Explanation Given information: Velocity of the shell, v m = 1500 f t s − 1 Weight of the shell, m g = 18 l b Velocity of the cannon, v M → Weight of the cannon, M g = 3000 l b The cannon rests on a horizontal surface and is free to move horizontally. Assume that the barrel of the cannon is rigidly attached to the frame (no recoil mechanism ) and the shell leaves the barrel 5 m s after firing. Concept used: General impulse-momentum principal, m v 1 + ∑ I m p 1 → 2 = m v 2 m v 1 − Initial momentum m v 2 − Final momentum ∑ I m p 1 → 2 − Impulse of the force during the time considered. Calculation: Impulse momentum diagram. g = 32.174 f t s − 2 M g = 3000 l b M = 3000 l b 32.174 f t s − 2 = 93.24 l b f t s − 2 m g = 18 l b m = 18 l b 32 To determine (b) The resultant of the vertical impulsive force exerted by the ground on the cannon.

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ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Chapter 13.3, Problem 13.153P

To determine

(a)

The recoil velocity of the canon.

To determine

(b)

The resultant of the vertical impulsive force exerted by the ground on the cannon.

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Chapter 13.3, Problem 13.152P

Chapter 13.3, Problem 13.154P

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(a) The recoil velocity of the canon.

Solution Summary: The author analyzes the recoil velocity of the canon. The cannon rests on a horizontal surface and is free to move.