Unit 6: Linear Momentum, Impulse, and Collisions

Unit 6: Linear Momentum, Impulse and Collisions

Introduction

• Quote by Neil deGrasse Tyson: "Physics is the only profession in which prophecy is not only accurate but routine."

Before We Begin

Power

• Definition:
  • Power is the rate of energy transfer.
• Mathematical Representation:
  • Average power ($ar{P}$) is given by:
    ar{P} = rac{W}{ riangle t}
  • Applying the definition of work:
    ar{P} = rac{W}{ riangle t} = rac{F riangle x}{ riangle t} = F ar{v} ext{cos} heta
• Units:
  • The unit of power is the Watt ($W$):
    1W = rac{J}{s} = kg rac{m^2}{s^3}
  • In the U.S. customary system, power is measured in horsepower ($hp$):
    $1hp = 746W$
  • Important Note: Do not confuse $[W]$ from Watts with “$W$” from work!
  • For electric power generation, the customary unit is kilowatt-hour ($kWh$) as a measure of energy:
    $1 kWh = 3.60 imes 10^6 J$

Example Problem

• Problem Statement:
  • A $1.00 imes 10^3 kg$ elevator car carries a maximum load of $8.00 imes 10^2 kg$. A constant frictional force of $4.00 imes 10^3 N$ retards its motion upward.
  • Question: What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator car at a constant speed ($v = 3.00 m/s$)?

Definition of Momentum

• Momentum is defined as the tendency to continue moving in some direction.
• Importance: Relevant in isolated systems, where two or more objects interact through collisions.
• Types of Momentum:
  1. Linear Momentum:
  • Movement in a straight line or the same direction (e.g., running at speed with direction).
  1. Angular Momentum:
  • Movement radially about an axis (e.g., ballerina spinning or tetherball moving around a pole).
• Focus: This unit concentrates on linear momentum.

Linear Momentum ($p$)

Mathematical Description

• Definition:
  • Linear Momentum ($p$) is mathematically described as:
    $p = m imes v$
  • Where:
  • $m$ = Mass of an object in kilograms
  • $v$ = Velocity in meters per second
• SI Unit:
  • The unit for momentum is $kg imes m/s$ (kilogram-meter per second).

Proportional Relationships

• Momentum is directly proportional to both mass and velocity:
  • If velocity is constant, then momentum increases with increasing mass:
    $p ext{ increases as } m ext{ increases, given } v ext{ is constant.}$
  • If mass is constant, then momentum increases with increasing velocity:
    $p ext{ increases as } v ext{ increases, given } m ext{ is constant.}$

Sample Problem

• Problem Statement: What velocity would a smaller mass ($m1$) need to achieve the same momentum as a bigger mass ($m2$) traveling at some speed ($v_2$)?
• Step 1: Set momentum of $m1$ equal to that of $m2$:
  • $p<em>1 = p</em>2$
  • Thus,
    $m<em>1 v</em>1 = m<em>2 v</em>2$
• To find $v_1$:
  • v1 = v2 rac{m2}{m1}

Example Calculation

• Given:
  • $m1 = 5 kg$, $m2 = 10 kg$, $v_2 = 30 m/s$
• Calculation:
  • v1 = v2 rac{m2}{m1} = 30 m/s imes rac{10 kg}{5 kg} = 60 m/s

Momentum-Force Relation

Changing Momentum

• Definition: The momentum ($p$) of an object can be changed ($ riangle p$) only when an external force ($F$) is applied to the object over a duration of time ($ riangle t = t{final} - t{initial}$).
• Change in momentum mathematically expressed:
  • $riangle p = m imes riangle v$
• Taking the ratio of momentum change and time gives:
  • rac{ riangle p}{ riangle t} = m imes rac{ riangle v}{ riangle t}
• By definition, acceleration ($a$) is given by:
  • a = rac{ riangle v}{ riangle t}
• This leads to:
  • $rac{ riangle p}{ riangle t} = m imes a$
• Note: The symbol $ riangle$ signifies “The change in [something] …” and is pronounced “delta [something].”

Newton's Second Law and Impulse Definition

• Newton’s second law states that the force exerted on an object is equal to the mass times acceleration:
  • $F = m imes a$
• Thus, we can express:
  • F = rac{ riangle p}{ riangle t}
• The impulse ($I$) can be defined as:
  • $I = F au = riangle p = m(v<em>f - v</em>i)$

Example Problem

• Problem Statement: A runaway train car with mass $15000 kg$ is moving at a velocity of $5.4 m/s$. Determine the time required for a force of $1500 N$ to bring the car to rest.

Impulse Calculation

• Definition of Impulse:
  • $I = riangle p = F imes riangle t$
• Change in momentum equals the average net external force multiplied by the time this force is applied.
• Example Problem:
  • Calculate the final speed of a $110 kg$ rugby player initially running at $8.00 m/s$ who collides head-on with a padded goalpost and experiences a backward force of $1.76 imes 10^4 N$ for $5.50 imes 10^{-2} s$.

Law of Conservation of Momentum

• Momentum is a conserved quantity stated as:
  • $ext{Σ}p<em>i = ext{Σ}p</em>f$
• The change in momentum for all agents can be expressed:
  • For object 1:
    $riangle p<em>1 = F</em>1 riangle t$
  • For object 2:
    $riangle p<em>2 = F</em>2 riangle t$
• According to Newton’s third law:
  • $F<em>1 = -F</em>2$
• This means:
  • $riangle p<em>1 = - riangle p</em>2$, leading to:
  • $p<em>f1 - p</em>i1 = -(p<em>f2 - p</em>i2)$
• Rearranging gives:
  • $p<em>f1 - p</em>i1 = p<em>i2 - p</em>f2$
  • Thus:
    $p<em>f1 + p</em>f2 = p<em>i2 + p</em>i1$
  • This establishes the law of conservation of momentum true in isolated systems where net external force is zero:
  • $F<em>1 + F</em>2 + … + F_N = 0$

Collisions

Types of Collisions

1. Elastic Collisions:

   • A collision that conserves kinetic energy:
     $K<em>i = K</em>f$
   • Expressed mathematically:
     rac{1}{2}m1v{1i}^2 + rac{1}{2}m2v{2i}^2 = rac{1}{2}m1v{1f}^2 + rac{1}{2}m2v{2f}^2
   • Objects bounce off one another after impact:
     $p<em>{1i} + p</em>{2i} = p<em>{1f} + p</em>{2f}$

2. Inelastic Collisions:

   • A collision that does not conserve kinetic energy
   • Objects stick together after impact, sharing final velocity:
     $m<em>1v</em>{1i} + m<em>2v</em>{2i} = (m<em>1 + m</em>2)v_f$

Strategic Steps for Solving Collision Problems

1. Coordinates:
   • Choose a coordinate axis that aligns with the direction of motion.
2. Diagram:
   • Sketch the problem to visualize forces and motion.
3. Conservation of Momentum:
   • Write expressions for total momentum in the initial and final states for both types of collisions.
4. Conservation of Energy:
   • Write total energy conservation expressions only for elastic collisions.
5. Solve Equations:

Example Problem (Elastic Collision)

• A block of mass $1.60 kg$ moves right at $4 m/s$ on a frictionless track and collides with a massless spring attached to a second block of mass $2.10 kg$ moving left at $-2.50 m/s$.
• Questions:
  1. Determine the velocity of the second block when the first block is moving at $3.00 m/s$.
  2. Find the compression of the spring at that moment.

Example Problem (Inelastic Collision)

• Problem Statement:
  • Two objects move in the same direction with velocities $v1 = 5 m/s$ and $v2 = 10 m/s$. Given $m1 = 10 kg$ and $m2 = 20 kg$, what is their final velocity after a perfectly inelastic collision?
• Solution Approach:
  • Inelastic objects stick together post-impact:
    $m<em>1v</em>{1i} + m<em>2v</em>{2i} = (m<em>1 + m</em>2)v_f$

Final Velocity Calculation

• Substitute and solve:
  • vf = rac{m1 v{1i} + m2 v{2i}}{m1 + m_2} = rac{10 kg imes 5 m/s + 20 kg imes 10 m/s}{10 kg + 20 kg} = 8.3 m/s

Collisions and Kinetic Energy

• In all systems, energy is conserved if all energy transfer mechanisms are considered.
• Kinetic Energy Formula:
  • ext{Total } K.E.: ext{Σ} rac{1}{2} mi vi^2 = ext{Σ} rac{1}{2} mf vf^2
• For elastic collisions:
  • $K<em>{K</em>i} = K<em>{K</em>f}$
• For inelastic collisions:
  • $K<em>{K</em>i} <br /> eq K<em>{K</em>f}$
  • Energy conservation may involve other forms:
  • $K<em>E</em>i = K<em>E</em>f + ext{Σ} U_{other}$
• Additional Energy Forms: Heat, sound, etc.