A. Key
Student
Date
Class
Momentum & Impulse
./
./
Momentum
Momentum
1
is a vector defined as the product of the mass and velocity of an object
p=mv
mass x length/time, and its 51 units are kilogram-meters
has dimensions
per second (kg.
m/s)
./
Momentum
./
A small object moving with a very high velocity (Le. a bullet) may have a larger
momentum
./
is a vector quantity,
so you must specify both its size and direction
than a more massive object that is moving slowly (Le. a bike) does
Small hailstones falling from very high clouds can have enough momentum
to hurt
someone or cause serious damages to cars and buildings
./
If you are coasting down a hill on your bike you will accelerate, in other words, you
are "picking up speed" or "gathering
./
momentum"
Impulse is a vector defined as the product of force and time interval over which the
] = F!:J.t
fforce acts on an object:
./
A change in momentum
./
Impulse-momentum
takes both force and time
says that: Force x time interval = Change in momentum
theorem
F!:J.t = !:J.p =
mv[ - mVi
./
Remember, we need a simple convention to describe the direction of vectors
./
We have been using negative speed for objects moving left or down (west or south)
and a positive
./
speed for objects moving right or up (east or north)
When a hitter hits the baseball coming from the pitcher he hits the ball as hard as
possible over a very small interval of time to change the momentum
./
of the ball
When a baseball player catches a ball in the field he travels with the ball as much as'
possible to increase the time interval and to decrease the force of impact
./
1.
Stopping times and distances depend on the impulse-momentum
A
2,25e kg pickup
What
is
the
truck
momentum of
has
a velocity
the
truck?
25 m/s to the
of
p ==
:=
\N'\
A deer
17 m/s.
with
a mass of
You are
going
f -:::"'"V
146 kg is
north.
= \Y b
Find
I£:6 (If
running
the
Wl/5 )
V
east.
= '2250 leg
5 b) '2 5 0
k~.
b)OC1O
k
~L5
2.
theorem
head-on
momentum
=- @7"4 g2
the
)ld.
a ~peed
Eot>+]
of
deer.
WI
I.>
")
>-
d. ~/s
with
of
Vv\ /
(2..5 ""'/s
S o~-f~
3.
A 21 kg child on a 5.9 kg bike is riding with a velocity of 4.5
m/s to the northwest.
a) What is total momentum of the child and the bike:together?
~ + oiC\ I
-:::. (\'V\
+ yYI
C
-=
V
k::> )
2. 6. q
/(8 (4. S ""l/.s
-==lt2.0
IL~.
b) What is the momentum of the child?
Pc '"
Me V
=
''3
21
(4.5
)
~/>
~7
__ ------------~
.,:;:{95
'"'/5)
J<tJ.~
1\1";]
c) What is the momentum of the bike?
f'b
4.
-=
Mb
V
-=
5.
ct
I~ ( Y. S M/$
)
A 1,400 kg car moving westward with a velocity of 15 m/s collides
with a utility pole and is brought to rest in 0.30 s. Find the
force exerted on the car during the collision.
[f"-== -
F". ~ t -:::
~F
F 6-e = Y'A (A\J"')
F'O.~OS
:::
70000
tV
0(2..
[!=
lL3(0 -IS M/5)
1400
70000
N
5.
I
boo~~
An 82 kg man drops from rest on a diving board 3.0 m above the
'-0 surface of the water and comes to rest 0.55 s after reaching the
v\- water. What is the net force on the diver as he is brought to
rest?
d.~.,.o~
,,~L :: v...,,'2. + 2-0. 01J
v~ -= 0
=-"';>
Vf = ,r
v 2(q.8/) '3.0-,"- 7. 672. "",/>
F. b. t
\J~
~61te("
"i t ..','
-::Ap
F(O.55'5)
=:'
=
A\l
VV\
F=.:... 1143. g3 N
gz..\~(o
-(.b12"",/~)
.- ... ~,'IJ -:::,0.
6.
A 0.40 kg soccer ball approaches a player horizontally with a
velocity of 18 m/s to the north. The player strikes the ball and
causes it to move in the opposite direction with a velocity of
22 m/s. What impulse was delivered to the ball by the player?
,
+
/"..F
~l
[F~ _\\00 NJ
U "f
J =- F.6t =: Ap
llr -==
1M (
::: O. 4 0
:::. - \ b
lJ:: \(,N.S
"f - Vl)
\L~ (- 2.2. WlI
k~
\WI
I~
>so:
I ~ •...
,1s )
lJ=-lbI'J
.
.s]
-----;>+
',--
F,
~
7.
,
~
i.
F2..
+-
I
- -----0 -- - - - - -0
.
tltl.
A e.se kg object is at rest. A 3.ee N force to the right acts on
the object during a time interval of l.ses.
a) What is the velocity of the object at the end of this
interval?
F.
I
At:::I
A0
't
-=
h1 ~
f! ( I. 55 )::: O. S
3
I"
V(
-==r.e
11"1
0 ~_
\l.~ ( ". - 0 ')
b) At the end of this interval, a constant force of 4.ee N to the
left is applied for 3.ee s. What is the velocity at the end of
the 3.ee s?
F ".L _ VV\ b,.\}
~.
=
_4"'(35)
8.
2-
L- L. -
LJ"
0.5
(Vz.-9"",/s)
\<:~
[\/2. = - 15 "'" 1$ f
A 2,24e kg car traveling to the west slows down uniformly from
2e.e m/s to s.ee mIse
20-. ~ __~
a) How long does it take the car to decelerate if the force on
the car is 8,41e N to the east?
F. b.t
=
8~lO' td:
VV'I/:::.'J
=
2.2...4C (-
~=
5 - (-20))
2.2.40 ( /5)
C7
-:::=
gY10
I'---Y.---
00
>
I
b) How far does the car travel during the deceleration? "
V f '"\
-.:::
V L'l +
d ==
9.
V
1..
V ,1..
f -
l.
2.(;(
cJ
Ac(eleflA
51.
_
2.
-
1..
0
2 ('3.75 )
LCil
Q=:£"=:
tlo,,", :
-::-150. 0
,--.
__
--.
VV'\
f-VL~3.(S"",15'2.
1: '.
M
[50.
f
0 R.
.'
~e>+~
0""
A 2,See kg car traveling to the north is slowed down uniformly
from an initial velocity of 2e.e m/s by a 6,2Se N braking force
acting opposite the car's motion.
1-''''''
V =O_I~
a) What is the car's velocity after 2.se s?
~ b:~~~2~S ~~~
500 (Vf - 20)
@'" [::=14~'»
1>.75 --:1,
b) How far does the car move during 2.se s?
F - (,2.50 N
et::: --:::
Vv't
'2.S DO /(4
~-2. S ""'/5 ~
(J
.J.
(,f -:::
~
Vf
I
L
=
<
1.
V f - Vi.. ~
2- '"
-
Vi.. +-
2(;\01
t.1.?
.h
-
2
Z. (- 2.. 5 )
c) How long does it take the car to come to a complete stop?
F. At
:::-M
_ ~'l.so'
6t
6. C ::- 'Z.Soo( -14~
~v
:: 2~OO(O-
- blS'O
14")
[bt=.5.6>/
@
I
'-:2- _
I.
'1
W\