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Chapter 8: Q8.5-34PE (page 290)

A battleship that is6.00×107kg and is originally at rest fires a1100-kg artillery shell horizontally with a velocity of575m/s . (a) If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ship’s recoil. Calculate its recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship and the shell). This energy is less than the energy released by the gun powder—significant heat transfer occurs.

Short Answer

Expert verified

(a) 0.0105m/s

(b) 1.82x108J

Step by step solution

01

Definition of kinetic energy

Kinetic energy is the energy that an item or particle has as a result of its movement. When work is done on an object by exerting a net force, the object accelerates and gains kinetic energy as a result.

02

Given data

ma=6x107kgmb=1100kgUa=0m/sUb=-575m/s

03

Recoil velocity

By putting all the value into the equation we get

MTVb=M1V1a+M2V2a0=M1V1a+M2V2aV1a=-M2V2aM1V1a=-1100-5756×107V1a=0.0105 m/s

Recoil velocity of the system is 0.0105m/s.

04

increase in internal Kinetic energy:

The kinetic energy increase will be the difference of the final kinetic energy to the initial kinetic energy

KEincrease=KEf-KEiKEincrease=12m1v1f2+12m2v2f2-12mivi2KEincrease=126×1070.01052+1100-5252-0KElost=1.82×108  J

The kinetic energy increased is 1.82x108J.

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