Exam 3B (2016).pdf

# M p v pi m p v pf m b v bf m p v pf m p v pi m b v bf

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m p v pi = m p v pf + m b v bf = ) m p v pf = m p v pi - m b v bf = ) v pf = m p v pi - m b v bf m p . Plugging in numbers, we have v p = (4 . 60 kg)(1 . 47422 m/s) - (2 . 50 kg)(1 . 30 kg) 4 . 60 kg = 0 . 767693 m/s ' 0 . 768 m/s . Copyright c 2016 University of Georgia. 6

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Physics 1111 Exam #3B-11:00 Solutions 17 November 2016 Last Name: IV: Pounce! (24 points) As Calvin runs due East at 3 . 30 m/s, Hobbes pounces on him from behind, moving at 6 . 90 m/s in a direction 50 . 0 North of East. Calvin has a mass of 22 . 8 kg and Hobbes has a mass of 13 . 1 kg. (a) What will their speed be immediately after the collision? (8 pts) This is a 2-D perfectly inelastic collision problem. We can set up conservation of momentum equations for both x (East) and y (North) directions. P x : m C v C + m H v H cos H = ( m C + m H ) v fx v fx = m C v C + m H v H cos H m C + m H = (22 . 8)(3 . 30) + (13 . 1)(6 . 90)(cos 50 . 0 ) 22 . 8 + 13 . 1 = 3 . 71425 m/s P y : m H v H sin H = ( m C + m H ) v fx v fy = m H v H sin H m C + m H = (13 . 1)(6 . 90)(sin 50 . 0 ) 22 . 8 + 13 . 1 = 1 . 92877 m/s Now that we have the components of the final velocity, we can find the magnitude: v f = q v 2 fx + v 2 fy = p (3 . 71425) 2 + (1 . 92877) 2 = 4 . 18519 m/s ' 4 . 19 m/s . (b) What direction will they move after the collision? (8 pts) We can also use the components to find the angle: f = tan - 1 v fy v fx = tan - 1 1 . 92877 3 . 71425 = 27 . 4423 ' 27 . 4 . Note that it would be fine to write the original equations in terms of v f and f instead of v fx and v fy . (c) What percentage of kinetic energy is lost in the collision? (8 pts) K i = 1 2 m C v 2 C + 1 2 m H v 2 H = (22 . 8)(3 . 30) 2 + (13 . 1)(6 . 90) 2 2 = 435 . 992 J K f = 1 2 ( m C + m H ) v 2 f = (22 . 8 + 13 . 1)(4 . 18519) 2 2 = 314 . 409 J K i - K f K i 100% ' 27 . 9% Copyright c 2016 University of Georgia. 7
Physics 1111 Exam #3B-11:00 Solutions 17 November 2016 Last Name: V: Potter’s Wheel (24 points) One type of potter’s wheel uses a large, heavy disk as a flywheel to keep the small disk that holds the pot turning smoothly for longer. The potter uses their foot to spin the big disk up to speed. One such disk has a diameter of 0 . 980 m and a mass of 38 . 0 kg. Assume the potter can give the wheel a constant angular acceleration of 0 . 340 rev/s 2 . (a) How many revolutions will the wheel make in reaching a speed of 1 . 05 rev/s, starting from rest? (8 pts) For this part, since the given values are in revolutions and the question asks for revolutions, we can keep working in revolutions. ! 2 f = ! 2 i + 2 ( Δ ) = ) Δ = ! 2 f - ! 2 i 2 Plugging in numbers: Δ = (1 . 05 rev/s) 2 - (0) 2(0 . 340 rev/s 2 ) ' 1 . 62 rev . It is also okay to convert all values to radians first, so long as you convert back to revolutions at the end. (b) To give this acceleration, the potter’s foot provides a tangential frictional force at 3 / 4 of the way from the center to the edge. Assuming this is the only force causing rotation, what is the magnitude of this force?
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