Chapter 7: Linear momentum

7.2 Collisions

In a collision:

  • Two or more objects strike each other
  • The total external force is either zero or negligible, and therefore momentum is conserved

Collisions are categorized by what happens to the total kinetic energy of the system. In an inelastic collision, the total kinetic energy is different before and after the collision. In an elastic collision, the total kinetic energy is the same before and after the collision.

On the scale of macroscopic objects, all collisions are inelastic. However, some are very close to elastic. For example, a collision between two pool balls is very nearly elastic. On the atomic and subatomic scale, collisions that are truly elastic occur regularly.

7.2.1 Inelastic collisions

Collisions in which objects permanently change shape (deform) are very common examples of inelastic collisions. Energy is required for deformation—this energy is provided by the kinetic energy the objects had before the collision.

Momentum is conserved (as in all collisions), but kinetic energy is different before and after:

\[
p_i = p_f \qquad\text{and}\qquad K_i \neq K_f \tag{7.3}
\]

As a special case of this, consider two objects that stick together after a collision. For two objects to stick together, there is generally some sort of deformation involved. When two objects stick together, it is often called a perfectly or completely inelastic collision.

However, objects do not need to stick together for the collision to be classified as inelastic. For example, a rock flying through and shattering a class window would be an inelastic collision: when the glass shatters, chemical bonds are broken, which requires energy. The energy required to break the glass comes from the kinetic energy of the rock.

Example

A 1200 kg car moves at 2.5 m/s, and collides with 2600 kg truck moving at 6.2 m/s in the opposite direction. The vehicles stick together. What is their speed after collision?

Say the car is traveling in the positive direction. Then, \(v_c = 2.5\ \textrm{m/s}\) and \(v_t = -6.5\ \textrm{m/s}\).

\[
\begin{align*}
p_i &= p_f \\
p_c + p_t &= p_f \\
m_c v_c + m_t v_t &= (m_c + m_t)v_f \\
\hookrightarrow v_f &= \frac{m_c v_c + m_t v_t}{m_c + m_t} \\
&= -3.45\ \textrm{m/s}
\end{align*}
\]

Remember speed is the magnitude of velocity. So the speed is 3.45 m/s.

Example

An object with mass \(m\) is traveling with a velocity of \(v_1\) when it collides with another object. The second object has the same mass, but is at rest when the collision occurs. They two objects stick together. Find the final velocity \(v_f\) of the two objects after the collision. Show that this is an inelastic collision. What is the ratio of final kinetic energy to initial?

\[
\begin{align*}
p_i &= p_f \\
p_1 + p_2 &= p_f \\
mv_1 + m(0) &= (m_\textit{tot})v_f \\
\hookrightarrow v_f &= \frac{mv_1}{2m} \\
&= \frac{1}{2}v_1
\end{align*}
\]

Note that the total mass is \(m_\textit{tot} = m + m = 2m\). To show this is an inelastic collision, look at the kinetic energy before and after the collision.

\[
\begin{align*}
K_f &= \frac{1}{2}(2m)v_f^2 & K_i &= K_1 + K_2 \\
&= \frac{1}{2}(2m)\left( \frac{1}{2}v_1 \right)^2 & &= \frac{1}{2}mv_1^2 + \frac{1}{2}m_2(0)^2 \\
&= \frac{1}{4}mv_1^2 & &= \frac{1}{2}mv_1^2
\end{align*}
\]

Since \( K_i \neq K_f \), this is an inelastic collision. The ratio of final kinetic energy to initial is

\[
\begin{align*}
\frac{K_f}{K_i} &= \left(K_f\right)\left(\frac{1}{K_i}\right) \\
&= \left( \frac{1}{4}mv_1^2 \right) \left( \frac{2}{mv_1^2} \right) \\
&= \frac{1}{2}
\end{align*}
\]

This suggests that for a perfectly inelastic collision between two objects of the same mass, exactly half of the initial kinetic energy is converted into other forms of energy!

On your own, see if this holds true if object 2 was initially moving. Try it for the following cases:

  1. \(v_2 = v_1\)
  2. \(v_2 = \frac{v_1}{2}\)
  3. \(v_2 = -v_1\)
  4. \(v_2 = -\frac{v_1}{2}\)

(Of course, you can try more than just these. You’ll need \(v_2\) to be some multiple of \(v_1\) for the math to work out nicely.)

Practice

Practice

Practice

Practice

7.2.2 Elastic collisions

As with all collisions, momentum is conserved in elastic collisions. Additionally, the total kinetic energy of a system is the same before and after an elastic collision:

\[
p_i = p_f \qquad\text{and}\qquad K_i = K_f \tag{7.4}
\]

Most collisions that you will encounter in daily life will be inelastic collisions, though some (such as collisions between pool balls) are very good approximations of elastic collisions.

Elastic collisions are very common on the atomic and subatomic scales. For example, if you’ve had chemistry you’re probably familiar with the Ideal Gas Law. This is a very accurate description of the interactions between certain atoms, and one of the primary assumptions that the ideal gas law is built on is that the collisions between atoms are elastic.

Example

Two carts are constrained to roll along a track. Cart \(A\) has a mass of 3 kg and initial velocity of 5 m/s. Cart \(B\) has a mass of 1 kg and an initial velocity of -5 m/s. After they collide, cart \(A\) is at rest. Determine the post-collision velocity of cart \(B\). Is this collision elastic or inelastic?

Analyze momentum to find the velocity of cart \(B\) after the collision:

\[
\begin{align*}
p_i &= p_f \\
m_av_{ai} + m_bv_{bi} &= m_av_{af} + m_bv_{bf} \\
m_av_{ai} + m_bv_{bi} &= m_a(0) + m_bv_{bf} \\
\hookrightarrow v_{bf} &= \frac{m_av_{ai} + m_bv_{bi}}{m_b} \\
&= \frac{(3\ \textrm{kg})(5\ \textrm{m/s}) + (1\ \textrm{kg})(-5\ \textrm{m/s})}{1\ \textrm{kg}} \\
&= 10\ \textrm{m/s}
\end{align*}
\]

To determine if this is elastic or inelastic, compare the initial and final kinetic energy:

\[
\begin{align*}
K_i &= K_{ai} + K_{bi} & K_f &= K_{af} + K_{bf} \\
&= \frac{1}{2}m_av_{ai}^2 + \frac{1}{2}m_bv_{bi}^2 & &= \frac{1}{2}m_a(0)^2 + \frac{1}{2}m_bv_{bf}^2 \\
&= 50\ \textrm{J} & &= 50\ \textrm{J}
\end{align*}
\]

Since \(K_i = K_f\), this was an elastic collision.

Example

Two two cars are constrained to roll along a track. Car 1 has a mass of 0.50 kg and is traveling to the right at 1.40 m/s. Car 2 has a mass of 1.50 kg and is traveling to the left at 0.70 m/s. The cars rebound off one another; after the collision, car 2 is traveling to the right at 0.22 m/s. Is this an elastic or an inelastic collision?

To determine if the collision is elastic or inelastic, we’ll need to analyze the total kinetic energy before and after the collision. To do that, we’ll need to know the speed of car 1 after the collision. So we begin by taking a look at momentum:

\[
\begin{align*}
p_i &= p_f \\
p_{1i} + p_{2i} &= p_{1f} + p_{2f} \\
m_1v_{1i} + m_2v_{2i} &= m_1v_{1f} + m_2v_{2f} \\
\hookrightarrow v_{1f} &= \frac{m_1v_{1i} + m_2v_{2i} – m_2v_{2f}}{m_1} \\
&= \frac{m_1v_{1i} + m_2(v_{2i} – v_{2f})}{m_1} \\
&= v_{1i} + \frac{m_2}{m_1}(v_{2i} – v_{2f}) \\
&= 1.40 \textrm{m/s} + \frac{1.50\ \textrm{kg}}{0.50\ \textrm{kg}}\left((-0.70\ \textrm{m/s}) – (0.22\ \textrm{m/s})\right) \\
&= -1.36\ \textrm{m/s}
\end{align*}
\]

Now we can calculate initial and final kinetic energies.

\[
\begin{align*}
K_i &= \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 & K_f &= \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2\\
&= 0.86\ \textrm{J} & &= 0.50\ \textrm{J}
\end{align*}
\]

Since \(K_i \neq K_f\), this is an inelastic collision. Even though this is not a completely inelastic collision (the objects do not stick together), it is still classified as being inelastic since some of the initial kinetic energy is converted into other forms of energy.