oldhw22 - This print-out should have 11 questions...

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Unformatted text preview: 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 0.414 kg bead slides on a straight friction- less wire with a velocity of 3.54 cm/s to the right, as shown. The bead collides elastically with a larger 0.649 kg bead initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.75 cm/s. . 414 kg 3 . 54 cm / s . 649 kg Find the distance the larger bead moves along the wire in the first 5.5 s following the collision. Correct answer: 15 . 0514 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. Δ x = v Δ t Given: Let to the right be positive: m 1 = 0 . 414 kg v 1 ,i = +3 . 54 cm / s m 2 = 0 . 649 kg v 1 ,f = − . 75 cm / s t = 5 . 5 s Solution: v 2 ,f = m 1 v 1 ,i − m 1 v 1 ,f m 2 = (0 . 414 kg)(3 . 54 cm / s) . 649 kg − (0 . 414 kg)( − . 75 cm / s) . 649 kg = 2 . 73661 cm / s to the right. Thus Δ x = (2 . 73661 cm / s)(5 . 5 s) = 15 . 0514 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 = 4 v 1 3 3. v ′ 2 = 3 v 1 4 4. v ′ 2 = v 1 3 5. v ′ 2 = 2 v 1 3 6. v ′ 2 = v 1 4 7. v ′ 2 = 0 8. v ′ 2 = 2 v 1 9. v ′ 2 = v 1 2 10. v ′ 2 = 5 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 . oldhomework 22 – Turner – (58185) 1 hinojosa (jlh3938) – oldhomework 22 – Turner – (58185) 2 So v ′ 2 = 2 parenleftBig v 1 2 parenrightBig − 0 = v 1 ....
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oldhw22 - This print-out should have 11 questions...

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