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A 12.0-kg shell is launched at an angle of 55.0o above the

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 110P Chapter 8

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 110P

A 12.0-kg shell is launched at an angle of 55.0o above the horizontal with an initial speed of 150 m/s. At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how much energy was released in the explosion?

Step-by-Step Solution:

Solution 110P Step 1: W can resolve the components of velocity as vertical and horizontal. Vertical component of velocity, v verticalv sin Provided, it is making an angle, = 55° with the horizontal Then, v vertical150 m/s × sin 55 = 150 m/s× 0.8192 = 122.87 m/s Since it is moving against the gravitational force, we can write, v - u = - 2 gS Where, v - final velocity, u - initial velocity, g - acceleration due to gravity And S - distance travelled In vertical motion, v = 0 m/s at a point 2 2 2 2 Therefore, 0 - 122.87 m /s = - 2 × 9.8 m/s × S The vertical displacement, S = 15098 m /s / 19.6 m/s = 770.26 m 2 Similarly, v = u - gt 0 = 122.87 m/s - 9.8 t Rearranging, t = 122.87 m/s / 9.8 m/s = 12.54 s 2 The time to reach maximum height will be, t = 12.54 s Therefore, the horizontal distance for explosion, H = v horizontal vhorizontal cos Provided, it is making an angle, = 55° with the horizontal Then, v horizontal50 m/s × cos 55 = 150 m/s× 0.5736 = 86.04 m/s Therefore, the horizontal distance for explosion, H = 86.04 m/s×12.54 s = 1078.9 m Step 2: During explosion, the shell breaks into two and one piece will be 3 time heavier than the other. Then the masses will be, m = 9 k1and m = 3 kg 2 Suppose, heavier mass is moving to the same position. According to the law of conservation of momentum, initial momentum = final momentum Speed of heavier mass is, v = - 86.04 m/s Therefore, (12 kg × 86.04 m/s) = (- 9 kg × 86.04 m/s) + (3 kg × v lighter 1032.48 kg m/s + 774.36 kg m/s = (3 kg × v lighter 1806.84 kg m/s = (3 kg × v lighter Rearranging, we get, v lighter806.84 kg m/s / 3 kg = 602.28 m/s Horizontal velocity of lighter mass is 602.28 m/s

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Chapter 8, Problem 110P is Solved
Step 4 of 4

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The answer to “A 12.0-kg shell is launched at an angle of 55.0o above the horizontal with an initial speed of 150 m/s. At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how much energy was released in the explosion?” is broken down into a number of easy to follow steps, and 80 words. This textbook survival guide was created for the textbook: University Physics, edition: 13. University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: shell, launched, point, fragment, heavier. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 110P from 8 chapter was answered, more than 276 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 110P from chapter: 8 was answered by , our top Physics solution expert on 05/06/17, 06:07PM.

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