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What gauge pressure in the water mains is necessary if a

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 40P Chapter 10

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 40P

Problem 40P

What gauge pressure in the water mains is necessary if a firehose is to spray water to a height of 15 m?

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ANSWER:Let’s take the respective masses and velocities of the two ball are, m , m , v = 200 m/s and v = 0. 1 2 1 2We know that if the collision is elastic, then the momentum and kinetic energy of the wholesystem will be conserved during the process.So, let’s impose these conditions.Momentum conservation says, m 1 1 m v =2 2v + m 1 1------2 2------------(1)And kinetic energy conservation says, 2 2 1/2 m 1 1 2 + 1/2 m v2 22 = 1/2 m v1 1 + 1/2 m v2 2 -------------(2)Here we know the initial velocities for both the objects and the masses are also known. Thefinal velocities are unknown. But, these are two equations with two unknowns which we canalways solve. a. The incoming ball is much more massive than the stationary ball:Then m >>1m , 2The solutions are, m1m 2 2m 2 v1= m + m × v 1 m + m × v 2--------------------------(3) 1 2 1 2 v = 2m 1 × v + m2 m 1× v -----------------------------(4) 2 m1+ m 2 1 m1+ m 2 2So, now let’s put the values in equation (3) and (4) to get the final velocities of the objects.\n m1 v1= m × 200 + 0 = 200 m/s. 1And 2×m v2= 1 × 200 + 0 = 400 m/s. m1CONCLUSION:So, if the incoming mass is much more massive than the stationary one, then the speeds of theball will be 200 m/s and 400 m/s. b. The stationary ball is much more massive than the incoming ball:Then, m >2 m and 1e can ignore the mass of the 1st ball. m2 2m 2 v1= m × 200 + m × 0 = 200 m/s 2 2 v = 2m1 × 200 + 0 = 0 2 m 2Here we have done a little approximation as, 2m1 m 2 goes to zero, as the the denominator is a very big number and the numerator isextremely small compared to the denominator. This approximation leads us to get the zeroover there.CONCLUSION:So, trivially the incoming ball will reverse it’s direction with the same magnitude and the ballat rest will not be affected at all.

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Chapter 10, Problem 40P is Solved
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Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

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