Linked List Implementation in Python

Lab 2: Linked List Implementation

• Objective is to implement a linked list data structure and common methods in Python.

Linked Lists

• Linked Lists are linear collections of data elements (nodes) with each node pointing to the next.
• Structure comprises a sequence of nodes.
• Singly-linked list nodes contain data and a reference to the next node.
• Doubly-linked list nodes contain data, a reference to the next node, and a reference to the previous node.

Common Operations

• Common linked list operations include insert, delete, search, len, and reverse.
• Insert: Add element to list.
• Delete: Remove element from list.
• Search: Find an element in list.
• Len: Get element count in list.
• Reverse: Reverse element order in list.

Implementation

• Implement a singly-linked list with methods: __init__, insert_head, insert_tail, delete_head, delete_tail, search, len, reverse, and print.

Node Class

• The Node class includes data and a pointer (next) to the next node.

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

LinkedList Class

• The LinkedList class has methods to manage the linked list.

class LinkedList:
    def __init__(self):
        self.head = None
        self.tail = None

Insert Head

• A new node is inserted at list's head.
• If the list is empty, the new node becomes both head and tail.
• Otherwise, the new node points to the current head, and it becomes the new head.

def insert_head(self, data):
    new_node = Node(data)
    if self.head is None:
        self.head = new_node
        self.tail = new_node
    else:
        new_node.next = self.head
        self.head = new_node

Insert Tail

• A new node is inserted at the list's tail.
• If list is empty, the new node becomes both head and tail.
• Otherwise, the current tail points to the new node, and the new node becomes the new tail.

def insert_tail(self, data):
    new_node = Node(data)
    if self.head is None:
        self.head = new_node
        self.tail = new_node
    else:
        self.tail.next = new_node
        self.tail = new_node

Delete Head

• The node at list's head is deleted.
• If the list is empty, returns None.
• The head is updated to the next node, and returns the data from the former head.

def delete_head(self):
    if self.head is None:
        return None
    else:
        temp = self.head
        self.head = self.head.next
        return temp.data

Delete Tail

• The node at the list's tail is deleted.
• If empty, returns None.
• If only one node, head and tail are set to None.
• Otherwise, iterates through the list to find the node before the tail, sets its next pointer to None, and updates the tail.

def delete_tail(self):
    if self.head is None:
        return None
    elif self.head == self.tail:
        temp = self.head
        self.head = None
        self.tail = None
        return temp.data
    else:
        temp = self.head
        while temp.next != self.tail:
            temp = temp.next
        temp.next = None
        self.tail = temp
        return self.tail.data

Search

• It searches for a node with matching data.
• Traverses list to find data and returns True if found, False if not.

def search(self, data):
    temp = self.head
    while temp is not None:
        if temp.data == data:
            return True
        temp = temp.next
    return False

Len

• To determine the number of nodes in list's.
• It then traverses the list, incrementing a counter, and returns the final count.

def len(self):
    count = 0
    temp = self.head
    while temp is not None:
        count += 1
        temp = temp.next
    return count

Reverse

• It reverses node order in the list.
• It uses three pointers: previous, current, and next, to rewire the next pointers of each node.

def reverse(self):
    prev = None
    current = self.head
    self.tail = self.head
    while(current is not None):
        next = current.next
        current.next = prev
        prev = current
        current = next
    self.head = prev

Print

• To display the data of each node.
• It traverses through the list, printing each node's data.

def print(self):
    temp = self.head
    while temp is not None:
        print(temp.data, end=" ")
        temp = temp.next
    print()

Testing

• Includes a main function provided to test implementation.

if __name__ == "__main__":
    ll = LinkedList()
    ll.insert_head(3)
    ll.insert_head(4)
    ll.insert_tail(5)
    ll.print() # 4 3 5
    ll.delete_head()
    ll.print() # 3 5
    ll.delete_tail()
    ll.print() # 3
    ll.insert_tail(10)
    ll.print() # 3 10
    ll.reverse()
    ll.print() # 10 3

Chapter 14: Acids and Bases

14.1 The Nature of Acids and Bases

Definitions of Acids and Bases

• Arrhenius: Acid produces $H^{+}$ in water, base produces $OH^{-}$ in water.
• Brønsted–Lowry: Acid is a proton ($H^{+}$) donor, base is a proton acceptor.
  • Broader definition than Arrhenius.
  • An acid and base always work together.
  • $HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$
    • $H_3O^+$ is the hydronium ion.
    • $A^-$ is the conjugate base of HA, and HA is the conjugate acid of $A^-$.
    • A conjugate acid-base pair consists of two substances related by the loss or gain of a single proton.
• Lewis: Acid is an electron-pair acceptor, base is an electron-pair donor.
  • Broadest definition.
  • All Brønsted–Lowry acids and bases are also Lewis acids and bases, but the converse is not true.
  • Lewis acids and bases explain the reactions of certain metal-containing compounds with electron-pair donors.

Acid Strength

• Strong acids completely transfer protons to water.
  • Their conjugate bases have negligible affinity for protons.
  • E.g., $HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)$
• Weak acids only partially transfer protons to water.
  • Their conjugate bases have a greater affinity for protons than water does.
  • E.g., $HF(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + F^-(aq)$
• Acid strength is the tendency to donate a proton.
• Strong bases react completely with water to form $OH^-$ ions.
  • The conjugate acids of strong bases have negligible acidity.
  • E.g., $O^{2-}(aq) + H_2O(l) \rightarrow 2OH^-(aq)$
• Weak bases react only partially with water to form $OH^-$ ions.
  • The conjugate acids of weak bases are weak acids.
  • E.g., $NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$
• Base strength is the tendency to accept a proton.

Factors Affecting Acid Strength

• Polarity of the H–A bond:
  • The greater the polarity, the more acidic the compound.
• Bond strength:
  • The weaker the bond, the more acidic the compound.
• Stability of the conjugate base, $A^-$:
  • The more stable the conjugate base, the more acidic the compound.
• For binary acids (H–A):
  • Acid strength increases from left to right across a period and increases down a group.
  • E.g., $HF < HCl < HBr < HI$
• Oxyacids (H–O–Y):
  • Acid strength increases with increasing electronegativity of Y and increases with increasing oxidation number of Y.
  • E.g., $HClO < HClO_2 < HClO_3 < HClO_4$
• Carboxylic acids (RCOOH):
  • Acid strength depends on the nature of the R group.
  • Electron-withdrawing groups increase acid strength.

Sample Exercise 14.1

• Arrange the following in order of increasing strength: $HClO_4(aq)$, $HBrO_4(aq)$, $HBrO_3(aq)$.

Solution

$HClO_4(aq) < HBrO_3(aq) < HBrO_4(aq)$

14.2 Autoionization of Water

The Ion-Product Constant

• Water is amphoteric; it can act as both an acid and a base.
• In pure water, a few molecules act as bases and take protons from other water molecules: $H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$
• This process is called the autoionization of water.
• The equilibrium constant for this process is called the ion-product constant, $K_w$: $K_w = [H_3O^+][OH^-] = [H^+][OH^-]$
• At 25 °C, $K_w = 1.0 \times 10^{-14}$
• In pure water, $[H^+] = [OH^-] = 1.0 \times 10^{-7} M$
• An acidic solution has $[H^+] > [OH^-]$.
• A basic solution has $[H^+] < [OH^-]$.
• A neutral solution has $[H^+] = [OH^-]$.
• The value of $K_w$ changes with temperature.

Sample Exercise 14.2

• Calculate the concentrations of $H^+(aq)$ or $OH^-(aq)$ in a solution at 25 °C in which $[OH^-] = 1.0 \times 10^{-9} M$.
• Identify the solution as acidic, basic, or neutral.

Solution

$[H^+] = 1.0 \times 10^{-5} M$ The solution is acidic.

14.3 The pH Scale

pH and pOH

• Because $[H^+]$ in an aqueous solution is usually quite small, it is often expressed as pH: $pH = -log[H^+]$
• Similarly, pOH is defined as: $pOH = -log[OH^-]$
• Because $[H^+][OH^-] = K_w = 1.0 \times 10^{-14}$, we can write: $pH + pOH = 14.00$
• $pH < 7.00$ indicates an acidic solution.
• $pH > 7.00$ indicates a basic solution.
• $pH = 7.00$ indicates a neutral solution.
• The pH scale is logarithmic, so a change of 1 pH unit represents a tenfold change in acidity.

Measuring pH

• pH can be measured using:
  • Acid-base indicators: Substances that change color as pH changes.
  • pH meters: Electronic instruments with a glass electrode that is sensitive to $H^+$ concentration.

Sample Exercise 14.3

• Calculate the pH of a solution in which $[H^+] = 6.0 \times 10^{-4} M$.

Solution

$pH = 3.22$

14.4 Strong Acids and Bases

Strong Acids

• Strong acids are those that completely dissociate in water.
• The common strong acids are $HCl, HBr, HI, HNO_3, H_2SO_4$, and $HClO_4$.
• In solutions of strong acids, $[H^+]$ is equal to the concentration of the acid.

Strong Bases

• Strong bases are those that completely dissociate in water to give $OH^-$ ions.
• The common strong bases are the hydroxides of the alkali metals ($NaOH, KOH, etc. ) and some alkaline earth metals ($Ca(OH)_2, Sr(OH)_2, Ba(OH)_2$).
• In solutions of strong bases, $[OH^-]$ is equal to the concentration of the base.

Sample Exercise 14.4

Calculate the pH of a 0.020 M solution of $HNO_3$.

Solution

$pH = 1.70$

14.5 Weak Acids

Acid-Dissociation Constant, $K_a$

• Weak acids only partially dissociate in water.
• The extent to which a weak acid dissociates is described by the acid-dissociation constant, $K_a$: $HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$ $K_a = \frac{[H_3O^+][A^-]}{[HA]} = \frac{[H^+][A^-]}{[HA]}$
• The larger the value of $K_a$, the stronger the acid.
• We can also express the strength of a weak acid as percent dissociation: Percent dissociation = $\frac{[H^+]_{eq}}{[HA]_0} \times 100%$

Calculating $K_a$

• To calculate $K_a$ from the pH of a solution of a weak acid:
  1. Write the equilibrium reaction and the expression for $K_a$.
  2. Use the pH to determine $[H^+]$ at equilibrium.
  3. Use stoichiometry to determine the equilibrium concentrations of all species.
  4. Substitute the equilibrium concentrations into the expression for $K_a$ and solve.

Sample Exercise 14.5

• A 0.10 M solution of formic acid (HCOOH) has a pH of 2.38 at 25 °C.
• Calculate $K_a$ for formic acid at this temperature.

Solution

$K_a = 1.8 \times 10^{-4}$

14.6 Weak Bases

Base-Dissociation Constant, $K_b$

• Weak bases only partially react with water to produce $OH^-$ ions.
• The extent to which a weak base reacts with water is described by the base-dissociation constant, $K_b$: $B(aq) + H_2O(l) \rightleftharpoons HB^+(aq) + OH^-(aq)$ $K_b = \frac{[HB^+][OH^-]}{[B]}$
• The larger the value of $K_b$, the stronger the base.

$K_a$ and $K_b$

• For a conjugate acid-base pair: $K_a \times K_b = K_w$
• Therefore: $pK_a + pK_b = pK_w = 14.00$
• The stronger an acid, the weaker its conjugate base, and vice versa.

Sample Exercise 14.6

• Calculate the pH of a 0.15 M solution of ammonia ($NH_3$).
• $K_b$ for ammonia is $1.8 \times 10^{-5}$.

Solution

$pH = 11.2$

14.7 Acid-Base Properties of Salts

Salts That Yield Neutral Solutions

• Salts are ionic compounds formed from the reaction of an acid and a base.
• Salts containing an alkali metal or alkaline earth metal cation (except $Be^{2+}$) and the conjugate base of a strong acid (e.g., $Cl^-, Br^-, I^-, NO_3^-, ClO_4^-$) will form neutral solutions.

Salts That Yield Acidic Solutions

• Salts containing a small, highly charged metal cation (e.g., $Al^{3+}, Fe^{3+}, Cu^{2+}$) or the conjugate acid of a weak base (e.g., $NH_4^+$) will form acidic solutions.

Salts That Yield Basic Solutions

• Salts containing the conjugate base of a weak acid (e.g., $F^-, CH_3COO^-, CO_3^{2-}$) will form basic solutions.

Sample Exercise 14.7

Predict whether aqueous solutions of the following salts will be acidic, basic, or neutral: (a) $KCl$, (b) $NH_4NO_3$, (c) $NaCH_3COO$.

Solution

(a) neutral, (b) acidic, (c) basic

14.8 Acid-Base Behavior and Chemical Structure

Factors That Affect Acid Strength

• Bond polarity: The greater the polarity of the H–A bond, the more acidic the compound.
• Bond strength: The weaker the H–A bond, the more acidic the compound.
• Stability of the conjugate base: The more stable the conjugate base, the more acidic the compound.

Hydrides

• For binary hydrides (H–A), acid strength increases from left to right across a period and increases down a group.
  • Acidity increases: $CH_4 < NH_3 < H_2O < HF$
  • Acidity increases: $HF < HCl < HBr < HI$

Oxyacids

• For oxyacids (H–O–Y), acid strength increases with increasing electronegativity of Y and increases with increasing oxidation number of Y.
  • Acidity increases: $HClO < HClO_2 < HClO_3 < HClO_4$

Carboxylic Acids

• For carboxylic acids (RCOOH), acid strength depends on the nature of the R group.
• Electron-withdrawing groups increase acid strength.

14.9 Lewis Acids and Bases

Lewis Acids

• Lewis acids are electron-pair acceptors.
• All Brønsted–Lowry acids are also Lewis acids, but the converse is not true.
• Molecules with an incomplete octet, such as $BF_3$ and $AlCl_3$, are Lewis acids.
• Metal cations, such as $Ag^+$ and $Fe^{3+}$, are Lewis acids.

Lewis Bases

• Lewis bases are electron-pair donors.
• All Brønsted–Lowry bases are also Lewis bases.
• Molecules with lone pairs of electrons, such as $NH_3$ and $H_2O$, are Lewis bases.

Sample Exercise 14.8

• Identify the Lewis acid and Lewis base in the following reaction: $BF_3 + NH_3 \rightarrow F_3BNH_3$.

Solution

• $BF_3$ is the Lewis acid.
• $NH_3$ is the Lewis base.

Thermodynamics

• First Law: Energy conservation.
• Second Law: Entropy increases in a closed system.

First Law Terms

• $U$: Internal energy.
• $Q$: Heat.
• $W$: Work.

First Law Equation

$\Delta U = U_{f} - U_{i} = Q - W$

Isobaric Process

• Constant pressure.
• $W = P\Delta V$
• $Q = nC_{p}\Delta T$
• $\Delta U = Q - W = nC_{p}\Delta T - P\Delta V = nC_{v}\Delta T$
• $C_{p} = C_{v} + R$
  • $C_p$: Molar specific heat at constant pressure.
  • $C_v$: Molar specific heat at constant volume.
  • $R$: Ideal gas constant.

Isochoric Process

• Constant volume.
• $W = 0$
• $\Delta U = Q = nC_{v}\Delta T$

Isothermal Process

• Constant temperature.
• $\Delta U = 0$
• $Q = W = nRT\ln(\frac{V_{f}}{V_{i}})$

Adiabatic Process

• No heat exchange.
• $Q = 0$
• $\Delta U = -W = nC_{v}\Delta T$
• $PV^{\gamma} = \text{constant}$
• $TV^{\gamma-1} = \text{constant}$
• $\gamma = \frac{C_{p}}{C_{v}}$

Second Law Equation

$\Delta S = \int_{i}^{f} \frac{dQ}{T}$ - $\Delta S$: Change in entropy.

Carnot Cycle

• $\varepsilon = 1 - |\frac{Q_{c}}{Q_{h}}|$
  • $\varepsilon$: Efficiency.
  • $Q_c$: Heat rejected to the cold reservoir.
  • $Q_h$: Heat absorbed from the hot reservoir.
• $\varepsilon_{c} = 1 - \frac{T_{c}}{T_{h}}$
  • $\varepsilon_c$: Carnot efficiency.
  • $T_c$: Temperature of the cold reservoir.
  • $T_h$: Temperature of the hot reservoir.

Entropy Change Equation

$\Delta S = nC_{v}\ln(\frac{T_{2}}{T_{1}}) + nR\ln(\frac{V_{2}}{V_{1}})$

Matrizen

Einführung

• Matrizen sind ein wichtiges Werkzeug in vielen Bereichen der Mathematik, Physik, Informatik und Ingenieurwissenschaften.
• Sie werden verwendet, um lineare Gleichungssysteme darzustellen, lineare Abbildungen zu beschreiben, Daten zu speichern und zu verarbeiten, und vieles mehr.

Definition

• Definition: Eine Matrix ist eine rechteckige Anordnung von Zahlen, die in Zeilen und Spalten angeordnet sind.
• Eine Matrix mit $m$ Zeilen und $n$ Spalten wird als $m \times n$-Matrix bezeichnet.
• Die Zahlen in der Matrix werden als Elemente oder Einträge der Matrix bezeichnet.
• Beispiel: $\qquad A = \begin{pmatrix} 1 & 2 & 3 & 4 \ 5 & 6 & 7 & 8 \ 9 & 10 & 11 & 12 \end{pmatrix}$
• Die Elemente der Matrix werden mit $a_{ij}$ bezeichnet, wobei $i$ die Zeilennummer und $j$ die Spaltennummer angibt.
• Im obigen Beispiel ist $a_{11} = 1$, $a_{23} = 7$ und $a_{34} = 12$.

Spezielle Matrizen

• Es gibt verschiedene Arten von speziellen Matrizen, die in der linearen Algebra eine wichtige Rolle spielen:
  • Quadratische Matrix: Anzahl der Zeilen und Spalten ist identisch ($m = n$)
  • Nullmatrix: Alle Elemente sind Null.
  • Einheitsmatrix: Diagonalelemente sind alle gleich Eins und deren alle anderen Elemente gleich Null sind; mit $I$ oder $I_n$ bezeichnet, wobei $n$ die Anzahl der Zeilen und Spalten angibt.
  • Diagonalmatrix: Elemente außerhalb der Diagonale gleich Null.
  • Symmetrische Matrix: Eine quadratische Matrix $A$, für die $A = A^T$.
  • Schiefsymmetrische Matrix: Eine quadratische Matrix $A$, für die $A = -A^T$.

Matrixoperationen

Addition und Subtraktion

• Definition: Zwei Matrizen $A$ und $B$ können addiert oder subtrahiert werden, wenn sie die gleiche Anzahl von Zeilen und Spalten haben.
• Die Summe oder Differenz von $A$ und $B$ ist eine Matrix, deren Elemente die Summe oder Differenz der entsprechenden Elemente von $A$ und $B$ sind. $\qquad (A + B){ij} = a{ij} + b_{ij}$ $\qquad (A - B){ij} = a{ij} - b_{ij}$
• Beispiel:* $\qquad A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}$ $\qquad A + B = \begin{pmatrix} 1+5 & 2+6 \ 3+7 & 4+8 \end{pmatrix} = \begin{pmatrix} 6 & 8 \ 10 & 12 \end{pmatrix}$ $\qquad A - B = \begin{pmatrix} 1-5 & 2-6 \ 3-7 & 4-8 \end{pmatrix} = \begin{pmatrix} -4 & -4 \ -4 & -4 \end{pmatrix}$

Skalarmultiplikation

• Definition: Eine Matrix $A$ kann mit einem Skalar $c$ multipliziert werden, indem jedes Element der Matrix mit dem Skalar multipliziert wird. $\qquad (cA){ij} = c \cdot a{ij}$
• Beispiel:* $\qquad A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}, \quad c = 2$ $\qquad cA = 2 \cdot \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix} = \begin{pmatrix} 2 & 4 \ 6 & 8 \end{pmatrix}$

Matrixmultiplikation

• Definition: Zwei Matrizen $A$ und $B$ können miteinander multipliziert werden, wenn die Anzahl der Spalten von $A$ gleich der Anzahl der Zeilen von $B$ ist.
• Das Produkt von $A$ und $B$ ist eine Matrix $C$, deren Elemente wie folgt berechnet werden: $\qquad C_{ij} = \sum_{k=1}^n a_{ik} \cdot b_{kj}$ wobei $n$ die Anzahl der Spalten von $A$ und die Anzahl der Zeilen von $B$ ist.
• Beispiel:* $\qquad A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}$ $\qquad A \cdot B = \begin{pmatrix} 1\cdot5 + 2\cdot7 & 1\cdot6 + 2\cdot8 \ 3\cdot5 + 4\cdot7 & 3\cdot6 + 4\cdot8 \end{pmatrix} = \begin{pmatrix} 19 & 22 \ 43 & 50 \end{pmatrix}$
• Achtung:* Die Matrixmultiplikation ist nicht kommutativ, d.h. im Allgemeinen gilt $A \cdot B \neq B \cdot A$.

Chapter 9: Momentum and Collisions

9.1 Linear Momentum

Definition of Linear Momentum

• The linear momentum $\vec{p}$ of an object of mass $m$ moving with a velocity $\vec{v}$ is defined as the product of the mass and the velocity: $\qquad \vec{p} = m\vec{v}$
  • It is a vector quantity.
  • Its direction is the same as the direction of $\vec{v}$.
  • The magnitude is $p = mv$.
  • Its SI unit is kg.m/s.

Example

A 7.0 kg bowling ball moves at 3.0 m/s. How big is its momentum? $p = mv = (7.0 kg)(3.0 m/s) = 21 kg.m/s$ Since it is a vector, we also need to specify the direction; for example, "in the forward direction."

Newton's Second Law Revisited

$\qquad \sum \vec{F} = \frac{\Delta \vec{p}}{\Delta t}$ - The time rate of change of the momentum of an object is equal to the net force acting on the object. $\qquad \sum \vec{F} = \frac{\Delta \vec{p}}{\Delta t} = \frac{\Delta (m\vec{v})}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ - Newton's Second Law says that a net force applied to an object changes its momentum. - The change in momentum is proportional to the magnitude of the force, and takes place in the direction of the force.

Example

• A hockey puck has a mass of 0.115 kg.
• Initially, it is at rest.
• A player applies a constant force of 6.0 N for a time of 0.021 s.
• What is the speed of the puck after the force is applied? $\Delta p = F\Delta t = (6.0 N)(0.021 s) = 0.13 kg.m/s$ $\Delta p = mv_f - mv_i = mv_f - 0 = mv_f$ $v_f = \frac{\Delta p}{m} = \frac{0.13 kg.m/s}{0.115 kg} = 1.1 m/s$

9.2 Impulse and Momentum

Impulse

• When a single constant force acts on the object, there is an impulse $\vec{I}$ defined as: $\qquad \vec{I} = \vec{F} \Delta t$
  • $\vec{F}$ is the constant force
  • $\Delta t$ is the time interval during which the force acts
• Impulse is a vector quantity
  • The direction of the impulse is the same as the direction of the force
  • The SI unit of impulse is N.s

Impulse – Momentum Theorem

$\qquad \vec{I} = \Delta \vec{p} = \vec{p_f} - \vec{p_i} = m\vec{v_f} - m\vec{v_i}$ - The impulse of the force acting on an object equals the change in momentum of the object. - This is equivalent to Newton's Second Law. - Impulse is a good way to determine the change in momentum of an object.

More About Impulse

• Impulse is a vector quantity with a magnitude equal to the area under the force-time curve.
• Assume the force varies in time. $\qquad \vec{I} = \int_{t_i}^{t_f} \sum \vec{F} dt$

Average Force

• The average force $\vec{\overline{F}}$ acting on an object during the time $\Delta t$ is the constant force that would give the same impulse to the object as the actual time-varying force gives during the time $\Delta t$. $\qquad \vec{\overline{F}} = \frac{\int_{t_i}^{t_f} \sum \vec{F} dt}{\Delta t} = \frac{\Delta \vec{p}}{\Delta t}$

Impulse Approximation

• In many cases, one force acting on an object for a short time is much greater than any other force present.
• In these cases, use the Impulse Approximation.
  • The net force acting on the object during the short time interval is the impulsive force.
  • The momentum of the object is changed only by the impulse.
  • Can ignore any other forces that may be present.

Example

A 3.0 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the ball by the wall? $p_i = mv_i = (3.0 kg)(10.0 m/s) = 30 kg.m/s$ $p_{ix} = p_i cos(60.0°) = (30 kg.m/s) cos(60.0°) = 15 kg.m/s$ $p_{iy} = -p_i sin(60.0°) = -(30 kg.m/s) sin(60.0°) = -26 kg.m/s$ $p_f = mv_f = (3.0 kg)(10.0 m/s) = 30 kg.m/s$ $p_{fx} = -p_f cos(60.0°) = -(30 kg.m/s) cos(60.0°) = -15 kg.m/s$ $p_{fy} = -p_f sin(60.0°) = -(30 kg.m/s) sin(60.0°) = -26 kg.m/s$ $I_x = \Delta p_x = p_{fx} - p_{ix} = -15 kg.m/s - (15 kg.m/s) = -30 kg.m/s$ $I_y = \Delta p_y = p_{fy} - p_{iy} = -26 kg.m/s - (-26 kg.m/s) = 0 kg.m/s$ $\overline{F_x} = \frac{\Delta p_x}{\Delta t} = \frac{-30 kg.m/s}{0.20 s} = -150 N$ $\overline{F_y} = \frac{\Delta p_y}{\Delta t} = \frac{0 kg.m/s}{0.20 s} = 0 N$

9.3 Conservation of Momentum

Isolated System

• In an isolated system, no external forces act on the system.

Conservation of Momentum

• Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant.
• The momentum of the system is conserved, but the momentum of an individual particle in the system may change. $\qquad \vec{p_1i} + \vec{p_2i} = \vec{p_1f} + \vec{p_2f}$ $\qquad m_1\vec{v_1i} + m_2\vec{v_2i} = m_1\vec{v_1f} + m_2\vec{v_2f}$
  • Momentum is conserved for