PHY 303K Homework 14

# PHY 303K Homework 14 - homework 14 – FIERRO JEFFREY –...

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Unformatted text preview: homework 14 – FIERRO, JEFFREY – Due: Mar 18 2008, 11:00 pm 1 Question 1, chap 10, sect 99. part 1 of 2 10 points A bullet of mass 7 g moving with an initial speed 400 m / s is fired into and passes through a block of mass 8 kg, as shown in the figure. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 786 N / m. 8 kg 400 m / s 7 g 786 N / m v 2 . 7 cm Assume the bullet’s speed to be so large for all practical purpose we can treat the problem in two steps: 1) the bullet passes through the block and the block picks up an speed 2) the moving block compresses the speed. Step 1 conserves the momentum, while step 2 conserves the energy. If the spring is compressed by the block with a distance 2 . 7 cm to the right after the bullet passed through it, find the speed v at which the bullet emerges from the block. Correct answer: 94 . 1405 m / s (tolerance ± 1 %). Explanation: Let : k = 786 N / m , x = 2 . 7 cm , m = 7 g , M = 8 kg , v = 400 m / s , and V = Block ′ s velocity From conservation of energy 1 2 M V 2 = 1 2 k x 2 . (1) Hence, V = radicalbigg k x 2 M = radicalBigg (786 N / m) (2 . 7 cm) 2 (8 kg) = 0 . 267627 m / s . Then from momentum conservation mv = M V + mv , or (2) v = v − M V m = (400 m / s) − (8 kg) (0 . 267627 m / s) (7 g) (0 . 001 kg / g) = 94 . 1405 m / s . Question 2, chap 10, sect 99. part 2 of 2 10 points Find the magnitude of the energy lost in the collision. Correct answer: 528 . 695 J (tolerance ± 1 %). Explanation: Using Eq. 1, the energy lost is Δ E = K f + U f − K i = mv 2 2 + k x 2 2 − mv 2 i 2 = (0 . 007 kg) (94 . 1405 m / s) 2 2 + (786 N / m) (0 . 027 m) 2 2 − (0 . 007 kg) (400 m / s) 2 2 = (31 . 0185 J) + (0 . 286497 J) − (560 J) = − 528 . 695 J | Δ E | = 528 . 695 J . Question 3, chap 10, sect 2. part 1 of 1 10 points homework 14 – FIERRO, JEFFREY – Due: Mar 18 2008, 11:00 pm 2 A circular hole of diameter 17 . 3 cm is cut out of a uniform square of sheet metal having sides 34 . 6 cm, as in the figure. d l l What is the distance between the center of mass and the center of the square? Correct answer: 2 . 98878 cm (tolerance ± 1 %). Explanation: Choose the center of the square as the ori- gin. Let the radius of the circle be r . Using the formula for center of mass, we have x cm = ∑ m i x i ∑ m i and likewise in y direction. Notice that the side of the square has length 4 r . Thinking of the absence of the circle as being a negative mass element there, we have, assuming mass density of ρ , x cm = ρ (4 r ) 2 (0) − r ρπ r 2 ρ (4 r ) 2 − ρπ r 2 = parenleftbigg − π 16 − π parenrightbigg r . The calculation is the same in y direction....
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PHY 303K Homework 14 - homework 14 – FIERRO JEFFREY –...

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