLUTIONS Page 2 of 6 CAVE RESCUE Q17. If the tension in the rope is T , then the net upward force is given by Fnet = T −W . By Newton’s second law, this equals ma, so the accceleration is a = T −W m = T −mg m = T m − g where m is the mass of the climber, m = 5209.8 = 53.1 kg (3 marks) Q18. The shortest time to get out of the cave will occur when the climber is accelerated at the maximum rate. Hence from (a), the maximum rate at which the climber can be accelerated amax occurs when the tension is maximum: amax = Tmax −mg m = 569− 520 53.1 = 0.923m s−2 Then we use s = 12at 2 to find the time taken: t = √ 2s a = √ 2× 35.1 0.923 = 8.7 s (3 marks) Q19. If the box is being lowered, then the net force on the box is F ′net = T ′ −mboxg The speed of the box is decreasing which means the acceleration is in the opposite direction to the velocity. The velocity is negative (it is moving downwards), so the acceleration must be positive. Hence F ′net = T ′ −mboxg = ma′ = 25× 4 Hence T ′ = m(a′ + g) = 25(4 + 9.8) = 345N (2 marks) Total: 8 marks FND Semester 1, 2021: SAMPLE SOLUTIONS Page 3 of 6 ARM Q20. Four forces: take clockwise torques to be positive, around the shoulder joint. • Weight of ball: force is weight of ball, acting at distance from shoulder. Force points down so torque is clockwise. – Fball = 9.8× 2 = 19.8N – dball = 0.38 + 0.45 = 0.83m – Direction of torque: + • Weight of arm: force is weight of arm, acting at CoM. Force points down so torque is clockwise. – Farm = 9.8× 8 = 78.4N – darm = 0.38m – Direction of torque: + • Force from deltoid: force is unknown, acting at attachment point. Force points up so torque is anticlockwise. – Fdeltoid = Fd – ddeltoid = 0.17m – Direction of torque: − • Force from shoulder joint: force is unknown (must exist to counteract horizontal component of force from deltoid), but distance is zero so produces no torque. – Fshoulder = Fs – dshoulder = 0 – Direction of torque: zero (4 marks) Q21. For equilibrium, the sum of the torques must be zero. Hence Fball × 0.83 + Farm × 0.38− Fd × 0.17 sin 15◦ = 0 so Fd = Fball × 0.83 + Farm × 0.38 0.17 sin 15◦ = 19.8× 0.83 + 78.4× 0.38 0.17× 0.2588 = 1050N (3 marks) Q22. Since Warm = 78.4N, this is roughly 13 times the weight of the arm. (1 mark) Total: 8 marks FND Semester 1, 2021: SAMPLE SOLUTIONS Page 4 of 6 TREE BRANCH Q23. Answer: C (1 mark) Q24. Since the branch is fixed to the tree at A, it cannot vibrate at that end so must be a node; it is free to vibrate at the other end, so B will be an antinode. (1 mark) Q25. The vibrating branch will behave like a closed pipe: the length of the branch is one-quarter of the wavelength for the fundamental mode, and three-quarter wavelengths for the second harmonic. Hence • Fundamental λ1 = 4L = 8m • Second harmonic: λ3 = 4L3 = 83 = 2.67m (2 marks) Q26. If v = 100m s−1, then • Fundamental f1 = vλ1 = 1008 = 12.5Hz • Second harmonic: f3 = vλ3 = 1002.67 = 37.5Hz (2 marks) Q27. You will hear the same frequencies as above: 12.5Hz and 37.5Hz. The sounds in air will have wavelength λ = vair/f , so λ′1 = 344 12.5 = 27.5m and λ ′ 3 = 9.2m (2 marks) Total: 8 marks FND Semester 1, 2021: SAMPLE SOLUTIONS Page 5 of 6 MUSICAL INSTRUMENT Q28. For a closed pipe, λ0 = 4L so we have λ0 = 4× 0.45 = 1.8m and so f0 = v λ = 344 1.8 = 191Hz (1 mark) Q29. A closed pipe can vibrate in many different normal modes. The normal modes must fulfill the boundary conditions, which for a closed pipe are that the open end must be a pressure node, and the closed end must be a pressure antinode. Such normal modes obey the relation λ = 4Lm for odd integers m = 1, 3, 5 . . . . (2 marks) Q30. Hence the third lowest frequency that can be produced corresponds to the m = 5 normal mode: λ5 = 4L 5 = 1.8 5 = 0.36m so f5 = 344 0.36 = 956Hz OR f5 = 5f0 = 5× 191 = 955Hz (1 mark) Q31. If the instrument was filled with helium instead of air, the wavelength of the note would stay the same (since it is fixed by the length of the instrument) but the wave speed would change (since it is a property of the medium, in this case helium). Helium has a much higher wave speed than air, so the new frequency will be higher than the same note played in air. (2 marks) Q32. If you filled a piano with helium the pitch of the notes would not change. The frequency of the note produced by the string is set by the length of the string (giving the wavelength) and the speed of the wave in the string. The string would therefore be vibrating at the same frequency, and so would produce a note of the same frequency in the helium. (2 marks) Total: 8 marks FND Semester 1, 2021: SAMPLE SOLUTIONS Page 6 of 6 EARTHQUAKE Q33. Loooking at the peaks between 760 and 780 s, there are two peaks in about 20 s, so T ≈ 10 s ⇒ f ≈ 0.1Hz (1 mark) Q34. (see above) (1 mark) Q35. If v = 3.7 km s−1, then λ = v/f = 3.7/0.1 = 37 km. (2 marks) Q36. The two waves are travelling the same distance but at different speeds, so they will arrive at different times. If the distance to the earthquake is d, the P-waves will take a time tp to arrive, given by tp = d/vp, while the S-waves will arrive in a time ts = d/vs. Hence ts − tp = d ( 1 vs − 1 vp ) or d = ts − tp( 1 vs − 1vp ) For the above seismogram, d = 760− 480 1 3.7 − 16.5 = 2400 km (2 marks: note that the question did not ask for a numerical value) Q37. If the frequency of the earthquake matches the natural frequency of the building, a resonance can be set up and the amplitude of the resulting vibration could get very large. To prevent this, the designer should ensure that the building does not have a natural frequency of vibration near 0.1Hz. If this is not possible, then the designer should ensure that any vibration is strongly damped before it can grow to destructive amplitude. (2 marks) Total: 8 marks























































































































































































































































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