Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a \(4800kg\)satellite uses this method to separate from the \(1500kg\)remains of its launcher, and that \(5000J\)of kinetic energy is supplied to the two parts. What are their subsequent velocities using the frame of reference in which they were at rest before separation?

The final velocity is \( - 0.704m/s\).The launcher here after the explosion moves towards right side with velocity \(2.25 m/s\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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.

The mass of first object is \(4800kg\). The mass of second object (Launcher) is \(1500kg\).The velocity before the explosion is zero.The velocity after the explosion is not known. The collision is elastic hence the total momentum will be conserved.

Step 2: Final velocity of the launcher and satellite

By putting all the value into the equation we get

\(\begin{aligned}{M_T}{V_i} &= {M_1}{V_{1a}} + {M_2}{V_{2a}}\\{\cancel{{{M_T}V}}_i} &= {M_1}{V_{1a}} + {M_2}{V_{2a}}\left( {{V_i} = 0} \right)\\ - {M_1}{V_{1a}} &= {M_2}{V_{2a}}\end{aligned}\)

This is the relation which we can use later on the increase in the kinetic energy is given as

\(\begin{aligned}K{E_{in}} &= K{E_f} - K{E_i}\\K{E_{in}} &= \left( {K{E_{1a}} + K{E_{2a}}} \right) - \cancel{{K{E_i}}}\\K{E_{in}} &= \left( {\dfrac{1}{2}{m_1}{v_{1a}}^2 + \dfrac{1}{2}{m_2}{v_{2a}}^2} \right) - 0\\K{E_{in}} &= \dfrac{1}{2}\left( {{m_1}{v_{1a}}^2 + {m_2}{{\left( {\dfrac{{ - {m_1}{v_{1a}}}}{{{m_2}}}} \right)}^2}} \right)....from\,eq\,(1)\end{aligned}\)

\(\begin{aligned}2K{E_{in}} &= {m_1}{v_{1a}}^2 + \dfrac{{{m_1}^2{v_{1a}}^2}}{{{m_2}}}\\2K{E_{in}} &= {v_{1a}}^2\left( {{m_1} + \dfrac{{{m_1}^2}}{{{m_2}}}} \right)\end{aligned}\)

\(\begin{aligned}{v_{1a}} &= \sqrt {\dfrac{{2K{E_{in}}}}{{\left( {{m_1} + \dfrac{{{m_1}^2}}{{{m_2}}}} \right)}}} \\{v_{1a}} &= \sqrt {\dfrac{{2\left( {5000} \right)}}{{\left( {4800 + \dfrac{{{{4800}^2}}}{{1500}}} \right)}}} \\{v_{1a}} &= \pm 0.704\,m/s\\{v_{1a}} &= - 0.704\,m/s\end{aligned}\)

Hence the final velocity is \( - 0.704m/s\)

To calculate for launcher, we will use equation (1)

\(\begin{aligned} - {M_1}{V_{1a}} &= {M_2}{V_{2a}}\\{V_{2a}} &= \dfrac{{ - {M_1}{V_{1a}}}}{{{M_2}}}\\{V_{2a}} &= \dfrac{{ - \left( {4800} \right)\left( { - 0.704} \right)}}{{\left( {1500} \right)}}\\{V_{2a}} &= 2.25\,m/s\end{aligned}\)

The launcher here after the explosion moves towards right side with velocity \(2.25 m/s\).

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College Physics (Urone)

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Short Answer

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Step 1: Definition of Kinetic Energy

Step 2: Final velocity of the launcher and satellite

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College Physics (Urone)

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Step 1: Definition of Kinetic Energy

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