FHSST Physics Collisions Explosions:Types of Collisions

Types of Collisions

We will consider two types of collisions in this section

• Elastic collisions
• Inelastic collisions

In both types of collision, total energy and total momentum is always conserved. Kinetic energy is conserved for elastic collisions, but not for inelastic collisions.

Elastic Collisions

Definition:

An elastic collision is a collision where total momentum and total kinetic energy are both conserved.

This means that the total momentum and the total kinetic energy before an elastic collision is the same as after the collision. For these kinds of collisions, the kinetic energy is not changed into another type of energy.

Before the Collision

In the following diagram,two balls are rolling toward each other, about to collide

Before the balls collide, the total momentum of the system is equal to all the individual momenta added together. The ball on the left has a momentum which we call ${\overrightarrow{p}}_1$ and the ball on the right has a momentum which we call ${\overrightarrow{p}}_2$, it means the total momentum before the collision is

$\begin{array}{c}{\overrightarrow{p}}_{\mathrm{B}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}}={\overrightarrow{p}}_1+{\overrightarrow{p}}_2\end{array}$

(8.1)

We calculate the total kinetic energy of the system in the same way. The ball on the left has a kinetic energy which we call K₁ and the ball on the right has a kinetic energy which we call K₂, it means that the total kinetic energy before the collision is

$\begin{array}{c}{K}_{\mathrm{B}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}}=K_1+K_2\end{array}$

(8.2)

After the Collision

The following diagram shows the balls after they collide

After the balls collide and bounce off each other, they have new momenta and new kinetic energies. Like before, the total momentum of the system is equal to all the individual momenta added together. The ball on the left now has a momentum which we call ${\overrightarrow{p}}_3$ and the ball on the right now has a momentum which we call ${\overrightarrow{p}}_4$, it means the total momentum after the collision is

$\begin{array}{c}{\overrightarrow{p}}_{\mathrm{A}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}}={\overrightarrow{p}}_3+{\overrightarrow{p}}_4\end{array}$

(8.3)

The ball on the left now has a kinetic energy which we call K₃ and the ball on the right now has a kinetic energy which we call K₄, it means that the total kinetic energy after the collision is

$\begin{array}{c}{K}_{\mathrm{A}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}}=K_3+K_4\end{array}$

(8.4)

Since this is an elastic collision, the total momentum before the collision equals the total momentum after the collision and the total kinetic energy before the collision equals the total kinetic energy after the collision

$\begin{array}{ccc}\text{Before}& & \text{After}\\ {\overrightarrow{p}}_{\mathrm{B}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}}& =& {\overrightarrow{p}}_{\mathrm{A}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}}\\ {\overrightarrow{p}}_1+{\overrightarrow{p}}_2& =& {\overrightarrow{p}}_3+{\overrightarrow{p}}_4\\ & \mathbf{\text{and}}& \\ K_{\mathrm{B}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}}& =& K_{\mathrm{A}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}}\\ K_1+K_2& =& K_3+K_4\end{array}$

File:Fhsst red line.png

Worked Example 46 An Elastic Collision

We will have a look at the collision between two pool balls. Ball 1 is at rest and ball 2 is moving towards it with a speed of 2 m·s⁻¹. The mass of each ball is 0.3 kg. After the balls collide elastically, ball 2 comes to a stop and ball 1 moves off. What is the final velocity of ball 1?

Step 1 : Draw the before diagram

Before the collision, ball 2 is moving we will call it's momentum P₂ and it's kinetic energy K₂. Ball 1 is at rest, so it has zero kinetic energy and momentum.

Step 2 : Draw the after diagram

After the collision, ball 2 is at rest but ball 1 has a momentum which we call P₃ and a kinetic energy which we call K₃.

RIAAN Note: second image on page 145 is missing

File:Fhsst expl4.png

Because the collision is elastic, we can solve the problem using momentum conservation or kinetic energy conservation. We will do it both ways to show that the answer is the same, whichever method you use.

Step 3 : Show the conservation of momentum

We start by writing down that the momentum before the collision ${\overrightarrow{p}}_{\mathrm{B}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}}$ is equal to the momentum after the collision ${\overrightarrow{p}}_{\mathrm{A}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}}$

$\begin{array}{ccc}\text{Before}& & \text{After}\\ {\overrightarrow{p}}_{\mathrm{B}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}}& =& {\overrightarrow{p}}_{\mathrm{A}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}}\\ {\overrightarrow{p}}_1+{\overrightarrow{p}}_2& =& {\overrightarrow{p}}_3+{\overrightarrow{p}}_4\\ 0+{\overrightarrow{p}}_2& =& {\overrightarrow{p}}_3+0\\ {\overrightarrow{p}}_2& =& {\overrightarrow{p}}_3\end{array}$

(8.7)

We know that momentum is just P = mv, and we know the masses of the balls, so we can rewrite the conservation of momentum in terms of the velocities of the balls

$\begin{array}{ccc}{\overrightarrow{p}}_2& =& {\overrightarrow{p}}_3\\ m_2{v}_2& =& m_3{v}_3\\ 0.3v_2& =& 0.3v_3\\ v_2& =& v_3\end{array}$

(8.8)

So ball 1 exits with the velocity that ball 2 started with!

$\begin{array}{c}{v}_3=2[\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}]\end{array}$

(8.9)

Step 4 : Show the conservation of kinetic energy

We start by writing down that the kinetic energy before the collision $K_{\mathrm{B}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}}$ is equal to the kinetic energy after the collision $K_{\mathrm{A}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}}$

$\begin{array}{ccc}\text{Before}& & \text{After}\\ K_{\mathrm{B}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}}& =& K_{\mathrm{A}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}}\\ K_1+K_2& =& K_3+K_4\\ 0+K_2& =& K_3+0\\ K_2& =& K_3\end{array}$

(8.10)

We know that kinetic energy is just $K=\frac{m{v}^2}{2}$, and we know the masses of the balls, so we can rewrite the conservation of kinetic energy in terms of the velocities of the balls

$\begin{array}{ccc}{K}_2& =& K_3\\ \frac{m_2{v}_2^2}{2}& =& \frac{m_3{v}_3^2}{2}\\ 0.15v_2^2& =& 0.15v_3^2\\ v_2^2& =& v_3^2\\ v_2& =& v_3\end{array}$

(8.11)

So ball 1 exits with the velocity that ball 2 started with, which agrees with the answer we got when we used the conservation of momentum.

$\begin{array}{c}{v}_3=2[\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}]\end{array}$

(8.12)

File:Fhsst red line.png

Worked Example 47 Elastic Collision 2

Question: Now for a slightly more difficult example. We have 2 marbles. Marble 1 has mass 50 g and marble 2 has mass 100&nbsp g. I roll marble 2 along the ground towards marble 1 in the positive x-direction. Marble 1 is initially at rest and marble 2 has a velocity of 3 $\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}$ in the positive x-direction. After they collide elastically, both marbles are moving. What is the final velocity of each marble?

Answer:

Step 1 :

Firstly, put all the quantities into S.I. units. So:

$\begin{array}{ccc}{m}_1=0.05\mathrm{k}\mathrm{g}& \mathrm{a}\mathrm{n}\mathrm{d}& m_2=0.1\mathrm{k}\mathrm{g}\end{array}$

Step 2 :

Draw the picture:

Before the collision:

After the collision:

Step 3 :

Since the collision is elastic, both momentum and kinetic energy are conserved in the collision. So:

$\begin{array}{ccc}{E}_{kBefore}& =& E_{kAfter}\\ & \mathrm{a}\mathrm{n}\mathrm{d}& \\ \overrightarrow{p_{Before}}& =& \overrightarrow{p_{After}}\end{array}$

There are two unknowns ($\overrightarrow{v_1}$ and $\overrightarrow{v_2}$) so we will need two equations to solve for them. We need to use both kinetic energy conservation and momentum conservation in this problem.

Step 4 : Let's start with energy conservation. Then:

$\begin{array}{ccc}{E}_{kBefore}& =& E_{kAfter}\\ \frac{1}{2}{m}_1{\overrightarrow{u_1}}^2+\frac{1}{2}{m}_2{\overrightarrow{u_2}}^2& =& \frac{1}{2}{m}_1{\overrightarrow{v_1}}^2+\frac{1}{2}{m}_2{\overrightarrow{v_2}}^2\\ m_1{\overrightarrow{u_1}}^2+m_2{\overrightarrow{u_2}}^2& =& m_1{\overrightarrow{v_1}}^2+m_2{\overrightarrow{v_2}}^2\end{array}$

But $\overrightarrow{u_1}$, and solving for ${\overrightarrow{v_2}}^2$

$\begin{array}{ccc}{\overrightarrow{v_2}}^2& =& {\overrightarrow{u_2}}^2-\frac{m1}{m2}{\overrightarrow{v_1}}^2\\ {\overrightarrow{v_2}}^2& =& (3{)}^2-\frac{(0.05)}{(0.10)}{\overrightarrow{v_1}}^2\\ {\overrightarrow{v_2}}^2& =& 9-\frac{1}{2}{\overrightarrow{v_1}}^2\mathrm{(}\mathbf{\text{A}}\mathrm{)}\end{array}$

Step 5 : Now we have simplified as far as we can, we move onto momentum conservation:

$\begin{array}{ccc}\overrightarrow{p_{Before}}& =& \overrightarrow{p_{After}}\\ m_1\overrightarrow{u_1}+m_2\overrightarrow{u_2}& =& m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}\end{array}$

But $\overrightarrow{u_1}$=0, and solving for $\overrightarrow{v_1}$

$\begin{array}{ccc}{m}_2\overrightarrow{u_2}& =& m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}\\ m_1\overrightarrow{v_1}& =& m_2\overrightarrow{u_2}-m_2\overrightarrow{v_2}\\ \overrightarrow{v_1}& =& \frac{m_2}{m_1}\overrightarrow{u_2}-\frac{m_2}{m_1}\overrightarrow{v_2}\\ \overrightarrow{v_1}& =& 2(3)-2\overrightarrow{v_2}\\ \overrightarrow{v_1}& =& 6-2\overrightarrow{v_2}\mathrm{(}\mathbf{\text{B}}\mathrm{)}\end{array}$

Step 6 :

Now we can substitute (B) into (A) to solve for $\overrightarrow{v_2}$:

$\begin{array}{ccc}{\overrightarrow{v_2}}^2& =& 9-\frac{1}{2}{\overrightarrow{v_1}}^2\\ {\overrightarrow{v_2}}^2& =& 9-\frac{1}{2}(6-2\overrightarrow{v_2}{)}^2\\ {\overrightarrow{v_2}}^2& =& 9-\frac{1}{2}(36-24\overrightarrow{v_2}+4{\overrightarrow{v_2}}^2)\\ {\overrightarrow{v_2}}^2& =& 9-18+12\overrightarrow{v_2}-2{\overrightarrow{v_2}}^2\\ 3{\overrightarrow{v_2}}^2& =& -9+12\overrightarrow{v_2}\\ {\overrightarrow{v_2}}^2& =& 4\overrightarrow{v_2}-3\\ {\overrightarrow{v_2}}^2-4\overrightarrow{v_2}+3& =& 0\\ (\overrightarrow{v_2}-3)(\overrightarrow{v_2}-1)& =& 0\\ \overrightarrow{v_2}=3& \mathbf{\text{or}}& \overrightarrow{v_2}=1\end{array}$

We were lucky in this question because we could factorise. If you can't factorise, then you can always solve using the formula for solving quadratic equations. Remember:

$\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$

So, just to check:

$\begin{array}{ccc}\overrightarrow{v_2}& =& \frac{4\pm \sqrt{4^2-4(1)(3)}}{2(1)}\\ \overrightarrow{v_2}& =& \frac{4\pm \sqrt{16-12}}{2}\\ \overrightarrow{v_2}& =& \frac{4\pm \sqrt{4}}{2}\\ \overrightarrow{v_2}& =& 2\pm 1\\ \overrightarrow{v_2}=3& \mathbf{\text{or}}& \overrightarrow{v_2}=1\mathrm{s}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{b}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}\end{array}$

Step 7 :

So finally, substituting into equation (B) to get $\overrightarrow{v_1}$:

$\begin{array}{c}\overrightarrow{v_1}=6-2\overrightarrow{v_2}\end{array}$

If $\overrightarrow{v_2}=3\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}$ then

$\begin{array}{c}\overrightarrow{v_1}=6-2(3)=0\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}\end{array}$

But, according to the question, marble 1 is moving after the collision. So $\overrightarrow{v_1}\ne 0$ and $\overrightarrow{v_2}\ne 3$. Therefore:

$\begin{array}{ccc}\overrightarrow{v_2}& =& \underset{\_}{1\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{p}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{x}\mathrm{-}\mathrm{d}\mathrm{i}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}\\ & \mathrm{a}\mathrm{n}\mathrm{d}& \\ \overrightarrow{v_1}& =& \underset{\_}{4\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{p}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{x}\mathrm{-}\mathrm{d}\mathrm{i}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}\end{array}$

File:Fhsst red line.png

Inelastic Collisions

Definition:

An inelastic collision is a collision in which total momentum is conserved but total kinetic energy is not conserved

the kinetic energy is transformed into other kinds of energy.

So the total momentum before an inelastic collisions is the same as after the collision. But the total kinetic energy before and after the inelastic collision is different. Of course this does not mean that total energy has not been conserved, rather the energy has been transformed into another type of energy.

As a rule of thumb, inelastic collisions happen when the colliding objects are distorted in some way. Usually they change their shape. To modify the shape of an object requires energy and this is where the missing kinetic energy goes. A classic example of an inelastic collision is a car crash. The cars change shape and there is a noticeable change in the kinetic energy of the cars before and after the collision. This energy was used to bend the metal and deform the cars. Another example of an inelastic collision is shown in the following picture.

Here an asteroid (the small circle) is moving through space towards the moon (big circle). Before the moon and the asteroid collide, the total momentum of the system is:

$\begin{array}{c}\overrightarrow{p_{Before}}=\overrightarrow{p_m}+\overrightarrow{p_a}\end{array}$

$\overrightarrow{p_m}$ stands for $\overrightarrow{p_{moon}}$ and $\overrightarrow{p_a}$ stands for $\overrightarrow{p_{asteroid}}$ and the total kinetic energy of the system is:

$\begin{array}{c}{E}_{Before}=E_{km}+E_{ka}\end{array}$

When the asteroid collides inelastically with the moon, its kinetic energy is transformed mostly into heat energy. If this heat energy is large enough, it can cause the asteroid and the area of the moon's surface that it hit, to melt into liquid rock! From the force of impact of the asteroid, the molten rock flows outwards to form a moon crater.

After the collision, the total momentum of the system will be the same as before. But since this collision is inelastic, (and you can see that a change in the shape of objects has taken place!), total kinetic energy is not the same as before the collision.

So:

$\begin{array}{ccc}\overrightarrow{p_{Before}}& =& \overrightarrow{p_{After}}\\ \overrightarrow{p_m}+\overrightarrow{p_a}& =& \overrightarrow{p_{After}}\\ & \mathbf{\text{but}}& \\ E_{kBefore}& \ne & E_{kAfter}\\ E_{km}+E_{ka}& \ne & E_{kAfter}\end{array}$

File:Fhsst red line.png

Worked Example 48 Inelastic Collision

Question: Let's consider the collision of two cars. Car 1 is at rest and Car 2 is moving at a speed of $2\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}$ in the negative x-direction. Both cars each have a mass of 500kg. The cars collide inelastically and stick together. What is the resulting velocity of the resulting mass of metal?

Answer:

Step 1 :

Draw the picture: Before the collision:

After the collision:

File:Fhsst expl11.png

Step 2 :

We know the collision is inelastic and there was a definite change in shape of the objects involved in the collision - there were two objects to start and after the collision there was one big mass of metal! Therefore, we know that kinetic energy is not conserved in the collision but total momentum is conserved.

So:

$\begin{array}{ccc}{E}_{kBefore}& \ne & E_{kAfter}\\ & \mathbf{\text{but}}& \\ \overrightarrow{p_{TBefore}}& =& \overrightarrow{p_{After}}\end{array}$

Step 3 :

So we must use conservation of momentum to solve this problem. Take the negative x-direction to have a negative sign:

$\begin{array}{ccc}\overrightarrow{p_{TBefore}}& =& \overrightarrow{p_{After}}\\ \overrightarrow{p_1}+\overrightarrow{p_2}& =& \overrightarrow{p_{After}}\\ m_1\overrightarrow{u_1}+m_2\overrightarrow{u_2}& =& (m_1+m_2)\overrightarrow{v}\\ 0+500(-2)& =& (500+500)\overrightarrow{v}\\ -1000& =& 1000\overrightarrow{v}\\ \overrightarrow{v}& =& -1\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}\end{array}$

Therefore,

$\begin{array}{c}\overrightarrow{v}=\underset{\_}{1\mathrm{m}\mathrm{.}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{g}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{x}\mathrm{-}\mathrm{d}\mathrm{i}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}\end{array}$

File:Fhsst red line.png