Homework 8 Solutions - homework 08 ALIBHAI, ZAHID Due: Mar...

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 08 ALIBHAI, ZAHID Due: Mar 21 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Two air blocks with masses 184 g and 184 g are equipped with identical springs ( k = 1560 N / m) . The blocks move toward each other with identical speeds of 1 m / s on a horizontal air track and collide, compressing the springs. 184 g 1 m / s 1560 N / m 184 g 1 m / s 1560 N / m Find the maximum compression of the spring attached to the 184 g mass. Correct answer: 1 . 08604 cm (tolerance 1 %). Explanation: Let : k = 1560 N / m , m 1 = 184 g , m 2 = 184 g , and bardbl vector v 1 bardbl = bardbl vector v 2 bardbl = v = 1 m / s . Note: Only the kinetic energy in the center- of-mass system is equivalent to the potential energy of the compressed springs. The center-of-mass velocity v cm is v cm = m 1 m 2 m 1 + m 2 v = (184 g) (184 g) (184 g) + (184 g) (1 m / s) = 0 m / s . The total kinetic energy of both masses is equal to the total potential of both com- pressed springs. 1 2 k x 2 1 + 1 2 k x 2 2 = 1 2 [ m 1 + m 2 ] [ v 2 v 2 cm ] = 1 2 [ m 1 + m 2 ] v 2 radicalBig x 2 1 + x 2 2 = radicalbigg [ m 1 + m 2 ] v 2 k = braceleftbigg [(184 g) + (184 g)] (1 m / s) 2 (1560 N / m) (1000 g / kg) bracerightbigg 1 / 2 (100 cm / m) = 1 . 5359 cm . Each of the two springs absorb the same amount of potential energy (one-half the total potential energy); i.e. , x = x 1 = x 2 , we have radicalBig x 2 1 + x 2 2 = 2 x, so x = radicalbigg x 2 1 + x 2 2 2 = (1 . 5359 cm) 2 = 1 . 08604 cm . Question 2 part 1 of 1 10 points An object of mass m is moving with speed v to the right on a horizontal frictionless surface, as shown, when it explodes into two pieces. Subsequently, one piece of mass 36 41 m moves with a speed v 36 / 41 = v 9 to the left. m v before 36 41 m v 9 5 41 m v 5 / 41 after What is the speed bardbl vectorv 5 / 41 bardbl of the other piece of the object? 1. bardbl vectorv 5 / 41 bardbl = 77 5 v . 2. bardbl vectorv 5 / 41 bardbl = 41 36 v . 3. bardbl vectorv 5 / 41 bardbl = 1476 5 v . homework 08 ALIBHAI, ZAHID Due: Mar 21 2007, 4:00 am 2 4. bardbl vectorv 5 / 41 bardbl = 205 36 v . 5. bardbl vectorv 5 / 41 bardbl = 41 v . 6. None of these 7. bardbl vectorv 5 / 41 bardbl = 41 5 v . 8. bardbl vectorv 5 / 41 bardbl = 46 36 v . 9. bardbl vectorv 5 / 41 bardbl = 9 v . correct 10. bardbl vectorv 5 / 41 bardbl = 5 v . Explanation: The horizontal component of the momen- tum is conserved, so 0 + mv = 36 41 mv 36 / 41 + 5 41 mv 5 / 41 0 + mv = 36 41 m parenleftBig v 9 parenrightBig + 5 41 mv 5 / 41 mv = 36 369 mv + 5 41 mv 5 / 41 5 41 v 5 / 41 = parenleftbigg 369 369 + 36 369 parenrightbigg v 5 41 v 5 / 41 = 405 369 v v 5 / 41 = 405 369 41 5 v bardbl vectorv 5 / 41 bardbl = 9 v ....
View Full Document

This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 16

Homework 8 Solutions - homework 08 ALIBHAI, ZAHID Due: Mar...

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