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Unformatted text preview: Bodet (ngb299) – oldhomework 19 – turner – (57340) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 0.374 kg bead slides on a straight friction less wire with a velocity of 3.11 cm/s to the right, as shown. The bead collides elastically with a larger 0.639 kg bead initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.79 cm/s. . 374 kg 3 . 11 cm / s . 639 kg Find the distance the larger bead moves along the wire in the first 5.2 s following the collision. Correct answer: 11 . 8697 cm. Explanation: Basic Concepts: m 1 v 1 ,i = m 1 v 1 ,f + m 2 v 2 ,f since v 2 ,i = 0 m/s. Δ x = v Δ t Given: Let to the right be positive: m 1 = 0 . 374 kg v 1 ,i = +3 . 11 cm / s m 2 = 0 . 639 kg v 1 ,f = . 79 cm / s t = 5 . 2 s Solution: v 2 ,f = m 1 v 1 ,i m 1 v 1 ,f m 2 = (0 . 374 kg)(3 . 11 cm / s) . 639 kg (0 . 374 kg)( . 79 cm / s) . 639 kg = 2 . 28263 cm / s to the right. Thus Δ x = (2 . 28263 cm / s)(5 . 2 s) = 11 . 8697 cm 002 (part 1 of 2) 10.0 points Consider the collision of two identical parti cles, with m 1 = m 2 = 10 g. The initial velocity of particle 1 is v 1 and particle 2 is initially at rest, v 2 = 0 m/s.. 1 2 v 1 After an elastic headon collision, the final velocity of particle 2 is v 2 and given by 1. v 2 = v 1 correct 2. v 2 = 0 3. v 2 = 3 v 1 4 4. v 2 = v 1 3 5. v 2 = 5 v 1 3 6. v 2 = v 1 2 7. v 2 = 2 v 1 3 8. v 2 = 4 v 1 3 9. v 2 = 2 v 1 10. v 2 = v 1 4 Explanation: For the final velocity of particle 2 after an elastic collision, we have v 2 = 2 v cm v 2 . For the present case, v cm = m 1 v 1 + m 2 v 2 m 1 + m 2 = v 1 2 . Bodet (ngb299) – oldhomework 19 – turner – (57340) 2 So v 2 = 2 v 1 2 0 = v 1 . 003 (part 2 of 2) 10.0 points Next replace particle 1 by a sledge hammer with mass m 1 = 10 kg, particle 2 by a golf ball with a mass m 2 = 10 g. Consider the elastic headon collision between the hammer and the ball. The initial velocity of the sledge hammer is v 1 , the golf ball is initially at rest, v 2 = 0 m/s. 1 2 v 1 Estimate the approximate final speed v 2 of the golf ball. 1. v 2 ≈ 5 v 1 2. v 2 ≈ v 1 3. v 2 ≈ 10 v 1 4. v 2 ≈ 20 v 1 5. v 2 ≈ 500 v 1 6. v 2 ≈ 1000 v 1 7. v 2 ≈ 50 v 1 8. v 2 ≈ 100 v 1 9. v 2 ≈ 200 v 1 10. v 2 ≈ 2 v 1 correct Explanation: Using again the same expression, the final velocity of the golf ball is v 2 = 2 v cm v 2 . Since m 1 m 2 , so v cm ∼ = v 1 . The approxi mate final speed of the ball is v 2 = 2 v 1 0 = 2 v 1 . 004 (part 1 of 2) 10.0 points A 2080 kg car moving east at 10.5 m/s collides with a 3190 kg car moving east. The cars stick together and move east as a unit after the collision at a velocity of 5.17 m/s....
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This note was uploaded on 10/06/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Friction, Work

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