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Chapter 8: Problem 17

The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A 30 -caliber bullet has mass 0.00720 \(\mathrm{kg}\) and a speed of 601 \(\mathrm{m} / \mathrm{s}\) relative to the muzzle when fired from a rifle that has mass 2.80 \(\mathrm{kg}\) . The loosely held rifle recoils at a speed of 1.85 \(\mathrm{m} / \mathrm{s}\) relative to the earth. Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle.

Short Answer

Expert verified
Momentum of gases is 0.8528 kg鈰卪/s.

Step by step solution

01

Define Conservation of Momentum

The conservation of momentum states that the total momentum before firing is equal to the total momentum after firing. Since initially everything is at rest, the total initial momentum is zero. Therefore, the total final momentum of the bullet, rifle, and gases must also be zero.
02

Calculate Momentum of the Bullet

The momentum of the bullet can be calculated using the formula: \( p_{ ext{bullet}} = m_{ ext{bullet}} \cdot v_{ ext{bullet}} \). Substituting the values: \( p_{ ext{bullet}} = 0.00720 \cdot 601 = 4.3272 \, \text{kg}\cdot\text{m/s} \).
03

Calculate Momentum of the Rifle

The momentum of the rifle is given by: \( p_{ ext{rifle}} = m_{ ext{rifle}} \cdot v_{ ext{rifle}} \). Substituting the values: \( p_{ ext{rifle}} = 2.80 \cdot (-1.85) = -5.18 \, \text{kg}\cdot\text{m/s} \). Note the negative sign indicates the rifle's momentum opposes the bullet's motion.
04

Apply Conservation of Momentum to Solve for Gas Momentum

According to conservation of momentum, the sum of the momenta of the bullet, rifle, and gases should be zero. Therefore, the momentum of the gases is given by: \( p_{ ext{gases}} = -(p_{ ext{bullet}} + p_{ ext{rifle}}) \). Substituting the momenta calculated earlier: \( p_{ ext{gases}} = -(4.3272 - 5.18) = 0.8528 \, \text{kg}\cdot\text{m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum
Momentum is a fundamental concept in physics that is defined as the product of an object's mass and its velocity. This concept is crucial because it allows us to understand how objects interact in motion. In our exercise, the momentum of various components鈥攂ullet, rifle, and gases鈥攚as analyzed. Each component has its momentum described by the formula:
  • Momentum (\( p \)) = Mass (\( m \)) \( \times \) Velocity (\( v \))
Different objects have different masses and velocities, thus contributing differently to the system's total momentum. It's important to remember:
  • Momentum is a vector quantity, meaning it has both magnitude and direction.
  • In a closed system, momentum is conserved, meaning the total momentum before an event equals the total momentum after.
In our example, since initially the system (rifle, bullet, and gases) was at rest, the initial momentum was zero. Hence, the final momentum also equals zero.
Recoil
Recoil is the backwards movement experienced by an object when it ejects another object forward. This phenomenon is commonly observed in firearms. Here, when the bullet is fired forward, the rifle experiences a backward recoil.
The concept of recoil can be explained through Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. When a bullet is fired:
  • The bullet moves forward with a certain momentum.
  • In reaction, the rifle moves backward with an equal and opposite momentum.
The backward velocity of the rifle is what we call recoil velocity. In this exercise, the rifle had a recoil speed of 1.85 m/s, highlighting how recoil is a direct result of the momentum conservation principle.
Propellant Gases
Propellant gases play a significant role in the firing mechanism of firearms. When the bullet is fired, these gases are expelled along with it. Their momentum contributes significantly to the backward recoil and needs to be accounted for in momentum calculations.
In our solution, the momentum of the propellant gases was calculated last, highlighting their role in ensuring the conservation of momentum. Despite not being visible, they have:
  • A notable mass, even though it is always less than the bullet.
  • A high velocity, which is why their momentum is not negligible.
The conservation of momentum allows us to calculate the momentum of these gases by accounting for the momentum of the rifle and bullet. This ensures that all components contribute accurately to the conservation equations.
Coordinate System
In analyzing physical problems, particularly in mechanics, it's critical to establish a reference frame or coordinate system. This allows observers to consistently analyze and interpret motion relative to a fixed point.
For this problem, the coordinate system is key in defining velocities and directions:
  • Since we consider a coordinate system attached to the earth, all velocities and momenta are measured relative to the earth.
  • The negative momentum of the rifle indicates that its recoil is in the opposite direction to the bullet's motion.
Using a consistent coordinate system allows physicists to apply laws of motion uniformly and derive accurate results, such as determining the direction and magnitude of the recoil or ensuring momentum is conserved throughout various events.

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Most popular questions from this chapter

A 45.0 -kg woman stands up in a 60.0 -kg canoe 5.00 \(\mathrm{m}\) long. She walks from a point 1.00 \(\mathrm{m}\) from one end to a point 1.00 \(\mathrm{m}\) from the other end (Fig. P8.106). If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?

At a classic auto show, a \(840-\mathrm{kg} 1955\) Nash Metropolitan motors by at 9.0 \(\mathrm{m} / \mathrm{s}\) , followed by a \(1620-\mathrm{kg} 1957\) Packard Clipper purring past at 5.0 \(\mathrm{m} / \mathrm{s}\) . (a) Which car has the greater kinetic energy? What is the ratio of the kinetic energy of the Nash to that of the Packard? (b) Which car has the greater magnitude of momentum? What is the ratio of the magnitude of momentum of the Nash to that of the Packard? (c) Let \(F_{\mathrm{N}}\) be the net force required to stop the Nash in time \(t,\) and let \(F_{\mathrm{P}}\) be the net force required to stop the Packard in the same time. Which is larger: \(F_{\mathrm{N}}\) or \(F_{\mathrm{P}} ?\) What is the ratio \(F_{\mathrm{N}} / F_{\mathrm{P}}\) of these two forces? (d) Now let \(F_{\mathrm{N}}\) be the net force required to stop the Nash in a distance \(d,\) and let \(F_{\mathrm{P}}\) be the net force required to stop the Packard in the same distance. Which is larger: \(F_{\mathrm{N}}\) or \(F_{\mathrm{P}} ?\) What is the ratio \(F_{\mathrm{N}} / F_{\mathrm{P}}\)?

Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 3.0 \(\mathrm{m}\) long and has mass 20.0 \(\mathrm{kg} .\) Burt has mass 30.0 \(\mathrm{kg}\) and Ernie has mass 40.0 \(\mathrm{kg} .\) Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it. Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log.

BIO Biomechanics. The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 \(\mathrm{ms}\) . (This number can also vary, depending on the racket and swing.) We shall assume a 30.0 -ms contact time for this exercise. The fastest-known served tennis ball was served by "Big Bill" Tilden in \(1931,\) and its speed was measured to be 73.14 \(\mathrm{m} / \mathrm{s}\) . (a) What impulse and what force did Big Bill exert on the tennis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of \(55 \mathrm{m} / \mathrm{s},\) what force and what impulse did he exert on the ball, assuming only horizontal motion?

A 1200 -kg station wagon is moving along a straight highway at 12.0 \(\mathrm{m} / \mathrm{s}\) . Another car, with mass 1800 \(\mathrm{kg}\) and speed \(20.0 \mathrm{m} / \mathrm{s},\) has its center of mass 40.0 \(\mathrm{m}\) ahead of the center of mass of the station wagon (Fig. E8.54). (a) Find the position of the center of mass of the system consisting of the two automobiles. (b) Find the magnitude of the total momentum of the system from the given data. (c) Find the speed of the center of mass of the system. (d) Find the total momentum of the system, using the speed of the center of mass. Compare your result with that of part (b).
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