Impulse, Momentum,
Work & Energy
Impulse




Force acting for a period of time
J = F∆t
Unit of measurement: N·s or kg·m/s
Describes how hard and how long we
push to change a motion
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Momentum





Impulse causes a change in
momentum.
Impulse-Momentum Theorem
F∆t = ∆p
Linear momentum is defined as the
product of mass x velocity
p = m∆v
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Unit of Measurement for ‘p’

MKS
N·s or kg·m/s

CGS dyn·s or g·cm/s

FPS
lb·s
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Vector Quantities
Both impulse and momentum are vector
quantities.
They have both magnitude and direction.
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Newton’s Second Law






F∆t = m∆v
F/m = ∆v/t
a=a
F = m ∆v/t
F = ma
Force is directly proportional to the change in
velocity if the mass is held constant.
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Exerting a given impulse
a)
A big force for a short time (F∆t)
b)
A smaller force for a longer time (F∆t)
c)
A force that changes while it acts
(F∆t)
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Direction of Impulse

Impulse points in the same direction as
the force and the change in
momentum.
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Impulse and Momentum
Sample Problems
1.
A baseball of mass 0.14 kg is moving at
+35 m/s.
a)
Find the momentum of the baseball.
m = 0.14 kg
v = +35 m/s
p=?
p = m∆v
= (0.14 kg)(+35 m/s)
p = +4.9 kg·m/s
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b.
Find the velocity at which a bowling ball, mass
7.26 kg, would have the same momentum as
the baseball.
p = +4.9 kg·m/s
m = 7.26 kg
∆V = ?
p = mv
∆v = p/m
= +4.9 kg·m/s
7.26 kg
∆v = +0.67 m/s
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c.
Find the average acceleration of the ball during
its contact with the bat if the average force is
-1.4 x 104 N and
m = 0.144 kg.
F = -1.4 x 104 N
m = 0.144 kg
a=?
F = ma
a = F/m
= -1.4 x 104N
0.144 kg
a = -9.7 x 104 m/s2
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2.
A 54 N·s impulse if given to a 6.0 kg
object. What is the change of
momentum for the object?
J = 54 N·s
m = 6.0 kg
p=?
p = F∆t
p=J
P = 54 N·s
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3.
Which quantities do not always occur in
equal and opposite pairs when an
interaction takes place within a
system?
a.
Impulses
b.
Acceleration
c.
Forces
d.
Momenta changes
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Answer to 3

accelerations
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Conservation of Momentum
Sum of Momenta before =
Sum of Momenta after

Any gain in momentum by an object
occurs only by the lost of a
corresponding amount of momentum by
another object.
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
In a system consisting of objects upon
which no external force is acting, the
momentum of the system is conserved.
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
Two objects of equal mass approach
each other head-on with the same
speed. The total momentum of the
system before the collision takes place
equal zero.
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Scenario 1
Before collision
After Collision
----->
at rest
---->
------>
m1v1 + m2v2 = m1v1’ + m2v2’
p1
+
p2
=
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p1’
+
p2’
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Scenario 2
Before collision
After Collision
----->
< ---<--------->
m1v1 - m2v2 = - m1v1’ + m2v2’
p1
-
p2
=
-p1’
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+
p2’
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Scenario 3
Before collision
After Collision
at rest
<---m1v1 + m2v2
<----< ---= m1v1’ + m2v2’
p1
=
+
p2
- p1’
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+ - p2’
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Scenario 4
Before collision
After Collision
----->
at rest
--- >------>
m1v1 + m2v2 = (m1 + m2) v’
p1
+
p2
=
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p’
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Examples
1.
Object A has a momentum of 60 N·s.
Object B, which has the same mass, is
standing motionless. Object A strikes
object B and stops. What is the velocity
of object B after the collision if the
mass of object B is 6 kg?
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Before
----->
m1v1
p1
60 N·s
collision
After Collision
at rest
stops
?
+ m2v2
=
m1v1’ + m2v2’
+ p2
= p1’
+ p2’
+ 0 N·s
= 0 N·s + (6.0 kg)v2’
v2’
= 10 m/s, to the right
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2.
Ball A of mass 0.355 kg moves along a
frictionless surface with a velocity of
+0.095 m/s. It collides with ball B of
mass 0.710 kg moving in the same
direction at a speed of +0.045 m/s.
After the collision, ball A continues in
the same direction with a velocity of
+0.035 m/s. What is the velocity and
direction of ball B after the collision?
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2.
Before collision
----->
----- >
m1v1 + m2v2
=
After Collision
------ >
?
m1v1’ + m2v2’
(0.355 kg)(0.095 m/s) + (0.710 kg)(0.045 m/s)
= (0.355 kg)(0.035 m/s) + (0.710 kg)V2’
v2’ = 0.075 m/s, to the right
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3.
A 0.105 kg hockey puck moving at
+48 m/s is caught by a 75 kg goalie at
rest. With what speed does the goalie
slide on the ice?
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3.
Before collision
----->
at rest
m1v1 + m2v2
=
After Collision
?
(m1 + m2)v’
(0.105 kg)(48 m/s) + 0 N·s = (0.105 kg + 75 kg)v’
v’ = 0.067 m/s
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4.
A 0.50 kg ball traveling at +6.0 m/s
collides head on with a 1.00 kg ball
moving in the opposite direction at a
velocity of -12.0 m/s. The 0.50 kg ball
moves away at -14 m/s in the opposite
direction after the collision. Find v2’.
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4.
Before collision
----->
< ---m1v1 + m2v2
=
After Collision
< ---?
m1v1’ + m2v2’
(0.50 kg)(+6.0 m/s) + (1.00 kg)(-12.0 m/s)
= (0.50 kg)(-14 m/s) + (1.00 kg)v2’
v2’ = -2.0 m/s
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Momentum at an Angle
Right Angle
Other Angles
1.
A 1325 kg car is moving due east at 27 m/s. It
collides with a 2165 kg car moving due
north at 17.0 m/s.
a.
Calculate the momentum of each car before
the collision.
b.
The two cars stick together and move off at
an angle. Calculate the resultant
momentum, speed and angle of the cars.
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a.
PN = mv
=(1325 kg)(27 m/s)
PN =3.6 x 104 N·s
PE= mv
= (2165 kg)(17.0 m/s)
PE = 3.7 x 104 N·s
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b.
p’ 2 = (3.6 x 104 N·s)(3.7 x 104 N·s)
p’ = 5.2 x 104 N·s
v’ = p’ / mt
= 5.2 x 104 N·s / (1325 kg + 2165 kg)
v’ = 14.9 m/s
tan Ө = PN / PE
= 3.6 x 104N·s / 3.7 x 104 N·s
Ө = 44o N of E
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2.
A 6.0 kg ball (A) is moving due east at 3.0 m/s. It
collides with another 6.0 kg ball (B) that is at rest.
Ball A moves off at an angle of 40.o north of east.
Ball B moves off at an angle of 50.o south of east.
a.
Calculate the momentum of ball A before the
collision.
b.
Calculate the x-component of ball A.
c.
Calculate the y-component of ball B.
d.
Calculate the velocity of ball A.
e.
Calculate the velocity of ball B.
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PA = mv
= (6.0 kg)(3.0 m/s)
PA = 18 kg·m/s
a)
b) PA’ = PA cos 40.o
= 18 kg·m/s (cos 40.o)
PA’ = 14 kg·m/s
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c)
d)
PB’ = PA sin 40.o
= 18 kg·m/s (sin 40.o)
PB’ = 12 kg·m/s
VA’ = PA’ / mA
= 14 kg·m/s / 6.0 kg
vA = 2.3 m/s
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e) vb’ = PB / mB
= 12 kg·m/s / 6.0 kg
vb’ = 2.0 m/s
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3.
Two objects of masses M1 = 1 kg and M2= 4 kg
are free to slide on a horizontal frictionless
surface. M1 is moving due east at 16 m/s. M2
is at rest. The objects collide and the
magnitudes and directions of the velocities of
the two objects before and after the collision
are M2 moves 37o N of E at 5 m/s; M1 movess
90o due S at 12 m/s.
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a.
b.
c.
d.
Calculate the x and y components of the
momenta of the balls before and after the
collision.
Show, using the calculations that
momentum is conserved.
Calculate the kinetic energy of the twoobject system before and after the collision.
Is kinetic energy conserved in the collision?
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Before Collision
Px = M1v1
= (1 kg)(16 m/s)
Px = 16 kg·m/s
Px = M2v2
=(4 kg)(0 m/s)
Px = 0 kg·m/s
Py = M1v1
= (1 kg)(0 m/s)
py = 0 kg·m/s
py = M2v2
= (4 kg)(0 m/s)
py = 0 kg·m/s
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After Collision
P’1x = M1v1cos 900
=(1kg)(12m/s)(0)
P’1x = 0 kg·m/s
P’2x = M2v2 cos 37o
p’1y = M1v1 sin 90o
=(1kg)(12 m/s)(1)
p’1y = 12 kg·m/s
P’2x = - M2v2 sin 37o
= (4 kg)(5 m/s)(0.80)
P’2x = 16 kg·m/s
= (4 kg)(5 m/s)(0.60)
P’2x = - 12 kg·m/s
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b.
Sum of Px before: (16 + 0) kg·m/s = 16 kg·m/s
Sum of Py before: (0 + 0) kg·m/s = 0 kg·m/s
Total = 16 kg·m/s
Sum of Px after = (0 + 16 kg·m/s) = 16 kg·m/s
Sum of Py after = [(+12) + (-12)] = 0 kg·m/s
Total = 16 kg·m/s
Px & py are conserved.
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c. K.E. = ½ mv2
Before: K.E. = ½ (1 kg)(16 m/s)2
= 128 J
After:
K.E. = ½ [(1 kg)(12 m/s)2 + (4 kg)(5m/s)2
K.E. = 122 J
d. K.E. isn’t conserved.
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4.
A fireworks rocket is moving at a speed of
50.0 m/s. The rocket suddenly breaks
into two pieces of equal mass, and they
fly off with velocities v1 and v2. v1 is
acting at an angle of 30.0o with the
original direction and v2 is acting at an
angle of 60.0o with the original
direction. What are the magnitudes of
v1 and v2?
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X - components
2mv1 = m1v1’ + m2v2’
2mv1 = m1v1’ cos 30.0o + m1v1’ cos 60.0o
2(50.0 m/s) = 0.866v1’ + 0.500 v2’
v1 = 1.73 v2
Y-components
2mv1 = m1v1’ sin 30.0o + m1v1’ sin 60.0o
0
= 0.500 v1 + 0.866v2
=(0.500)(1.73 v2) + 0.866 v2
V2 = 50.0 m/s
V1 = (1.73)(50.0 m/s)
V1 = 86.5 m/s
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Work
Work is done only when a force moves an
object.
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The distance the object moves must be in
the same direction as the force applied
to the object.
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Work = force x distance
when both the force and displacement
are in the same direction
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Work = force (cos Ө) x distance
When both the force and displacement
are not in the same direction
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If W = F cos Ө d, then the units of
measurements are
MKS ---- > N·m
or J
CGS ---- > dyn·cm or erg
FPs ---- > ft·lb
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(Joule)
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Example

A force of 10,000 N is applied to a
stationary wall. How much work is
performed?
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
0. The wall didn’t move.
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Example 2

A 950 N skydiver jumps from an
altitude of 3000 m. What is the total
work performed on the skydiver?
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Answer:
W = F cos Ө d
= (950 N) (3000 m)
W = 2.9 x 106 N·m
Work is done by gravity.
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Example 3
A 200 N sled is pulled a distance of 20. m
with a rope held at an angle of 30o.
Calculate the force exerted by the rope
if W = 2.0 x 104J.
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Answer:
W = F cos Ө d
2.0 x 104 J = F (cos 30o) (20. m)
F = 1.2 x 103 N
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Mechanical Energy
Mechanical energy is what an object has
due to its motion or position.
Potential energy
Kinetic energy
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Kinetic Energy
K.E.
Energy due to the motion of an object
Work can increase or decrease K.E.
Equation: K.E. = ½ mv2
Units of measurements:
MKS: N·m or J
(J = Joule)
CGS: dyn·cm or erg
FPS: ft·lb
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Potential Energy
P.E.




Stored in an object because of its state or
position
Gravitational P.E. depends on an object’s
position above the earth surface. (base level)
Equation: P.E. = mgh
Units of measurements
MKS: N·m or J
(J = Joule)
CGS: dyn·cm or erg
FPS: ft·lb
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Gravitational Potential Energy
Newton’s Law of Univ. Grav.
=
P.E.
Fg
mgh
= Gm1m2
r2
= m x Gm x h
r2
=
Gm1m2
r
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Blue Book Reviewing Concepts
P. 236
Type or form of energy
2.
wound-up watch spring = P.E.
work was done to wind up watch
running mechanical watch use
= P.E. converted to K.E.
watch runs down
= energy converted to heat and sound
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3.
Types of mechanical energy in the
Earth-Sun system.
Both P. E. and K.E. Both changing. K.E.
is greatest when earth is closest to
sun.
P. E. is greatest when earth is farthest
from the sun.
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B.B. = Applying Concepts
5.
Can the kinetic energy of a baseball
ever have a negative value?
No. Mass is positive. The square of the
velocity is positive.
The change in K.E. can be negative.
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B.B. P. 257
Applying Concepts
1.
Which has more kinetic energy
Compact car or b. semi-truck. Both
are traveling at the same velocity.
Answer:
Semi-truck. It has more mass. K.E. is
directly proportional to the mass.
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4.
Is it possible to exert a force and yet
not cause a change in kinetic energy?
Answer:
Yes. The force cannot be applied through
a distance. No work is done and there
is no change in kinetic energy
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Example 1
1.
A 1600-kg car travels at a speed of
12.5 m/s. What is its kinetic energy?
Answer:
K.E. = ½ mv2
= ½ (1600 kg)(12.5 m/s)2
K.E. = 1.3 x 105 J
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Example Problem 2
12.
How much potential energy does Tim,
mass 60.0 kg gain when he climbs a
gymnasium rope a distance of 3.5 m?
Answer:
P.E. = mgh
= (60.0 kg)(9.80 m/s2)(3.5 m)
P.E. = 2.1 x 103 J
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Applications of Work and
Energy Equations
Work-Energy Theorem
Work causes a change in either potential
or kinetic energy.
W = ∆P.E.
W = ∆K.E.
= mg(h1 – h2)
= ½ m (v22 –v12)
or
or
Fd = mg(h1 – h2)
Fd = ½ m (v22 –v12)
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Conservation of Energy
∆P.E. = ∆K.E.
There is no lost or gain of energy. P.E. is
converted to K.E. or K.E. is converted to
P.E.
Energy can also be converted to heat,
sound or other forms of energy.
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Conservation of Energy
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/ie.html
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Nonconservative (external)
Conservative (internal) Forces
Internal Forces
External Forces
Fgrav
Fspring
Fapplied
Ffriction
Fair
Ftension
Fnormal
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Nonconservative Forces
Total mechanical energy of an object is
changed
Total mechanical energy = K.E. + P.E.
Work is positive - object will gain energy
Work is negative - object will lose energy.
The gain or loss in energy can be in the form of
P. E., K.E. or both.
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Conservative Forces
Internal forces that change the form of energy
without changing the total amount of
mechanical energy.
Energy changes forms from kinetic to potential
(or vice versa)
The total amount of mechanical is conserved.
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Example
1.
a)
b)
A 98-N sack of grain is hoisted to a storage
room 50 m above the ground floor or a
grain elevator.
W = Fd cosӨ
= (98 N)(50 m)(cos Oo)
W = 5 x 103 J
P.E. = mg∆h
or ∆P.E. = W
=(98 N)(50 m)
= 5 x 103 J
P.E. = 5 x 103 J
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1 c) ∆P.E. = ∆K.E.
= 5 x 103 J
Using free-fall equation
V22= V12 + 2 g d
= (0 m/s)2 + 2(9.80 m/s2)(50 m)
V2 = 31 m/s
m = W/g
= 98 N / 9.8 m/s2 = 10 kg
K.E. = ½ (10 kg)(31 m/s)2 = 5 x 103 J
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Example
23.
a.
A 2.0 kg rock initially at rest loses 400 J of potential energy
while falling to the ground.
Calculate the kinetic energy that the rock gains while falling.
∆K.E. = ∆P.E.
∆K.E. = 400 J
b.
V2 = 2(∆K.E.) / m
= (2)(400 J) / 2.0 kg
V = 20 m/s2
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Example: Roller Coaster
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Starting from rest
a)
When is the P.E. at its maximum?
b)
When is K.E. at its maximum?
c)
What stops the roller coaster?
d)
Did the wall do work on the roller
coaster?
e)
Explain how you can determine the
velocity at each point?
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1.
2.
3.
4.
5.
At the top. Greatest height
At the bottom. Greatest change in height
from beginning.
the wall
Yes. The wall applied a force through a
distance.
Use conservation of energy equation.
∆K.E. = ∆P.E. in the forward direction.
W = ∆K.E. + ∆P.E.
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Example. The Pendulum and
Conservation of Energy
animation: http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/pe.html
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1.
2.
3.
4.
5.
Maximum P.E.
Maximum K.E.
Equation used to calculate the maximum
speed.
Do you need to know the mass of the
pendulum bob? Why?
Equation used to calculate the speed
between the maximum height and bottom
of the swing.
animation: http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/pe.html

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1.
At beginning and end of swing
2.
At the bottom of the swing
3.
4.
5.
∆K.E. = ∆P.E.
m(v22 – v12) = m(h1 – h2)
No. Mass cancels out.
h = L – L cosӨ
V2 = (L)(g) tan Ө
or
T sin Ө = mv2/L
animation: http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/pe.html
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Power
Rate at which work is done by the net force
Power = work / time
Units in MKS system
= J/s or Watt (W)
Unit in FPS system is the horsepower (hp)
1 horsepower = 550 ft·lb/sec = 745 watts
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Since it is measured in J/s and Joule is a
unit of measurement for energy
Power can also be defined as
P = change in energy/ time
or
P = ∆E/t
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Deriving another equation for
Power
work
W
t
= force x displacement
= Fd
t
Or
P = Fv
v = d/t
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60.
a) P = Fv
= (2.00 x 102N)(20.0 m/s)
P = 4.00 x 103 W
b) F = Fa + mg sin 37.0o
P = Fv
= (Fa + mg sin 37.0o)v
=[2.00 x 102 N +
(2.0 x 102 kg)(9.80 m/s2)sin 37.0o](20.0 m/s)
P = 3.35 x 104 W
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61.
P1
=
Fwater
P2
Fwater + T
Fwater = 7.50 x 104 W = 6250 N
2 m/s
T = Fwater [(P2 /P1)] – 1]
= 6250 N[(8.30 x 104 W / 7.50 x 104 W) – 1]
T = 6.7 x 102 N
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61.
Constant speed: FBoat(1) = FWater
Pulling a skier: FBoat(2) = Fwater + Trope
P = Fv
V is the same in both cases
P1 =
P2
F1(boat)
F2(boat)
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64.
a.
W01 = ½ (+6.0 N)(1.0 m – 0 m) = 3.0 J
b.
No area. Work12 = 0 J
c.
W23 = ½ (-6.0 N)(3.0 m – 2.0 m) = -3.0 J
W34 = (-6.0 N)(4.0 m – 3.0 m) = - 6.0 J
Total Work24 = (-3.0 J) + (-6.0 J) = - 9.0 J
Work is negative because force is negative.
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65.
a) W03 = (31 N)(3.0 m – 0 m) = 93 J
b.
c.
W36 = 0 J No area under curve
W = ½ m (v22 - v12)
93 J = ½ (65 kg) [(v2)2 - (1.5 m/s)2]
v2 = 2.3 m/s
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67.
W0,10= ½ (10.0 N)(10.0 m – 0 m) =5.00 x 101 J
W10,20=(10.0 m)(20.0 m–10.0 m) = 1.00 x 102 J
Total Work = 1.50 x 102 J
b.
W = ½ m (v22 - v12)
1.50 x 102 J = ½ (6.00 kg) [(v2)2 - ( 0 m/s)2]
v2 = 7.07 m/s
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