9.4 Types of Collisions

Explosions

The first possibility is that a single object may break apart into two or more pieces. An example of this is a firecracker, or a bow and arrow, or a rocket rising through the air toward space. These can be difficult to analyze if the number of fragments after the collision is more than about three or four; but nevertheless, the total momentum of the system before and after the explosion is identical.

Note that if the object is initially motionless, then the system (which is just the object) has no momentum and no kinetic energy. After the explosion, the net momentum of all the pieces of the object must sum to zero (since the momentum of this closed system cannot change). However, the system will have a great deal of kinetic energy after the explosion, although it had none before. Thus, we see that, although the momentum of the system is conserved in an explosion, the kinetic energy of the system most definitely is not; it increases. This interaction—one object becoming many, with an increase of kinetic energy of the system—is called an explosion.

Where does the energy come from? Does conservation of energy still hold? Yes; some form of potential energy is converted to kinetic energy. In the case of gunpowder burning and pushing out a bullet, chemical potential energy is converted to kinetic energy of the bullet, and of the recoiling gun. For a bow and arrow, it is elastic potential energy in the bowstring.

Inelastic

The second possibility is the reverse: that two or more objects collide with each other and stick together, thus (after the collision) forming one single composite object. The total mass of this composite object is the sum of the masses of the original objects, and the new single object moves with a velocity dictated by the conservation of momentum. However, it turns out again that, although the total momentum of the system of objects remains constant, the kinetic energy doesn’t; but this time, the kinetic energy decreases. This type of collision is called inelastic.

Any collision where the objects stick together will result in the maximum loss of kinetic energy (i.e., $K_{\text{f}}$ will be a minimum).

Such a collision is called perfectly inelastic. In the extreme case, multiple objects collide, stick together, and remain motionless after the collision. Since the objects are all motionless after the collision, the final kinetic energy is also zero; therefore, the loss of kinetic energy is a maximum.

• If $0<K_{\text{f}}<K_{\text{i}}$, the collision is inelastic.
• If $K_{\text{f}}$ is the lowest energy, or the energy lost by both objects is the most, the collision is perfectly inelastic (objects stick together).
• If $K_{\text{f}}=K_{\text{i}}$, the collision is elastic.

Elastic

The extreme case on the other end is if two or more objects approach each other, collide, and bounce off each other, moving away from each other at the same relative speed at which they approached each other. In this case, the total kinetic energy of the system is conserved. Such an interaction is called elastic.

Example 9.10

Formation of a Deuteron

A proton (mass $1.67\times {10}^{\mathrm{-27}}\text{kg}$) collides with a neutron (with essentially the same mass as the proton) to form a particle called a deuteron. What is the velocity of the deuteron if it is formed from a proton moving with velocity $7.0\times {10}^6\text{m/s}$ to the right and a neutron moving with velocity $4.0\times {10}^6\text{m/s}$ to the left?

Strategy

Define the system to be the two particles. This is a collision, so we should first identify what kind. Since we are told the two particles form a single particle after the collision, this means that the collision is perfectly inelastic. Thus, kinetic energy is not conserved, but momentum is. Thus, we use conservation of momentum to determine the final velocity of the system.

Solution

Treat the two particles as having identical masses M. Use the subscripts p, n, and d for proton, neutron, and deuteron, respectively. This is a one-dimensional problem, so we have

$$M{v}_{\text{p}}-M{v}_{\text{n}}=2M{v}_{\text{d}}.$$

The masses divide out:

$$\begin{array}{ccc}\hfill v_{\text{p}}-v_{\text{n}}& =\hfill & 2v_{\text{d}}\hfill \\ \hfill 7.0\times {10}^6\text{m/s}-4.0\times {10}^6\text{m/s}& =\hfill & 2v_{\text{d}}\hfill \\ \hfill v_{\text{d}}& =\hfill & 1.5\times {10}^6\text{m/s.}\hfill \end{array}$$

The velocity is thus ${\overrightarrow{v}}_{\text{d}}=\left(1.5\times {10}^6\text{m/s}\right)\widehat{i}$.

Significance

This is essentially how particle colliders like the Large Hadron Collider work: They accelerate particles up to very high speeds (large momenta), but in opposite directions. This maximizes the creation of so-called “daughter particles.”

Example 9.11

Ice Hockey 2

(This is a variation of an earlier example.)

Two ice hockey pucks of different masses are on a flat, horizontal hockey rink. The red puck has a mass of 15 grams, and is motionless; the blue puck has a mass of 12 grams, and is moving at 2.5 m/s to the left. It collides with the motionless red puck (Figure 9.20). If the collision is perfectly elastic, what are the final velocities of the two pucks?

Figure 9.20 Two different hockey pucks colliding. The top diagram shows the pucks the instant before the collision, and the bottom diagram show the pucks the instant after the collision. The net external force is zero.

Strategy

We’re told that we have two colliding objects, and we’re told their masses and initial velocities; we’re asked for both final velocities. Conservation of momentum seems like a good strategy; define the system to be the two pucks. There is no friction, so we have a closed system. We have two unknowns (the two final velocities), but only one equation. The comment about the collision being perfectly elastic is the clue; it suggests that kinetic energy is also conserved in this collision. That gives us our second equation.

The initial momentum and initial kinetic energy of the system resides entirely and only in the second puck (the blue one); the collision transfers some of this momentum and energy to the first puck.

Solution

Conservation of momentum, in this case, reads

$$\begin{array}{ccc}\hfill p_{\text{i}}& =\hfill & p_{\text{f}}\hfill \\ \hfill m_2{v}_{\text{2,i}}& =\hfill & m_1{v}_{\text{1,f}}+m_2{v}_{\text{2,f}}.\hfill \end{array}$$

Conservation of kinetic energy reads

$$\begin{array}{ccc}\hfill K_{\text{i}}& =\hfill & K_{\text{f}}\hfill \\ \hfill \frac{1}{2}{m}_2{v}_{\text{2,i}}^2& =\hfill & \frac{1}{2}{m}_1{v}_{\text{1,f}}^2+\frac{1}{2}{m}_2{v}_{\text{2,f}}^2.\hfill \end{array}$$

There are our two equations in two unknowns. The algebra is tedious but not terribly difficult; you definitely should work it through. The solution is

$$\begin{array}{ccc}\hfill v_{\text{1,f}}& =\hfill & \frac{\left(m_1-m_2\right)v_{\text{1,i}}+2m_2{v}_{\text{2,i}}}{m_1+m_2}\hfill \\ \hfill v_{2\text{f}}& =\hfill & \frac{\left(m_2-m_1\right)v_{\text{2,i}}+2m_1{v}_{\text{1,i}}}{m_1+m_2}.\hfill \end{array}$$

Substituting with the given numbers where a positive direction is to the left, we obtain

$$\begin{array}{ccc}\hfill v_{\text{1,f}}& =\hfill & 2.22\frac{\text{m}}{\text{s}}\hfill \\ \hfill v_{\text{2,f}}& =\hfill & \mathrm{-0.28}\frac{\text{m}}{\text{s}}.\hfill \end{array}$$

Significance

Notice that after the collision, the blue puck is moving to the right; its direction of motion was reversed. The red puck is now moving to the left.

Subatomic Collisions and Momentum

Conservation of momentum is crucial to our understanding of atomic and subatomic particles because much of what we know about these particles comes from collision experiments.

At the beginning of the twentieth century, there was considerable interest in, and debate about, the structure of the atom. It was known that atoms contain two types of electrically charged particles: negatively charged electrons and positively charged protons. (The existence of an electrically neutral particle was suspected, but would not be confirmed until 1932.) The question was, how were these particles arranged in the atom? Were they distributed uniformly throughout the volume of the atom (as J.J. Thomson proposed), or arranged at the corners of regular polygons (which was Gilbert Lewis’ model), or rings of negative charge that surround the positively charged nucleus—rather like the planetary rings surrounding Saturn (as suggested by Hantaro Nagaoka), or something else?

The New Zealand physicist Ernest Rutherford (along with the German physicist Hans Geiger and the British physicist Ernest Marsden) performed the crucial experiment in 1909. They bombarded a thin sheet of gold foil with a beam of high-energy (that is, high-speed) alpha-particles (the nucleus of a helium atom). The alpha-particles collided with the gold atoms, and their subsequent velocities were detected and analyzed, using conservation of momentum and conservation of energy.

If the charges of the gold atoms were distributed uniformly (per Thomson), then the alpha-particles should collide with them and nearly all would be deflected through many angles, all small; the Nagaoka model would produce a similar result. If the atoms were arranged as regular polygons (Lewis), the alpha-particles would deflect at a relatively small number of angles.

What actually happened is that nearly none of the alpha-particles were deflected. Those that were, were deflected at large angles, some close to $180\text{°}$—those alpha-particles reversed direction completely (Figure 9.21). None of the existing atomic models could explain this. Eventually, Rutherford developed a model of the atom that was much closer to what we now have—again, using conservation of momentum and energy as his starting point.

Figure 9.21 The Thomson and Rutherford models of the atom. The Thomson model predicted that nearly all of the incident alpha-particles would be scattered and at small angles. Rutherford and Geiger found that nearly none of the alpha particles were scattered, but those few that were deflected did so through very large angles. The results of Rutherford’s experiments were inconsistent with the Thomson model. Rutherford used conservation of momentum and energy to develop a new, and better model of the atom—the nuclear model.