the “big mo”. momentum is the product of __________ x the _____________ of an object
DESCRIPTION
The “Big MO” Momentum is the product of mass X the velocity of an objectTRANSCRIPT
![Page 1: The “Big MO”. Momentum is the product of __________ X the _____________ of an object](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/1.jpg)
The “Big MO”
![Page 2: The “Big MO”. Momentum is the product of __________ X the _____________ of an object](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/2.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/3.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/4.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/5.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/6.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/7.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/8.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/9.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/10.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/11.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/12.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/13.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/14.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/15.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/16.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/17.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/18.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/19.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/20.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/21.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/22.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/23.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/24.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/25.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/26.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/27.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/28.jpg)
Alternate Form of Newton’s Second Law
![Page 29: The “Big MO”. Momentum is the product of __________ X the _____________ of an object](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/29.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/30.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/31.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/32.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/33.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/34.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/35.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/36.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/37.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/38.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/39.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/40.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/41.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/42.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/43.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/44.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/45.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/46.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/47.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/48.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/49.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/50.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/51.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/52.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/53.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/54.jpg)
Impulse
![Page 55: The “Big MO”. Momentum is the product of __________ X the _____________ of an object](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/55.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/56.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/57.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/58.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/59.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/60.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/61.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/62.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/63.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/64.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/65.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/66.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/67.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/68.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/69.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/70.jpg)
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](https://reader036.vdocuments.site/reader036/viewer/2022062401/5a4d1b437f8b9ab0599a227b/html5/thumbnails/71.jpg)
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