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Chapter 8: Problem 19

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 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
The momentum of the propellant gases is 0.8528 kg路m/s.

Step by step solution

01

Understand the Concept of Conservation of Momentum

The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act upon it. In this problem, we need to use this principle to find the momentum of the propellant gases. The system consists of the bullet, rifle, and gases.
02

Calculate the Bullet's Momentum

To calculate the bullet's momentum, use the formula:\[ p_{bullet} = m_{bullet} \times v_{bullet} \]where \( m_{bullet} = 0.00720 \, \text{kg} \) and \( v_{bullet} = 601 \, \text{m/s} \). Substitute these values to find \( p_{bullet} \).\[ p_{bullet} = 0.00720 \, \text{kg} \times 601 \, \text{m/s} = 4.3272 \, \text{kg} \cdot \text{m/s} \].
03

Calculate the Rifle's Momentum

Similarly, calculate the rifle's momentum using its mass and recoil speed:\[ p_{rifle} = m_{rifle} \times v_{rifle} \]where \( m_{rifle} = 2.80 \, \text{kg} \) and \( v_{rifle} = -1.85 \, \text{m/s} \) (the negative sign indicates the opposite direction to the bullet's motion).\[ p_{rifle} = 2.80 \, \text{kg} \times (-1.85 \, \text{m/s}) = -5.18 \, \text{kg} \cdot \text{m/s} \].
04

Set Up the Conservation of Momentum Equation

According to conservation of momentum:\[ p_{initial} = p_{final} \]Initially, the system is at rest, so \( p_{initial} = 0 \). Thus,\[ 0 = p_{bullet} + p_{rifle} + p_{gases} \]Solve for the momentum of the gases, \( p_{gases} \):\[ p_{gases} = -(p_{bullet} + p_{rifle}) \].
05

Solve for the Momentum of the Propellant Gases

Substitute the known values into the equation from Step 4:\[ p_{gases} = - (4.3272 \, \text{kg} \cdot \text{m/s} - 5.18 \, \text{kg} \cdot \text{m/s}) \]\[ p_{gases} = - (4.3272 \, \text{kg} \cdot \text{m/s} + (-5.18 \, \text{kg} \cdot \text{m/s})) \]\[ p_{gases} = - (-0.8528 \, \text{kg} \cdot \text{m/s}) = 0.8528 \, \text{kg} \cdot \text{m/s} \].

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Key Concepts

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

Physics Problem Solving
Approaching physics problems like the one analyzed here demands keen understanding and application of fundamental principles. First, identify what is given and what needs to be found. The problem states that a rifle and bullet system involves gases that need momentum calculation. By analyzing the scenario, one can recognize it's governed by the conservation of momentum.

Using a systematic approach, break down the problem into manageable parts:
  • Understand the key principles or laws involved - here, conservation of momentum.
  • Accurately note all given quantities such as masses and velocities.
  • Apply the relevant equations methodically, as done with momentum calculations for each part of the system: rifle, bullet, and gases.

Being systematic helps ensure each aspect of the problem is thoroughly considered, reducing errors and improving understanding.
Kinematics
Kinematics focuses on the motion of objects without considering the forces that cause this motion. In the scenario of the bullet fired from a rifle, kinematics helps calculate how fast and in which direction each object is moving.

Key kinematic quantities in this problem include:
  • Velocity, which is how fast and in what direction an object is moving.
  • Relative velocity, which relates the speed of the bullet to the rifle and to the earth.

For example, the bullet's speed is given relative to the muzzle, highlighting the importance of reference frames in kinematics. Understanding the difference in these frames is crucial for solving this problem accurately.
Impulse and Momentum
Impulse and momentum are closely related concepts in physics. In your exercise, momentum plays a core role in understanding how the rifle and bullet move upon firing.

Momentum (\( p \)) is the product of an object's mass and velocity (\( p = m imes v \)). Conservation of momentum states that within a closed system like the rifle and bullet scenario, the total momentum before and after an event must remain unchanged if no external forces act on it.

Impulse can change an object's momentum. It鈥檚 the force applied over time (\( J = F imes ext{time} \)). In this scenario, the focus is primarily on how the gases' momentum complements the bullet and rifle momenta to satisfy the conservation law. By understanding how impulse can change momentum, recognizing the effects of the propellant gases becomes more accessible.

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Most popular questions from this chapter

A soldier on a firing range fires an eight-shot burst from an assault weapon at a full automatic rate of 1000 rounds per minute. Each bullet has a mass of 7.45 \(\mathrm{g}\) and a speed of 293 \(\mathrm{m} / \mathrm{s}\) relative to the ground as it leaves the barrel of the weapon. Calculate the average recoil force exerted on the weapon during that burst.

A blue puck with mass 0.0400 \(\mathrm{kg}\) , sliding with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}\) on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass \(m\) , initially at rest. After the collision, the velocity of the blue puck is 0.050 \(\mathrm{m} / \mathrm{s}\) in the same direction as its initial velocity. Find (a) the velocity (magnitude and dircction) of the red puck after the collision; and (b) the mass \(m\) of the red puck.

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a \(0.400-\mathrm{kg}\) ball that is traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s}\) . Your mass is 70.0 \(\mathrm{kg}\) . (a) If you catch the ball, with what speed do you and the ball move after-ward? (b) If the ball hits you and bounces off your chest, so after-ward it is moving horizontally at 8.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, what is your speed after the collision?

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass \(70.0 \mathrm{kg},\) is moving to the right at 2.00 \(\mathrm{m} / \mathrm{s}\) , while the other. of mass 65.0 \(\mathrm{kg}\) , is moving to the left at 2.50 \(\mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the velocity of these skaters just after they collide?

An astronaut in space cannot use a scale or balance to weigh objects because there is no gravity. But she does have devices to measure distance and time accurately. She knows her own mass is 78.4 \(\mathrm{kg}\) , but she is unsure of the mass of a large gas canister in the airless rocket. When this canister is approaching her at 3.50 \(\mathrm{m} / \mathrm{s}\) , she pushes against it, which slows it down to 1.20 \(\mathrm{m} / \mathrm{s}\) (but does not reverse it) and gives her a speed of 2.40 \(\mathrm{m} / \mathrm{s} .\) What is the mass of this canister?
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