271_PartUniversity Physics Solution

271_PartUniversity Physics Solution - 8-30 Chapter 8 The...

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8-30 Chapter 8 The result of Problem 8.96 part (d) applies here since the collision is elastic This says that 25.0 90 , B θ °+ = ° so that 65.0 . B ( A and B move off in perpendicular directions.) EXECUTE: x P is conserved so 11 2 2 AAx BBx mv +=+ . But AB mm = so 12 2 cos25.0 cos65.0 AA B vv v = ° . y P is conserved so AAy BBy . 22 0 Ay By =+ . 0 sin25.0 sin65.0 −° . () sin25.0 /sin65.0 B A =°° . This result in the first equation gives 2 sin25.0 cos65.0 cos25.0 sin65.0 A v °° ⎛⎞ + ⎜⎟ ° ⎝⎠ . 1.103 = . 21 /1.103 (15.0 m/s)/1.103 13.6 m/s == = . And then ( ) 2 sin25.0 /sin65.0 13.6 m/s 6.34 m/s. B v = EVALUATE: We can use our numerical results to show that K K = and that x x PP = and . y y = 8.98. IDENTIFY: Since there is no friction, the horizontal component of momentum of the system of Jonathan, Jane and the sleigh is conserved. SET UP: Let + x be to the right. 800 N A w = , 600 N B w = and 1000 N C w = . EXECUTE: x x = gives 222 0 CCx =++ . 22 22 2 Cx CC wv v mw ++ . 2 (800 N)( [5.00 m/s]cos30.0 (600 N)( [7.00 m/s]cos36.9 ) 0.105 m/s 1000 N v −+ + °) ° . The sleigh&s velocity is 0.105 m/s, to the left. EVALUATE: The vertical component of the momentum of the system consisting of the two people and the sleigh is not conserved, because of the net force exerted on the sleigh by the ice while they jump. 8.99. IDENTIFY: In Eq. 8.28 treat each straight piece as an object in the system. SET UP: The center of mass of each piece of length L is at its center. EXECUTE: (a) From symmetry, the center of mass is on the vertical axis, a distance ( /2)cos( /2) L α below the apex. (b) The center of mass is on the vertical axis of symmetry, a distance 2( /2)/3 /3 L L = above the center of the horizontal segment. (c) Using the wire frame as a coordinate system, the coordinates of the center of mass are equal and each is equal to ( /2)/2 /4 L L = . The center of mass is along the bisector of the angle, a distance /8 L from the corner. (d) By symmetry, the center of mass is at the center of the equilateral triangle, a distance (/ 3 ) s i n 6 0 /1 2 LL = ° above the center of the horizontal segment. EVALUATE: The center of mass need not lie on any point of the object, it can be in empty space. 8.100. IDENTIFY: There is no net horizontal external force so cm v is constant. SET UP: Let + x be to the right, with the origin at the initial position of the left-hand end of the canoe. A 45.0 kg m = , 60.0 kg B m = . The center of mass of the canoe is at its center. EXECUTE: Initially, cm 0 v = , so the center of mass doesn&t move. Initially, 11 cm1 BB mx x + = + . After she walks, cm2 x + = + . cm1 cm2 x x = gives . She walks to a point 1.00 m from the right-hand end of the canoe, so she is 1.50 m to the right of the center of mass of the canoe and 1.50 m xx . (45.0 kg)(1.00 m) (60.0 kg)(2.50 m) (45.0 kg)( 1.50 m) (60.0 kg) B B x x += + + . 2 (105.0 kg) 127.5 kg m B x =⋅ and 2 1.21 m B x = . 1.21 m 2.50 m 1.29 m −= = . The canoe moves 1.29 m to the left.
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Momentum, Impulse, and Collisions 8-31 EVALUATE: When the woman walks to the right, the canoe moves to the left. The woman walks 3.00 m to the right relative to the canoe and the canoe moves 1.29 m to the left, so she moves 3.00 m 1.29 m 1.71 m −= to the right relative to the water. Note that this distance is (60.0 kg/45.0 kg)(1.29 m) .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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271_PartUniversity Physics Solution - 8-30 Chapter 8 The...

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