oldhomework 19-solutions

# oldhomework 19-solutions - Bodet(ngb299 oldhomework 19...

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Bodet (ngb299) – oldhomework 19 – turner – (57340) 1 This print-out should have 15 questions. Multiple-choice 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. 0 . 374 kg 3 . 11 cm / s 0 . 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 = - 0 . 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) 0 . 639 kg - (0 . 374 kg)( - 0 . 79 cm / s) 0 . 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 head-on 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 .

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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 head-on 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. a) What is the velocity of the 3190 kg car before the collision? Correct answer: 1 . 69464 m / s. Explanation: Basic Concept: The cars have the same final speed, so m 1 v i, 1 + m 2 v i, 2 = ( m 1 + m 2 ) v f Given: Let east be positive: m 1 = 2080 kg v 1 ,i = +10 . 5 m / s m 2 = 3190 kg v f = +5 . 17 m / s Solution: v 2 ,i = ( m 1 + m 2 ) v f - m 1 v i, 1 m 2 = (2080 kg + 3190 kg)(5 . 17 m / s) 3190 kg - (2080 kg)(10 . 5 m / s) 3190 kg = 1 . 69464 m / s to the east.
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