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Unformatted text preview: yoon (jy4326) – HW07 – markert – (58840) 1 This printout should have 22 questions. Multiplechoice 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 13 cm along a circu lar arc with a 23 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 . 02 kg mass. The acceleration of gravity is 9 . 8 m / s 2 . 2 3 cm 5 . 02 kg 30 g v bullet 13 cm Calculate the total energy of the composite system at any time after the collision. Correct answer: 6 . 4337 J. Explanation: Let : r = 23 cm = 0 . 23 m , h = 13 cm = 0 . 13 m , m block = 5 . 02 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 = 6 . 4337 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: 268 . 701 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 . 13 m) = 1 . 59625 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 . 02 kg) (0 . 03 kg) × (1 . 59625 m / s) = 268 . 701 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 bullet + m block ) v bullet 2. p f = 1 2 ( m bullet + m block ) radicalbig g h 3. p f = m block radicalbig g h 4. p f = m block v bullet 5. p f = m bullet v bullet correct yoon (jy4326) – HW07 – markert – (58840) 2 6. p f = √ m bullet + m block g h 7. p f = 1 2 ( m bullet + m block ) v bullet 8. p f = m bullet radicalbig g h 9. p f = ( m bullet + m block ) radicalbig g h 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 8 kg mass slides to the right on a surface having a coefficient of friction 0 . 27 as shown in the figure. The mass has a speed of 4 m / s when contact is made with a spring that has a spring constant 157 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 01/08/2010 for the course CH 53675 taught by Professor Bocknack during the Fall '09 term at University of Texas.
 Fall '09
 BOCKNACK
 Organic chemistry

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