midterm 04-solutions

# midterm 04-solutions - Version 075/ABACD midterm 04 turner...

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Unformatted text preview: Version 075/ABACD midterm 04 turner (57340) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A(n) 23 kg object, initially at rest in free space, explodes into three segments. The masses of two of these segments are both 8 kg and their velocities are 5 . 1 m / s. The an- gle between the direction of motion of these segments is 49 . What is the speed of the third segment? 1. 4.28294 2. 8.00292 3. 6.8048 4. 10.6075 5. 19.16 6. 11.8862 7. 18.9 8. 4.90434 9. 3.8011 10. 4.52156 Correct answer: 10 . 6075 m / s. Explanation: The mass of the third segment is m 3 = m 123- m 1- m 2 = (23 kg)- (8 kg)- (8 kg) = 7 kg . The momentum of the segments are p 1 = m 1 v 1 = (8 kg) (5 . 1 m / s) = 40 . 8 kg m / s p 2 = m 2 v 2 = (8 kg) (5 . 1 m / s) = 40 . 8 kg m / s p 3 = m 3 v 3 = (7 kg) v 3 . Momentum is conserved. 0 = &amp;p 1 + &amp;p 2 + &amp;p 3 p p p 1 2 3 = 180 - = 180 - 49 = 131 The law of cosines gives p 2 3 = p 2 1 + p 2 2- 2 p 1 p 2 cos(180 - ) ( 1 ) = (40 . 8 kg m / s) 2 + (40 . 8 kg m / s) 2- 2 (40 . 8 kg m / s) (40 . 8 kg m / s) cos(131 ) = (3329 . 28 kg 2 m 2 / s 2 )- (- 2184 . 2 kg 2 m 2 / s 2 ) = (5513 . 48 kg 2 m 2 / s 2 ) p 3 = 5513 . 48 kg 2 m 2 / s 2 = 74 . 2528 kg m / s . Since p = mv , we have v 3 = p 3 m 3 = 74 . 2528 kg m / s 7 kg = 10 . 6075 m / s . Alternative Solution: Symmetry greatly simplifies this calculation. v 3 = 2 m 1 m 3 v 1 cos 2 = 2 8 kg 7 kg (5 . 1 m / s) cos 49 2 = 10 . 6075 m / s . 002 10.0 points If r A = 6 . 5 m, F = 3 N, and = 68 , what is the magnitude of the torque about location A? 1. 21.941 2. 5.1683 3. 18.08 Version 075/ABACD midterm 04 turner (57340) 2 4. 4.4375 5. 13.775 6. 19.234 7. 27.17 8. 2.9089 9. 10.392 10. 0.65118 Correct answer: 18 . 08 Nm. Explanation: Let : r A = 6 . 5 m , = 68 , and F = 3 N . The value of torque is given by &amp; A = &amp;r A &amp; F Thus, the magnitude of torque is A = r A F sin ( ) A = (6 . 5 m)(3 N) sin(68) A = 18 . 08 Nm 003 10.0 points A 1 kg, 3 . 6 m long rod is supported on a knife edge at its midpoint. A 2 . 1 kg ball of clay is dropped from rest from a height of 1 . 8 m and makes a perfectly inelastic collision with the rod 1 . 1 m from the point of support. 1 . 8 m 3 . 6 m 1 . 1 m 1 kg 2 . 1 kg Find the magnitude of the angular momen- tum of the rod and clay system about the point of support immediately after the inelas- tic collision. The acceleration of gravity is 9 . 81 m / s 2 . 1. 67.9486 2. 23.1958 3. 7.14364 4. 22.5313 5. 50.5132 6. 38.8925 7. 78.5683 8. 13.7277 9. 127.486 10. 25.8852 Correct answer: 13 . 7277 J s....
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## This note was uploaded on 10/06/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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midterm 04-solutions - Version 075/ABACD midterm 04 turner...

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