the “big mo”. momentum is the product of __________ x the _____________ of an object

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The “Big MO”

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The “Big MO” Momentum is the product of mass X the velocity of an object

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Page 1: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

The “Big MO”

Page 2: 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

Page 3: 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

Page 4: 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

• Symbol:

Page 5: 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

• Symbol: p

Page 6: 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

• Symbol: p• Formula: ?

Page 7: 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

• Symbol: p• Formula: p = mv

Page 8: 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

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

Page 9: 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

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

Page 10: 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

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

Page 11: 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

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

Page 12: 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

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

Page 13: 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

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

Page 14: 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

• 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

Page 15: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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?

Page 16: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =?

Page 17: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =? Formula: p = ?

Page 18: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =? Formula: p = mv

Page 19: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =? Formula: p = mv Sub: p = ?

Page 20: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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

Page 21: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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

Page 22: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =? 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 = ?

Page 23: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =? 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 = ?

Page 24: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =? 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

Page 25: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =? 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 = ?

Page 26: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =? 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 = ?

Page 27: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 =? 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]

Page 28: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

Page 29: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• What is Newton’s second law equation?

Page 30: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• F net = m a

Page 31: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

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

for acceleration?

Page 32: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• F net = m a• a = ∆v / ∆t

Page 33: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

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

Page 34: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

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

Page 35: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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?

Page 36: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

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

Page 37: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

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

Page 38: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

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

Page 39: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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 = ?

Page 40: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

Page 41: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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 = ?

Page 42: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

Page 43: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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 = ?

Page 44: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

• m v1 = initial momentum or p1

Page 45: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

• m v1 = initial momentum or p1

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

Page 46: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

• m v1 = initial momentum or p1

• F net = ( p2 – p1 ) / ∆t

Page 47: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

• m v1 = initial momentum or p1

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

Page 48: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

• m v1 = initial momentum or p1

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

Page 49: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

• m v1 = initial momentum or p1

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

Page 50: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

• m v1 = initial momentum or p1

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

Page 51: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

• m v1 = initial momentum or p1

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

Page 52: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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

• m v1 = initial momentum or p1

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

Page 53: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Alternate Form of Newton’s Second Law

• 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.

Page 54: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

Impulse

Page 55: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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

law:

Page 56: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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

law: F net = ∆ p / ∆t

Page 57: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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

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

get?

Page 58: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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

law: F net = ∆ p / ∆t F net ∆t = ∆ p

Page 59: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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”.

Page 60: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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”. Why is “J” a vector ?

Page 61: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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.

Page 62: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 = ?

Page 63: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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

Page 64: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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.

Page 65: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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 = ?

Page 66: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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

Page 67: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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:

Fnet

Time of contact during collisionArea

Page 68: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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.

Fnet

Time of contact during collisionArea

Page 69: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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

Page 70: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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]

Page 71: The “Big MO”. Momentum is the product of __________ X the _____________ of an object

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