Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The USS Montana is a massive battleship with a weight of \(136,634,000 \mathrm{lb}\). It has twelve 16 -inch guns, which are capable of firing \(2700 .-\) lb projectiles at a speed of \(2300 . \mathrm{ft} / \mathrm{s}\). If the battleship fires three of these guns (in the same direction), what is the recoil velocity of the ship?

Short Answer

Expert verified
Answer: The recoil velocity of the USS Montana battleship is approximately 0.136 ft/s.

Step by step solution

01

Calculate the initial momentum

Initially, both the battleship and the projectiles are at rest, so their initial momentum is 0.
02

Calculate the mass and velocity of three projectiles

Each projectile has a mass of \(2700\,\mathrm{lb}\) and is fired at a velocity of \(2300\,\mathrm{ft/s}\). Since there are three guns, we need to find the total mass and momentum of the three projectiles. Total mass of projectiles: \(3 \times 2700\,\mathrm{lb}=8100\,\mathrm{lb}\) Velocity of each projectile is given as \(2300\,\mathrm{ft/s}\).
03

Calculate the final momentum of three projectiles

As per the law of conservation of momentum, the momentum before firing will be equal to the momentum after firing. The final momentum of the three projectiles can be calculated using the formula: Final momentum of projectiles = mass of projectiles × velocity of projectiles Final momentum of projectiles = \(8100\,\mathrm{lb} \times 2300\,\mathrm{ft/s} = 18,630,000\, \mathrm{lb\cdot ft/s}\)
04

Calculate the recoil velocity of the battleship

Now we equate the initial and final momenta and solve for the recoil velocity of the battleship (v). Initial momentum = Final momentum \(136,634,000\,\mathrm{lb} \times v = 18,630,000\,\mathrm{lb\cdot ft/s}\) Solve for v: v = \(\frac{18,630,000\,\mathrm{lb\cdot ft/s}}{136,634,000\,\mathrm{lb}}\) v \(\approx 0.136\,\mathrm{ft/s}\)
05

State the final answer

The recoil velocity of the USS Montana battleship when it fires three of its guns in the same direction is approximately \(0.136\,\mathrm{ft/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Conservation of Momentum
The principle of conservation of momentum is a fundamental concept in physics, asserting that the total momentum of an isolated system remains constant if no external forces act on it. Momentum is a vector quantity, described as the product of an object's mass and velocity. In the case of the USS Montana battleship, when the guns are fired, the momentum of the projectiles moving forward is balanced by the recoil of the ship in the opposite direction, ensuring the system's total momentum remains unchanged. This is because the forces involved are internal to the battleship-projectile system.

Example: When the battleship's guns fire, the momentum gained by the projectiles is equal and opposite to the momentum the battleship gains in the recoil, according to the conservation of momentum.
Momentum Calculation
Calculating momentum is straightforward: simply multiply the mass of the object by its velocity. Momentum has both magnitude and direction, making it a vector quantity. In physics problems, it's essential to consider the direction when performing momentum calculations, as direction affects how momentum values are combined.

Example: For a projectile, its momentum is the product of its mass and its velocity in the direction of motion. If multiple projectiles are moving in the same direction, their momenta are additive. The combined momentum of multiple objects, like the projectiles fired by the battleship, can be found by summing the individual momenta when their directions are the same.
Projectile Motion
Projectile motion pertains to objects thrown or launched into the air and are subject to gravity. The motion of projectiles is typically two-dimensional and can be split into horizontal and vertical components. However, in some physics problems, such as with the recoil velocity calculation for the battleship, only the horizontal component is relevant because there is no vertical motion involved.

Example: The projectiles fired from the battleship's guns exhibit projectile motion, but since the recoil velocity is in the opposite direction of the horizontal component of the projectiles' velocity, only this horizontal direction is considered in the conservation of momentum.
Physics Problem Solving
Tackling physics problems often requires a systematic approach: identify the relevant principles, translate given information into mathematical terms, apply equations correctly, and solve for unknowns. For problems involving recoil velocity, the conservation of momentum is a key principle. It's critical to organize information, such as masses and velocities, calculate the momenta of moving objects, and set up equations that reflect the conservation law.

Example: In solving for the battleship's recoil velocity, steps include calculating the initial momentum (which is zero as the ship and projectiles start at rest), computing the combined momentum of the fired projectiles, and then solving for the unknown recoil velocity of the battleship to satisfy the conservation of momentum.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A garden hose is used to fill a 20 - \(L\) bucket in 1 min. The speed of the water leaving the hose is \(1.05 \mathrm{~m} / \mathrm{s}\). What force acts on the person holding the hose? a) \(0.35 \mathrm{~N}\) b) \(2.1 \mathrm{~N}\) c) \(9.8 \mathrm{~N}\) d) \(12 \mathrm{~N}\) e) \(21 \mathrm{~N}\)

A one-dimensional rod has a linear density that varies with position according to the relationship \(\lambda(x)=c x,\) where \(c\) is a constant and \(x=0\) is the left end of the rod. Where do you expect the center of mass to be located? a) the middle of the rod b) to the left of the middle of the rod c) to the right of the middle of the rod d) at the right end of the rod e) at the left end of the rod

A soda can of mass \(m\) and height \(L\) is filled with soda of mass \(M .\) A hole is punched in the bottom of the can to drain out the soda. a) What is the center of mass of the system consisting of the can and the soda remaining in it when the level of soda in the can is \(h,\) where \(0

A proton with mass \(1.673 \cdot 10^{-27} \mathrm{~kg}\) is moving with a speed of \(1.823 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) toward an alpha particle with mass \(6.645 \cdot 10^{-27} \mathrm{~kg}\), which is at rest. What is the speed of the center of mass of the proton-alpha particle system?

Two point masses are located in the same plane. The distance from mass 1 to the center of mass is \(3.0 \mathrm{~m} .\) The distance from mass 2 to the center of mass is \(1.0 \mathrm{~m}\). What is \(m_{1} / m_{2}\), the ratio of mass 1 to mass 2 ? a) \(3 / 4\) b) \(4 / 3\) c) \(4 / 7\) d) \(7 / 4\) c) \(4 / 7\) d) \(7 / 4\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free