Lecture 10

Impulse and Linear Momentum

Introduction to Linear Momentum

Linear momentum is a fundamental concept in physics that describes the quantity of motion an object possesses. It is a vector quantity, dependent on both the mass and velocity of an object.

• Definition: Linear momentum (p) is defined as the product of an object's mass (m) and its velocity (v).

• Formula: $\vec{p} = m\vec{v}$

• Vector Nature: Momentum has both magnitude and direction, following the direction of velocity.

• Example: A 3 kg object moving at 2 m/s has a momentum of $6\ \mathrm{kg\cdot m/s}$ in the direction of motion.

Total Linear Momentum of a System

The total linear momentum of a system is the vector sum of the individual momenta of all objects within the system.

• Formula: $\vec{p}_{\text{net}} = \sum_i m_i \vec{v}_i$

• Application: Used to analyze systems of multiple objects, such as in collisions.

Mass: m	Velocity:		Momentum:
3 kg	$\vec{v}_1 = (-6\ \mathrm{m/s}, 0)$	$\vec{p}_1 = (-18\ \mathrm{kg\cdot m/s}, 0)$
9 kg	$\vec{v}_2 = (-2\ \mathrm{m/s}, 0)$	$\vec{p}_2 = (-18\ \mathrm{kg\cdot m/s}, 0)$
3 kg	$\vec{v}_3 = (+4\ \mathrm{m/s}, -2\ \mathrm{m/s})$	$\vec{p}_3 = (+12\ \mathrm{kg\cdot m/s}, -6\ \mathrm{kg\cdot m/s})$

• Net Momentum Calculation:

  • $p_{\text{net},x} = -18\ \mathrm{kg\cdot m/s} - 18\ \mathrm{kg\cdot m/s} + 12\ \mathrm{kg\cdot m/s} = -24\ \mathrm{kg\cdot m/s}$

  • $p_{\text{net},y} = -6\ \mathrm{kg\cdot m/s}$

Impulse and the Impulse-Momentum Theorem

Impulse

Impulse is the effect of a force acting over a time interval, resulting in a change in momentum.

• Definition: Impulse (J) is the product of the average net force and the time interval over which it acts.

• Formula: $\vec{J} = \vec{F}_{\text{net}} \Delta t$

• Units: $\mathrm{N\cdot s}$ or $\mathrm{kg\cdot m/s}$

• Graphical Representation: The area under a force vs. time graph represents impulse.

Impulse-Momentum Theorem

The impulse-momentum theorem relates the impulse delivered to an object to its change in momentum.

• Statement: The net impulse on an object equals its change in momentum.

• Formula: $\vec{J} = \Delta \vec{p}$

• Expanded: $\vec{F}_{\text{net}} \Delta t = m\vec{v}_f - m\vec{v}_i$

• Example: If a 4 N force acts for 2 s, $J = 4\ \mathrm{N} \times 2\ \mathrm{s} = 8\ \mathrm{N\cdot s}$

Applications: Solving Problems with Impulse and Momentum

Example 1: Collision Force Calculation

A pool stick strikes a motionless 0.50 kg cue ball for $\Delta t = 0.00003\ \mathrm{s}$, causing the ball to roll left at $3.0\ \mathrm{m/s}$. Find the average collision force.

• Step 1: Calculate change in momentum:

  • $\Delta p_x = m v_{f,x} - m v_{i,x} = (0.50\ \mathrm{kg})(-3.0\ \mathrm{m/s}) - 0 = -1.5\ \mathrm{kg\cdot m/s}$

• Step 2: Relate to impulse:

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

  • $-1.5\ \mathrm{N\cdot s} = F_{\text{net},x}(0.00003\ \mathrm{s})$

  • $F_{\text{net},x} = -50,000\ \mathrm{N}$

• Comparison: The collision force is much larger than gravity acting on the ball.

Concept Check: Stopping an Object with Impulse

• A 0.5 kg egg tossed with $2\ \mathrm{kg\cdot m/s}$ of momentum can be stopped by different combinations of force and time, as long as $J = \Delta p$.

• For example, a strong force over a short time or a gentle force over a longer time can both stop the egg.

Conservation of Momentum

Momentum Conservation Principle

Momentum is conserved in an isolated system, meaning the total momentum before an interaction equals the total momentum after, provided no external forces act.

• Formula: $\vec{p}_{\text{total, i}} = \vec{p}_{\text{total, f}}$

• Application: Used to analyze collisions and explosions.

• Equal and Opposite Exchange: Forces between interacting objects cause equal but opposite changes in momentum.

Collisions: Momentum Exchange

• During a collision, each object experiences a change in momentum equal in magnitude and opposite in direction.

• Formula: $m_1 \Delta v_1 = -m_2 \Delta v_2$

• Example: Two carts collide and exchange momentum, but the total system momentum remains constant.

Example 2: Conservation of Momentum in an Isolated System

A 150 kg crate sits on a motionless 500 kg boat. The crate slides off at 3 m/s. Find the boat's final speed.

• Step 1: Initial momentum is zero: $\vec{p}_{\text{sys,i}} = m_B \vec{v}_{B,i} + m_C \vec{v}_{C,i} = 0$

• Step 2: Apply conservation of momentum:

  • $0 = (500\ \mathrm{kg})v_{B,f} + (150\ \mathrm{kg})(+3\ \mathrm{m/s})$

  • $v_{B,f} = -0.9\ \mathrm{m/s}$

• Comments:

  • Equal amounts of momentum in opposite directions.

  • High mass boat travels slower than the lighter crate.

Summary Table: Key Equations and Concepts

Concept	Equation	Description
Linear Momentum	$\vec{p} = m\vec{v}$	Product of mass and velocity
Impulse	$\vec{J} = \vec{F}_{\text{net}} \Delta t$	Force applied over time interval
Impulse-Momentum Theorem	$\vec{J} = \Delta \vec{p}$	Impulse equals change in momentum
Conservation of Momentum	$\vec{p}_{\text{total, i}} = \vec{p}_{\text{total, f}}$	Total momentum remains constant in isolated system

Additional info: These notes expand on the original lecture slides by providing full definitions, formulas, and step-by-step problem solutions, ensuring a self-contained study guide for exam preparation.