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Chapter 8: Problem 73

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.

Short Answer

Expert verified
The average recoil force exerted is 29.2 N.

Step by step solution

01

Determine Firing Rate

Since the firing rate is 1000 rounds per minute, for an 8-shot burst, it will take a time of \(t = \frac{8}{1000} \times 60\) seconds. Calculate this time to understand the duration of the burst.
02

Calculate Total Bullet Momentum

The momentum of a bullet can be found using \(p = mv\), where \(m\) is the mass of the bullet (7.45 g = 0.00745 kg) and \(v\) is the speed (293 m/s). Calculate the momentum of one bullet, then multiply by 8 for the total burst momentum.
03

Calculate Average Recoil Force

The average force can be determined using the impulse-momentum theorem, \(F_{avg} = \frac{\Delta p}{\Delta t}\), where \(\Delta p\) is the change in momentum and \(\Delta t\) is the time duration of the burst determined in Step 1. Use these values to calculate the average recoil force.

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Key Concepts

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

Impulse-Momentum Theorem
The Impulse-Momentum Theorem is a fundamental principle in physics that connects impulse and momentum. It states that the change in momentum of an object is equal to the impulse applied to it. The formula for this theorem is \( F_{avg} \cdot \Delta t = \Delta p \), where \( F_{avg} \) is the average force applied, \( \Delta t \) is the time over which the force acts, and \( \Delta p \) is the change in momentum. In other words, this theorem tells us that to change the momentum of an object, a force must be applied over a period of time.

For the soldier's weapon firing bullets, we focus on calculating the change in momentum \( \Delta p \), which we've done by determining the momentum for all bullets fired in one burst. Once the bullet's momentum is calculated, using the theorem helps us find the average force (recoil force) exerted by the gun on the soldier by dividing the change in momentum by the time it takes to shoot the burst.
Bullet Momentum
Momentum is a key concept in understanding the motion of objects. It is defined as the product of an object's mass and its velocity. The formula for momentum \( p \) is given by \( p = m \times v \), where \( m \) is the mass of the object and \( v \) is the velocity. In our context, each bullet shot from the weapon has momentum due to its mass and speed.

For this exercise, we first convert the mass of the bullet from grams to kilograms to keep the units consistent with the velocity, which is in meters per second. Using the formula, we find the momentum of one bullet, and then multiply by the number of bullets in the burst (8 bullets in this case) to find the total momentum. Bullet momentum directly contributes to the recoil experienced by the weapon, as it represents the force exerted to propel the bullets outward.
Firing Rate Calculation
The firing rate is crucial for determining the time duration for which a force is exerted. It is expressed in terms of rounds fired per minute. In this exercise, we have a firing rate of 1000 rounds per minute. However, for an 8-shot burst, we need to convert this rate into the time it takes to fire these 8 rounds to use in further calculations.

The time \( \Delta t \) can be calculated using the formula \( \Delta t = \frac{8}{1000} \times 60 \). This gives us the time duration in seconds. The calculated time helps us in using the Impulse-Momentum Theorem to find the average recoil force by providing the duration over which the force acts. Firing rate impacts not only how quickly a gun can shoot but also the succession in which the recoil force is applied.

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Most popular questions from this chapter

On a very muddy football field, a \(110-\mathrm{kg}\) linebacker tackles an \(85-\mathrm{kg}\) halfback. Immediately before the collision, the line-backer is slipping with a velocity of 8.8 \(\mathrm{m} / \mathrm{s}\) north and the halfback is sliding with a velocity of 7.2 \(\mathrm{m} / \mathrm{s}\) east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t}$$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k v\) . This gives, since \(F_{\text { ext }}=m g .\) $$m g=m \frac{d v}{d t}+v(k v)$$ Or, dividing by \(k\) $$x g=x \frac{d v}{d t}+v^{2}$$ This is a differential equation that has a solution of the form \(v=a t,\) where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v,\) find the acceleration \(a\) . (b) Find the distance the raindrop has fallen in \(t=3.00 \mathrm{s}\) . (c) Given that \(k=2.00 \mathrm{g} / \mathrm{m}\) , find the mass of the raindrop at \(t=3.00 \mathrm{s}\) . For many more intriguing aspects of this problem, see \(\mathbf{K} .\) S. Krane, Amer Jour. Phys, Vol. 49\((1981)\) pp. \(113-117\)

A ball with mass \(M,\) moving horizontally at 5.00 \(\mathrm{m} / \mathrm{s}\) , collides elastically with a block with mass 3 \(\mathrm{M}\) that is initially hanging at rest from the ceiling on the end of a 50.0 -m wire. Find the maximum angle through which the block swings after it is hit.

A \(20.00-\mathrm{kg}\) lead sphere is hanging from a hook by a thin wire 3.50 \(\mathrm{m}\) long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a \(5.00-\mathrm{kg}\) steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

(a) What is the magnitude of the momentum of a \(10,000-k g\) truck whose speed is 12.0 \(\mathrm{m} / \mathrm{s} ?\) (b) What speed would a \(2,000-\mathrm{kg}\) SUV have to attain in order to have (i) the same momentum? (ii) the same kinetic energy?
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