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Unformatted text preview: Quach, Phuc Homework 8 Due: Oct 16 2006, 5:00 pm Inst: Mark Rupright 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Sand from a stationary hopper falls on a mov ing conveyor belt at the rate of 7 . 36 kg / s, as shown in the figure. The belt is supported by frictionless rollers and moves at 0 . 985 m / s un der the action of a horizontal external force supplied by the motor that drives the belt. F ext Find the frictional force exerted by the belt on the sand. Correct answer: 7 . 2496 N. Explanation: Let : m = 7 . 36 kg / s and v = 0 . 985 m / s . The momentum of the sand on the belt is p x = mv . Therefore, the rate of change of momentum in the horizontal direction is p x = d dt ( mv ) = dm dt v = mv = (7 . 36 kg / s)(0 . 985 m / s) = 7 . 2496 N . From Newtons second law, the net force act ing on the sand in the horizontal direction is equal to the change of momentum of the sand in the horizontal direction. But, the only horizontal force on the sand is belt friction, therefore, the frictional force is f = p x = 7 . 2496 N . keywords: 002 (part 1 of 1) 10 points You want to enlarge a skating surface so you stand on the ice at one end and aim a hose horizontally to spray water on the schoolyard pavement. Water leaves the hose at 2 . 8 kg / s with a speed 11 . 4 m / s. If your mass is 42 kg, what is your recoil acceleration? Neglect friction and the mass of the hose. Correct answer: . 76 m / s 2 . Explanation: Let : dm w dt = 2 . 8 kg / s , m = 42 kg , and v w = 11 . 4 m / s . Applying conservation of momentum, mv + m w v w = 0 . Differentiating with respect to t , m dv dt + v dm dt + m w dv w dt + v w dm w dt = 0 ma + v dm dt + m w a w + v w dm w dt = 0 v w dm w dt + ma = 0 . Since a w = 0 and dm w dt = 0 , so a = v w m dm w dt a = 11 . 4 m / s 42 kg (2 . 8 kg / s) = . 76 m / s 2 . keywords: 003 (part 1 of 1) 10 points Quach, Phuc Homework 8 Due: Oct 16 2006, 5:00 pm Inst: Mark Rupright 2 The graph below shows the force on an object of mass M as a function of time. Time (s) Force(N) 1 2 3 4 10 10 For the time interval 0 to 4 s, the total change in the momentum of the object is 1. p = 0 kgm / s . correct 2. p = 20 kgm / s . 3. Indeterminable unless the mass M of the object is known 4. p = 40 kgm / s . 5. p = 20 kgm / s . Explanation: The Newtons second law of motion, in one dimension, is F = M a = M dv dt . From this, we obtain F dt = M dv , = ( M v ) = Z F dt, where M v is the momen tum of the object. So from the graph above, the change in the momentum is zero ....
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 Fall '08
 Guzman
 Physics, Work

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