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CP A 20.00-kg lead sphere is hanging from a hook by a thin

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

Solution for problem 88P 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 88P

CP? A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.00-kg steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

Step-by-Step Solution:

Solution 88P Step 1: Data given Mass of sphere m =120.0 kg Mass of dart m 2 5.0 kg Length of wire d = 2.80 m We need to find the initial speed of the dart From the above explanation of the problem we know it is an inelastic collision We shall use the law of conservation of energy to find the velocity of the dart We have KE = PE 1/2 mv = mgh After the collision the mass will m = m 1 m = 20.0 kg + 5.0 kg = 25.0 kg Substituting this we get 1/2 × 25.0 kg v = 25.0 kg × 9.8 × 2 × 2.80 m Solving this we get v = 9.8×2×2.80 m 1/2 v = 10.47 m/s Hence we have the velocity of the sphere and dart after the collision We have the momentum of the sphere and dart after collision as = mv = 25.0 kg × 10.4 m/s = 260 m/s kg Since we need to find the initial velocity of the dart to hit the sphere we have 260 m/s kg = m v + m v 2 2 1 1 Since the sphere is at rest while collision v = 0 1 Hence we have 260 m/s kg = m v 2 2 \ 260 m/s kg = 5.0 kg × v 2 v 2 ( 260 m/s kg) /5.0 kg v 2 52.0 m/s Hence the dart should hit the sphere with velocity 52.0 m/sto make a complete circle

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Chapter 8, Problem 88P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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CP A 20.00-kg lead sphere is hanging from a hook by a thin