/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 66 Old naval ships fired 10 kg cann... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Q. 66 (page 290)

Old naval ships fired 10 kg cannon balls from a 200 kg cannon. It was very important to stop the recoil of the cannon, since otherwise the heavy cannon would go careening across the deck of the ship. In one design, a large spring with spring constant 20,000 N/m was placed behind the cannon. The other end of the spring braced against a post that was firmly anchored to the ship鈥檚 frame. What was the speed of the cannon ball if the spring compressed 50 cm when the cannon was fired?

Short Answer

Expert verified

Speed of the cannon ball is 100 m/s.

Step by step solution

01

Step 1. Given information is :Mass of cannon M = 200 kgMass of the cannon ball m = 10 kgSpring constant k = 2000 N/mDistance by which spring is compressed∆x=50 cm=0.5m

We need to find out the speed of the cannon ball if the spring if the compressed by 50 cm when the cannon was fired

02

Step 2. Using the Law of conservation of momentum and Law of conservation of Energy

Let velocity of cannon = V

Velocity of cannon ball = v

Using Law of conservation of Momentum,

MV=mvv=MmV

Using Law of conservation of Energy,

12k(鈭�x)2=12MV220000(0.5)2=200V2V2=25V=5m/sAndweknowthat,v=MmVv=20010脳5=100m/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 91影视!

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

You have been asked to design a 鈥渂allistic spring system鈥� to measure the speed of bullets. A spring whose spring constant is k is suspended from the ceiling. A block of mass M hangs from the spring. A bullet of mass m is fired vertically upward into the bottom of the block and stops in the block. The spring鈥檚 maximum compression d is measured.

a. Find an expression for the bullet鈥檚 speed vB in terms of m, M, k, and d.

b. What was the speed of a data-custom-editor="chemistry" 10gbullet if the block鈥檚 mass is data-custom-editor="chemistry" 2.0kgand if the spring, with data-custom-editor="chemistry" k=50N/m, was compressed by data-custom-editor="chemistry" 45cm?

A spaceship of mass 2.0 脳 106 kg is cruising at a speed of 5.0 脳 106 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.0 脳 105 kg, is blown straight backward with a speed of 2.0 脳 106 m/s. A second piece, with mass 8.0 脳 105 kg, continues forward at 1.0 脳106 m/s. What are the direction and speed of the third piece?

A 300gbird flying along at 6.0m/ssees a 10g insect heading straight toward it at a speed of 30m/s. The bird opens its mouth wide and enjoys a nice lunch. What is the bird鈥檚 speed immediately after swallowing?

At what speed do a bicycle and its rider, with a combined mass of 100kg, have the same momentum as a 1500kg car traveling at 5.0m/s?

What is the impulse on a 3.0kgparticle that experiences the force shown in FIGURE EX11.4?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Energy Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Engineering Physics

Read Explanation

Dynamics

Read Explanation

Force

Read Explanation
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.