Linear Momentum, Center of Mass, and Collisions: Study Notes

Linear Momentum, Center of Mass, and Collisions

Center of Mass

The center of mass of a system is a special point that represents the average position of all the mass in the system. For a system of particles, the center of mass moves as if all the external forces act at that point.

• Position Vector of Center of Mass (RCM): The position of the center of mass relative to an origin O is given by the vector $\vec{R}_{CM}$.

• Coordinates of Center of Mass: For N particles with masses $m_i$ and positions $x_i$, $y_i$, $z_i$:

  • $x_{CM} = \frac{\sum m_i x_i}{\sum m_i}$

  • $y_{CM} = \frac{\sum m_i y_i}{\sum m_i}$

  • $z_{CM} = \frac{\sum m_i z_i}{\sum m_i}$

• Newton's Second Law for Center of Mass: The motion of the center of mass is governed by the net external force:

  • $M \vec{a}_{CM} = \vec{F}_{ext}$

  • Where $M$ is the total mass, $\vec{a}_{CM}$ is the acceleration of the center of mass, and $\vec{F}_{ext}$ is the net external force.

• Internal vs. External Forces: Internal forces (forces between particles within the system) do not affect the motion of the center of mass; only external forces do.

Example: If a projectile explodes mid-air, the center of mass continues to follow the same parabolic trajectory as if the explosion had not occurred, because the explosion involves only internal forces.

Application: Exploding Projectile Problem

When an object breaks into fragments (e.g., a rocket at the top of its trajectory), the center of mass continues its motion as if the object had not broken apart.

• For two identical fragments, if one lands a distance D from the explosion point and the center of mass lands at 2D, the other fragment must land at 3D.

• Calculation:

  • Let $x_1 = D$, $x_2 = x$, both masses $m$.

  • $x_{CM} = \frac{m x_1 + m x_2}{2m} = \frac{D + x}{2}$

  • If $x_{CM} = 2D$, then $2D = \frac{D + x}{2} \implies x = 3D$

Linear Momentum

Definition and Properties

Linear momentum is a vector quantity defined as the product of an object's mass and its velocity.

• Formula: $\vec{p} = m \vec{v}$

• Direction of momentum is the same as the direction of velocity.

• For a system of particles, total momentum is the vector sum of the momenta of all particles.

Newton's Second Law (General Form)

• Net force equals the rate of change of momentum:

  • $\vec{F}_{net} = \frac{d\vec{p}}{dt}$

  • For a finite time interval: $\vec{F}_{net} = \frac{\Delta \vec{p}}{\Delta t}$

Conservation of Linear Momentum

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

• Condition: $\vec{F}_{ext} = 0 \implies \Delta \vec{p} = 0$

• Conservation Law: $\vec{p}_{initial} = \vec{p}_{final}$

• Momentum conservation applies separately to each component (x, y, z):

  • $\Delta p_x = F_{ext, x} \Delta t$

  • $\Delta p_y = F_{ext, y} \Delta t$

• Even if there is an external force, if it acts only in one direction, momentum may be conserved in the perpendicular direction.

Example: In projectile motion, gravity acts vertically, so horizontal momentum is conserved.

Classic Example: Two People on Ice

• Two people push off each other on frictionless ice.

• Initial momentum is zero (both at rest).

• After pushing, they move in opposite directions with momenta $m_1 v_1$ and $m_2 v_2$.

• By conservation: $m_1 v_1 + m_2 v_2 = 0 \implies v_2 = -\frac{m_1}{m_2} v_1$

Example: Bullet Embedding in a Block

• Bullet of mass $m$ and velocity $v$ embeds in a block of mass $M$ at rest.

• After collision, both move together with velocity $V$.

• By conservation:

  • $m v = (M + m) V \implies V = \frac{m}{M + m} v$

Momentum Conservation in Two Dimensions

When a particle breaks into fragments moving in different directions, momentum conservation must be applied to each component.

• For a particle of mass $M$ moving with velocity $V$ that breaks into two fragments:

  • Let fragment 1: mass $m_1$, velocity $\vec{v}_1$ at angle $\theta_1$

  • Fragment 2: mass $m_2$, velocity $\vec{v}_2$ at angle $\theta_2$

Component	Equation
X-direction	$M V = m_1 v_1 \cos \theta_1 + m_2 v_2 \cos \theta_2$
Y-direction	$0 = m_1 v_1 \sin \theta_1 - m_2 v_2 \sin \theta_2$

These equations allow solving for unknown velocities or angles after the explosion.

Collisions

Types of Collisions

• Elastic Collision: Both momentum and kinetic energy are conserved.

• Inelastic Collision: Momentum is conserved, but kinetic energy is not.

• Completely Inelastic Collision: The colliding objects stick together after collision.

Impulse and Average Force

During a collision, the force between objects can be large and act over a short time. The impulse is the product of force and the time interval over which it acts, and equals the change in momentum.

• Impulse: $J = F_{avg} \Delta t = \Delta p$

• The area under the force vs. time curve represents the impulse.

Collisions and Conservation Laws

• During a collision, internal forces (action-reaction pairs) do not change the total momentum of the system.

• Momentum conservation can be applied to the system of colliding objects if external forces are negligible or cancel out.

Completely Inelastic Collision Example

• Two cars collide and stick together.

• Let car 1: mass $m_1$, velocity $v_1$; car 2: mass $m_2$, at rest.

• After collision, combined mass $(m_1 + m_2)$ moves with velocity $V$.

• By conservation:

  • $m_1 v_1 = (m_1 + m_2) V \implies V = \frac{m_1}{m_1 + m_2} v_1$

• Kinetic Energy Loss:

  • Initial: $KE_i = \frac{1}{2} m_1 v_1^2$

  • Final: $KE_f = \frac{1}{2} (m_1 + m_2) V^2$

  • Fraction remaining: $\frac{KE_f}{KE_i} = \frac{m_1}{m_1 + m_2}$ (less than 1, so kinetic energy is lost)

Summary Table: Types of Collisions

Type	Momentum Conserved?	Kinetic Energy Conserved?	Example
Elastic	Yes	Yes	Billiard balls
Inelastic	Yes	No	Car crash (not sticking)
Completely Inelastic	Yes	No	Objects stick together

Key Points

• Momentum is always conserved in isolated systems (no net external force).

• Kinetic energy is only conserved in elastic collisions.

• Impulse is the change in momentum and can be used to estimate average force during a collision.

• In two-dimensional problems, apply conservation laws to each component separately.

Additional info: Some explanations and context have been expanded for clarity and completeness, including the summary tables and explicit formulas for two-dimensional momentum conservation.