Question

A 500.0 kg module is attached to a 400.0 kg shuttle craft, which moves at $\mathbf{1000}\frac{\mathbf{m}}{\mathbf{s}}$ relative to the stationary main spaceship. Then a small explosion sends the module backward with speed $\mathbf{100}\mathbf{.}\mathbf{0}\frac{\mathbf{m}}{\mathbf{s}}$ relative to the new speed of the shuttle craft. As measured by someone on the main spaceship, by what fraction did the kinetic energy of the module and shuttle craft increase because of the explosion?

Short answer

Fraction of kinetic energy of the module and shuttlecraft increase is $\frac{\left(∆K.E\right)}{K.E_i}=2.5\times {10}^{-3}$.

Step-by-step solution

Understanding the given information

1. Mass of module,$M\mathrm{is}500.0\mathrm{kg}$.
2. Mass of shuttle craft,$m\mathrm{is}500.0\mathrm{kg}$.
3. Speed of module and shuttle,$V_i\mathrm{is}1000\frac{\mathrm{m}}{\mathrm{s}}$.
4. Speed of module relative to shuttle, $V_R\mathrm{is}100.0\frac{\mathrm{m}}{\mathrm{s}}$.

Concept and formula used in the given question

Consider by conservation of momentum and kinetic energy find the velocity of the shuttle. From the kinetic energies of the module and shuttle and find the fraction. The formula used are stated below.

$$\begin{gathered} m_1{v}_{i1}+m_2{v}_{i2}=m_1{v}_{f1}+m_2{v}_{f2} \\ K.E=\frac{1}{2}m{v}^2 \end{gathered}$$

Calculate the fraction for which the kinetic energy of the module and shuttle craft increase because of the explosion

Determine the speed of the shuttle:

$m{v}_{i1}+m{v}_{i2}=m_1{v}_{f1}+m_2{v}_{f2}$

Initially, both the module and the shuttle had the same speed solve as:

$$\begin{gathered} m{v}_{i1}+m{v}_{i1}=m\left(V_{shuttle}-100.0\right)+m{v}_{shuttle} \\ \left(500.0+400.0\right)\left(1000\right)=500.0\left(V_{shuttle}-100.0\right)+\left(400.0\right)V_{shuttle} \\ 900000=900.0v_{shuttle}-50000 \\ 900.0v_{shuttle}=950000 \end{gathered}$$

Solve further as:

$$\begin{gathered} V_{shuttle}=\frac{950000}{900.0} \\ {V}_{shuttle}=1055.6\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

New speed for shuttle,

$$\begin{gathered} V_{new}=V_{shuttle}-100.0 \\ =1055.6-100 \\ =955.6\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Finally, the module and shuttle have different speeds.

Now the change in kinetic energy of the module and shuttle is determined as:

$$\begin{gathered} K.E_f-K.E_i=\left[\frac{1}{2}\left(500.0\right){\left(955.6\right)}^2+\frac{1}{2}\left(400\right){\left(1055.6\right)}^2\right]-\frac{1}{2}\left(900\right){\left(1000\right)}^2 \\ K.E_f-K.E_i=1151112 \end{gathered}$$

So, the fraction is,

$$\begin{gathered} \frac{\left(K.E_f-K.E_i\right)}{K.E_i}=\frac{1151112}{4.5\times {10}^8} \\ =2.5\times {10}^{-3} \end{gathered}$$