hw 22 - homework 22 – HETTIGER, CHRISTOF – Due: Mar 22...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: homework 22 – HETTIGER, CHRISTOF – Due: Mar 22 2008, 4:00 am 1 Question 1, chap 11, sect 2. part 1 of 1 0 points A 0.376 kg bead slides on a straight fric- tionless wire with a velocity of 4.08 cm/s to the right, as shown. The bead collides elas- tically with a larger 0.628 kg bead initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.66 cm/s. . 376 kg 4 . 08 cm / s . 628 kg Find the distance the larger bead moves along the wire in the first 5.5 s following the collision. Correct answer: 15 . 6088 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. Δ x = v Δ t Given: Let to the right be positive: m 1 = 0 . 376 kg v 1 ,i = +4 . 08 cm / s m 2 = 0 . 628 kg v 1 ,f = − . 66 cm / s t = 5 . 5 s Solution: v 2 ,f = m 1 v 1 ,i − m 1 v 1 ,f m 2 = (0 . 376 kg)(4 . 08 cm / s) . 628 kg − (0 . 376 kg)( − . 66 cm / s) . 628 kg = 2 . 83796 cm / s to the right. Thus Δ x = (2 . 83796 cm / s)(5 . 5 s) = 15 . 6088 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 head-on collision, the final velocity of particle 2 v ′ 2 is given by 1. v ′ 2 = 2 v 1 2. v ′ 2 = 4 v 1 3 3. v ′ 2 = 0 4. v ′ 2 = v 1 correct 5. v ′ 2 = v 1 3 6. v ′ 2 = 2 v 1 3 7. v ′ 2 = v 1 4 8. v ′ 2 = 3 v 1 4 9. v ′ 2 = 5 v 1 3 10. v ′ 2 = v 1 2 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 – HETTIGER, CHRISTOF – 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 head-on collision between the hammer and the ball....
View Full Document

This note was uploaded on 05/09/2010 for the course PHYSICS PHY 301k taught by Professor Turner during the Spring '10 term at University of Texas-Tyler.

Page1 / 6

hw 22 - homework 22 – HETTIGER, CHRISTOF – Due: Mar 22...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online