Impulse and Momentum

Introduction – A Simple Collision

Impulse and momentum are useful when discussing collisions.
A collision involves a short-duration interaction between two objects.
Collisions occur over small time intervals but are not instantaneous.
Example: A racket hitting a tennis ball exerts a force on the ball.
- This force is large but short-lived, referred to as an impulsive force.
- The ball compresses and expands during the impact.

Introduction – A Simple Collision

During a collision with a wall, an impulsive force acts on an object.
- This force varies with time and reaches its maximum at maximum compression.
A graph of $F_x$ vs. $t$ helps visualize the interaction.
The object deforms during the collision, indicating it's an elastic object rather than a particle.

Introduction – A Simple Collision

The object's velocity changes as a result of the collision.
Newton's 2nd Law can quantify the change in velocity:
- $\vec{F} = m\vec{a} = m \frac{d\vec{v}}{dt}$
Rearranging gives:
- $m d\vec{v} = \vec{F} dt$
Integrating both sides:
- $\int{vi}^{vf} m d\vec{v} = \int{ti}^{tf} \vec{F}(t) dt$
- $m\vec{v}f - m\vec{v}i = \int{ti}^{t_f} \vec{F}(t) dt$

Momentum & Impulse

Momentum ($\vec{p}$) is defined as mass times velocity:
- $\vec{p} = m\vec{v}$
- It's a vector quantity pointing in the direction of velocity.
- Units: kg ⋅ m/s
- Can be broken into x and y components for problem-solving.
Impulse ( $J_x$ ) describes the effect of force over a time interval:
- $Jx = \int{ti}^{tf} F_x(t) dt$
- Impulse is not a vector; it describes the action of a single force component.
- Units: N ⋅ s (equivalent to kg ⋅ m/s)

An Alternate Version of Newton’s 2nd Law

Newton's 2nd Law can be expressed in terms of momentum:
- $\vec{F} = \frac{d}{dt} m\vec{v} = \frac{d\vec{p}}{dt}$
- Force is the rate of change of momentum with respect to time.
This is a more general form of the law.
- The earlier version assumes constant mass, which isn't always the case (e.g., rockets burning fuel).

Relating Momentum and Impulse

From previous expressions, we have:
- $mv{xf} - mv{xi} = \int{ti}^{t_f} \vec{F}(t) dt$
- $p{xf} - p{yi} = J_x$
The momentum principle states that an impulse delivered to an object causes a change in the object's momentum.
- $\Delta px = Jx$
- A force in the x-direction only changes the x-component of momentum.

The Momentum Principle

Consider a rubber ball hitting a wall.
- The initial momentum is positive (to the right).
- The final momentum is negative (to the left after rebound).
- The force exerted is to the left, so the force curve is inverted, and the impulse is negative.
- The impulse causes a change in the ball's momentum.

Impulse – One More Thing…

The force during a collision can be complex, making impulse computation difficult.
Instead of $F(t)$ , we can use the average force ( $F_{avg}$ ) exerted during the collision.
$Jx = F{avg} \Delta t$
The area under the $F_{avg}$ curve (a rectangle) is the same as the area under the original $F(t)$ curve.

Momentum Principle – Analogous to Energy Principle

The momentum principle is similar to the energy principle from Chapter 9.
When a force acts on an object, it does work and creates an impulse.
The choice of which principle to use depends on the context.
- Energy Principle: $\Delta K = W = \int{xi}^{xf} Fx dx$
- Momentum Principle: $\Delta px = Jx = \int{ti}^{tf} Fx dt$

Conservation of Momentum

Consider two objects headed toward each other.
- Assumptions:
  - The objects are elastic.
  - No external forces act on either object during the interaction.
- When they collide, each exerts a force on the other (action/reaction pair).
- After the collision, they move away from each other, and their momenta have changed.

Conservation of Momentum

Objects have momenta $p{x1}$ and $p{x2}$ .
Total momentum of the system: $p{x1} + p{x2}$ .
The change in momentum of the system over time interval $dt$ is:
- $\frac{d}{dt} (p{x1} + p{x2}) = \frac{d\vec{p}{x1}}{dt} + \frac{d\vec{p}{x2}}{dt}$
Applying Newton's Laws:
- Newton's 2nd Law: $\frac{d\vec{p}{x1}}{dt} = F{12}$ , $\frac{d\vec{p}{x2}}{dt} = F{21}$
- Newton's 3rd Law: $F{21} = -F{12}$

Conservation of Momentum

$\frac{d\vec{p}{x1}}{dt} + \frac{d\vec{p}{x2}}{dt} = F{12} + F{21} = F{12} + (-F{12}) = 0$
- $\frac{d}{dt} (p{x1} + p{x2}) = 0$
- The total momentum doesn't change with time.
- $p{x1} + p{x2} = constant$
$p{x1,i} + p{x2,i} = p{x1,f} + p{x2,f}$
The expressions don't depend on force.
- We can analyze the interaction using momentum even if the force is unknown.

Momentum of a System

Consider a system of $N$ interacting particles with external forces also acting on the system.
The total momentum ( $P$ ) of the system is:
- $P = \vec{p}1 + \vec{p}2 + \cdots + \vec{p}N = \sumk \vec{p}_k$
Applying Newton's 2nd Law to each particle:
- $\frac{dP}{dt} = \sumk \sum{j \neq k} \vec{F}{jk} + \sumk \vec{F}{ext, k} = \sumk \vec{F}_{ext, k}$
- Interaction forces cancel out; the change in momentum is due to external forces.

Law of Conservation of Momentum

An isolated system has a net external force of zero ( $\vec{F}_{ext} = 0$ ).
If we have an isolated system:
- $\frac{dP}{dt} = 0$
The Law of Conservation of Momentum states:
- The total momentum ( $P$ ) of an isolated system does not change.
- Interactions within the system can change individual momenta, but the total remains constant.
- $Pf = Pi$

Problem Solving Strategy: Conservation of Momentum

MODEL: Clearly define the system.
- If possible, choose a system that is isolated ( $F_{net} = 0$ ).
- If it's not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion.
- Other segments of the motion can be analyzed using Newton's laws or conservation of energy.
VISUALIZE Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you're trying find.
SOLVE The mathematical representation is based on the law of conservation of momentum: $Pf = Pi$ . In component form, this is
- $(P{fx})1 + (P{fx})2 + (P{fx})3 + … = (P{ix})1 + (P{ix})2 + (P{ix})3 + …$
- $(P{fy})1 + (P{fy})2 + (P{fy})3 + … = (P{iy})1 + (P{iy})2 + (P{iy})3 + …$
REVIEW Check that your result has correct units and significant figures, is reasonable, and answers the question.

Problem Solving Strategy: Conservation of Momentum

Example 11.3: Rolling Away

Conservation of Momentum – Choosing a System

Problem-solving with momentum conservation depends on system choice.
Consider a ball dropped from a height above the Earth.
- If the system is just the ball, gravity is an external force.
  - Impulse changes momentum; momentum is not conserved.
- If the system includes the Earth, gravitational interaction forces are internal.
  - The system is isolated, and momentum is conserved.

Collisions – A Detailed Look

Momentum conservation is a major tool in studying collisions.
Two types of collisions:
- Perfectly Inelastic Collisions:
  - Objects collide and stick together, moving with the same final velocity.
  - Momentum is conserved.
  - Mechanical energy is not conserved (transformed into thermal energy).
  - Example: A dart hitting a dartboard.
- Perfectly Elastic Collisions:
  - Objects collide and bounce off each other.
  - Momentum is conserved.
  - Kinetic energy is conserved if $\Delta U = 0$ .
  - Example: Billiard balls colliding.

A Perfectly Elastic Collision – Special Case

Consider a perfectly elastic collision between two balls with masses $m1$ and $m2$ .
Initially, Ball 1 moves with velocity $v_{ix,1}$ and Ball 2 is at rest.
- Only Ball 1 has nonzero initial momentum and kinetic energy.
Afterward, both balls are moving.
- Both have nonzero final momentum and kinetic energy.
- Both energy and momentum are conserved.

A Perfectly Elastic Collision – Special Case

Using conservation laws, we can find expressions for the final velocities.
Momentum conservation:
- $P{ix} = P{fx}$
- $m1v{ix,1} = m1v{fx,1} + m2v{fx,2}$
Energy conservation:
- $K{1i} = K{1f} + K_{2f}$
- $\frac{1}{2} m1v{ix,1}^2 = \frac{1}{2} m1v{fx,1}^2 + \frac{1}{2} m2v{fx,2}^2$
We now have two equations and two unknowns.

A Perfectly Elastic Collision – Special Case

Result: In a perfectly elastic collision where Ball 2 is initially at rest, the final velocities are:
- $v{f1} = \frac{m1 - m2}{m1 + m2} v{i1}$
- $v{f2} = \frac{2m1}{m1 + m2} v_{i1}$

Perfectly Elastic Collision – Special Case

Case A: Let $m1 = m2$
- Substituting into the expressions:
 - $v_{fx,1} = 0$
 - $v{fx,2} = v{ix,1}$
- Ball 1 transfers its momentum to Ball 2.

Perfectly Elastic Collision – Special Case

Case B: Let $m1 >> m2$
- Substituting into the expressions, we can make the approximations:
 - $v{fx,1} \approx v{ix,1}$
 - $v{fx,2} \approx 2v{ix,1}$
- Ball 1 essentially stays on course while Ball 2 flies ahead of it.

Perfectly Elastic Collision – Special Case

Case C: Let $m1 << m2$
- Substituting into the expressions, we can make the approximations:
 - $v{fx,1} \approx -v{ix,1}$
 - $v_{fx,2} \approx 0$
- Ball 1 rebounds, flying off with about the same speed it had initially, while Ball 2 remains at rest.

Model: Inelastic & Elastic Collisions

In a perfectly inelastic collision, the objects stick and move together. Kinetic energy is transformed into thermal energy.
- Mathematically: $(m1 + m2)v{fx} = m1(v{ix})1 + m2(v{ix})_2$
In a perfectly elastic collision, the objects bounce apart with no loss of energy.
- Mathematically: If object 2 is initially at rest, then
 - $(v{fx})1=\frac{m1 - m2}{m1 + m2}(v{ix})1$
 - $(v{fx})2 = \frac{2m1}{m1 + m2}(v{ix})_1$

Explosions

Explosions involve a brief, intense interaction where objects move away from each other.
An explosion can be seen as the opposite of a collision.
Explosive forces are internal, so momentum is conserved if there are no external forces.
Examples:
- Radioactivity: A uranium atom ejecting an alpha particle (daughter nucleus recoils).
- Rocket Propulsion: Burning fuel expels exhaust gas downward, pushing the rocket up.

Momentum in Two Dimensions

Momentum is a vector quantity; total momentum is the vector sum of individual momenta.
Each component has its own conservation equation:
- $P{xi} = P{xf} \Rightarrow \sumk p{xi,k} = \sumk p{xf,k}$
- $P{yi} = P{yf} \Rightarrow \sumk p{yi,k} = \sumk p{yf,k}$
- Total momentum is conserved only if each component is conserved.
Treat each component independently.

Example 11.9: A Three Piece Explosion

APPENDIX

A Perfectly Elastic Collision – Special Case

Using conservation laws, we can find expressions for the final velocities.
Momentum conservation:
- $P{ix} = P{fx}$
- $m1v{ix,1} = m1v{fx,1} + m2v{fx,2}$
Energy conservation:
- $K{1i} = K{1f} + K_{2f}$
- $\frac{1}{2} m1v{ix,1}^2 = \frac{1}{2} m1v{fx,1}^2 + \frac{1}{2} m2v{fx,2}^2$
We now have two equations and two unknowns.

A Perfectly Elastic Collision – Special Case

Solve the system to find the final velocities of the two balls:
- $m1v{ix,1}^2 = m1v{fx,1}^2 + m2v{fx,2}^2$
- $m1v{ix,1} = m1v{fx,1} + m2v{fx,2}$
Simplify by dropping the “x” subscripts since the motion is one dimension and solve each equation for $v_{i1}^2$ .
- $v{i1}^2 = v{f1}^2 + \frac{m2}{m1} v_{f2}^2$
- $v{f1} = \frac{1}{m1}(m1v{f1} + m2v{f2})$
- $v{f1}^2 = \frac{1}{m1^2}(m1v{f1} + m2v{f2})^2 = \frac{1}{m1^2}(m1^2v{f1}^2 + 2m1m2v{f1}v{f2} + m2^2v{f2}^2) = v{f1}^2 + 2\frac{m2}{m1}v{f1}v{f2} + \frac{m2^2}{m1^2}v_{f2}^2$

A Perfectly Elastic Collision – Special Case

Solve the system to find the final velocities of the two balls:

A Perfectly Elastic Collision – Special Case

Solve the system to find the final velocities of the two balls:
Now we can solve for $v{f2}$ in terms of $v{f1}$
- $v{f2} = 2v{f1} + \frac{m2}{m1} v{f2} \Rightarrow v{f2} \left(1 - \frac{m2}{m1}\right) = 2v{f1} \Rightarrow v{f2} \left(\frac{m1 - m2}{m1}\right) = 2m1v{f1} \Rightarrow v{f2} = \frac{2m1}{m1 - m2} v{f1}$
We can substitute this expression into the original momentum conservation expression, Eq (b), to get an expression for $v_{f1}$ . Start by doing the substitution:
- $m1v{i1} = m1v{f1} + m2v{f2} = m1v{f1} + m2 \frac{2m1}{m1 - m2} v_{f1}$
- $m1v{i1} = m1v{f1} + \frac{2m1m2}{m1 - m2} v_{f1}$
- $v{i1} = v{f1} + \frac{2m2}{m1 - m2} v{f1}$

A Perfectly Elastic Collision – Special Case

Solve the system to find the final velocities of the two balls:
It’s now time to solve for $v{f1}$ . First, multiply everything by $m1 - m_2$ to get rid of the fraction.
- $(m1 - m2)v{i1} = (m1 - m2)v{f1} + 2m2v{f1} = (m1 - m2 + 2m2) v{f1} = (m1 + m2) v_{f1}$
- $v{f1} = \frac{m1 - m2}{m1 + m2} v{i1}$
Our last step is to substitute this into our $v_{f2}$ expression so that both of our final velocities are in terms of the initial velocity.
- $v{f2} = \frac{2m1}{m1 - m2} v{f1} = \frac{2m1}{m1 - m2} \frac{m1 - m2}{m1 + m2} v_{i1}$
- $v{f2} = \frac{2m1}{m1 + m2} v_{i1}$

A Perfectly Elastic Collision – Special Case

Result: In a perfectly elastic collision where Ball 2 is initially at rest, the final velocities are:
- $v{f1} = \frac{m1 - m2}{m1 + m2} v{i1}$
- $v{f2} = \frac{2m1}{m1 + m2} v_{i1}$