06_Momentum

Conservation of Momentum

1. Invariance of the Work-Kinetic Energy Theorem

The Work-Kinetic Energy Theorem states: $K_f - K_i = W$.

We can demonstrate that this theorem holds across different inertial frames under a Galilean transformation.

Galilean Transformation:

Consider a frame moving with constant velocity $\vec{v}_0$. The transformations for position, velocity, and acceleration are:

$\vec{r}' = \vec{r} - \vec{v}_0 t$
$\vec{v}' = \vec{v} - \vec{v}_0$
$\vec{a}' = \vec{a}$

Proof (Assuming 1D motion and constant force):

In the original frame, work is $W = Fd$.

In the moving frame, displacement is $d' = d - v_0 t$, so the work is:

\[ W' = Fd' = F(d - v_0 t) \]

The kinetic energy in the moving frame is $K' = \frac{1}{2}m(v')^2 = \frac{1}{2}m(v - v_0)^2 = K - v_0mv + \frac{1}{2}mv_0^2$.

The change in kinetic energy in the moving frame is:

\[ K_f' - K_i' = K_f - K_i - v_0m(v_f - v_i) = Fd - v_0m(v_f - v_i) \]

Rewriting $Fd$ in terms of $d'$:

\[ K_f' - K_i' = F(d' + v_0 t) - v_0m(v_f - v_i) = W' + v_0[Ft - m(v_f - v_i)] \]

From kinematics, under constant acceleration, $v_f = v_i + at \Rightarrow m(v_f - v_i) = mat = Ft$. Therefore, the bracketed term is zero.

\[ \Rightarrow K_f' - K_i' = W' \]

Conclusion: The Work-Kinetic Energy Theorem always holds, regardless of the inertial reference frame.

2. Linear Momentum and Impulse

Linear Momentum: Defined as $\vec{p} \equiv m\vec{v}$
Impulse: Defined as the integral of force over time: $\vec{I} \equiv \int_{t_i}^{t_f} \vec{F} dt$

Impulse-Momentum Theorem: $\vec{I} = \vec{p}_f - \vec{p}_i$

Proof:

Starting from Newton's Second Law: $\vec{F} = m\frac{d\vec{v}}{dt} \Rightarrow \vec{F} = \frac{d\vec{p}}{dt}$ (differential form).

Integrating both sides: $\int_{t_i}^{t_f} \vec{F} dt = \int_{\vec{p}_i}^{\vec{p}_f} d\vec{p} \Rightarrow \vec{I} = \vec{p}_f - \vec{p}_i$.

(Example: For a constant force, $\vec{F}t = m(\vec{v}_f - \vec{v}_i)$, as used in the Galilean proof.)

3. Conservation of Linear Momentum

Isolated Two-Particle System:

Consider particles $m_1$ and $m_2$ interacting with forces $\vec{F}_{12}$ (on 1 by 2) and $\vec{F}_{21}$ (on 2 by 1).

\[ \vec{F}_{21} = \frac{d\vec{p}_1}{dt}, \quad \vec{F}_{12} = \frac{d\vec{p}_2}{dt} \]

By Newton's Third Law, $\vec{F}_{21} = -\vec{F}_{12}$, which means:

\[ \frac{d\vec{p}_1}{dt} + \frac{d\vec{p}_2}{dt} = 0 \Rightarrow \vec{p}_1 + \vec{p}_2 = \text{constant} \Rightarrow \vec{p}_{1i} + \vec{p}_{2i} = \vec{p}_{1f} + \vec{p}_{2f} \]

Generalization:

For any system, if the system is closed (no particles enter/leave) and isolated (no net external force), the total linear momentum is conserved:

\[ \vec{P} = \sum_i \vec{p}_i = \text{constant} \]

Origin of Conservation Laws (Symmetry):

Energy conservation stems from time translation invariance.
Linear momentum conservation stems from spatial translation invariance.

(Formal proofs are covered in advanced Theoretical Mechanics).

4. Collisions

During a collision, internal forces ($\vec{F}_{12}$ and $\vec{F}_{21}$) are equal and opposite, resulting in equal and opposite impulses. Therefore, the total momentum of the isolated system is conserved: $\vec{P} = \vec{p}_1 + \vec{p}_2 = \text{constant}$.

Kinetic energy is NOT necessarily conserved in collisions.

Elastic collision: Kinetic energy is conserved ($K_i = K_f$).
Inelastic collision: Kinetic energy is lost ($K_i > K_f$).

Perfectly Inelastic Collisions

Objects stick together and share a final velocity: $\vec{v}_{1f} = \vec{v}_{2f} = \vec{v}_f$.

Momentum Conservation: $\vec{v}_f = \frac{m_1\vec{v}_{1i} + m_2\vec{v}_{2i}}{m_1 + m_2}$
Loss of Kinetic Energy:

\[ K_i - K_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 - \frac{1}{2}(m_1+m_2)v_f^2 = \frac{m_1 m_2}{2(m_1+m_2)} (\vec{v}_{1i} - \vec{v}_{2i})^2 \]

This value is strictly $>0$ unless $\vec{v}_{1i} = \vec{v}_{2i}$ (in which case no collision occurs).

Example Applications:

Ballistic Pendulum: A mass $m_1$ at $v_{1i}$ embeds into $m_2$ at rest. The system swings to height $h$.
```
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v_f = \frac{m_1 v_{1i}}{m_1 + m_2} \Rightarrow h = \frac{v_f^2}{2g} = \frac{m_1^2 v_{1i}^2}{2(m_1+m_2)^2 g} $$
Spring Deformation: Mass $m_1$ ($v_{1i}$) strikes $m_2$ ($v_{2i}$) through a spring $k$, where $v_{1i} > v_{2i}$.

Maximal deformation $x$ occurs when velocities are momentarily equal (acting as a perfectly inelastic collision).

\[ \Delta K = \frac{1}{2}kx^2 \Rightarrow x = \sqrt{\frac{m_1 m_2}{k(m_1+m_2)}} (v_{1i} - v_{2i}) \]

Elastic Collisions

1D Elastic Collision (Head-to-head):

Solving energy and momentum conservation equations yields:

\[ v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} + \frac{2m_2}{m_1 + m_2}v_{2i} \]

\[ v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} + \frac{m_2 - m_1}{m_1 + m_2}v_{2i} \]

Special Cases (assuming target $m_2$ is initially at rest, $v_{2i} = 0$):
- $m_1 \ll m_2$: $v_{1f} \approx -v_{1i}$, $v_{2f} \approx 0$ (Light object bounces off).
- $m_1 \gg m_2$: $v_{1f} \approx v_{1i}$, $v_{2f} \approx 2v_{1i}$ (Heavy object plows through).
- $m_1 = m_2$: $v_{1f} = v_{2i}$, $v_{2f} = v_{1i}$ (Velocities exchange).
2D Elastic Collision:

Requires considering momentum in x and y directions separately alongside energy conservation. Yields 3 equations for 4 unknowns ($v_{1f}, v_{2f}, \theta, \phi$), requiring an extra constraint to solve.

Special case: If $m_1 = m_2$ and target is at rest, the final velocity vectors are always orthogonal ($\vec{v}_{1f} \cdot \vec{v}_{2f} = 0$).

5. Center of Mass (CM)

The Center of Mass represents the mean position of the mass distribution.

Many-particle system: $\vec{r}_{CM} \equiv \frac{\sum m_i \vec{r}_i}{\sum m_i}$
Extended object: $\vec{r}_{CM} = \frac{\int \vec{r} dm}{M} = \frac{\int \vec{r} \rho dV}{M}$

Examples of calculating CM:

Uniform Rod (Length $L$): $x_{CM} = \frac{\int_0^L \lambda x dx}{\int_0^L \lambda dx} = \frac{1}{2}L$
Non-uniform Rod ($\lambda(x) = \alpha x$): $x_{CM} = \frac{\int_0^L (\alpha x) x dx}{\int_0^L (\alpha x) dx} = \frac{2}{3}L$
Uniform Right Triangle (Base $a$, height $b$): Utilizing double integration over area, $x_{CM} = \frac{2}{3}a$, $y_{CM} = \frac{1}{3}b$.

Motion of a System of Particles:

Taking derivatives of $\vec{r}_{CM} = \frac{\sum m_i \vec{r}_i}{M}$:

Velocity: $\vec{v}_{CM} \equiv \frac{d\vec{r}_{CM}}{dt} = \frac{\sum \vec{p}_i}{M} = \frac{\vec{P}_{tot}}{M}$
Acceleration: $\vec{a}_{CM} \equiv \frac{d\vec{v}_{CM}}{dt} = \frac{\sum \vec{F}_i}{M}$

This leads to Newton's Second Law for a system: $\sum \vec{F}_{ext} = M\vec{a}_{CM} = \frac{d\vec{P}_{tot}}{dt}$. (Note: All internal forces cancel out).

6. The Center of Mass Frame

König's Theorem

The total kinetic energy of a system is the sum of the kinetic energy of the Center of Mass (bulk motion) and the internal kinetic energy relative to the CM.

\[ E_K = E_{k, cm} + E_K' \]

Proof: Let particle velocities in the CM frame be $\vec{v}_i' = \vec{v}_i - \vec{v}_{CM}$.

$E_K = \sum \frac{1}{2}m_i (\vec{v}_{CM} + \vec{v}_i') \cdot (\vec{v}_{CM} + \vec{v}_i') = \frac{1}{2}M v_{CM}^2 + \frac{1}{2}\sum m_i(v_i')^2 + \vec{v}_{CM} \cdot \sum m_i \vec{v}_i'$.

Since $\sum m_i \vec{r}_i' = 0$ by definition of the CM frame, its derivative $\sum m_i \vec{v}_i'$ is also zero. This eliminates the cross term, proving the theorem.

Momentum in the CM Frame

The total momentum of any system evaluated within its own Center of Mass frame is always zero.

\[ \vec{P}' = \sum \vec{p}_i' = 0 \]

Collisions in the CM Frame

Analyzing collisions in the CM frame simplifies the math significantly.

1D Elastic Collision:

Particles approach with initial momenta $\vec{p}_{1i}' = -\vec{p}_{2i}'$.

After an elastic collision, their velocities simply reverse direction:

\[ \vec{v}_{1f}' = -\vec{v}_{1i}' \quad \text{and} \quad \vec{v}_{2f}' = -\vec{v}_{2i}' \]
2D Elastic Collision:

In the original frame, particles scatter at complex angles. In the CM frame, particles always approach head-on and recede back-to-back. Energy conservation dictates that the magnitudes of their velocities remain unchanged; the collision merely rotates the axis of their back-to-back trajectory.