25 - fryer(vdf96 Conservation Of Energy graves(6 This...

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fryer (vdf96) – Conservation Of Energy – graves – (6) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 (part 1 of 2) 10.0 points A student performs a ballistic pendulum experiment using an apparatus similar to that shown in the Fgure. Initially the bullet is Fred at the block while the block is at rest (at its lowest swing point). After the bullet hits the block, the block rises to its highest position, see dashed block in the Fgure, and continues swinging back and forth. The following data is obtained: the maximum height the pendulum rises is 4 . 3 cm, the mass of the bullet is 70 g, and the mass of the pendulum bob is 904 kg. The acceleration of gravity is 9 . 8 m / s 2 . 904 kg 70 g v i v f 44 4 . 3 cm ±ind the Fnal velocity of the system ( m 1 + m 2 ) immediately after the collision and before the pendulum starts to swing upwards. Correct answer: 0 . 918041 m / s. Explanation: Let : θ = 44 . 4932 , m 1 = 70 g = 0 . 07 kg , m 2 = 904 kg , and h = 4 . 3 cm = 0 . 043 m . ±rom the setup, the Fnal velocity of the col- lision is the initial velocity of the subsequent swing. During the swinging process the total energy is conserved E i = E f 1 2 ( m 1 + m 2 ) v 2 f = ( m 1 + m 2 ) g h. Therefore v f = r 2 g h = R 2 (9 . 8 m / s 2 ) (0 . 043 m) = 0 . 918041 m / s . 002 (part 2 of 2) 10.0 points ±ind v 1 , the initial speed of m 1 . Correct answer: 11856 . 8 m / s. Explanation: The linear momentum is conserved in a collision p i = p f m 1 v 1 = ( m 1 + m 2 ) v f Therefore v 1 = m 1 + m 2 m 1 v f = (0 . 07 kg) + (904 kg) (0 . 07 kg) (0 . 918041 m / s) = 11856 . 8 m / s . 003 10.0 points Water ²ows over a section of Niagara ±alls at a rate of 1 . 21 × 10 6 kg / s and falls 69 m. The acceleration of gravity is 9 . 8 m / s 2 . How many 40 W bulbs can be lit with this power? Correct answer: 2 . 0455 × 10 7 . Explanation: Let : r = 1 . 21 × 10 6 kg / s , and h = 69 m .
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fryer (vdf96) – Conservation Of Energy – graves – (6) 2 The power of the water is equal to the change in potential energy per unit time, so P = U t = mg h t = m t g h = (1 . 21 × 10 6 kg / s) (9 . 8 m / s 2 ) (69 m) = 8 . 18202 × 10 8 W . The power required to light
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This note was uploaded on 12/11/2011 for the course PHYSICS 101 taught by Professor Graves during the Fall '10 term at San Jose City College.

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25 - fryer(vdf96 Conservation Of Energy graves(6 This...

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