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Homework 8 Solutions

Homework 8 Solutions - homework 08 ALIBHAI ZAHID Due 4:00...

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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 0 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 0 9 to the left. m v 0 before 36 41 m v 0 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 0 . 2. bardbl vectorv 5 / 41 bardbl = 41 36 v 0 . 3. bardbl vectorv 5 / 41 bardbl = 1476 5 v 0 .

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homework 08 – ALIBHAI, ZAHID – Due: Mar 21 2007, 4:00 am 2 4. bardbl vectorv 5 / 41 bardbl = 205 36 v 0 . 5. bardbl vectorv 5 / 41 bardbl = 41 v 0 . 6. None of these 7. bardbl vectorv 5 / 41 bardbl = 41 5 v 0 . 8. bardbl vectorv 5 / 41 bardbl = 46 36 v 0 . 9. bardbl vectorv 5 / 41 bardbl = 9 v 0 . correct 10. bardbl vectorv 5 / 41 bardbl = 5 v 0 . Explanation: The horizontal component of the momen- tum is conserved, so 0 + m v 0 = 36 41 m v 36 / 41 + 5 41 m v 5 / 41 0 + m v 0 = 36 41 m parenleftBig v 0 9 parenrightBig + 5 41 m v 5 / 41 m v 0 = 36 369 m v 0 + 5 41 m v 5 / 41 5 41 v 5 / 41 = parenleftbigg 369 369 + 36 369 parenrightbigg v 0 5 41 v 5 / 41 = 405 369 v 0 v 5 / 41 = 405 369 41 5 v 0 bardbl vectorv 5 / 41 bardbl = 9 v 0 . Question 3 part 1 of 1 10 points A gadget of mass 20 . 24 kg floats in space without motion. Because of some internal malfunction, the gadget violently breaks up into 3 fragments flying away from each other. The first fragment has mass m 1 = 6 . 45 kg and speed v 1 = 4 . 38 m / s while the second fragment has mass m 2 = 5 . 62 kg and speed v 2 = 2 . 87 m / s. The angle between the veloc- ity vectors vectorv 1 and vectorv 2 is θ 12 = 80 .
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