What is momentum?

Momentum is a word that we hear used colloquially in everyday life. We are often told that sports teams and political candidates have "a lot of momentum". In this context, the speaker usually means to imply that the team or candidate has had a lot of recent success and that it would be difficult for an opponent to change their trajectory. This is also the essence of the meaning in physics, though in physics we need to be much more precise.

Momentum is a measurement of mass in motion: how much mass is in how much motion. It is usually given the symbol \[\mathbf{p}\].

By definition, \[\boxed{\mathbf{p} = m \cdot \mathbf{v}}.\]

Where \[m\] is the mass and \[\mathbf{v}\] is the velocity. The standard units for momentum are \[\mathrm{kg \cdot m/s}\], and momentum is always a vector quantity. This simple relationship means that doubling either the mass or velocity of an object will simply double the momentum.

The useful thing about momentum is its relationship to force. You might recall from the kinematic equations that change in velocity \[\Delta v\] can also be written as \[a\cdot \Delta t\].

We can then see that any change in momentum following an acceleration can be written as

\[\begin{aligned} \Delta \mathbf{p} &= m\cdot \Delta v\\
&= m\cdot \mathbf{a}\cdot \Delta t
\\ &= \mathbf{F}\cdot \Delta t\end{aligned}\]

What is impulse?

Impulse is a term that quantifies the overall effect of a force acting over time. It is conventionally given the symbol \[\text{J}\] and expressed in Newton-seconds.

For a constant force, \[\mathbf{J} = \mathbf{F} \cdot \Delta t\].

As we saw earlier, this is exactly equivalent to a change in momentum \[\Delta \mathbf{p}\]. This equivalence is known as the impulse-momentum theorem. Because of the impulse-momentum theorem, we can make a direct connection between how a force acts on an object over time and the motion of the object.

One of the reasons why impulse is important and useful is that in the real world, forces are often not constant. Forces due to things like people and engines tend to build up from zero over time and may vary depending on many factors. Working out the overall effect of all these forces directly would be quite difficult.

When we calculate impulse, we are multiplying force by time. This is equivalent to finding the area under a force-time curve. This is useful because the area can just as easily be found for a complicated shape—variable force—as for a simple rectangle—constant force. It is only the overall net impulse that matters for understanding the motion of an object following an impulse.

Explain, please.

The concept of impulse that is both external and internal to a system is also fundamental to understanding conservation of momentum.

Momentum in space

Most people are familiar with seeing astronauts working in orbit. They appear to effortlessly push around freely floating objects. Because astronauts and the objects they are working with are both in free-fall, they do not have to contend with the force of gravity. However, heavy moving objects still possess the same momentum that they do on earth, and it can be just as difficult to change this momentum.

Suppose that an emergency occurs on a space station and an astronaut needs to manually move a free-floating 4,000 kg space capsule away from a docking area. On earth, the astronaut knows she can hold a 50 kg weight above herself for 3 seconds. How quickly could she get the capsule moving?

We first calculate the total impulse that the astronaut can apply. Note that the astronaut is pushing vertically in both cases so we don't need to keep track of the direction of the force.

\[\begin{aligned} J &= (mg)\cdot \Delta t\\ &= 50~\mathrm{kg} \cdot 9.81~\mathrm{m/s^2} \cdot 3~\mathrm{s} = 1471.5 ~\mathrm{Ns} \end{aligned}\]

And, by the impulse-momentum theorem, we can find the velocity of the spacecraft:

\[\frac{1471.5~\mathrm{Ns}}{4000~\mathrm{kg}} = 0.37 ~\mathrm{m/s}\]

What is specific impulse?

Specific impulse—\[I_{SP}\]—is a specification commonly given to engines which produce a thrust force. Jet engines and rocket engines are two common examples. In this context, specific impulse is a measure of the efficiency of using fuel to produce thrust and is one of the most important specifications of such an engine.

When the prefix specific is used in physics, it means "relative to" a particular quantity. Specific gravity and specific heat are two examples of where you may have seen this prefix used. Specific impulse is impulse measured relative to the weight of fuel—on earth—used to produce the impulse.

\[I_\mathrm{SP} = \frac{F \cdot \Delta t}{m_\mathrm{fuel}g}\]

Because we are dividing an impulse by a force—the force on the fuel due to earth's gravity—the force units cancel out, and the units for specific impulse are simply seconds.

A rocket might have a specific impulse of 300 s. This means that it could use fuel weighing 1 N to produce 1 N of thrust for 300 s. In practice, the rocket might have some minimum thrust, say 100 N. In this case it could use fuel weighing 1 N to produce the 100 N thrust for 3 s.

Impulse of an aircraft

A Boeing 747 aircraft has four engines, each of which can produce a thrust force of up to 250 kN. It takes around 30 s for the aircraft to get up to take-off speed. The thrust produced by the engines during take off is approximated by the force-time curve shown below.

CF6 Engine thrust during take off of Boeing 747 [1]

Exercise 1a: What is the total impulse produced by the aircraft in getting up to take-off speed?

Exercise 1b: The specific impulse of the jet engines is known to be around 6000 s. How many kilograms of fuel were burned in getting the aircraft up to take-off speed?

Attributions

1. Data from (a) W.A. Fasching 9/1979 NASA-CR-159564 CF6 Jet Engine Performance Improvement Program (b) Project for the Sustainable Development of Heathrow, Ch 3 – Emission Sources. 7/2006.

What is the principle of conservation of momentum?

In physics, the term conservation refers to something which doesn't change. This means that the variable in an equation which represents a conserved quantity is constant over time. It has the same value both before and after an event.

There are many conserved quantities in physics. They are often remarkably useful for making predictions in what would otherwise be very complicated situations. In mechanics, there are three fundamental quantities which are conserved. These are momentum, energy, and angular momentum. Conservation of momentum is mostly used for describing collisions between objects.

Just as with the other conservation principles, there is a catch: conservation of momentum applies only to an isolated system of objects. In this case an isolated system is one that is not acted on by force external to the system—i.e., there is no external impulse. What this means in the practical example of a collision between two objects is that we need to include both objects and anything else that applies a force to any of the objects for any length of time in the system.

If the subscripts \[i\] and \[f\] denote the initial and final momenta of objects in a system, then the principle of conservation of momentum says

\[\mathbf{p}_\mathrm{1i} + \mathbf{p}_\mathrm{2i} + \ldots = \mathbf{p}_\mathrm{1f} + \mathbf{p}_\mathrm{2f} + \ldots\]

Why is momentum conserved?

Conservation of momentum is actually a direct consequence of Newton's third law.

Consider a collision between two objects, object A and object B. When the two objects collide, there is a force on A due to B—\[F_\mathrm{AB}\]—but because of Newton's third law, there is an equal force in the opposite direction, on B due to A—\[F_\mathrm{BA}\].

\[F_\mathrm{AB} = - F_\mathrm{BA}\]

The forces act between the objects when they are in contact. The length of time for which the objects are in contact—\[t_\mathrm{AB}\] and \[t_\mathrm{BA}\]—depends on the specifics of the situation. For example, it would be longer for two squishy balls than for two billiard balls. However, the time must be equal for both balls.

\[t_\mathrm{AB} = t_\mathrm{BA}\]

Consequently, the impulse experienced by objects A and B must be equal in magnitude and opposite in direction.

\[F_\mathrm{AB}\cdot t_\mathrm{AB} = – F_\mathrm{BA}\cdot t_\mathrm{BA}\]

If we recall that impulse is equivalent to change in momentum, it follows that the change in momenta of the objects is equal but in the opposite directions. This can be equivalently expressed as the sum of the change in momenta being zero.

\[\begin{aligned}m_\mathrm{A} \cdot \Delta v_\mathrm{A} &= -m_\mathrm{B} \cdot \Delta v_\mathrm{B} \\
m_\mathrm{A} \cdot \Delta v_\mathrm{A} + m_\mathrm{B} \cdot \Delta v_\mathrm{B} &= 0\end{aligned}\]

What is interesting about conservation of momentum?

There are at least four things that are interesting—and sometimes counter-intuitive—about momentum conservation:

Momentum is a vector quantity, and therefore we need to use vector addition when summing together the momenta of the multiple bodies which make up a system. Consider a system of two similar objects moving away from each other in opposite directions with equal speed. What is interesting is that the oppositely-directed vectors cancel out, so the momentum of the system as a whole is zero, even though both objects are moving.

Collisions are particularly interesting to analyze using conservation of momentum. This is because collisions typically happen fast, so the time colliding objects spend interacting is short. A short interaction time means that the impulse, \[F\cdot \Delta t\], due to external forces such as friction during the collision is very small.

It is often easy to measure and keep track of momentum, even with complicated systems of many objects. Consider a collision between two ice hockey pucks. The collision is so forceful that it breaks one of the pucks into two pieces. Kinetic energy is likely not conserved in the collision, but momentum will be conserved.

Provided we know the masses and velocities of all the pieces just after the collision, we can still use conservation of momentum to understand the situation. This is interesting because by contrast, it would be virtually impossible to use conservation of energy in this situation. It would be very difficult to work out exactly how much work was done in breaking the puck.

• Collisions with "immovable" objects are interesting. Of course, no object is really immovable, but some are so heavy that they appear to be. Consider the case of a bouncy ball of mass \[m\] traveling at velocity \[v\] towards a brick wall. It hits the wall and bounces back with velocity \[-v\]. The wall is well attached to the earth and doesn't move, yet the momentum of the ball has changed by \[2mv\] since velocity went from positive to negative.

If momentum is conserved, then the momentum of the earth and wall also must have changed by \[2mv\]. We just don't notice this because the earth is so much heavier than the bouncy ball.

What kind of problems can we solve using momentum conservation?

Exercise 1a: The recoil of a cannon is probably familiar to anyone who has watched pirate movies. This is a classic problem in momentum conservation. Consider a wheeled, 500 kg cannon firing a 2 kg cannonball horizontally from a ship. The ball leaves the cannon traveling at 200 m/s. At what speed does the cannon recoil as a result?

Show solution.

Exercise 1b: Suppose the cannon was raised to fire at an angle \[\alpha=30^\circ\] to the horizontal. What is the recoil speed in this case? Where did the additional momentum go?

Exercise 2a: A golf club head of mass \[m_c=0.25~\mathrm{kg}\] is swung and collides with a stationary golf ball of mass \[m_b=0.05~\mathrm{kg}\]. High speed video shows that the club is traveling at \[v_c = 40~\mathrm{m/s}\] when it touches the ball. It remains in contact with the ball for \[t=0.5~\mathrm{ms}\]; after that, the ball is traveling at a speed of \[v_b=40~\mathrm{m/s}\]. How fast is the club traveling after it has hit the ball?

Exercise 2b: What is the average force on the club due to the golf ball in the previous problem?

Exercise 3: Suppose a 100 kg football player is at rest on an ice rink. A friend throws a 0.4 kg football towards him at a speed of 25 m/s. In a smooth motion he receives the ball and throws it back in the same direction at a speed of 20 m/s. What is the speed of the player after the throw?

How can we solve 2-dimensional collision problems?

In other articles, we have looked at how momentum is conserved in collisions. We have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy. We have applied these principles to simple problems, often in which the motion is constrained in one dimension.

If two objects make a head on collision, they can bounce and move along the same direction they approached from (i.e. only a single dimension). However, if two objects make a glancing collision, they'll move off in two dimensions after the collision (like a glancing collision between two billiard balls).

For a collision where objects will be moving in 2 dimensions (e.g. x and y), the momentum will be conserved in each direction independently (as long as there's no external impulse in that direction).

In other words, the total momentum in the x direction will be the same before and after the collision.

\[\Large \Sigma p_{xi}=\Sigma p_{xf}\]

Also, the total momentum in the y direction will be the same before and after the collision.

\[\Large \Sigma p_{yi}=\Sigma p_{yf}\]

In solving 2 dimensional collision problems, a good approach usually follows a general procedure:

Identify all the bodies in the system. Assign clear symbols to each and draw a simple diagram if necessary.

Write down all the values you know and decide exactly what you need to find out to solve the problem.

Select a coordinate system. If many of the forces and velocities fall along a particular direction, it is advisable to use this direction as your x or y axis to simplify calculation; even if it makes your axes not parallel to the page in your diagram.

Identify all the forces acting on each of the bodies in the system. Make sure that all impulse is accounted for, or that you understand where external impulses can be neglected. Remember that conservation of momentum only applies in cases where there is no external impulse. However, conservation of momentum can be applied separately to horizontal and vertical components. Sometimes it is possible to neglect an external impulse if it is not in the direction of interest.

Write down equations which equate the momentum of the system before and after the collision. Separate equations can be written down for momentum in the x and y directions.

Solve the resulting equations to determine an expression for the variable(s) you need.

Substitute in the numbers you know to find the final value. Should this require adding vectors, it is often useful to do this graphically. A vector diagram can be drawn and the method of adding vectors head-to-tail used. Trigonometry can then be used to find the magnitude and direction of all the vectors you need to know.

Billiard ball problem

Figure 1 describes the geometry of a collision of a white and a yellow billiard ball. The yellow ball is initially at rest. The white ball is played in the positive-x direction such that it collides with the yellow ball. The collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis.

The mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg. A sound recording reveals that the collision happens 0.25 s after the player has struck the white ball. The yellow ball falls into the pocket 0.35 s after the collision.

Figure 1: Collision of white and yellow billiard balls.

Bouncing baseball

Consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine. The machine is set to deliver the 0.145 kg ball at 10 m/s. The ball hits the board, making an angle of 45° from the surface of the board. The board is slightly flexible and the collision is inelastic. The ball bounces back at an angle of 40° from the surface of the board as shown in figure 3.

Hint: When a ball bounces off a surface, the impulse responsible for the bounce is always directed normal to the surface.

Figure 3: Trajectory of a baseball bouncing off a back-board.

Exercise 2a: What is the velocity of the ball after the collision?

Solution