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Midterm 02 - Bautista Aldo Midterm 2 Due 10:00 pm Inst...

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Bautista, Aldo – Midterm 2 – Due: Oct 19 2005, 10:00 pm – Inst: Maxim Tsoi 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 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 17 cm along a circu- lar arc with a 31 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 . 88 kg mass. The acceleration of gravity is 9 . 8 m / s 2 . 31 cm 5 . 88 kg 30 g v bullet 17 cm Calculate the total energy of the composite system at any time after the collision. Correct answer: 9 . 84606 J. Explanation: Let : r = 31 cm = 0 . 31 m , h = 17 cm = 0 . 17 m , m block = 5 . 88 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 = 9 . 84606 J , which is the total energy after collision. 002 (part 2 of 3) 10 points Taking the same parameter values as those in Part 1, determine the initial velocity of the bullet. Correct answer: 359 . 599 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 = p 2 g h = q 2 (9 . 8 m / s 2 ) (0 . 17 m) = 1 . 82538 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 . 88 kg) (0 . 03 kg) × (1 . 82538 m / s) = 359 . 599 m / s . 003 (part 3 of 3) 10 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 = m bullet + m block v bullet 3. p f = 1 2 ( m bullet + m block ) p g h 4. p f = m bullet p g h 5. p f = m block p g h 6. p f = 1 2 ( m bullet + m block ) v bullet
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Bautista, Aldo – Midterm 2 – Due: Oct 19 2005, 10:00 pm – Inst: Maxim Tsoi 2 7. p f = m bullet v bullet correct 8. p f = ( m bullet + m block ) p g h 9. p f = m block v bullet 10. p f = m bullet + m block g h Explanation: As in part 2, due to the conservation of linear momentum, p f = p i = m bullet v bullet . 004 (part 1 of 3) 10 points Consider the general case of collisions of two masses m 1 = m with m 2 = 2 m along a fric- tionless horizontal surface. Denote the initial and the final center of mass momenta to be p i cm = p 1 + p 2 , and p f cm = p 0 1 + p 0 2 . And the initial and final kinetic energies to be K i = K 1 + K 2 , and K f = K 0 1 + K 0 2 . v 1 m v 2 2 m For an elastic collision which pair of state- ments is correct? 1. p i cm < p f cm , K i = K f 2. p i cm = p f cm , K i > K f 3. p i cm < p f cm , K i > K f 4. p i cm > p f cm , K i < K f 5. p i cm > p f cm , K i = K f 6. p i cm > p f cm , K i > K f 7. p i cm = p f cm , K i = K f correct 8. p i cm = p f cm , K i < K f 9. p i cm < p f cm , K i < K f Explanation: Momentum and energy are always con- served in elastic collision.
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