homework 27 ans

# homework 27 ans - homework 27 NOORANI ZOHEB Due Apr 3 2008...

This preview shows pages 1–2. Sign up to view the full content.

homework 27 – NOORANI, ZOHEB – Due: Apr 3 2008, 4:00 am 1 Question 1, chap 10, sect 99. part 1 of 2 10 points Two particles of masses m 1 = 6 . 8 kg and m 2 = 18 . 6 kg are moving toward each other along the x axis with equal speeds 7 . 22 m / s. Specifically, v 1 x = +7 . 22 m / s (particle 1 moves to the right) and v 2 x = - 7 . 22 m / s (particle 2 moves to the left). The particles collide elastically. After the collision, the first particle moves at θ 1 = 90 to its original direction while the second par- ticle is deflected through a smaller angle θ 2 < 90 . x y vectorv 1 vectorv 2 vectorv 1 vectorv 2 θ 1 = 90 θ 2 Find the final speed | v 1 | of the first particle. Correct answer: 10 . 0281 m / s (tolerance ± 1 %). Explanation: In elastic collision, the net momentum vector P and the net kinetic energy K are both con- served. In components, P x = m 1 v 1 - m 2 v 2 = 0 - m 2 v 2 cos θ 2 , (1) P y = 0 = m 1 v 1 - m 2 v 2 sin θ 2 , (2) K = m 1 v 2 1 2 + m 2 v 2 2 2 = m 1 v 2 1 2 + m 2 v 2 2 2 . (3) Given v 1 = v 0 , v 2 = - v 0 , eq. (1) gives v 2 cos θ 2 = parenleftbigg 1 - m 1 m 2 parenrightbigg v 0 (4) while eq. (2) implies v 2 sin θ 2 = m 1 m 2 v 1 ; (5) therefore v 2 2 = ( v 2 cos θ 2 ) 2 + ( v 2 sin θ 2 ) 2 = parenleftbigg 1 - m 1 m 2 parenrightbigg 2 v 2 0 + m 2 1 m 2 2 v 2 1 . (6) Substituting this formula into the energy con- servation eq. (3), we arrive at m 1 + m 2 2 v 2 0 = m 1 2 v 2 1 + m 2 2 bracketleftBigg parenleftbigg 1 - m 1 m 2 parenrightbigg 2 v 2 0 + m 2 1 m 2 2 v 2 1 bracketrightBigg = m 1 ( m 2 + m 1 ) 2 m 2 v 2 1 + ( m 2 - m 1 ) 2 2 m 2 v 2 0 (7) and hence (after a bit of algebra) v 2 1 v 2 0 = 3 m 2 - m 1 m 2 + m 1 . (8) Consequently, v 1 = v 0 × radicalbigg 3 m 2 - m 1 m 2 + m 1 = 10 . 0281 m / s . (9) Question 2, chap 10, sect 99. part 2 of 2 10 points Find the deflection angle θ 2 of the second particle. Correct answer: 38 . 6738 (tolerance ± 1 %).

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern