Momentum, Collisions in Two Dimensions, and Angular Momentum: Study Notes

Momentum and Collisions in Two Dimensions

Conservation of Momentum in Two Dimensions

When collisions occur in two dimensions, the law of conservation of momentum applies to each component of the momentum vector. This means that both the x- and y-components of the total momentum before the collision must equal the respective components after the collision.

• Momentum Conservation Equations:

• For the x-component: $(p_{1x})_i + (p_{2x})_i + (p_{3x})_i + ext{...} = (p_{1x})_f + (p_{2x})_f + (p_{3x})_f + ext{...}$

• For the y-component: $(p_{1y})_i + (p_{2y})_i + (p_{3y})_i + ext{...} = (p_{1y})_f + (p_{2y})_f + (p_{3y})_f + ext{...}$

• Total Momentum: The total momentum of the system is conserved, even though individual momenta may change during the collision.

• Application: This principle is used to analyze billiard ball collisions, explosions, and other multi-object interactions.

Example Problems: Momentum Vectors

Problems often involve drawing or calculating the momentum vector of a fragment or object after a collision or explosion, using the conservation of momentum.

• Explosions: If an object at rest explodes into fragments, the vector sum of all fragment momenta must be zero.

• Collisions: After a collision, the vector sum of the momenta of all objects must equal the total initial momentum.

Worked Example: Peregrine Falcon Strike

This example demonstrates the application of conservation of momentum in two dimensions for a perfectly inelastic collision.

• Scenario: A falcon (mass 0.80 kg, velocity 18 m/s at 45° below horizontal) collides with a pigeon (mass 0.36 kg, velocity 9.0 m/s horizontally).

• Strategy: Treat the system as falcon + pigeon. External forces are negligible; total momentum is conserved. After collision, both move together at a common velocity.

• Calculations:

  • Find x- and y-components of initial momentum: $(P_x)_i = m_f v_{fx} + m_p v_{px} = (0.80)(-18 \cos 45^\circ) + (0.36)(-9.0) = -13.4 \text{ kg m/s}$ $(P_y)_i = m_f v_{fy} + m_p v_{py} = (0.80)(-18 \sin 45^\circ) + 0 = -10.2 \text{ kg m/s}$

  • After collision, both birds move with velocity $\vec{v}$: $(P_x)_f = (m_f + m_p) v_x$ $(P_y)_f = (m_f + m_p) v_y$

  • Solve for $v_x$ and $v_y$: $v_x = \frac{-13.4}{0.80 + 0.36} = -11.6 \text{ m/s}$ $v_y = \frac{-10.2}{0.80 + 0.36} = -8.79 \text{ m/s}$

  • Find direction: $\alpha = \tan^{-1} \left( \frac{v_y}{v_x} \right) = \tan^{-1} \left( \frac{-8.79}{-11.6} \right) = 37^\circ$ below horizontal

  • Find speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-11.6)^2 + (-8.79)^2} = 15 \text{ m/s}$

• Assessment: The falcon slows down after catching the pigeon, and the final direction is more horizontal, consistent with the initial conditions.

Angular Momentum

Definition and Properties

Angular momentum is the rotational analog of linear momentum and is conserved in the absence of external torque.

• Angular Momentum ($L$): For an object with moment of inertia $I$ rotating at angular velocity $\omega$:

• $L = I \omega$

• SI Units: $\text{kg} \cdot \text{m}^2/\text{s}$

• Newton's Second Law for Rotation: $\alpha = \frac{\tau_{net}}{I}$

• Angular Acceleration: $\alpha = \frac{\Delta \omega}{\Delta t}$

• Impulse-Momentum Theorem (Rotational): $\tau_{net} \Delta t = I \Delta \omega$

Comparison: Rotational vs. Linear Dynamics

Rotational Dynamics	Linear Dynamics
Torque $\tau_{net}$	Force $F_{net}$
Moment of inertia $I$	Mass $m$
Angular velocity $\omega$	Velocity $v$
Angular momentum $L = I\omega$	Linear momentum $\vec{p} = m\vec{v}$

Conservation of Angular Momentum

Law of Conservation

If the net external torque on a system is zero, the total angular momentum of the system remains constant.

• Mathematical Statement: $\tau_{net} \Delta t = \Delta L$

• If $\tau_{net} = 0$, then $L_f = L_i$

• For multiple objects:

• $(I_1 \omega_1)_i + (I_2 \omega_2)_i + \cdots = (I_1 \omega_1)_f + (I_2 \omega_2)_f + \cdots$

Varying Moment of Inertia

The moment of inertia of an object can change if the distribution of mass changes, affecting angular velocity to conserve angular momentum.

• Example: Figure skaters spin faster by pulling in their arms, reducing their moment of inertia.

• Isolated Rotating Object: Can experience a change in angular velocity if its moment of inertia changes.

Worked Example: Period of a Merry-Go-Round

This example illustrates conservation of angular momentum when a person moves on a rotating platform.

• Scenario: Joey (36 kg) stands at the center of a 200 kg merry-go-round (radius 2.0 m, period 2.5 s). Joey walks to the edge. What is the new period?

• Strategy: System is Joey + merry-go-round. No external torque; angular momentum is conserved.

• Moment of Inertia: For disk: $I_{disk} = \frac{1}{2}MR^2$; for Joey at edge: $I_{Joey} = mR^2$

• Initial Angular Momentum: $L_i = \left( \frac{1}{2}MR^2 \right) \omega_i$

• Final Angular Momentum: $L_f = \left( mR^2 + \frac{1}{2}MR^2 \right) \omega_f$

• Conservation: $L_i = L_f$

• Solve for $\omega_f$: $\omega_f = \frac{M}{M + 2m} \omega_i$

• Calculate new period: $T_f = \frac{2\pi}{\omega_f}$

• Result: The merry-go-round rotates more slowly after Joey moves to the edge.

Worked Example: Angular Momentum of a Bicycle

Bicycle wheels have angular momentum due to their rotation, which helps riders stay upright.

• Moment of Inertia (rim): $I = mr^2$

• Angular Velocity: $\omega = \frac{v}{r}$

• Angular Momentum per Wheel: $L = I\omega$

• Example Calculation: For $m = 1.5 \text{ kg}, r = 0.33 \text{ m}, v = 4.5 \text{ m/s}$: $I = 1.5 \times 0.33^2 = 0.16 \text{ kg m}^2$ $\omega = 4.5 / 0.33 = 14 \text{ rad/s}$ $L = 0.16 \times 14 = 2.2 \text{ kg m}^2/\text{s}$ (per wheel) $L_{total} = 4.4 \text{ kg m}^2/\text{s}$ (for two wheels)

Summary: General Principles and Applications

Conservation Laws

• Conservation of Momentum: The total momentum of an isolated system remains constant if no net external force acts.

• Conservation of Angular Momentum: The total angular momentum of a system remains constant if no net external torque acts.

Solving Conservation Problems

• Strategize: Choose an isolated system.

• Prepare: Draw before-and-after diagrams.

• Solve: Apply conservation equations to each component.

• Assess: Check if the result is reasonable.

Important Concepts

• Momentum: $\vec{p} = m\vec{v}$

• Impulse: $J = \int F dt$

• Impulse-Momentum Theorem: $\Delta \vec{p} = J$

• Angular Momentum: $L = I\omega$

Applications

• Collisions: Objects stick together (perfectly inelastic) or bounce apart (elastic), but total momentum is conserved.

• Explosions: Objects move apart; total momentum remains conserved.

• Two Dimensions: Both x- and y-components of momentum must be conserved, leading to simultaneous equations.