# Collisions

Edited by msridhar30 (Spring 2018)

A **collision** is the act in which two or more entities exert forces on each other over a time period, often expressed in a short period. Collisions behave according to the fundamental principles of physics. The nature of collisions allow for assumptions, as discussed later, to solve for unknowns.

## The Main Idea

If we take the colliding objects as the system, what assumption can we make about the total momentum and energy of this system?

In our definition of a collision, we've said that interactions preceding and succeeding the process are tiny compared to the ones we see during the process. The forces that contribute to this sudden spike of interactions is internal to the system: they are between the colliding objects. Since we take them as the system, we can make the assumption that the outside forces that act on the system are negligible.

By this assumption, the change in the total momentum and energy of the system (of both the colliding objects) is negligible.

For [math]\displaystyle{ \Delta p = F * \Delta t }[/math], [math]\displaystyle{ F }[/math] is negligible, thus [math]\displaystyle{ \Delta p }[/math] is negligible.

For [math]\displaystyle{ \Delta E = W_{surr} + Q }[/math], [math]\displaystyle{ W_{surr} = F_{surr} * \Delta x }[/math] and [math]\displaystyle{ Q = m * C * \Delta t }[/math]. [math]\displaystyle{ F_{surr} }[/math] and [math]\displaystyle{ \Delta t }[/math] are negligible. Thus [math]\displaystyle{ \Delta E }[/math] is negligible.

Therefore, we can assume that the total momentum and energy of the system (of both the colliding objects) don't change during the collision. Furthermore, the assumptions of conservation of momentum and conservation of kinetic energy allow us to find the final velocities of each entity in a two-body collision.

Example: Swinging Masses

The conservation of momentum and conservation of energy can be identified in a system which features steel spherical masses that hang side by side in a horizontal line from each other. If a single mass is pulled to the side and released, it follows in the opposing motion and strikes the line. The outer mass swings in the line of movement. Similarly, if two balls are pulled to the side and released, the other two balls swing out in the line of movement.

p_initial = p_final KE_initial = KE_final

Case 1:momentum in = momentum out mv = momentum out Kinetic Energy in = Kinetic Energy out (1/2)*mv^2 = kinetic energy outCase 2:momentum in = momentum out 2mv = momentum out Kinetic Energy in = Kinetic Energy out (1/2)*2*mv^2 = kinetic energy out

Both the conservation of momentum and conservation of energy are respected in both cases. The conservation of momentum and conservation of energy allow for the understanding of the relationships between internal and external forces. Realistically, there are very few perfectly elastic collisions. Often, kinetic energy is converted into internal energy, or dissipates as sound or heat energy.

**Types of Collisions**

In Elastic Collisions, the total kinetic energy is conserved. None of the kinetic energy of the system is transformed from or into [math]\displaystyle{ E_{internal} }[/math], such as thermal energy and spring potential energy.

Elastic collisions are regarded as one-dimensional collisions where the impact is along the line of movement.

An example for such a process is the Newton's Cradle, where the kinetic energy of the colliding ball gets transformed into the kinetic energy of the last ball in the cradle. The total kinetic energy is conserved.

In Inelastic Collisions, the total kinetic energy is not conserved. Some/all of the kinetic energy of the system can be transformed from or into [math]\displaystyle{ E_{internal} }[/math].

Inelastic collisions are regarded as two-dimensional collisions where the impact is not along the line of movement.

An example for such a process is a car crash, where the kinetic energy of the colliding car gets transformed into thermal energy and internal energy contained in the crumbled, tense parts of the car chassis.

(Inelastic Collisions where there's maximum kinetic energy dissipation
are called Maximally Inelastic Collision. This doesn't necessarily mean
that *all* of the kinetic energy gets transformed into [math]\displaystyle{ E_{internal} }[/math],
since the total momentum is conserved. Collisions where the objects stick together are
Maximally Inelastic Collision.)

### Mathematical model

By our assumptions, [math]\displaystyle{ p_f = p_i }[/math] and [math]\displaystyle{ E_f = E_i }[/math], where subscripts [math]\displaystyle{ f }[/math] and [math]\displaystyle{ i }[/math] mean final and initial respectively. These equations hold true for all collisions.

In an elastic collision, [math]\displaystyle{ K_{total, f} = K_{total, i} }[/math].

In an inelastic collision, this equation doesn't hold, since [math]\displaystyle{ \Delta E = \Delta K + \Delta E_{internal} = 0 }[/math], and [math]\displaystyle{ \Delta E_{internal} }[/math] might be nonzero.

### Computational model

In this model, two objects of different masses are moving towards each other with equal magnitude, but opposite direction momenta.

The objects collide, and after the collision, it is observed that the objects are embedded within each other with final velocity zero.

This is an example of a Maximally Inelastic Collision: none of the kinetic energy is conserved.

Note: The behavior of the colliding objects are hardcoded in the model to embed (vs. bouncing, etc.) This is an assumption we make about the materials/internal structure of the balls. Correctly modeling deforming objects without simplifying assumptions is a very complex task.

## Examples

### Simple

**Question**

A [math]\displaystyle{ 0.5 kg }[/math] soccer ball is moving with a speed of [math]\displaystyle{ 5 m/s }[/math] directly towards a [math]\displaystyle{ 0.7 kg }[/math] basketball, which is at rest.

The two balls collide and stick together. What will be their final speed?

**Answer**

Let [math]\displaystyle{ m_1 = 0.5 kg }[/math], [math]\displaystyle{ v_1 = 5 m/s }[/math], [math]\displaystyle{ m_2 = 0.7 kg }[/math], [math]\displaystyle{ v_2 = 0 m/s }[/math], where subscripts [math]\displaystyle{ 1 }[/math] and [math]\displaystyle{ 2 }[/math] signify the soccer ball and basketball respectively.

We're going to compute [math]\displaystyle{ v_f }[/math] for the collided mass.

We know that momentum is conserved. [math]\displaystyle{ p_f = p_i \implies p_f = p_1 + p_2 \implies p_f = (m_1 * v_1) + (m_2 * v_2) }[/math].

We know that the objects stick together, hence their mass [math]\displaystyle{ m_1 + m_2 }[/math]. Then [math]\displaystyle{ p_f = (m_1 + m_2) * v_f \implies (m_1 + m_2) * v_f = (m_1 * v_1) + (m_2 * v_2) }[/math].

If we solve for [math]\displaystyle{ v_f }[/math], we get

[math]\displaystyle{ \begin{aligned} v_f &= \frac{(m_1 * v_1) + (m_2 * v_2)}{m_1 + m_2} \\ &= \frac{(0.5 * 5) +(0.7 * 0)}{0.5 + 0.7} \\ &= 2.083 m/s \end{aligned} }[/math]

This is a Maximally Inelastic Collision, as the collided objects stuck together. [math]\displaystyle{ K_{tot, f} \lt K_{tot, i} }[/math].

### Middling

**Question**

A [math]\displaystyle{ 10 kg }[/math] asteroid with a velocity of [math]\displaystyle{ \lt 10, 0, 0\gt m/s }[/math] crashes into the Earth in the Sahara Desert and bounces back into the space with the same speed and opposite direction.

Nobody is hurt, but some say that they felt the Earth recoil during this unusual collision. Find the recoil velocity of the earth to see if it was significant enough for people to have felt it. (Keep in mind that the human sensory facilities aren't very precise.)

**Answer**

Let [math]\displaystyle{ m_a = 10 kg }[/math], [math]\displaystyle{ m_e = 6 * 10^{24} kg }[/math], [math]\displaystyle{ v_{a, i} = \lt 10, 0, 0\gt m/s }[/math], [math]\displaystyle{ v_{a, f} = \lt -10, 0, 0\gt m/s }[/math], and [math]\displaystyle{ v_{e, i} = \lt 0, 0, 0\gt m/s }[/math], where subscripts [math]\displaystyle{ a }[/math] and [math]\displaystyle{ e }[/math] signify the asteroid and the earth respectively.

We're going to compute [math]\displaystyle{ v_{e, f} }[/math]

We know that momentum is conserved.

[math]\displaystyle{ \begin{aligned} &p_f = p_i \\ \implies &p_{a, f} + p_{e, f} = p_{a, i} + p_{e, i} \\ \implies &(m_a * v_{a, f}) + (m_e * v_{e, f}) = (m_a * v_{a, i}) + (m_e * v_{e, i}) \end{aligned} }[/math]

Solving for [math]\displaystyle{ v_{e, f} }[/math], we get

[math]\displaystyle{ \begin{aligned} v_{e, f} &= \frac{(m_a * v_{a, i}) + (m_e * v_{e, i}) - (m_a * v_{a, f})}{m_e} \\ &= \frac{(10 * \lt 10, 0, 0\gt ) + (6 * 10^{24} * \lt 0, 0, 0\gt ) - (10 * \lt -10, 0, 0\gt )}{6 * 10^{24}} \\ &= \frac{\lt 100, 0, 0\gt + \lt 100, 0, 0\gt }{6 * 10^{24}} \\ &= \lt 3.33 * 10^{-23}, 0, 0\gt m/s \end{aligned} }[/math]

The recoil velocity of the earth is too small for anyone to feel it.

### Difficult

**Question**

Two rocks in space are approaching a third, stationary rock. Rock 1 has a mass of [math]\displaystyle{ 30 kg }[/math] and moves with a velocity [math]\displaystyle{ \lt 13, -10, -2\gt m/s }[/math]. Rock 2 has a mass of [math]\displaystyle{ 60 kg }[/math], and moves with a velocity [math]\displaystyle{ \lt -3, 23, 3\gt m/s }[/math]. Rock 3 has a mass of [math]\displaystyle{ 35 kg }[/math] and is stationary. The rocks collide at Rock 3's location.

After the collision, it is observed that Rock 1 has embedded into Rock 3, which is now moving with a new velocity of [math]\displaystyle{ \lt 3, 5, 2\gt m/s }[/math], and that a chunk of mass [math]\displaystyle{ 15 kg }[/math] has broken off of the combined rock.

Rock 2, in the meantime, has a new velocity of [math]\displaystyle{ \lt -2, 12, 2\gt m/s }[/math].

What is the change in the internal energy [math]\displaystyle{ \Delta E_{internal} }[/math] in the system as a result of this three-way collision?

**Answer**

Due to the complexity of this question, we are going to use a computer to solve this problem. Conceptually, everything remains the same, but instead of computing things by hand, we're getting help from a computer. Feel free to solve it by hand if you want to!

# Define the initial masses/velocities of the rocks. mass_1_i = 30 mass_2_i = 60 mass_3_i = 35 v_1_i = vector(13, -10, -2) v_2_i = vector(-3, 23, 3) v_3_i = vector(0, 0, 0) # Calculate the initial total momentum. total_p_i = (mass_1_i * v_1_i) + (mass_2_i * v_2_i) + (mass_3_i * v_3_i) # Calculate the masses/velocities of the rocks after the collision. mass_chunk = 15 mass_2_f = 60 mass_3_f = mass_1_i + mass_3_i - mass_chunk v_2_f = vector(-2, 12, 2) v_3_f = vector(3, 5, 2) # Calculate the velocity of the chunk. Remember that the momentum is conserved. v_chunk = (total_p_i - ((v_2_f * mass_2_f) + (v_3_f * mass_3_f))) / mass_chunk # Calculate the initial total kinetic energy. def kinetic_energy(mass, velocity): return 1/2 * mass * mag(velocity) ** 2 k_tot_i = kinetic_energy(mass_1_i, v_1_i) \ + kinetic_energy(mass_2_i, v_2_i) \ + kinetic_energy(mass_3_i, v_3_i) # Calculate the final total kinetic energy. k_tot_f = kinetic_energy(mass_2_f, v_2_f) \ + kinetic_energy(mass_3_f, v_3_f) \ + kinetic_energy(mass_chunk, v_chunk) # Calculate the change in internal energy. d_e_internal = k_tot_f - k_tot_i

This Python snippet, when evaluated, will show that [math]\displaystyle{ \Delta E_{internal} = 5175 J }[/math].

## Connectedness

Collisions are everywhere in our daily lives. Some examples are: a bat colliding with a baseball in your neighbor's backyard, two cars crashing in the highway in front of you, your finger hitting the keyboard, and so on.

Depending on the nature of the collision we can usually guess whether it's an Elastic Collisions or an Inelastic Collisions, and reason about it. (If it's a car crash, things crumble, so there's increase in thermal energy, etc.)

Implications of collisions may be included in the understanding of the ideal gas law. As temperatures rise within a given sector, the movement of molecules become excited. Molecules of ideal gases will move with perfect elasticity in relation to one another. In other words, the kinetic energy before the initial collision will remain the same as the final kinetic energy after the collision. However, most gases do not behave ideally. More accurately, moments of kinetic energy may dissipate or be absorbed as heat energy and sound energy, respectively.

Another very interesting application of collisions is in the field of astronomy.

The universe is ever-changing and we never know what might come our way. A large meteor could be flying at the Earth with a very high speed and we could use our knowledge of collisions to find out how the Earth will move after the collision, what the temperature change of the area would be, and how the rest of the Earth would be effected by the collision. By using our collision skills, we could evacuate anyone who is prone to harm during the collision.

Since collisions are ubiquitous processes, knowledge about them is instrumental for any engineer who deals with the physical world.

## History

We've mentioned that collisions are not special case events in Physics-- the fundamental principles apply fully for them, and we reason about them using these principles. Then it won't surprise us that the history of collisions is tangled with the history of the fundamental principles.

In 1661, Christiaan Huygens, a Dutch physicist, concluded that when a moving body struck a stationary object of equal mass, the initial moving body would lose all of its momentum while the second object would pick up the same amount of velocity that the moving body had before the collision. He also stated that the total "quantity of motion" should be the same before and after the collision. This was the first thought of the conservation of momentum.

In 1687, Sir Isaac Newton went a little further and stated in his third law of motion that the forces of action and reaction between two bodies are opposite and equal. Because of this law; we know that the total momentum of a closed system is constant. Every force in the system has a reciprocal pair with an opposing direction, which implies that their momenta cancel out, keeping the total momentum constant.

In Ernest Rutherford's famous Gold Foil Experiment of 1899, he found out that the plum pudding model of the atom was inaccurate, and that there was a concentrated core at the center of the atom (nucleus) and a lot of empty space around. He didn't have access to the sophisticated tools/knowledge that we have now to assist in his discovery: he did this through investigating atomic collisions between alpha particles and gold foil. He shot alpha particles to a piece of thin gold foil, and observed that some of the particles were scattered through angles larger than 90 degrees during the collisions. Using collision principles, he was then able to reason that the atom had most of its mass concentrated in a dense core, rather than spread out evenly throughout the atom as the plum pudding model suggested.

## See also

Other pages related to collisions:

- Maximally Inelastic Collision
- Inelastic Collisions
- Elastic Collisions
- Linear Momentum
- Energy Principle,
- Kinetic Energy
- Net Force.

### Further reading

Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. 4th ed. Vol. 1. Hoboken, NJ: Wiley, 2015. Print. Matter and Interactions.

Ehrhardt, H., and L. A. Morgan. Electron Collisions with Molecules, Clusters, and Surfaces. New York: Plenum, 1994. Print.

### External links

Simulation for collisions: Collision Simulation

Video on different collision types: video

## References

Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. 4th ed. Vol. 1. Hoboken, NJ: Wiley, 2015. Print. Matter and Interactions.

"The Gold Foil Experiment." The Gold Foil Experiment. N.p., n.d. Web. 17 Apr. 2016.