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Unformatted text preview: burgin (lcb626) – HW07 – markert – (56480) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus the bullet rises to a height of 6 cm along a circular arc with a 11 cm radius. Assume: The entire track is frictionless. A bullet with a m 1 = 30 g mass is fired horizontally into a block of wood with m 2 = 5 . 95 kg mass. The acceleration of gravity is 9 . 8 m / s 2 . 1 1 cm 5 . 95 kg 30 g v bullet 6 cm Calculate the total energy of the composite system at any time after the collision. Correct answer: 3 . 51624 J. Explanation: Let : r = 11 cm = 0 . 11 m , h = 6 cm = 0 . 06 m , m block = 5 . 95 kg , and m bullet = 30 g = 0 . 03 kg . The mechanical energy is conserved after collision. Choose the position when the sys- tem stops at height h , where the kinetic en- ergy is 0 and the potential energy is given by ( m bullet + m block ) g h = 3 . 51624 J , which is the total energy after collision. 002 (part 2 of 3) 10.0 points Taking the same parameter values as those in Part 1, determine the initial velocity of the bullet. Correct answer: 216 . 164 m / s. Explanation: During the rising process the total energy is conserved E i = 1 2 ( m bullet + m block ) v 2 f and E f = ( m bullet + m block ) g h, so v f = radicalbig 2 g h = radicalBig 2 (9 . 8 m / s 2 ) (0 . 06 m) = 1 . 08444 m / s . The linear momentum is conserved in a colli- sion. p i = m bullet v i p f = ( m bullet + m block ) v f . Therefore v i = m bullet + m block m bullet v f = (0 . 03 kg) + (5 . 95 kg) (0 . 03 kg) × (1 . 08444 m / s) = 216 . 164 m / s . 003 (part 3 of 3) 10.0 points Denote v bullet to be the initial velocity, find the momentum of the compound system im- mediately after the collision. 1. p f = m block radicalbig g h 2. p f = m bullet radicalbig g h 3. p f = √ m bullet + m block g h 4. p f = 1 2 ( m bullet + m block ) radicalbig g h 5. p f = m bullet v bullet correct burgin (lcb626) – HW07 – markert – (56480) 2 6. p f = 1 2 ( m bullet + m block ) v bullet 7. p f = √ m bullet + m block v bullet 8. p f = ( m bullet + m block ) radicalbig g h 9. p f = m block v bullet 10. p f = ( m bullet + m block ) v bullet Explanation: As in part 2, due to conservation of linear momentum, p f = p i = m bullet v bullet . 004 10.0 points A 6 kg mass slides to the right on a surface having a coefficient of friction 0 . 64 as shown in the figure. The mass has a speed of 6 m / s when contact is made with a spring that has a spring constant 165 N / m. The mass comes to rest after the spring has been compressed a distance d . The mass is then forced toward the left by the spring and continues to move in that direction beyond the unstretched posi- tion. Finally the mass comes to rest a distance D to the left of the unstretched spring....
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This note was uploaded on 10/25/2010 for the course PHY PHYSICS taught by Professor Swinney during the Fall '08 term at University of Texas.
- Fall '08