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Version 001 – Summer Review #5 Circular Motion, Gravity, Energy and Momentum – tubman – (IBII20 s This print-out should have 12 questions. gR 10. v = correct Multiple-choice questions may continue on µ the next column or page – find all choices Explanation: before answering. The maximum frictional force due to friction is fmax = µ N , where N is the inward Barrel of Fun 04 directed normal force of the wall of the cylin001 (part 1 of 2) 10.0 points der on the person. To support the person An amusement park ride consists of a large vertically, this maximal friction force fsmax vertical cylinder that spins about its axis fast must be larger than the force of gravity m g enough that any person inside is held up so that the actual force, which is less than against the wall when the floor drops away µN , can take on the value m g in the positive (see figure). The coefficient of static friction vertical direction. Now, the normal force supbetween the person and the wall is µ and the v2 radius of the cylinder is R. plies the centripetal acceleration on the R person, so from Newton’s second law, R m v2 N = . R Since µ m v2 ≥ mg, R the minimum speed required to keep the person supported is at the limit of this inequality, which is fsmax = µ N = ω What is the minimum tangential velocity needed to keep the person from slipping downward? p 1. v = g R p 2. v = µ g R p 3. v = 2 g R p 4. v = 2 µ g R p 5. v = µ 2 π g R 1p gR µ 1p 7. v = gR 2 p 8. v = 2 µ g R p 9. v = µ 2 g R 6. v = 2 µ m vmin = mg R 1/2 gR vmin = . µ 002 (part 2 of 2) 10.0 points Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimum necessary. Find the magnitude of the frictional force between the person and the wall. µ m v2 − mg R m v2 2. F = − µmg R µ m v2 3. F = m g + R m v2 4. F = µ m g + R 1. F = 5. F = m g correct Version 001 – Summer Review #5 Circular Motion, Gravity, Energy and Momentum – tubman – (IBII20 m v2 6. F = µR mg 7. F = µ 8. F = µ m g m v2 R µ m v2 10. F = R Explanation: The vertical friction force must equal m g in order to balance the force of gravity and not have any acceleration in the vertical direction. 9. F = Conceptual 05 11 003 (part 1 of 2) 10.0 points Calculate the period of a ball tied to a string of length 1.3 m making 4.4 revolutions every second. An air puck of mass 0.031 kg is tied to a string and allowed to revolve in a circle of radius 1.5 m on a frictionless horizontal surface. The other end of the string passes through a hole in the center of the surface, and a mass of 1.4 kg is tied to it, as shown. The suspended mass remains in equilibrium while the puck revolves on the surface. 0.031 kg 1.5 m 1.4 kg What is the magnitude of the force that maintains circular motion acting on the puck? The acceleration due to gravity is 9.81 m/s2 . Correct answer: 13.734 N. Explanation: Correct answer: 0.227273 s. Explanation: Let : T = Let : f = 4.4 rev/s , R = 1.3 m . and g = 9.81 m/s2 , mp = 0.031 kg , m = 1.4 kg , and r = 1.5 m . 1 1 = = 0.227273 s . f 4.4 rev/s m r 004 (part 2 of 2) 10.0 points Calculate the speed of the ball. Correct answer: 35.9398 m/s. mp Explanation: D 2πR v= = T T 2 π (1.3 m) = 0.227273 s = 35.9398 m/s . Holt SF 07Rev 52 005 (part 1 of 2) 10.0 points Fc = Fg = m g = (1.4 kg) (9.81 m/s2 ) = 13.734 N . 006 (part 2 of 2) 10.0 points What is the linear speed of the puck? Correct answer: 25.7788 m/s. Explanation: Version 001 – Summer Review #5 Circular Motion, Gravity, Energy and Momentum – tubman – (IBII20 v2 Fc = mp t s s r Fc r (13.734 N) (1.5 m) = vt = mp (0.031 kg) Correct answer: 2778.3 J. Explanation: The kinetic energy of the bullet is 1 1 mb vb2 = (56 g) (315 m/s)2 2 2 = 2778.3 J . Kb = = 25.7788 m/s . Firing a Gun 007 (part 1 of 4) 10.0 points A 30 kg gun is standing on a frictionless surface. The gun fires a 56 g bullet with a muzzle velocity of 315 m/s. The positive direction is that of the bullet. Calculate the momentum of the bullet immediately after the gun was fired. 010 (part 4 of 4) 10.0 points Calculate the kinetic energy of the gun immediately after the gun was fired. Correct answer: 17.64 kg · m/s. The recoil velocity of the gun comes from its momentum: pg = mg vg pg vg = mg Explanation: Let : mb = 56 g and vb = 315 m/s . The momentum of the bullet is pb = mb vb = (56 g) (315 m/s) = 17.64 kg · m/s . 008 (part 2 of 4) 10.0 points Calculate the momentum of the gun immediately after the gun was fired. Correct answer: 5.18616 J. Explanation: Let : mg = 30 kg . Thus the kinetic energy of the gun is p2g 1 (−17.64 kg · m/s)2 2 Kg = mg vg = = 2 2 mg 2 (30 kg) = 5.18616 J . Drag Racer 02 011 (part 1 of 2) 10.0 points A 901 kg drag race car accelerates from rest to 113 km/h in 1.03 s. What change in momentum does the force produce? Correct answer: −17.64 kg · m/s. Explanation: By conservation of momentum, pbo + pgo = pbf + pgf 0 + 0 = pb + pg pg = −pb = −17.64 kg · m/s . 009 (part 3 of 4) 10.0 points Calculate the kinetic energy of the bullet immediately after the gun was fired. Correct answer: 28281.4 kg · m/s. Explanation: Let : m = 901 kg and v = 113 km/h = 31.3889 m/s . Momentum is defined as ∆p = m ∆v = (901 kg) (31.3889 m/s − 0 m/s) = 28281.4 kg · m/s . Version 001 – Summer Review #5 Circular Motion, Gravity, Energy and Momentum – tubman – (IBII20 012 (part 2 of 2) 10.0 points What average force is exerted on the car? Correct answer: 27457.7 N. Explanation: Let : t = 1.03 s . Momentum and impulse are related by ∆p = F t , F = so ∆p 28281.4 kg · m/s = = 27457.7 N . t 1.03 s