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Unformatted text preview: arellano (aa39398) – Practice Homework Chapter 9 Solutions – weathers – (17101) 1 This printout should have 74 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The amount of momentum an object has de pends on its 1. mass and height from the ground. 2. only its speed. 3. mass and speed. correct 4. only its height from the ground. 5. only its mass. Explanation: 002 10.0 points What is the formula for momentum? 1. p = Δ v Δ t 2. p = mv correct 3. p = 1 2 mv 2 4. p = mg h 5. p = Δ d Δ t Explanation: 003 10.0 points Which of these quantities must be a scalar? 1. linear momentum 2. kinetic energy correct 3. acceleration 4. velocity 5. force Explanation: Kinetic energy is a measured quantity with out direction. 004 (part 1 of 2) 10.0 points An object is moving so that its kinetic energy is 229 J and the magnitude of its momentum is 36 . 2 kg · m / s. Determine the speed of the object. Correct answer: 12 . 6519 m / s. Explanation: Let : K = 229 J and p = 36 . 2 kg · m / s . The kinetic energy is given by K = 1 2 mv 2 , and the magnitude of the momentum is p = mv. Dividing one equation by the other yields K p = v 2 . v = 2 K p = 2 (229 J) 36 . 2 kg · m / s = 12 . 6519 m / s . 005 (part 2 of 2) 10.0 points Determine the mass of the object. Correct answer: 2 . 86122 kg. Explanation: m = p v = 36 . 2 kg · m / s 12 . 6519 m / s = 2 . 86122 kg . 006 10.0 points A bullet of mass 0 . 7 kg is headed at hori zontal speed 150 m / s toward a block of mass 14(0 . 7 kg) = 9 . 8 kg resting on a horizontal frictionless surface. The bullet strikes the block and comes to rest relative to the block somewhere deeply embedded inside the block. arellano (aa39398) – Practice Homework Chapter 9 Solutions – weathers – (17101) 2 9 . 8 kg . 7 kg v v f What average force did the block exert on the bullet if it took 0 . 1 s to stop the bullet? 1. 70 N 2. 980 N 3. 70 N 4. Not enough information is provided. 5. 1050 N 6. 1050 N 7. 980 N correct 8. Zero, since the bullet comes to rest rela tive to the block. Explanation: During the completely inelastic collision momentum is conserved, so p = p f mv = ( m + 14 m ) v f v f = 1 15 v and the impulse on the bullet is F avg Δ t = Δ p b = p b,f p b,i = mv f mv = m parenleftbigg 1 15 v parenrightbigg mv = 14 15 mv F avg = 14 mv 15Δ t = 14(0 . 7 kg)(150 m / s) 15(0 . 1 s) = 980 N . 007 10.0 points A bullet of mass m is headed at horizontal speed v toward a block of mass 12 m resting on a horizontal frictionless surface. The bullet strikes the block and comes to rest relative to the block somewhere deeply embedded inside the block. 12 m m v v f What average force did the block exert on the bullet if it took a time Δ t to stop the bullet?...
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This note was uploaded on 10/31/2010 for the course PHYSICS 1710 taught by Professor Weathers during the Fall '10 term at North Texas.
 Fall '10
 Weathers
 Momentum, Work

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