Momentum
Linear Linear momentum is the
product of mass (scalar) and
velocity (vector)
Physics for Scientists &
Engineers 1
p = mv
Momentum vector and
velocity vector are parallel
to each other
Magnitude of the linear
momentum
Spring Semester 2012
Lecture 17
Momentum and Collisions, Impulse, Conservation of Momentum
p = mv
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Physics for Scientists&Engineers 1
Momentum and Force
dv
d
p=m
= ma = F
dt
dt
K = 1 mv 2
2
Use p=mv and then obtain
K=
term is zero
February 14, 2012
K=
3
Impulse
tf
and therefore:
Then the impulse is simply:
J = Fave t = m v
This result seems rather trivial (the integral is still
in the definition of the average force), but is very
useful for practical purposes
J = p
ti
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Physics for Scientists&Engineers 1
4
i
p
x ,f
dpx
dt = dpx = px ,f px ,i px
dt
ti
ti
p
x ,i
tf
tf
pf
dp
F dt = dt = dp = pf pi p
dt
pi
ti
ti
tf
Impulse: J F dt
Physics for Scientists&Engineers 1
Define the average force acting over a time
interval t = tf - ti
f
Fdt = 1 f Fdt = 1 f Fdt
Fave = i f
t f t i i
t i
dt
Obtain an expression for momentum change by
going back to the relationship between momentum
and force, and integrating both sides over time
Fx dt =
February 14, 2012
Impulse and Average Force
Change in momentum (f = final, i = initial)
p pf pi
tf
p2
2m
We can reformulate all of mechanics we have studied thus
far in terms of momentum instead of velocity
dpy
dp
dpx
; Fy =
; Fz = z
dt
dt
dt
Physics for Scientists&Engineers 1
mv 2 m 2 v 2
p2
=
=
2
2m
2m
Find relationship between momentum and kinetic energy
dp
This Newtons Second Law for momentum F =
dt
In components: Fx =
2
We already know
dv dm
d d
p = (mv ) = m +
v
dt
dt dt
dt
If the mass is constant in time,
Physics for Scientists&Engineers 1
Momentum and Kinetic Energy
Take the time derivative of the definition of the
momentum
2nd
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Physics for Scientists&Engineers 1
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1
Example: A positive force and then a negative
force acting on a mass
Example: A positive force and then a negative
force acting on a mass (2)
A 0.442 kg object is at rest at the origin of a coordinate
system. A 3.38 N force in the positive x direction acts on
the object for 1.57 s. What is the velocity at the end of
this interval?
Use the relationship between impulse and linear
momentum
J = p
F t = mv = m ( v f vi )
A 0.442 kg object is at rest at the origin of a coordinate
system. A 3.38 N force in the positive x direction acts on
the object for 1.57 s. What is the velocity at the end of
this interval? ANS: +12.0m/s
At the end of this interval, a constant force of 5.92 N is
applied in the negative x direction for 3.65 s. What is the
velocity at the end of the 3.65 s?
F t =m ( v f vi )
F t
F t
v f vi =
vf =
+ vi
m
m
In this case, vi=+12.0 m/s
F t 5.92 N 3.65s
m
m
vf =
+ vf =
+ 12.0 = 36.9
m
0.442kg
s
s
Starts from rest, so vi=0
F t 3.38 N 1.57 s
m
vf =
=
= +12.0
m
0.442kg
s
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Physics for Scientists&Engineers 1
Baseball Home Run
9
Physics for Scientists&Engineers 1
( 80.35 m/s)2 + ( 31.71 m/s )2
Start with the change in velocity in the x-direction
vx = ( 49.17 m/s ) ( cos 35.0 ) ( 40.23 m/s ) ( cos185) = 80.35 m/s
Now in the y-direction
vy = ( 49.17 m/s ) ( sin 35.0 ) ( 40.23 m/s ) ( sin185) = 31.71 m/s
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Why can one hit for more power with less tension on the
tennis racket strings?
Answer: relationship between impulse, force, and time
= 86.38 m/s
p = mv = ( 0.145 kg ) ( 86.38 m/s ) = 12.5 kg m/s
PROBLEM 2
The ball-bat contact lasts 1.2 ms
What is the average force exerted by the bat?
SOLUTION 2
The force can be calculated in terms of the impulse
p = J = Fave t
weight of baseball team:
p 12.5 kg m/s
Fave =
=
= 10.4 kN 9 ( 90 kg ) ( 9.81 m/s2 ) = 8 kN
t
0.0012 s
Physics for Scientists&Engineers 1
Physics for Scientists&Engineers 1
Impulse and Sports
The change in momentum is then
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Lets make a sketch
Baseball Home Run
The change in velocity of the baseball is
2
2
v = vx + vy =
Physics for Scientists&Engineers 1
Baseball Home Run
A pitcher throws a fastball, which crosses the plate with a
speed of 90.0 mph (= 40.23 m/s) and an angle of 5.0
relative to the horizontal
The batter hits it for a home run, launching it with 110.0
mph (= 49.17 m/s) at an angle of 35.0 to the horizontal
Mass of the baseball = 5.10 ounces (= 0.145 kg)
PROBLEM 1:
What is the magnitude of the impulse the baseball
receives from the bat?
SOLUTION 1:
The impulse is equal to the momentum change that the
baseball receives
Need to calculate the change in velocity then multiply by
the mass
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11
J = Fave t
If the average force is
fixed, increasing the
contact time
between racket and
ball increases
impulse given to ball
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Physics for Scientists&Engineers 1
12
2
Impulse and Car Safety Devices
Collisions
Car collision: very large impulse (cannot be changed)
Typical collision time is very short
J
Since Fave =
, this means a very large force acting on the
t
driver
All passive safety devices belt, (seat air bag, crash crumple
zones) are designed to extend the time interval and thus to
reduce the average force
Suppose two objects collide with each other
They might rebound away from each other like two
billiard balls
Elastic collision
They might stick together like a subcompact car colliding
with an 18-wheeler
Totally inelastic collision
Looking at the momenta, p1 and p2, of two objects during a
collision, we find that
pf1 + pf2 = pi1 + pi 2
Conservation of total momentum
The most important result of this chapter
Our second conservation law
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Physics for Scientists&Engineers 1
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Momentum Conservation Derivation
Use definition of impulse and its relationship to momentum
change we get
tf
F12 dt = p2 = pf 2 pi2 ( i means initial, f means final)
Integration of this equation gives us
tf
tf
tf
0 = ( F21 + F1 2 ) dt = F21 dt + F1 2 dt = pf1 pi1 + pf2 pi2
ti
During a collision object 2 exerts a force on object 1
F21
So we get
2 1
ti
dt = p1 = pf1 pi1
Physics for Scientists&Engineers 1
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Momentum Conservation Derivation (3)
F
ext
dt << p1 = p2
ti
Physics for Scientists&Engineers 1
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Totally Elastic Collisions in 1d
Conversely, the forces between these two collision partners
are very large, compare home run example
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ti
pf1 + pf2 = pi1 + pi2
Only condition used for derivation is Newtons Third Law
What about other, external forces?
Neglected them in the derivation
Why is this justified?
Collision times are very short, and during these short times
the impulse due to external forces can be neglected
tf
ti
Collecting the initial momentum vectors on one side and final
momentum vectors on the other side gives us
ti
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14
Newtons Third Law tells us that these forces are equal and
opposite to each other
F12 = F21 F12 + F21 = 0
F 2
1
F
Physics for Scientists&Engineers 1
Momentum Conservation Derivation (2)
During a collision object 1 exerts a force on object 2
tf
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Video sequence shows
collision between two
carts on a
(approximately)
frictionless track.
Equal time intervals
Velocities before and
after collision are
marked by colored lines
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Physics for Scientists&Engineers 1
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3
Totally Elastic Collisions in 1d (2)
Totally Elastic Collisions in 1d (3)
A totally elastic collision is an ideal case in which two
conditions are fulfilled:
Total momentum is conserved (valid for all collisions)
We can also obtain expression for velocities instead of
momenta, by using pi1,x=m1vi1,x,
Expression for final velocities from initial ones
pf1,x + pf2,x = pi1,x + pi2,x
Total kinetic energy is conserved (only elastic collisions)
2
2
2
2
pf1,x pf2,x pi1,x pi2,x
+
=
+
2 m1 2 m2 2 m1 2 m2
Result: can calculate momenta after collision
m m2
2 m1
pi1,x +
pi2,x
pf1,x = 1
m1 + m2
m1 + m2
2 m2
m m1
pi1,x + 2
pi2,x
pf2,x =
m1 + m2
m1 + m2
February 14, 2012
vf2, x
Go through
Derivation 7.2
in the book
Physics for Scientists&Engineers 1
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Totally Elastic Collisions in 1d (4)
m1 ( vf1,x vi1,x ) = m2 ( vi2,x vf2,x )
m1vf1,x + m2 vf2,x = m1vi1,x + m2 vi2,x
Kinetic energy result
m1 ( vf1,x vi1,x )( vf1,x + vi1,x ) = m2 ( vi2,x vf2,x )( vi2,x + vf2,x )
m1 ( vf1,x vi1,x ) = m2 ( vi2,x vf2,x )
Divide kinetic energy result by momentum result
Total kinetic energy is conserved
vf1,x + vi1,x = vi2,x + vf2,x
1
1
1
1
2
2
2
2
m1vf1,x + m2 vf2,x = m1vi1,x + m2 vi2,x
2
2
2
2
Rearrange to get
Rearrange and simplify
2
2
2
2
2
2
2
2
m1vf1,x + m2 vf2,x = m1vi1,x + m2 vi2,x m1 ( vf1,x vi1,x ) = m2 ( vi2,x vf2,x )
m1 ( vf1,x vi1,x )( vf1,x + vi1,x ) = m2 ( vi2,x vf2,x )( vi2,x + vf2,x )
Physics for Scientists&Engineers 1
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Physics for Scientists&Engineers 1
Momentum result
pf1,x + pf2,x = pi1,x + pi2,x
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Totally Elastic Collisions in 1d (5)
Total momentum is conserved
Rearrange
m1 m2
2 m2
vi1, x +
vi2.x
m1 + m2
m1 + m2
2 m1
m m1
=
vi1,x + 2
vi2, x
m1 + m2
m1 + m2
vf1, x =
21
vf1, x vf2, x = vi1, x + vi2, x
Collisions change
the sign of the
relative velocity!
vf1, x vf2, x = (vi1, x vi2, x )
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Physics for Scientists&Engineers 1
Example: P024
Example: P024 (2)
y
x
p = mv
v1, x = v1 sin 1 v1, y = v1 cos 1
v2, x = v2 sin 2 v2, y = v2 cos 2
p = p2 p1
px = mv2, x mv1, x = m ( v2 sin 2 v1 sin 1 )
p y = mv2, y mv1, y = m ( v2 cos 2 ( v1 cos 1 ) ) = m ( v2 cos 2 + v1 cos 1 )
px = ( 0.250 kg ) ( ( 2.20 m/s ) sin 57.0 ( 2.40 m/s ) sin 42.0 ) = 0.05979 kg m/s
p y = ( 0.250 kg ) ( ( 2.20 m/s ) cos 57.0 + ( 2.40 m/s ) cos 42.0 ) = 0.7454 kg m/s
p =
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February 14, 2012
( p x ) + ( p x )
2
2
=
( 0.05979 kg m/s ) + ( 0.7454 kg m/s )
2
Physics for Scientists&Engineers 1
2
= 0.748 kg m/s
24
4
Example: P024 (3)
p = p2 p1
p2 = p1 + p
2
px = 0.05979 kg m/s
p y = 0.7454 kg m/s
p y
1 0.7454 kg m/s
= tan
= 85.4
0.05979 kg m/s
p x
= tan 1
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Physics for Scientists&Engineers 1
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5