Momentum And Collisions

Newton's 2nd Law Revisited:

Momentum is the measure of a body’s tendency to remain in motion. This quantity is proportional to both the mass and the velocity of the object. Mathematically, it is the product of the mass and the velocity.

(1)
\begin{align} \vec p = m \vec v \end{align}

Notice that this is a vector equation! In this way, momentum is different from kinetic energy, which is a scalar. When Newton wrote his 2nd Law of motion, he was really meaning something more general than $\vec F = m \vec a$. Here’s a excerpt from Newton’s famous book Philosophiae Naturalis Principia Mathematica where Newton describes his 2nd Law of Motion:

“The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.”

We can use this notion of momentum to rewrite our previous notion of Newton’s 2nd Law as

(2)
\begin{align} \vec F = \frac { \Delta \vec p } { \Delta t } \end{align}

Impulse

Impulse, from a physics perspective, is the change in momentum that results in applying a force over a given period of time. Looking at our alteration of Newton's 2nd Law, we can derive a mathematical expression for impulse by multiplying both sides of Eq.2 by $\Delta t$.

(3)
\begin{align} \vec {I} = \vec {F} \Delta t = \Delta \vec {p} = \Delta \left( m \vec {v} \right) \end{align}

Notice this, too, is a vector equation. Impulse does have direction and it is important! During a collision, the impulse can be used to resolve what the final velocity for an object will be. Let's look at a simple example of applying a force for a given amount of time, and using the impulse to determine the final velocity of a spacecraft.

Example:
A 135-kg spacecraft moving with an initial velocity, $\vec {v_i}$ = <40, 20, -5> m/s, fires its thrusters,
producing a force of $\vec F$ = <150, -120, 40> N, for 18 seconds. Determine the final velocity.

$\vec {F} \Delta t = \Delta \left( m \vec {v} \right)$
Since the mass remains constant, we can rewrite the above equation as
$\vec {F} \Delta t = m \Delta \left( \vec {v} \right)$
Divide both sides by mass.
$\vec {F} \Delta t / m = \vec {v_f} - \vec {v_i}$
Now we can solve for the final velocity.
$\vec {v_f} = \vec {v_i} + \vec {F} \Delta t / m$
$\vec {v_f} = <40, 20, -5> + \left( 18 / 135 \right) * <150, -120, 40> = <60, 4.0, 0.33> m/s$

Very often in in game programming, the term Impulse refers to not a change in momentum, but a change in velocity. Although the definition and application are similar to the formal physical definition and applications, don't confuse the two. If it helps, think of the game development usage as assuming a unit mass. Let's take a look at an example using the classic coin-op game, Asteroids.

Example:
The player's ship is moving with a velocity of $\vec {v_i} = <12, 5> px/step$ and
experiences a thrust of $\vec {F} = <-0.3, -0.5> px/step^2$ over a period of 60 steps,
where the mass has a value of 1 and is unitless. Find the final velocity of the ship.

$\vec {I} = \vec {F} t$
$\vec {I} = <-0.3, -0.5> px/step^2 * 24 steps$
$\vec {I} = <-0.3*24, -0.5*24> px/step$
$\vec {I} = <-7.2, -12> px/step$

Since we're assuming a unit mass, $\vec {I}$ is numerically the same as $\Delta \vec {v}$
$\Delta \vec {v} = \vec {v_f} - \vec {v_i}$
$\vec {v_f} = \vec {v_i} + \Delta \vec {v}$
$\vec {v_f} = <12, 5> px/step + <-7.2, -12> px/step$
$\vec {v_f} = <12-7.2, 5-12> px/step$
$\vec {v_f} = <4.8, -7> px/step$

Recoil

One of the most easily demonstrated affects of momentum and impulse is the feeling of recoil or thrust. When you fire a shotgun, for instance, the detonation of the charge pushes the shot at high velocity. This represents a change in momentum for the shot, but if the shot, gun, and shooter are considered as a single system, the force applied to the shot by the charge is an internal force and therefore there should be no change in momentum for the system as a whole. This means that if the shot is carrying momentum in one direction, out of the barrel, then the rest of the system, namely the gun and shooter, will carry that same amount of momentum in the opposite direction. Let's take a look at an example.

Example
A 75-kg person fires a 28.35-g slug from a 2.5-kg 12-gauge shotgun. The shooter holds the
shotgun tightly to their shoulder so that the shooter and the gun recoil as a unit from the shot.
If the slug leaves the gun with a velocity of 549 m/s, with what velocity do the shooter and
gun recoil?

$\left( m_{shooter} + m_{gun} \right) v_{recoil} = m_{slug} v_{slug}$
$v_{recoil} = \left( m_{slug} v_{slug} \right) / \left( m_{shooter} + m_{gun} \right)$
$v_{recoil} = \left( 0.02835 kg * 549 m/s \right) / \left( 75 kg + 2.5 kg \right) = 0.20 m/s$

That's a stout recoil and why a shotgun should always be fired held tight to the shoulder. Let's examine the recoil velocity if the shotgun is held off of the shoulder so that the shooter's mass is no longer part of the system.

$m_{gun} v_{recoil} = m_{slug} v_{slug}$
$v_{recoil} = \left( m_{slug} v_{slug} \right) / m_{gun}$
$v_{recoil} = \left( 0.02835 kg * 549 m/s \right) / \left( 2.5 kg \right) = 6.2 m/s$

That's a significant difference! Keep in mind that 6.2 m/s is equal to 14 mph. Holding a shotgun in this fashion is a good way to end up with a broken shoulder.

Rocket Propulsion

Recoil is an impulsive event and encapsulated in a finite time interval, but what if mass were continuously being ejected? There would be a continual change in the momentum of the remaining system. This is exactly how jet and rocket motors function to generate thrust. Let's examine the recoil effect of exhausting a parcel of combustion gas out of an engine at a specific speed.

(4)
\begin{align} m_{vehicle} \Delta v_{vehicle} = m_{exhaust} v_{exhaust} \end{align}

Dividing both sides by the time interval over which the combustion gases are exhausted from the vehicle yields a statement of the force applied.

(5)
\begin{align} \frac {m_{vehicle} \Delta v_{vehicle}} {\Delta t} = \frac {m_{exhaust} v_{exhaust}} {\Delta t} \\ m_{vehicle} \left( \frac {\Delta v_{vehicle}} {\Delta t} \right) = \left( \frac {m_{exhaust}} {\Delta t} \right) v_{exhaust} \end{align}

Notice on the left side of the equation, we have $\Delta v / \Delta t$. This is an acceleration! Therefore, the left side of the above equation represents the force on the vehicle, or the thrust. On the right side of the equation, we have $\Delta m / \Delta t$. This is the rate at which fuel is being combusted and is called the "Burn Rate". Now we can rewrite the above equation in terms of Thrust, Burn Rate, and Exhaust Velocity.

(6)
\begin{align} \text{Thrust} = \text{Burn Rate} * \text{Exhaust Velocity} \end{align}

1D Collisions

When two objects collide, momentum may be transferred from one to the other. The total momentum of the system, however, will always remain the same. If one object loses momentum, the other object must gain momentum. Furthermore, the amount that one loses and the other gains must be equal. This is the foundation for the Conservation of Linear Momentum. This conservation law states that the total momentum of a system before a collision must equal the total momentum of a system afterwords. In the language of algebra, this is

(7)
\begin{equation} m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} . \end{equation}

There are two different types of collisions, elastic and inelastic. The most important difference between the two is that elastic collisions conserve kinetic energy. Elastic collisions involve two objects that collide and bounce off of each other while retaining their original shapes. Two billiard balls colliding would be an example of an elastic collision. Inelastic collisions involve deformation of one or both of the colliding objects. This is why inelastic collisions do not conserve kinetic energy. An example of this type of collision would be the collision between a two pieces of putty. Part of the system’s original kinetic energies used to deform the shape of the objects or to increase the temperature of the two objects.

Perfectly Inelastic Collisions

A perfectly inelastic collision is a special case in which the two colliding bodies stick together after the collision. They then move off as a single unit with the same velocity. Mathematically, this means the final velocities in Eq.(7) are equal and the equation can be rewritten as

(8)
\begin{align} m_1 v_{1i} + m_2 v_{2i} = \left( m_1 + m_2 \right) v_{f} . \end{align}

Real Collisions

Real collisions are typically somewhere between the two cases of a perfectly elastic and perfectly elastic collision. The degree to which the collision is elastic can be described by the coefficient of restitution, $\varepsilon$. For a perfectly elastic collision, the total amount of kinetic energy is conserved. In a real collision, some of this kinetic energy is lost. By using the conservation of momentum and kinetic energy for an elastic collision, one can derive what’s called the Velocity Equation

(9)
\begin{equation} v_{2f} - v_{1f} = v_{1i} - v_{2i} . \end{equation}

When the collision is not precisely elastic, the right side of Eq.(9) is multiplied by coefficient of restitution. If $\varepsilon = 1$, the collision is perfectly elastic. If $\varepsilon = 0$, the collision is perfectly inelastic.

(10)
\begin{align} v_{2f} - v_{1f} = \varepsilon \left( v_{1i} - v_{2i} \right) . \end{align}

This means that one can use the same equation for both elastic and inelastic collisions, and therefore the same function block. You won’t need two separate routines by using the coefficient of restitution.

Reflections

So far, only one-dimensional collisions have been detailed, but multi-dimensional collisions are simply a superposition of the two or three orthogonal directions. Eqs. (7) and (10) can be used to evaluate the final velocities in the x, y, and z directions separately. So long as the components of the initial velocities are known, finding the three-dimensional final velocities is simply a matter of applying Eqs. (7) and (10) three times, once for each direction.
One special type of collision is the reflection. Imagine a billiard ball colliding with a rail and bouncing off. Figure 1 shows a diagram of this type of collision. Here we assume the mass of the rail is vastly greater than the mass of the ball. In the case of the rail being aligned with a coordinate axis, as is shown in Figure 1, then Eq. (10) in the direction perpendicular to the rail becomes

(11)
\begin{align} v_{1i} = - \varepsilon \cdot v_{1f} \end{align}

The parallel velocity will remain the same. This is a special case. The surface with which an object collides is not always going to be aligned with a coordinate axis. To address the more general problem of an object bouncing off of a surface with any given orientation, we will need to do some more geometry, see Figure 2. As with the ball-rail problem, the component of the velocity parallel to the surface remains unchanged, while the velocity perpendicular to the surface is inverted as described in Eq. (11). Using this notion of parallel and perpendicular velocities, we can write the initial and final velocities as such,

(12)
\begin{align} \vec v_{i} = \left( \begin{array}{cc} v_{\parallel} \\ v_{\perp} \end{array} \right) , \end{align}

and

(13)
\begin{align} \vec v_{f} = \left( \begin{array}{cc} v_{\parallel} \\ - \varepsilon \cdot v_{\perp} \end{array} \right) . \end{align}

The change in velocity, given the above, is

(14)
\begin{align} \Delta \vec v_{f} = \left( \begin{array}{cc} 0 \\ - \left( \varepsilon + 1 \right) \cdot v_{\perp} \end{array} \right) , \end{align}

such that

(15)
\begin{align} \vec v_f = \vec v_i + \Delta \vec v \end{align}

The more general case of Eqs. (14) and (15) involve using the normal vector to the surface to find the perpendicular projection of the initial velocity vector. The general form of Equation 10 now becomes

(16)
\begin{align} \vec v_f = \vec v_i - \left( \varepsilon + 1 \right) \left( \vec v_i \cdot \hat n \right) \hat n \end{align}

where $\hat n$ is the vector normal to the surface. This equation will work for either the 2D or 3D case since it’s only the perpendicular velocity that is changing.

Questions and Exercises

Concept Questions

1. What are the two types of collisions? What do they have in common, and how do they differ?
2. A hollywood stunt coordinator will use a large airbag so that a stunt double can "fall" from a building and land safely. Using our revised form of Newton's 2nd Law, describe what in the equation is changing that allows the stunt double to survive the fall unharmed.
3. Both momentum and kinetic energy are computed using an object’s mass and its velocity. What is the primary difference between the two quantities?

Problems

1. A 700-kg Indy-style race car decelerates from 93 m/s to 31 m/s in 0.137 s as it crashes into the outside wall of a turn. What is the total impulse on the car and what is the average force on the car during the crash.
2. A 85-kg person running at 5.4 m/s jumps into a 11-kg canoe. What is the final velocity of the person and canoe?
3. How much kinetic energy is gained or lost during the collision described in Problem #2? What type of collision is this? Defend your answer.
4. A 3.3-kg ball moving to the right with a velocity of 7.0 m/s collides elastically with 7.9-kg ball moving to the left with a velocity of 4.9 m/s. What are the final velocities of the two balls?
5. A ball moving with a velocity of $\vec v = \left( 16.5 \hat x + 1.2 \hat y \right) m/s$ collides with a wall described by the vector, $\vec B = 5.4 \hat x - 10.8 \hat y$ where $\vec B$ is parallel to the wall. What is the velocity of the ball after the collision assuming the coefficient of restitution is 0.75?
6. Use Equations 7 and 10 to write a general expression for the final velocity of Object 2. You will need this to create your collision function.
7. Write a general expression for the final velocity of Object 1 using Equation 10 and the results from the above question.
page revision: 17, last edited: 21 Oct 2018 14:01