oldhw 18 - alexander (jra2623) oldhomework 18 Turner...

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Unformatted text preview: alexander (jra2623) oldhomework 18 Turner (92510) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 0.499 kg bead slides on a straight friction- less wire with a velocity of 4.68 cm/s to the right, as shown. The bead collides elastically with a larger 0.600 kg bead initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.68 cm/s. . 499 kg 4 . 68 cm / s . 6 kg Find the distance the larger bead moves along the wire in the first 4.3 s following the collision. Correct answer: 19 . 1683 cm. Explanation: Basic Concepts: m 1 vectorv 1 ,i = m 1 vectorv 1 ,f + m 2 vectorv 2 ,f since v 2 ,i = 0 m/s. x = v t Given: Let to the right be positive: m 1 = 0 . 499 kg v 1 ,i = +4 . 68 cm / s m 2 = 0 . 600 kg v 1 ,f = . 68 cm / s t = 4 . 3 s Solution: v 2 ,f = m 1 v 1 ,i m 1 v 1 ,f m 2 = (0 . 499 kg)(4 . 68 cm / s) . 6 kg (0 . 499 kg)( . 68 cm / s) . 6 kg = 4 . 45773 cm / s to the right. Thus x = (4 . 45773 cm / s)(4 . 3 s) = 19 . 1683 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 = 2 v 1 2. v 2 = v 1 correct 3. v 2 = v 1 4 4. v 2 = 2 v 1 3 5. v 2 = 0 6. v 2 = 4 v 1 3 7. v 2 = 5 v 1 3 8. v 2 = v 1 2 9. v 2 = 3 v 1 4 10. v 2 = v 1 3 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 . alexander (jra2623) oldhomework 18 Turner (92510) 2 So v 2 = 2 parenleftBig v 1 2 parenrightBig 0 = v 1 ....
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oldhw 18 - alexander (jra2623) oldhomework 18 Turner...

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