The “Big MO”. Momentum is the product of X the ___ of an object

The “Big MO”

• Momentum is the product of __________ X the _____________ of an object

The “Big MO”

• Momentum is the product of mass X the velocity of an object

The “Big MO”

• Symbol:

The “Big MO”

• Symbol: p

The “Big MO”

• Symbol: p• Formula: ?

The “Big MO”

• Symbol: p• Formula: p = mv

The “Big MO”

• Symbol: p• Formula: p = mv• Vector or Scalar ???

The “Big MO”

• Symbol: p• Formula: p = mv• Vector or Scalar

The “Big MO”

• Symbol: p• Formula: p = mv• Vector or Scalar• Unit: ?

The “Big MO”

• Symbol: p• Formula: p = mv• Vector or Scalar• Unit: kg m/s

The “Big MO”

• Symbol: p• Formula: p = mv• Vector or Scalar• Unit: kg m/s• Other Forms of the momentum equation: v = ? m = ?

The “Big MO”

• Symbol: p• Formula: p = mv• Vector or Scalar• Unit: kg m/s• Other Forms of the momentum equation: v = p / m m = ?

The “Big MO”

• Symbol: p• Formula: p = mv• Vector or Scalar• Unit: kg m/s• Other Forms of the momentum equation: v = p / m m = ǀ p ǀ / ǀ v ǀ

Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum?b) A 2.0 X 103 kg car is moving with a momentum of 8.0 X 104 kg m/s [S]. What is its velocity?

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =?

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = ?

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = ?

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = ?

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W]

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W]b) Given: m = 2.0 X 103 kg p = 8.0 X 104 kg m/s [S] v = ?

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W]b) Given: m = 2.0 X 103 kg p = 8.0 X 104 kg m/s [S] v = ? Formula: v = ?

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W]b) Given: m = 2.0 X 103 kg p = 8.0 X 104 kg m/s [S] v = ? Formula: v = p / m

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W]b) Given: m = 2.0 X 103 kg p = 8.0 X 104 kg m/s [S] v = ? Formula: v = p / m Sub: v = ?

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W]b) Given: m = 2.0 X 103 kg p = 8.0 X 104 kg m/s [S] v = ? Formula: v = p / m Sub: v = 8.0 X 104 kg m/s [S] / 2.0 X 103 kg = ?

a) Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W]b) Given: m = 2.0 X 103 kg p = 8.0 X 104 kg m/s [S] v = ? Formula: v = p / m Sub: v = 8.0 X 104 kg m/s [S] / 2.0 X 103 kg = 40.0 m/s [S]

Alternate Form of Newton’s Second Law

• What is Newton’s second law equation?

• F net = m a

• F net = m a• In terms of delta notation, what is the defining equation

for acceleration?

• F net = m a• a = ∆v / ∆t

• F net = m a• a = ∆v / ∆t So what is F net = ?

• F net = m a• a = ∆v / ∆t therefore F net = m ∆v / ∆t

• F net = m a• a = ∆v / ∆t therefore F net = m ∆v / ∆t• What is the equation for ∆v in terms of final velocity and

initial velocity?

• F net = m a• a = ∆v / ∆t therefore F net = m ∆v / ∆t• ∆v = v2 – v1

• F net = m a• a = ∆v / ∆t therefore F net = m ∆v / ∆t• ∆v = v2 – v1 So what is F net = ?

• F net = m a• a = ∆v / ∆t therefore F net = m ∆v / ∆t• ∆v = v2 – v1 therefore F net = m ( v2 – v1 ) / ∆t•

• F net = m a• a = ∆v / ∆t therefore F net = m ∆v / ∆t• ∆v = v2 – v1 therefore F net = m ( v2 – v1 ) / ∆t• Expand the numerator F net = ?

• F net = m a• a = ∆v / ∆t therefore F net = m ∆v / ∆t• ∆v = v2 – v1 therefore F net = m ( v2 – v1 ) / ∆t• Expand the numerator F net = ( m v2 – mv1 ) / ∆t

• F net = m a• a = ∆v / ∆t therefore F net = m ∆v / ∆t• ∆v = v2 – v1 therefore F net = m ( v2 – v1 ) / ∆t• Expand the numerator F net = ( m v2 – mv1 ) / ∆t• But what is m v2 = ?

• F net = m a• a = ∆v / ∆t therefore F net = m ∆v / ∆t• ∆v = v2 – v1 therefore F net = m ( v2 – v1 ) / ∆t• Expand the numerator F net = ( m v2 – mv1 ) / ∆t• m v2 = final momentum or p2

• But what is m v1 = ?

• m v1 = initial momentum or p1

• What is F net in terms of final and initial momentum?

• F net = ( p2 – p1 ) / ∆t

• F net = ( p2 – p1 ) / ∆t• What is ( p2 – p1 ) in terms of delta notation?

• F net = ( p2 – p1 ) / ∆t• ( p2 – p1 ) = ∆ p

• F net = ( p2 – p1 ) / ∆t• ( p2 – p1 ) = ∆ p So what is F net = ?

• F net = ( p2 – p1 ) / ∆t• ( p2 – p1 ) = ∆ p therefore F net = ∆ p / ∆t

• F net = ( p2 – p1 ) / ∆t• ( p2 – p1 ) = ∆ p therefore F net = ∆ p / ∆t• Using the term “rate” , define net force in words.

• F net = ( p2 – p1 ) / ∆t• ( p2 – p1 ) = ∆ p therefore F net = ∆ p / ∆t• The net force is the rate of change of momentum.

• F net = ∆ p / ∆t • The net force is the rate of change of momentum.• This is another and better form or equation for Newton’s

second law than F net = m a . This equation for Newton’s second law takes into consideration the case when mass changes as well as velocity, such as rocket motion.

Impulse

Impulse• Let’s start with the alternate equation for Newton’s second

law: F net = ∆ p / ∆t

law: F net = ∆ p / ∆t • If we multiply both sides of this equation by ∆t , what do we

law: F net = ∆ p / ∆t F net ∆t = ∆ p

law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”.

law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. Why is “J” a vector ?

law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p.

law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p.• Units of J = ?

law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p.• Units of J = Newton X second = Ns

law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p.• Units of J = Newton X second = Ns• We use impulse for short time interval collisions, like cricket

bat hitting ball, baseball bat hitting baseball etc.

Impulse• Let’s start with the alternate equation for Newton’s second law:

F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p.• Units of J = Newton X second = Ns• We use impulse for short time interval collisions, like cricket bat

hitting ball, baseball bat hitting baseball etc.• Equation for J = ?

Impulse• Let’s start with the alternate equation for Newton’s second law:

F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p.• Units of J = Newton X second = Ns• We use impulse for short time interval collisions, like cricket bat

hitting ball, baseball bat hitting baseball etc.• Equation J = ∆ p or F net ∆t

Impulse• Let’s start with the alternate equation for Newton’s second law: F net = ∆

p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p.• Units of J = Newton X second = Ns• We use impulse for short time interval collisions, like cricket bat hitting ball,

baseball bat hitting baseball etc.• Equation J = ∆ p or F net ∆t• We can also calculate the impulse as the area under a “net force” vs “time of

collision” curve:

Time of contact during collisionArea

Impulse• Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p.• Units of J = Newton X second = Ns• We use impulse for short time interval collisions, like cricket bat hitting ball,

baseball bat hitting baseball etc.• Equation J = ∆ p or F net ∆t• We can also calculate it as the area under a “net force” vs “time of collision” curve:• The area can be approximated as two triangles as shown.

Time of contact during collisionArea

Example : A 50.00 g ping-pong ball moving horizontally at 6.00 m/s [right] strikes a wall. The ball makes contact with the wall for 0.0600 seconds. The ping-pong ball rebounds off the wall with a horizontal velocity of 4.00 m/s [left]. Ignoring any vertical forces, find: a) change in momentum of the ball b) impulse of the wall on the ball c) the average net horizontal force on the ball

a) ∆ p = p2 – p1 = m v2 – mv1 = 0.05(-4.00) - 0.05(+6.00) ∆ p = - 0.200 – 0.300 = - 0.500 kg m/s or 0.500 kg m/s [L]b) J = ∆ p = 0.500 Ns [left]c) F net = ∆ p / ∆t = 0.500 kg m/s [r] /0.0600 s = 8.33 N [left]

Homework

• Try the Impulse and momentum problems handout and check answers

• New textbook: Read p222 – p225 Try q1,2 p223 q1,q2 p226 Q1-Q12 p227 check answers p717• Old Textbook: Read p232 – pp236 Try 6-10 p237 check same page

The “Big MO”. Momentum is the product of X the ___ of an object

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The “Big MO”. Momentum is the product of __________ X the _____________ of an object

The “Big MO”. Momentum is the product of X the ___ of an object