3-4: Impulse and Momentum
Introduction
In today’s class, we define impulse and momentum, and see how those principles are useful for collisions.
At the end of today’s class, you will be able to:
- Define impulse and momentum
- Describe the impulse-momentum principle
- Describe what a perfectly inelastic collision is
- Apply Conservation of Momentum to perfectly inelastic collision problems
Impulse and Momentum
Today, we will introduce a powerful concept called Conservation of Momentum. This is useful for problems involving collisions.
Momentum and Impulse
$\overline{F} =m\overline{a}$
Multiply each side by dt and integrate between the initial time, ti, and the final time, tf, to get Equation 1:
$\int_{t_{i}}^{t_{f}}\overline{F}dt = \int_{t_{i}}^{t_{f}}m\overline{a}dt$
The mass can be taken outside of the integral because it is constant. The integral of adt is the area under the acceleration-time curve, or the change in velocity.
$m\int_{t_{i}}^{t_{f}}\overline{a}dt = m\Delta \overline{v} = m \overline{v} _{f} - m \overline{v} _{i}$
Momentum is defined as mass multiplied by velocity.
The left hand side of Equation 1 is the integral of Fdt. This integral is defined as the impulse , and is the area under the force-time curve. Another way of expressing the impulse is the average force multiplied by a change in time.
$Impulse = \int_{t_{i}}^{t_{f}}\overline{F}dt = \overline{F}_{avg}\Delta t$
The area under the force-time graph, as shown below, is Favgt.
This results in the impulse-momentum principle, or the change in momentum is equal to the impulse. Note that both impulse and momentum are vectors.
$\int_{t_{i}}^{t_{f}}\overline{F} dt = m\overline{v} _{f} - m\overline{v} _{i}$
Another way of writing this is that the final momentum is equal to the initial momentum plus the impulse.
$m\overline{v} _{f} = m\overline{v} _{i} + \int_{t_{i}}^{t_{f}}\overline{F} dt$
Units of Impulse and Momentum
At first glance, it may seem that the units of impulse and momentum could not be the same. Let’s look at dimensional analysis that compares momentum and impulse.
$mv = [m][\frac{L}{T}] = \frac{[m][L]}{[T]} \frac{[T]}{[T]} = \frac{[m][L]}{[T]^{2}}[T]=[F][T]$
Thus, momentum has the same units as impulse.
In the SI system, momentum and impulse have the following units: $[kg][\frac{m}{s}]=[N][s]$
In the USC system, momentum and impulse have the following units: $[slug][\frac{ft}{s}]=[lb][s]$
Example: Hitting a Tennis Ball
In this example we will determine the aveage force applied to a tennis ball during a serve.
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Check out the video of the super slow motion of the serve.
Impulse and Momentum Textbook Pages
Calculate the momentum for the following situations.
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Conservation of Momentum
System of Particles
If we have a system of particles, such as in a game of pool, we can write the impulse-momentum equation for a pool ball, or we can add up all the impulses and momentums, and write the impulse-momentum equation for all the pool balls.
$\sum_{j=1}^{n} m\overline{v} _{f} = \sum_{j=1}^{n} m\overline{v} _{i} + \int_{t_{i}}^{t_{f}}\overline{F} dt$
where n is the number of pool balls.
Here is what is happening when we break in a game of pool. The initial momentum is zero; everything is at rest. We use the cue stick to apply an impulse to the cue ball. This first creates momentum in the cue ball.
Now what happens? If we look at the system of the cue ball and all the other pool balls, there will be no external force, and hence no impulse. Thus, the momentum of all the pool balls right after the cue ball strikes them is equal to the momentum of the cue ball before the collision. This almost seems impossible, but remember that momentum is a vector. Some pool balls are going in one direction, and some in the opposite direction, with the vector sum of the momentum components perpendicular to the original momentum will equal 0.
Conservation of Momentum
The principle of conservation of momentum states that if there is no external force applied to a system, then momentum does not change, or is conserved. The final momentum is equal to the initial momentum.
$\sum_{j=1}^{n} m\overline{v} _{f} = \sum_{j=1}^{n} m\overline{v}_{i}$
Conservation of momentum is very useful in some types of problems, particularly collisions.
Answer the questions using impulse-momentum.
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Collisions
We can classify collisions as follows.
Perfectly Inelastic Collision
A perfectly inelastic collision occurs when two objects collide and stick together. Momentum is conserved but mechanical energy is lost.
Examples of a perfectly inelastic collision include a train car coupling and ‘Velcro Letterman’.
Perfectly Elastic Collision
A perfectly elastic collision occurs when both momentum and mechanical energy are conserved.
An example of a perfectly elastic collision is two pool balls colliding.
Inelastic Collisions
In between a perfectly elastic and perfectly inelastic collision is an inelastic collision. The objects bounce off of one another, but mechanical energy is lost.
An example of an inelastic collision is a ball bouncing off a floor where it does not bounce to its original height.
Explosion
An explosion is similar to a collision in the sense that momentum is conserved. An example is the fireworks after a UT touchdown at Neyland Stadium.
Types of Collisions Textbook Pages
Answer the following concept questions on impulse, momentum, and collisions.
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Summary
- Momentum
- Single object: $m\overline{v}$
- System of objects: $\sum_{j=1}^{n} m\overline{v}$
- Impulse
- $\int_{t_{i}}^{t_{f}}\overline{F} dt$
- area under F vs t graph
- F~avg~ * t
- Units
- [force][time] = [mass][velocity]
- N-s = kg-m/s
- lb-s = slug-ft/s
- Impulse - Momentum
- Single object: $\int_{t_{i}}^{t_{f}}\overline{F} dt = m\overline{v} _{f} - m\overline{v} _{i}$
- System of objects: $\sum_{j=1}^{n} m\overline{v} _{f} = \sum_{j=1}^{n} m\overline{v} _{i} + \int_{t_{i}}^{t_{f}}\overline{F} dt$
- Conservation of Momentum
- No external forces
- $\sum_{j=1}^{n} m\overline{v}_{f} = \sum_{j=1}^{n} m\overline{v}_{i}$
- Collisions
- Perfectly Inelastic
- Inelastic
- Perfectly Elastic
- Perfectly Inelastic Collisions
- Objects stick together, final velocity is the same
- Momentum conserved, mechanical energy lost
- $m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2}){v}^\prime$
- Inelastic Collisions
- Objects DO NOT stick together
- Momentum conserved, mechanical energy lost
- $m_{1}v_{1} + m_{2}v_{2} = m_{1}{v_1}^\prime + m_{2}{v_2}^\prime$
- Explosions
- Objects start together, initial velocity is the same
- Momentum convserved, mechanical energy added
- $(m_{1} + m_{2}){v} = m_{1}v_{1}^\prime + m_{2}v_{2}^\prime$