In a collision, two particles come together for a short time and thereby produce impulsive forces on each other.

Remember that all forces are field forces when viewed at small enough scale.

Even an object in space pulled around by the gravitational forces of the planets or Sun may be considered as collisions.

Consider two objects with mass $m_{1}$ and $m_{2}$ undergoing a collision. During the collision, they will exert a force on each other, as given by $\vec{F}_{12}$ and $\vec{F}_{21}$. The forces will vary according to the graph given below:

Hence, the total change in momentum for object 1 with mass $m_{1}$ is:

$$\Delta \vec{p}_{1} = \int\limits_{t_{i}}^{t_{f}} \vec{F}_{21} \, dt$$

For object 2 with mass $m_{2}$:

$$\Delta \vec{p}_{2} = \int\limits_{t_{i}}^{t_{f}} \vec{F}_{12} \, dt$$

From Newton’s Third Law,

$$\begin{aligned} \Delta \vec{p}_{2} &=-\Delta \vec{p}_{2} \\ \Delta \vec{p}_{1} + \Delta \vec{p}_{2} &= 0 \\ \vec{p}_{\text{tot}} &= \vec{p}_{1} + \vec{p}_{2} = \text{const.} \end{aligned}$$

### Conservation Of Linear Momentum

The total momentum of an isolated system just before a collision equals the total momentum of the system just after the collision.

- It is a direct result from Newton’s Second and Third Laws of Motion.
- It provides a convenient way to calculate some parameters, such as final velocities, in collisions without the need to know the exact forms of the forces involved.

### Elastic & Inelastic Collisions

Momentum is always conserved.

If the total kinetic energy is also the same before and after the collision, the collision is elastic.

If the total kinetic energy is not the same before and after the collision, the collision is inelastic.

#### Elastic Collision

Consider the elastic collision as shown in the diagram above.

From momentum conservation:

$$\begin{aligned} \text{Initial Momentum} &= \text{Final Momentum} \\ m_{1}v_{1i} + m_{2}v_{2i} &= m_{1}v_{1f}+m_{2}v_{2f} \end{aligned}$$

From energy conservation:

$$\begin{aligned} \text{Initial Kinetic Energy} &= \text{Final Kinetic Energy} \\ \frac{1}{2}m_{1}v_{1i}^{2} + \frac{1}{2}m_{2}v_{2i}^{2} &= \frac{1}{2} m_{1}v_{1f}^{2} + \frac{1}{2} m_{2} v_{2f}^{2} \end{aligned}$$

Let’s find the final velocities in terms of the initial velocities.

Let’s start with the energy conservation equation,

$$\begin{aligned} \frac{1}{2}m_{1}v_{1i}^{2} + \frac{1}{2}m_{2}v_{2i}^{2} &= \frac{1}{2} m_{1}v_{1f}^{2} + \frac{1}{2} m_{2} v_{2f}^{2} \\ m_{1} \left( v_{1i}^{2}-v_{1f}^{2} \right) &= m_{2} \left( v_{2f}^{2}-v_{2i}^{2} \right) \\ m_{1} \left( v_{1i}-v_{1f} \right) \left( v_{1i} + v_{1f} \right) &= m_{2} \left( v_{2f}-v_{2i} \right) \left( v_{2f} + v_{2i} \right) \, \, – \text{Eqn 1}\end{aligned}$$

From the conservation of momentum equation,

$$\begin{aligned} m_{1}v_{1i} + m_{2}v_{2i} &= m_{1}v_{1f}+m_{2}v_{2f} \\ m_{1} \left( v_{1i}-v_{1f} \right) &= m_{2} \left( v_{2f}-v_{2i} \right) \, \, – \text{Eqn 2} \end{aligned}$$

Dividing Equation 1 by Equation 2, we have:

$$\begin{aligned} v_{1i} + v_{1f} &= v_{2f} + v_{2i} \\ v_{1i}-v_{2i} &= v_{2f}-v_{1f} \, \, – \text{Eqn 3}\end{aligned}$$

Notice that we can get $\frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}} = 1$ from Equation 3. The fraction $\frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}}$ is known as the coefficient of restitution, $e$. We will look into that later.

Continuing from Equation 3,

$$\begin{aligned} v_{1i}-v_{2i} &= v_{2f}-v_{1f} \\ v_{2f} &= v_{1i} + v_{1f}-v_{2i} \, \, – \text{Eqn 4} \end{aligned}$$

Sub. Equation 4 into momentum conservation equation – $m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}$, we have:

$$\begin{aligned} m_{1}v_{1i} + m_{2}v_{2i} &= m_{1}v_{1f} + m_{2} \left( v_{1i} + v_{1f}-v_{2i} \right) \\ \left( m_{1}-m_{2} \right) v_{1i} + \left( m_{2} + m_{2} \right) v_{2i} &= \left( m_{1} + m_{2} \right) v_{1f} \end{aligned}$$

Making $v_{1f}$ the subject, we obtain:

$$v_{1f} = \left( \frac{m_{1}-m_{2}}{m_{1}+m_{2}} \right) v_{1i} + \left( \frac{2m_{2}}{m_{1}+m_{2}} \right) v_{2i} $$

Using the same method, one can find the final velocity for object 2:

$$v_{2f} = \left( \frac{2m_{1}}{m_{1}+m_{2}} \right) v_{1i} + \left( \frac{m_{2}-m_{1}}{m_{1}+m_{2}} \right) v_{2i}$$

Armed with the result, let’s look at specific cases.

**Case 1:** If $m_{1} = m_{2}$, we have:

$$\begin{aligned} v_{1f} &= v_{2i} \\ v_{2f} &= v_{1i} \end{aligned}$$

The particles exchange velocities.

**Case 2:** If $m_{2}$ is initially at rest $\left( v_{2i} = 0 \right)$, then

$$\begin{aligned} v_{1f} &= \left( \frac{m_{1}-m_{2}}{m_{1}+m_{2}} \right) v_{1i} \\ v_{2f} &= \left( \frac{2 m_{1}}{m_{1} + m_{2}} \right) v_{1i} \end{aligned}$$

When $m_{1} \gg m_{2}$, we have $v_{1f} \approx v_{1i}$ and $v_{2f} \approx 2 v_{1i}$.

When $m_{2} \gg m_{1}$, we have $v_{1f} \approx-v_{1i}$ and $v_{2f} \approx 0$.

##### Transfer Of Kinetic Energy During Elastic Collision

Often, we want to know how much kinetic energy is transferred to a stationary target $\left( v_{2i} = 0 \right)$. Using $v_{2f} = \left( \frac{2 m_{1}}{m_{1} + m_{2}} \right) v_{1i}$, we have

$$\begin{aligned} \text{KE}_{2f} &= \frac{1}{2} m_{2} \left( \frac{2m_{1}}{m_{1}+m_{2}} \right)^{2} v_{1i}^{2} \\ &= \frac{4m_{2}m_{1}}{\left( m_{1} + m_{2} \right)^{2}} \text{KE}_{1i} \end{aligned}$$

Let’s look at two cases,

**Case 1:** $m_{2} \gg m_{2}$, we have:

$$\text{KE}_{2f} \approx 4 \frac{m_{1}}{m_{2}} \text{KE}_{1i}$$

Case 2: $m_{1} \gg m_{2}$, we have:

$$\text{KE}_{2f} \approx 4 \frac{m_{2}}{m_{1}} \text{KE}_{1i}$$

From the above, you can see that maximum kinetic energy transfer will occur when $m_{1} = m_{2}$

#### Perfectly (Totally) Inelastic Collisions

In a totally inelastic collision, two particles with masses $m_{1}$ and $m_{2}$ collide head-on and stick together after the collision. The combined particle will move with some common velocity $\vec{v}_{f}$.

From momentum conservation:

$$\begin{aligned} m_{1} \vec{v}_{1i} + m_{2} \vec{v}_{2i} &= \left( m_{1} + m_{2} \right) \vec{v}_{f} \\ \vec{v}_{f} &= \frac{m_{1}\vec{v}_{1i} + m_{2} \vec{v}_{2i}}{m_{1}+m_{2}} \end{aligned}$$

#### Coefficient Of Restitution

Coefficient of restitution is defined as:

$$\begin{aligned} e &= \frac{\text{Relative speed of separation}}{\text{Relative speed of approach}} \\ &= \frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}} \end{aligned}$$

For a perfectly inelastic collision, e is 0. For inelastic collision, $e < 1$. For elastic collision, $e > 1$.

#### 2-D Collisions

For 2D collisions, the momentum in each direction is conserved. The conservation equations for the above is:

$$\begin{aligned} m_{1}v_{1ix} + m_{2}v_{2ix} &= m_{1}v_{1fx} + m_{2}v_{2fx} \\ m_{1}v_{1iy}+m_{2}v_{2iy} &= m_{1}v_{1fy} + m_{2} v_{2fy} \end{aligned}$$

In two dimensions, there are **four unknowns**, either $v_{1fx}$, $v_{1fy}$, $v_{2fx}$, $v_{2fy}$, or if we use the magnitudes and angles, $v_{1f}$, $v_{2f}$, $\theta$ and $\phi$.

We can only write down **two** equations for conservation of linear momentum in their component directions.

We may be able to add **one more** equation relating energy if the collision is perfectly elastic.

We will need **at least another parameter** to determine the final outcome. The final parameter will be one of the unknowns as stated above.

The **exception** to this rule will be perfectly inelastic collision. In perfectly inelastic collision, there are only two unknowns and the final outcome is uniquely determined by conservation of linear momentum along.

##### Elastic Glancing Collision

The diagram above shows the case of elastic glancing collision. From the conservation of momentum, we have:

In the horizontal direction,

$$m_{1}v_{1i} = m_{1}v_{1f} \cos{\theta} + m_{2}v_{2f} \cos{\theta}$$

In the vertical direction,

$$0 = m_{1}v_{1f}\sin{\theta}-m_{2}v_{2f}\sin{\theta}$$

Since it is an elastic collision, the kinetic energy is conserved, which is given by:

$$\frac{1}{2}m_{1}v_{1i}^{2} = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2}$$

You can analyze the rest of the equations with the same methods that has been shown above. Let’s look at a special case of the elastic glancing collision, whereby $m_{1} = m_{2} = m$.

From the conservation of momentum, we have:

$$\vec{v}_{1i} = \vec{v}_{1f} + \vec{v}_{2f}$$

From the conservation of energy, we have:

$$v_{1i}^{2} = v_{1f}^{2} + v_{2f}^{2}$$

From the above 2 equations, we can see that $\vec{v}_{1i}$, $\vec{v}_{1f}$ and $\vec{v}_{2f}$ makes a triangle, as shown by the figure below.

Hence, we can conclude that:

$$\theta + \phi = 90^{\circ}$$

The objects leave each other at a $90^{\circ}$ angle.

##### Problem Solving Strategy For 2-D Collision Problems

- Set up an appropriate coordinate system. It will be useful to choose x-axis to coincide with one of the initial velocity.
- Sketch and label all velocity vectors.
- Write down the momentum conservation equations in the 2 component directions.
- If the collision is elastic, write down the kinetic energy conservation equation.
- If the collision is inelastic, additional information may be required. If totally inelastic, the final velocities of both particles are equal.
- Use other information given. E.g. Direction of one of the final velocities.