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Q37PE

Expert-verifiedFound in: Page 290

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 \(2.25 m/s\).

**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.

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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