Download Unit 6 MOMENTUM AND ITS Conservation 1

Principles of
Technology/Physics in
Context (PT/PIC)
Chapter 6
Momentum and Its Conservation 1
Text p. 111-113
Principles of Technology/Physics in Context
(PT/PIC)
• During the lecture assessment
questions will be asked.
• The assessment questions will be
collected at the end of the lecture for
grading.
Key Objectives
At the conclusion of this chapter you’ll be able
to:
• Define the term momentum, and state its SI
unit.
• Solve problems involving mass, velocity, and
momentum.
• Define the term impulse, and state its SI unit.
Key Objectives
At the conclusion of this chapter you’ll be able
to:
• Relate impulse to change in momentum.
• Solve impulse-momentum problems.
6.1 MOMENTUM
• Suppose we wanted to measure the
“tendency” of an object to remain in
motion.
• We would find that two factors are
necessary to describe this tendency:
• the mass and the velocity of the object.
6.1 MOMENTUM
• The more mass an object has, the more
force is required to bring it to rest.
• Similarly, the greater the velocity of the
object, the more force is necessary to
bring it to rest.
• These two factors, mass and velocity, are
combined into a single quantity that we
call momentum.
6.1 MOMENTUM
• We define momentum, symbolized as p.
as the product of mass and velocity:
• p=mv
6.1 MOMENTUM
• Momentum is a vector quantity and its
direction is the direction of the velocity of
the object. Its unit is the kilogram meter
per second (kg∙m/s).
• p=mv
Assessment Question 1
• All of the following are true EXCEPT:
A.Momentum is the “tendency” of an object to
remain in motion is.
B.The more mass an object has, the more force is
required to bring it to rest.
C.The greater the velocity of the object, the more
force is necessary to bring it to rest.
D.Mass and velocity, are combined into a single
quantity that we call momentum.
E.Momentum is a scalar quantity and its magnitude
is the magnitude of the velocity of the object.
6.1 MOMENTUM
• PROBLEM
• An object whose mass is 3.5 kilograms is
traveling at 20. meters per second [east]
Calculate the momentum of the object.
6.1 MOMENTUM
• SOLUTION
Assessment Question 2
• An object whose mass is 35 kilograms is
traveling at 42 meters per second [west]
Calculate the momentum of the object.
• p = mv = 35 kg ∙ 42 m/s [west] =
A.0.833 kg∙m/s [west]
B.1.2 kg∙m/s [west]
C.7 kg∙m/s [west]
D.77 kg∙m/s [west]
E.1500 kg∙m/s [west]
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• When Newton developed his second law
of motion, he recognized that the
unbalanced force on an object caused a
change in the object momentum:
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• Since mass is usually constant:
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• We can rewrite Newton’s second law in
the following form:
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• We call the quantity F∙t, the impulse
delivered to the object.
• Impulse is symbolized by the letter J, and
its unit is the newton second (N∙s).
• Impulse is a vector quantity, and its
direction is the direction of the net force.
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• From Newton’s second law it follows that
the impulse delivered to the object
changes its momentum, and the unit
newton second is equivalent to the unit
kilogram meter per second.
Assessment Question 3
• All of the following are true EXCEPT
A. The unbalanced force on an object caused a
change in the object momentum
B. The quantity F∙t, the impulse delivered to the
object.
C. Impulse is symbolized by the letter J, and its
unit is the newton second (N∙s).
D. Impulse is a vector quantity, and its direction is
the direction of the net force.
E. The impulse delivered to an object does not
affect its momentum.
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• PROBLEM
• A 5.0-kilogram object traveling at 3.0
meters per second [east] is subjected to a
force that increases its velocity to 7.0
meters per second [east]
• Calculate: (a) the initial momentum of the
object,
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• SOLUTION
• Calculate: (a) the initial momentum of the
object,
Assessment Question 4
•
•
A.
B.
C.
D.
E.
A 15 kg (m) object traveling at 13 m/s
[west] (vi) is subjected to a force that
increases its velocity to 27 m/s [west] (vf)
Calculate the initial momentum of the
object. pi = mvi = 15 kg∙13 m/s [west] =
0.087 kg∙m/s [west]
1.2 kg∙m/s [west]
28 kg∙m/s [west]
200 kg∙m/s [west]
350 kg∙m/s [west]
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• PROBLEM
• A 5.0-kilogram object traveling at 3.0
meters per second [east] is subjected to a
force that increases its velocity to 7.0
meters per second [east]
• Calculate: (b) the final momentum of the
object,
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• SOLUTION
• Calculate: (b) the final momentum of the
object,
Assessment Question 5
•
•
A.
B.
C.
D.
E.
A 15 kg (m) object traveling at 13 m/s
[west] (vi) is subjected to a force that
increases its velocity to 27 m/s [west] (vf)
Calculate the final momentum of the
object, pf = mvf = 15 kg∙27 m/s [west] =
0.55 kg∙m/s [west]
1.8 kg∙m/s [west]
54 kg∙m/s [west]
240 kg∙m/s [west]
410 kg∙m/s [west]
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• PROBLEM
• A 5.0-kilogram object traveling at 3.0
meters per second [east] is subjected to a
force that increases its velocity to 7.0
meters per second [east]
• Calculate: (c) the change in momentum of
the object,
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• SOLUTION
• Calculate: (c) the change in momentum of
the object,
Assessment Question 6
•
A 15 kg (m) object traveling at 13 m/s
[west] (vi) is subjected to a force that
increases its velocity to 27 m/s [west] (vf)
• Calculate the change in momentum of the
object Δp = pf - pi = mvf - mvi =
Δp = 15 kg∙27 m/s [W] - 15 kg∙13 m/s [W] =
A.
B.
C.
D.
E.
55 kg∙m/s [west]
130 kg∙m/s [west]
210 kg∙m/s [west]
340 kg∙m/s [west]
420 kg∙m/s [west]
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• PROBLEM
• A 5.0-kilogram object traveling at 3.0
meters per second [east] is subjected to a
force that increases its velocity to 7.0
meters per second [east]
• Calculate: (d) the impulse delivered to the
object.
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• SOLUTION
• Calculate: (d) the impulse delivered to the
object.
Assessment Question 7
•
A 51 kg (m) object traveling at 31 m/s [west] (vi)
is subjected to a force that increases its velocity
to 72 m/s [west] (vf).
• Calculate the impulse delivered to the object.
• J = F∙t = Δp = pf - pi = mvf - mvi =
Δp = 51 kg∙72 m/s [W] - 51 kg∙31 m/s [W] =
A. 75 kg∙m/s [west]
B. 210 kg∙m/s [west]
C. 510 kg∙m/s [west]
D. 1300 kg∙m/s [west]
E. 2100 kg∙m/s [west]
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• PROBLEM
• A 5.0-kilogram object traveling at 3.0
meters per second [east] is subjected to a
force that increases its velocity to 7.0
meters per second [east]
• Calculate: (e) If the force acts for 0.20
second, what are its magnitude and its
direction?
6.2 NEWTON’S SECOND LAW
AND MOMENTUM
• SOLUTION
• Calculate: (e) If the force acts for 0.20
second, what are its magnitude and its
direction?
Assessment Question 8
•
•
A.
B.
C.
D.
E.
If an impulse of 55 kg∙m/s [w] acts for 15
second, what are its magnitude and its direction
of the force of the impulse?
F = J/t = 55 kg∙m/s [W] / 15 s =
0.27 N [w]
3.7 N [w]
40 N [w]
70 N [w]
825 N [w]
Assessment Question 9
•
•
A.
B.
C.
D.
E.
If an impulse of 85 kg∙m/s [W] (J) acts with a
force of 98 N [W](F) how long is the impulse (t)?
t = J/F = 85 kg∙m/s [W] / 98 N [W] =
0.87 s
1.2 s
13 s
290 s
830 s
Conclusion
• Momentum is the product of mass and
velocity.
• It is a vector quantity that measures the
tendency of an object to remain in motion.
• An impulse is the product of force and time
and is responsible for changing the
momentum of an object.
• An impulse is also a vector quantity.
Assessment Question 10
• All of the following are true EXCEPT:
A. The product of mass and velocity is
momentum.
B. Momentum has magnitude and direction.
C. Momentum measures the tendency of an
object to remain in motion.
D. An impulse is the product of force and
time and is responsible for changing the
momentum of an object.
E. An impulse has magnitude only.