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Unformatted text preview: homework 22 PAPAGEORGE, MATT Due: Mar 22 2008, 4:00 am 1 Question 1, chap 11, sect 2. part 1 of 1 0 points A 0.479 kg bead slides on a straight fric tionless wire with a velocity of 4.62 cm/s to the right, as shown. The bead collides elas tically with a larger 0.600 kg bead initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.69 cm/s. . 479 kg 4 . 62 cm / s . 6 kg Find the distance the larger bead moves along the wire in the first 4.8 s following the collision. Correct answer: 20 . 3479 cm (tolerance 1 %). 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 . 479 kg v 1 ,i = +4 . 62 cm / s m 2 = 0 . 600 kg v 1 ,f = . 69 cm / s t = 4 . 8 s Solution: v 2 ,f = m 1 v 1 ,i m 1 v 1 ,f m 2 = (0 . 479 kg)(4 . 62 cm / s) . 6 kg (0 . 479 kg)( . 69 cm / s) . 6 kg = 4 . 23915 cm / s to the right. Thus x = (4 . 23915 cm / s)(4 . 8 s) = 20 . 3479 cm Question 2, chap 11, sect 2. part 1 of 2 10 points Consider the collision of two identical par ticles, where the initial velocity of particle 1 is v 1 and particle 2 is initially at rest. 1 2 v 1 After an elastic headon collision, the final velocity of particle 2 v 2 is given by 1. v 2 = v 1 4 2. v 2 = 4 v 1 3 3. v 2 = v 1 correct 4. v 2 = v 1 2 5. v 2 = 3 v 1 4 6. v 2 = 5 v 1 3 7. v 2 = 2 v 1 8. v 2 = 2 v 1 3 9. v 2 = 0 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 . So v 2 = 2 parenleftBig v 1 2 parenrightBig 0 = v 1 . homework 22 PAPAGEORGE, MATT Due: Mar 22 2008, 4:00 am 2 Question 3, chap 11, sect 2. part 2 of 2 10 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....
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 Spring '09
 KLEINMAN
 Physics, Friction, Work

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