# Explosions: Energy and Heat

In explosions, you have seen that kinetic energy is not conserved. But remember that total energy is always conserved. Let's look at what happens to the energy in some more detail. If a given amount of energy is released in an explosion it is not necessarily all transformed into kinetic energy. Due to the deformation of the exploding object, often a large amount of the energy is used to break chemical bonds and heat up the pieces.
Energy is conserved but some of it is transferred through non-conservative processes like heating. This just means that we cannot get the energy back. It will be radiated into the environment as heat energy but it is all still accounted for.
Now we can start to mix the ideas of momentum conservation with energy transfer to make longer problems. These problems are not more complicated just longer. We will start off short and then combine the different ideas later on.
   Long problems should be treated like a number of smaller problems. Focus on them one at a time.

## Worked Example 51 Energy Accounting 1

Question: An object with a mass of m$_t$ = 17 kg explodes into two pieces of mass m$_1$ = 7 kg and m$_2$ = 10 kg; m$_1$ has a velocity of 9 m·s$^{−1}$ in the negative x-direction and m$_2$ has a velocity of 6.3 m·s$^{−1}$ in the positive x-direction. If the explosion released a total energy of 2000 J, how much was used in a non-conservative way?
Answer:
Step 1 :
Draw the picture. Before the collision:
Riaan Note: first image on page 157 is missing
After the collision:
Step 2 :
We are asked how much energy was used in a non-conservative fashion. This is the difference between how much energy was used in a conservative fashion and how much was used in total. We are lucky because we have everything we need to determine the kinetic energy of both pieces. The kinetic energy of the pieces is energy that was used in a conservative way.
Step 3 :
The sum of the kinetic energy for the two blocks is the total kinetic energy of the pieces. So:
${\begin{matrix}E_{kTot}&=&E_{k1}+E_{k2}\\&=&{\frac {1}{2}}m_{1}{\overrightarrow {v_{1}}}^{2}+{\frac {1}{2}}m_{2}{\overrightarrow {v_{2}}}^{2}\\&=&{\frac {1}{2}}(7)(9)^{2}+{\frac {1}{2}}(10)(6.3)^{2}\\&=&283.5+198.45\\E_{kTot}&=&481.95{\rm {\ J}}\end{matrix}}$
Step 4 :
The total energy that was transformed into kinetic energy is 481.95 J. We know that 2000 J of energy were released in total. the question makes no statements about other types of energy so we can assume that the difference was lost in a non-conservative way. Thus the total energy lost in non-conservative work is:
${\begin{matrix}E-E_{kTot}&=&2000-481.95\\&=&1518.05{\rm {\ J}}\end{matrix}}$

## Worked Example 52 Energy Accounting 2

Question: An object at rest, with mass m$_{Tot}$ = 4 kg, explodes into two pieces (m$_1$, m$_2$) with m$_1$ = 2.3 kg. m$_1$ has a velocity of 17 m·s$^{−1}$ in the negative x-direction. If the explosion released a total energy of 800 J,
* What is the velocity of m$_2$?
* How much energy does it carry?
* And how much energy was used in a non-conservative way?
Answer:
Step 1 :
Draw the picture. Before the collision:
After the collision:
Step 2 :
Now we know that in an explosion, total kinetic energy is not conserved. There is a definite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So:
${\begin{matrix}{\overrightarrow {p_{Before}}}&=&{\overrightarrow {p_{After}}}\\{\overrightarrow {p_{Before}}}&=&{\overrightarrow {p_{1}}}+{\overrightarrow {p_{2}}}\end{matrix}}$
But the object was initially at rest so:
${\begin{matrix}0&=&{\overrightarrow {p_{1}}}+{\overrightarrow {p_{2}}}\\0&=&m_{1}{\overrightarrow {v_{1}}}+m_{2}{\overrightarrow {v_{2}}}{\rm {\qquad ({\textbf {A}})}}\end{matrix}}$
Step 3 :
Now we know that m$_1$= 2.3 kg but we do not know what the mass of m$_2$ is. However, we do know that:
${\begin{matrix}m_{Tot}&=&m_{1}+m_{2}\\m_{2}&=&m_{Tot}-m_{1}\\&=&4{\rm {{\ kg}-2.3{\rm {\ kg}}}}\\&=&1.7{\rm {\ kg}}\end{matrix}}$
Step 4 :
Now we can substitute all the values we know into equation (A) and solve for ${\overrightarrow {v_{2}}}$. Let's choose the positive x-direction to have a positive sign and the negative x-direction to have a negative sign:
${\begin{matrix}0&=&m_{1}{\overrightarrow {v_{1}}}+m_{2}{\overrightarrow {v_{2}}}{\rm {\qquad ({\textbf {A}})}}\\0&=&(2.3)(-17)+1.7{\overrightarrow {v_{2}}}\\0&=&-39.1+1.7{\overrightarrow {v_{2}}}\\1.7{\overrightarrow {v_{2}}}&=&39.1\\{\overrightarrow {v_{2}}}&=&23{\rm {\ \mathrm {m.s^{-1}} }}\end{matrix}}$
* ${\overrightarrow {v_{2}}}=23\mathrm {m.s^{-1}}$ in the positive x-direction
Step 5 :
Now we need to calculate the energy that the second piece carries:
${\begin{matrix}E_{k2}&=&{\frac {1}{2}}m_{2}{\overrightarrow {v_{2}}}^{2}\\&=&{\frac {1}{2}}(1.7)(23)^{2}\\&=&449.65{\rm {\ J}}\end{matrix}}$
* The kinetic energy of the second piece is $E_{k2}=449.65J$
Step 6 :
Now the amount of energy used in a non-conservative way in the explosion, is the difference between the amount of energy released in the explosion and the total kinetic energy of the exploded pieces:
${\begin{matrix}E-E_{kTot}&=&800-E_{kTot}\\\end{matrix}}$
We know that:
${\begin{matrix}E_{kTot}&=&E_{k1}+E_{k2}\\&=&{\frac {1}{2}}m_{1}{\overrightarrow {v_{1}}}^{2}+449.65\\&=&{\frac {1}{2}}(2.3)(17)^{2}+449.65\\&=&332.35+449.65\\&=&782{\rm {\ J}}\end{matrix}}$
Step 7 :
So going back to:
${\begin{matrix}E-E_{kTot}&=&800-E_{kTot}\\&=&800-782\\&=&18{\rm {\ J}}\end{matrix}}$
* 18 J of energy was used in a non-conservative way in the explosion