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Chapter 7: Problem 18

A rifle has a mass of \(4.5 \mathrm{kg}\) and it fires a bullet of mass $10.0 \mathrm{g}\( at a muzzle speed of \)820 \mathrm{m} / \mathrm{s} .$ What is the recoil speed of the rifle as the bullet leaves the gun barrel?

Short Answer

Expert verified
Answer: The recoil speed of the rifle as the bullet leaves the gun barrel is approximately -1.82 m/s.

Step by step solution

01

Identify the given information

We have been given the following information: - Mass of the rifle, \(m_{rifle} = 4.5 \mathrm{kg}\) - Mass of the bullet, \(m_{bullet} = 10.0 \mathrm{g} = 0.01 \mathrm{kg}\) (converted from grams to kilograms) - The velocity of the bullet after being fired, \(v_{bullet} = 820 \mathrm{m} / \mathrm{s}\) - The initial total momentum of the system is 0, as both the bullet and the rifle are at rest.
02

Apply the conservation of momentum principle

According to the conservation of momentum principle, the total momentum before the bullet is fired should equal the total momentum after the bullet is fired. Mathematically, we can represent this as: \(m_{rifle} v_{rifle_{initial}} + m_{bullet} v_{bullet_{initial}} = m_{rifle} v_{rifle_{final}} + m_{bullet} v_{bullet_{final}}\) Since both the rifle and the bullet are initially at rest, \(v_{rifle_{initial}} = v_{bullet_{initial}} = 0\). Therefore, the equation becomes: \(m_{rifle} v_{rifle_{final}} = -m_{bullet} v_{bullet_{final}}\)
03

Solve for the final velocity of the rifle

We have the equation: \(m_{rifle} v_{rifle_{final}} = -m_{bullet} v_{bullet_{final}}\) To find the final velocity of the rifle, \(v_{rifle_{final}}\), we need to rearrange the equation: \(v_{rifle_{final}} = -\dfrac{m_{bullet} v_{bullet_{final}}}{m_{rifle}}\) Now, let's plug in the given values: \(v_{rifle_{final}} = -\dfrac{(0.01 \mathrm{kg})(820 \mathrm{m}/\mathrm{s})}{4.5 \mathrm{kg}}\) \(v_{rifle_{final}} = -\dfrac{8.2 \mathrm{m}/\mathrm{s}}{4.5}\) \(v_{rifle_{final}} \approx -1.82 \mathrm{m}/\mathrm{s}\).
04

Interpret the result

The recoil speed of the rifle as the bullet leaves the gun barrel is approximately \(-1.82 \mathrm{m}/\mathrm{s}\). The negative sign indicates that the rifle is moving in the opposite direction of the fired bullet.

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