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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# 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 $$2.25 m/s$$.

See the step by step solution

## 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 $$2.25 m/s$$.