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AP PHYSICS B
Linear Momentum & Impulse
Teacher Packet
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involved in the production of this material.
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Linear Momentum & Impulse
Objective
To review the student on the concepts, processes and problem solving strategies necessary to
successfully answer questions on linear momentum and impulse.
Standards
Linear momentum and impulse are addressed in the topic outline of the College Board AP*
Physics Course Description Guide as described below.
I. Newtonian Mechanics
D. Systems of particles, linear momentum
1. Center of mass
2. Impulse and momentum
3. Conservation of linear momentum, collisions
AP Physics Exam Connections
Topics relating to linear momentum and impulse are tested every year on the multiple choice and
in most years on the free response portion of the exam in conjunction with work and energy. The
list below identifies free response questions that have been previously asked over linear
momentum and impulse. These questions are available from the College Board and can be
downloaded free of charge from AP Central. http://apcentral.collegeboard.com.
2008
2005
2002
2001
1999
Free Response Questions
Question 1
2008 Form B
Question 2
2006 Form B
Question 1
2005 Form B
Question 2
2002 Form B
Question 1 (ex a)
Question 1
Question 2
Question 2
Question 1
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involved in the production of this material.
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Linear Momentum & Impulse
What I Absolutely Have to Know to Survive the AP* Exam
Momentum (p) is a vector quantity which has the same direction as the velocity. Momentum can be
thought of as inertia in motion. It is the product of mass times velocity. To have momentum an object
must be moving.
Impulse (J) equals the change in momentum of an object and is a vector that has the same direction as
the net Force. It is derived from Newton’s second law.
For a collision, the area under a force versus time graph yields the impulse applied during the collision.
There are two types of collisions, elastic and inelastic. In both types of collisions momentum is
conserved. In elastic collisions, kinetic energy is conserved as well.
Key Formulas and Relationships
p = mv
Momentum
Units: kg ⋅ m / s
Impulse J = FΔt
Units: N ⋅ s
Note: kg ⋅ m / s and N ⋅ s are equivalent units
Impulse – momentum theorem
J = Δp
FΔt = mv f − mv i
Conservation of linear momentum pa + pb = p'a + p'b
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involved in the production of this material.
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Linear Momentum & Impulse
Important Concepts
•
•
Momentum (p) is a vector quantity which has the same direction as the velocity. Momentum can be
thought of as inertia in motion. To have momentum an object must be moving.
Impulse (J) equals the change in momentum of an object and is a vector that has the same direction
as the net Force. It is derived from Newton’s second law.
F = ma
F = m( v f − v i )
Δt
FΔt = m( v f − v i )
FΔt = mv f − mv i
FΔt = Δp
J = FΔt is used on the AP exam for impulse. The impulse is the applied force multiplied by the time over
which it acts. Impulse is equal to the change in the momentum of a system. Hence, if we apply a large force
for a short time we can generate a momentum change of the same magnitude as having a small force for a
long time.
F
•
=
t
F
t
The area under a Force vs. Time curve = the change in momentum of the object
Force
(N)
Time
(s)
•
The impulse—momentum theorem says that Impulse (J) is the product of the average force acting
on an object and the time interval during which the force acts. This impulse produces a change in an
object’s momentum J = Ft = Δp and can be used to derive the law of conservation of linear
momentum.
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Linear Momentum & Impulse
Derivation:
Imagine two objects traveling in opposite directions that collide. According to Newton’s
F1 = − F2
third law they exert equal and opposite forces upon each other
Since the forces act upon each other for the same time,
F1t = − F2t
Since Ft = Δp we can write Δp1 = −Δp 2
p1' − p1 = −( p '2 − p 2 )
therefore
m1 v1' − m1 v1 = − m2 v '2 + m2 v 2
this gives us the law of conservation of momentum
m1 v1 + m2 v 2 = m1 v1' + m2 v '2
•
•
When no external forces act on a system, the total momentum remains the same.
o Example 1: Two objects collide. The momentum of object one before the collision plus the
momentum of object two before the collision equals the momentum of objects one and two
after the collision.
m1v1 + m2v2 = m1 v1' + m2 v2'
o Example 2: A gun is fired and recoils. The sum of the momentum of the gun and bullet
before it is fired (which is 0) equals the sum of the momentum of the gun and bullet after it is
fired (also 0). The gun kicks and has a negative velocity and therefore a negative
momentum. The bullet has a positive velocity and momentum, and when added together, the
resulting momentum is zero.
0 + 0 = − p a + pb = 0
When objects collide, there are two types of collisions that can occur-- elastic and inelastic.
o Elastic: In this type both kinetic energy and momentum are conserved.
If an equation is classified as elastic, both the conservation of linear momentum and
conservation of KE equations can be used
pa + pb = p'a + p'b
KEa + KEb = KEa' + KEb'
When two objects collide and bounce off each other, there are two possible energy
interactions. Kinetic energy can be conserved or not conserved. If Kinetic energy is
conserved, then we have an elastic collision. Momentum will always be conserved.
o Inelastic: In these collisions only momentum is conserved
When two objects collide and bounce off each other, some of the kinetic energy is
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Linear Momentum & Impulse
transformed into other forms of energy. Real world collisions are inelastic. On the AP exam,
you may be asked whether kinetic energy is conserved and, if it isn’t, then what happened to
that energy or how much kinetic energy is transferred elsewhere.
When two objects collide and they stick together after they collide, then this is called a
perfectly inelastic collision. The key thing to remember here is that after the collision both
objects stick together and have the same velocity. Only momentum is conserved in this type
of collision.
A third example of an inelastic collision on the AP Exam is the two body explosion. The
idea of an explosion is that you have two bodies that are at rest – this means that they have
no momentum. After the explosion takes place, the bodies end up moving away from each
other. Because momentum is conserved, their final momentum, when added together, must
still equal zero (the momentum before the explosion). Remember momentum is a vector.
•
Momentum is conserved in two (and three) dimensions as well. Whenever we analyze a collision in
two dimensions the momentum has to be split up into components. The sum of the momenta of all
components in the x and y directions before a collision must equal the sum of the momenta after the
collision.
∑ px = ∑ px' and ∑ p y = ∑ p 'y
o The diagram below shows the momentum vector of ball A colliding with ball B which is at
rest. Ball A has momentum in the x direction but none in the y while ball B has no
momentum at all.
Ball A
Ball B
After the collision:
Ball A
Ball B
The components of the x and y momentum vectors after the collision must equal the components before the
collision.
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Linear Momentum & Impulse
=
In the x direction
=
+
In the y direction since the momentum before the collision was zero, the y components of the momentum
after the collision must be equal to zero. Since they act in opposite directions, their magnitudes must be
equal to each other.
=
On the AP Exam, collisions in two dimensions are frequent. Momentum still has to be conserved, but the
problems can become fairly complex. Note: keep track of what’s going on by breaking things into x and y
components: momentum is a vector, not a scalar.
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Linear Momentum & Impulse
Free Response
Question 1 (15 pts)
A 15.0 kg wagon with frictionless wheels travels to the right at 5.0 m/s on a horizontal
frictionless surface as shown in the diagram above. When it reaches point A, a 5.0 kg
mass is placed upon it.
v = 5.0 m/s
5.0 kg mass
A
A. Determine the magnitude of the momentum of the wagon before the mass is placed
upon it.
1 point for the correct expression for
momentum
(2 points max)
p=mv
p= (15kg)(5m/s)= 75
1 point for the correct answer including
correct units and reasonable number of
significant digits
kgm
s
B. Determine the speed of the wagon after the mass is placed upon it.
1 point for any statement that
momentum is conserved
(3 points max)
Pwagon+ pmass=pwagon+mass
1 point for adding the masses
75 + 0 = (15 kg + 5kg)v
1 point for the correct answer
including correct units and reasonable
number of significant digits
m
v = 3.75
s
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Linear Momentum & Impulse
The wagon and mass continue moving until reaching another frictionless wagon of mass
15.0 kg at point B. This wagon is attached to a spring of negligible mass which has a
spring constant of 250 N/m. The other end of the spring is fixed to the wall. Upon
impact, the two wagons stick together and compress the spring.
B
C. Determine the speed of the two wagons immediately after they stick together.
(3 points max)
1 point for a correct application of
conservation of momentum
pwagon1+pwagon2=pwagon1+wagon2
20kg(3.75m/s) + 0 = (20kg+15kg)vwagon1&2
2.14
1 point for using the speed found in
part b
1 point for the correct answer
including correct units and reasonable
number of significant digits
m
= vwagon1&2
s
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Linear Momentum & Impulse
D. Determine the kinetic energy of the two wagons immediately after they stick together.
(3 points max)
1 point for a correct expression for
Kinetic energy
K = ½ mv2
K= ½(35)(2.14)
1 point for correct substitution using
answer in part c
2
1 point for the correct answer
including correct units and reasonable
number of significant digits
K=80 J
E. Determine the distance the wagons travel until the spring is fully compressed.
1 point for a correct expression for
conversation of energy
(4 points max)
K=U
1 point for the correct expression for
the potential energy of a spring
U = ½ kx2
80 J = ½ (250 N/m)(x2)
1 point for correct substitution using
answer in part d
x = 0.80 m
1 point for the correct answer
including correct units and reasonable
number of significant digits
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Linear Momentum & Impulse
Question 2 (15 pts)
An arrow with a mass of 0.20 kg attached to a bowstring experiences a variable force for
0.85 s as it is shot from a bow as shown in the diagram below.
50
Force
(N)
0
Time
(s)
.85
A. Determine the Impulse applied to the arrow by the bowstring during the 0.85 s that
they are in contact.
(3 points max)
J = F Δt = mΔ v
1
J = F Δt = bh
2
1
J = ( 0.85s )( 50N )
2
J = 21.25 N is
1 point for any statement indicating that
impulse equals change in momentum
1 point for using the area under the curve
of the F vs. t graph to determine the
impulse or change in momentum of the
arrow
1 point for the correct answer including
correct units and reasonable number of
significant digits
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Linear Momentum & Impulse
B. Determine the speed of the arrow at the instant it leaves the bow.
(2 points max)
J = F Δ t = mΔ v
1 point for the correct application of
the impulse momentum theorem
kgm
s
0.2 kg
m
Δv = 106.25
s
1 point for the correct answer or an
answer consistent with part (A),
including correct units and reasonable
number of significant digits
J
Δv = =
m
21.25
The arrow then travels in a parabolic path until it hits an apple atop a man’s head.
When the arrow hits the apple, it is at the same height as it was when it left the bow.
The arrow and apple stick together and fly off his head landing a certain distance
behind the man. The mass of the apple is 0.4 kg.
C. Determine the speed of the apple/arrow as they leave the man’s head.
1 point for any statement that
momentum is conserved
(4 points max)
P1 + P2 = P1'+ 2
m1v1 + 0 = ( m1 + m2 ) v '
m⎞
⎛
0.2 kg ) ⎜ 106.25 ⎟
(
m1v1
s⎠
⎝
=
v' =
m1 + m2
( 0.2 kg + 0.4 kg )
v ' = 35.4
m
s
1 point for a correct expression of
momentum p=mv
1 point for adding the masses of the
arrow and the apple
1 point for the correct answer or an
answer consistent with part (B),
including correct units and reasonable
number of significant digits
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Linear Momentum & Impulse
D. Calculate the kinetic energy of the two-object system before and after the collision.
1 point for a correct expression of
kinetic energy
(4 points max)
1
KE = mv 2
2
Before
2
1
m⎞
⎛
KE= ( 0.2 kg ) ⎜ 106.25 ⎟ = 1130 J
2
s⎠
⎝
After
2
1
m⎞
⎛
KE= ( 0.2 kg + 0.4 kg ) ⎜ 35.4
⎟ = 376 J
2
s⎠
⎝
1 point for a correct quantity or
answer consistent with part B
1 point for a correct quantity or
answer consistent with part C
1 point for a correct identification of
before and after
E. Is the collision elastic? Justify your answer.
(2 points max)
No
1 point for indicating the collision is
NOT elastic
KEBefore ≠ KE After
1 point for a statement that kinetic
energy is not conserved in the collision
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Linear Momentum & Impulse
Multiple Choice
1. Two objects of mass 3m and 4m are initially at rest on a frictionless horizontal surface.
An ideal spring is compressed between the objects which are held together by a thread of
negligible mass. After the thread is cut and the spring is released, the object with a mass
of 3m has a speed v. What is the speed of the object with the mass of 4m?
A)
1/7 v
B)
3/4 v
C)
v
D)
4/3 v
E)
7v
Conservation of p
(explosion)
p1 + p 2 = p1' + p'2
B
0 + 0 = 3mv + 4mV
−3mv = 4mV
−3/ 4 = V
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Linear Momentum & Impulse
question 2
4 m/s
10 kg
2 m/s
5 kg
2. A 10 kg block moving to the right with a speed of 4 m/s on a frictionless surface
collides with a 5 kg block also moving to the right with a speed of 2 m/s. If the
blocks stick together upon impact, what is their new speed?
A) 3
3
m/s
4
B) 3
1
m/s
2
C) 3
1
m/s
3
D) 3 m/s
E) 2
1
m/s
2
Conservation of p (objects in
same direction stick
together)
p1 + p 2 = p1' + p '2
C
40 + 10 = 15v
50 /15 = v
3 13 = v
3. In a demonstration a physics professor throws an egg against a blanket and it does
not break. When the same egg is thrown against a brick wall, it breaks. Which of
the following best describes the egg’s momentum change?
A)
B)
C)
D)
The brick wall provides a greater momentum change than the blanket.
The blanket provides a greater momentum change than the brick wall.
The momentum changes in both cases are the same.
The momentum change between the egg and the brick wall occurs over more time
with more force.
E) The momentum change between the egg and the blanket occurs over less time
with less force.
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Linear Momentum & Impulse
Momentum change
The momentum change depends only on
mass as well as initial and final velocities.
C
4. If the momentum of an object changes and its mass remains the same, what else must
be happening to the object?
I. it is accelerating
II. its velocity is changing
III. there is a net force acting on it
A)
B)
C)
D)
E)
I only
II only
III only
I and II only
I , II, and III
What is momentum
If p is changing and m stays the same,
then v must be changing. Therefore it
must be accelerating, and to accelerate
and net force must be acting on it
E
5. A 2 kg ball traveling eastward at 3.0 m/s is kicked in such a way that it is now
traveling westward with a speed of 8.0 m/s. The impulse applied to the ball is
A)
B)
C)
D)
E)
- 6 Ns
- 10 Ns
- 18 Ns
- 22 Ns
- 48 Ns
Impulse = Δp
Δp = p 2 − p1
D
Δp = (2)( −8) − (2)(3)
Δp = −16 − 6
Δp = −22
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Linear Momentum & Impulse
Question 6
M
2v
v
m
6. An object with mass M and speed v collides with a second object of mass m traveling
to the left with a speed 2v. They are both traveling on a frictionless surface. If they
collide and stick together upon impact, they will move off together with a new speed of
(M ≠ m)
A)
B)
1
2
v
M +m
( M + 2m ) v
C)
M +m
( M − 2m )v
D)
( M + 2m ) v
M +m
E)
( M − 2m )v
M +m
Mv + m( −2v ) = ( M + m )V
( M − 2m )v = ( M + m )V
Conservation of p
(objects in opposite
directions stick together)
( M −2 m ) v
( M +m )
®
E
=V
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Linear Momentum & Impulse
Questions 7-9
A net force that varies as a function of time is applied to a mass of 7.0 kg according to the
graph as shown below:
Force
(N)
4
0
10
25
Time
(s)
7. Calculate the impulse applied to the mass during the first 25 seconds
A) 70 Ns
B) 40 Ns
C) 30 Ns
D) - 2/25 Ns
E) - 4/25 Ns
Impulse = area under F vs. t
curve
Area = 4(10)+1/2(15)(4)
Area = 40 + 30
Area = 70
A
8. If the object had an initial velocity of 3.0 m/s at time t = 0, it’s velocity after 25 s
would be
A)
B)
C)
D)
E)
7 m/s
10 m/s
13 m/s
15 m/s
25 m/s
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Linear Momentum & Impulse
Impulse = Δp
Using data from graph
Δp = m( v 2 − v1 )
C
70 = 7( v 2 − 3)
10 = ( v 2 − 3)
13 = v 2
9. What is the acceleration of the object during the first 10 seconds?
0 m/s2
4/7 m/s2
7/4 m/s2
28 m/s2
40 m/s2
A)
B)
C)
D)
E)
a=F/m
using data from graph
a=F/m
a=4/7
B
10. In an elastic collision
I.
II.
III.
A)
B)
C)
D)
E)
Linear momentum is conserved
Kinetic Energy is conserved
Elastic Potential Energy is conserved
I only
II only
III only
I & II only
II & III only
Definition of an elastic
collision
In an elastic collision, both momentum
and Kinetic Energy are conserved
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Linear Momentum & Impulse
11. Two objects, one of which is at rest, collide. The arrows below represent the
momenta of the two objects after the collision.
Which of the following may represent the initial momentum of the moving object before
the collision?
A)
B)
C)
D)
E)
zero
Two dimensional momentum Adding the two vectors tip-to-tail gives a
resultant which points in the direction of
the vector arrow D.
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D
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Linear Momentum & Impulse
12. A circus performer is shot out of a cannon and flies through the air until landing
harmlessly in a net. Which of the following best describes why the performer is not
injured after this “death defying” event?
A) The net decreases the amount of time the force is acting on the person and
increases the change in momentum.
B) The net increases the amount of time the force is acting on the person and
decreases the change in momentum.
C) The net decreases the amount of time the momentum is changing and
increases the average force acting on the person.
D) The net increases the amount of time the momentum is changing and
decreases the average force acting on the person.
E) The net increases the amount of time the momentum is changing and
decreases the impulse acting on the person.
The nature of impulse
Ft = Δp by increasing the time of the
momentum change, the force applied to
the circus performer is reduced
D
13. A 0.1 kg ball traveling horizontally westward at 20 m/s strikes a vertical wall and
bounces off the wall eastward with a speed of 20 m/s. What is the impulse acting on the
ball?
A)
B)
C)
D)
E)
zero
2 Ns westward
2 Ns eastward
4 Ns westward
4 Ns eastward
Impulse = Δp
Assume east is positive
I = Δp
Δp = p 2 − p1
E
Δp = (.1)(20) − (.1)( −20)
Δp = 4
The positive sign indicates eastward
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Linear Momentum & Impulse
14. Object A has a mass m and a velocity v. Object B has the same mass but a velocity
V. If the magnitude of object A’s momentum equals the magnitude of object B’s kinetic
energy, what is V in terms of v?
A) v
B) 2v
C) 2 v
D) 1 2 v 2
E) v 2
Relating momentum to KE
mv = 1/ 2mV 2
D
2v = V 2
V2
v=
2
Their magnitudes are equal (obviously p
does not equal KE)
15. Which of the following have equivalent SI units?
(I)
(II)
(III)
A)
B)
C)
D)
E)
Force
Momentum
Impulse
I only
II only
III only
I and II only
II and III only
Units of Momentum and
Impulse
Momentum’s unit Kg m/s is equivalent to
Impulse’s unit Ns
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E
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