Chapter 9 concepts
J Momentum
J Conservation of momentum
J Impulse
J Inelastic collisions
Physics 151
1
PRS Jocko
Jocko, who has a mass of 60 kg, catches a 20 kg
ball that was traveling horizontally at 10 m/s.
After Jocko catches it, how fast will he and the
ball move across the ice. (Friction with the ice is
negligible.)
1) 30 m/s
2) 10 m/s
3) 5.0 m/s
4) 2.5 m/s
5) Insuff. info.
Physics 151
2
Jocko soln
Jocko, who has a mass of 60 kg, catches a 20 kg
ball that was traveling
at 10 m/s.
mi x vi =horizontally
20 x 10 = 200
After Jocko catches
how
fast
will
he and the
mf x vit,
=
80
x
?
=
200
f
ball move across—>
thev ice.
(Friction
with the ice is
=
2.5
m/s
f
negligible.)
1) 30 m/s
2) 10 m/s
3) 5.0 m/s
4) 2.5 m/s
5) Insuff. info.
Momentum is
conserved!
Physics 151
3
MOMENTUM
P
definition
and
force
1. Definition: p = mv
2. Relationship to force:
dp d( mv)
dv
dm
=
=m +v
dt
dt
dt
dt
= ma
= åF
Improved form of
2nd
Mass usually doesn’t
change with time
Law:
dp
åF =
dt
Probably was Newton’s original form of 2nd Law.
Works even when mass does change, e.g., rockets.
Physics 151
4
MOMENTUM
Conservation of p
3. Momentum is conserved in the
absence of a net external force.
dp
0 = Fnet = å F =
Þ
dt
p = constant
Conservation
of Momentum
Physics 151
5
masses
Momentum is stillTwo
conserved
even if
there are internal forces on a system.
e.g., Consider two interacting
masses. By Newton’s 3rd law:
F21 = -F12
F21 + F12 = 0
dp1 dp 2 d (p1 + p 2 )
+
=
= 0
dt
dt
dt
i.e., Total momentum Sp = p1 + p2 remains constant
Physics 151
6
Professor on ice
An 80 kg professor standing on a frozen pond flings a 2kg
textbook at 4 m/s. If friction with the ice is negligible, find
the recoil velocity of the professor.
Spi = Spf = 0
0 = pprof i + pbook i = pprof f + pbook f
pprof f = – pbook f
mprofvprof f = – mbookvbook f
v prof = -
professor &
book initially
at rest
mbook v book
2 kg × 4 m/s
== - 0.1 m/s (to the left)
m prof
80 kg
Physics 151
7
PRS exploding shell
A shell, fired from a gun, explodes into two fragments of
equal mass at the top of its parabolic trajectory, which is a
horizontal distance X from the gun. After the explosion, one
fragment is momentarily at rest. Where does the other
fragment land?
1) x = X
2) x = 2X
3) x = 3X
4) Elsewhere
5) No Idea
Physics 151
8
Exploding
shell
solution
The total momentum
the instant after the
collision is equal to
the total momentum
the instant before.
m v0
m v0
=
m
m
,v1=0
,v 2 =?
2
2
m
m
×0 +
× v2
2
2
Þ v 2 = 2v 0
Velocity v2= 2v0 is in the x-direction, and because ax=0,
the forward-going fragment retains this horizontal velocity
component until it lands on the ground.
After the explosion, the forward fragment travels an
additional horizontal distance 2X: Ans: x = 3X
Both pieces hit the ground at the same time.
Physics 151
9
IMPULSE
Impulse
If the net (external) force is zero, momentum is conserved:
0 = Fnet
dptot
= åF =
Þ p = constant
dt
If the net external force is NOT zero, the momentum changes:
This integral is known
as the impulse, J
Physics 151
10
IMPULSE
Golf ball impulse
The golf club is in contact with the ball for a short time interval, in
which the force on the ball varies from zero to something very large.
(Note the distortion of the ball.)
The impulse is the
momentum transferred
between the objects.
The area under the force-time
curve is the impulse
J = Fdt = Favgt = p
Physics 151
11
Spit wad A thought
Force (Newtons)
A student fires a 1-gram spit wad
out of a straw at his professor. The
horizontal force on the wad while it
is moving through the straw is
shown below. What is the velocity
of the wad after it leaves the straw?
Physics 151
experiment
Area =
0.014 N-s
time (seconds)
12
Example car & van collision
A 1500 kg car traveling east at 25 m/s collides with a 2500 kg van
moving north at 20 m/s. If the car and van stick together following
the collision, what is their velocity (magnitude and direction)
immediately after the impact?
Although we know nothing
about the details of the
collision, we can (and will)
solve this problem using
conservation of momentum.
= å pi
å p xf = å pxi
å p yf = å p yi
å pf
Physics 151
13
MOMENTUM
Momentum review
 Momentum p = mv
Usually a good approximation if
the interacting objects have short
impact times, and/or if the
impulsive forces they exert on
each other are much larger than
external forces.
dp
 Relationship to net force: S F =
dt
 The total momentum Sp of a system of objects is
conserved provided there is no net external force.
 Impulse J = Fdt = Favg t = p
Physics 151
14
PRS bowling pin
You seek to knock down a bowling pin by throwing a ball at it.
You have two balls of equal size and mass, one rubber and the
other putty. The rubber ball bounces back, while the putty sticks
to the pin. Which is most likely to topple the bowling pin?
1)
2)
3)
4)
5)
Rubber ball
Putty ball
No difference
Insuff info
No idea
Physics 151
15
bowling pin soln
You seek to knock down a bowling pin by throwing a ball at it.
You have two balls of equal size and mass, one rubber and the
other putty. The rubber ball bounces back, while the putty sticks
to the pin. Which is most likely to topple the bowling pin?
1)
2)
3)
4)
5)
Rubber ball
Putty ball
No difference
Insuff info
No idea
Impulse
= mv
Physics 151
Impulse
 2mv
16
Inelastic collisions
Perfectly elastic collision:
The participating objects
recoil without loss of
energy.
Physics 151
Perfectly inelastic collision:
Objects stick together.
17
Magic table cloth
Physics 151
18
PRS Sticking masses
Two masses moving on a frictionless surface as shown have a
totally inelastic, head-on collision. Following impact, the
velocity of the 4 kg mass is...
1) +2.33 m/s
2) +1.00 m/s
3) +0.50 m/s 4) –0.33 m/s
Physics 151
19
Sticking masses soln
m1 v1
m1v1
m2
+
m2v2
(m1v1 +m2v2)
vf =
(m1 + m2)
v2
(m1 + m2)
=
vf
(m1 + m2)vf
(2 kg)(3 m/s) + (4 kg)(-2 m/s)
=
(2 kg + 4 kg)
= (-2 kg m/s) / (6 kg) = – 0.33 m/s
Physics 151
The 2 masses move to
the left after collision
20
IMPULSE REVIEW
Impulse review
The area under the force-time
curve is the impulse
J = Fdt = Favgt = p
Physics 151
In order to make a
small impulse
a) use a small force
b) use a short time
21
PRS balls & cart
Barry is on a cart that is initially at rest on a horizontal track
with negligible friction. What happens between the time he
throws the ball and when it hits the rigid partition? While the
ball is in flight, the cart is...
1) moving to the left
2) moving to the right
3) stationary
Physics 151
22
PRS balls & cart II
Barry is on a cart that is initially at rest on a horizontal track
with negligible friction. He throws balls at a rigid partition and
the balls bounce back as shown. After the last ball has
bounced from the partition, the cart is...
1) moving to the left
2) moving to the right
3) stationary
Physics 151
23
PRS balls & cart III
Barry is on a cart that is initially at rest on a horizontal track with
negligible friction. He throws balls at a padded partition and the
balls drop down. After the last ball has come to rest, the cart
is...
1) moving to the left
2) moving to the right
3) stationary, at its original
position
4) stationary, displaced to the
left of its original position
5) stationary, displaced to the
right of its original position
Physics 151
24
Billiard balls
vf
vi

A cue ball strikes a stationary
eight ball with speed vi. It
v8 = (vi vf cos vf sin
continues at angle  with
speed vf. What is the velocity
of the eight ball?
Physics 151
25
Rocket
problem
A rocket traveling with
momentum Mv throws gas out
the back at a constant rate
dM/dt with a fixed speed
vexhaust relative to the engine.
Is the acceleration constant?
Is the thrust constant?
The thrust may seem constant as viewed by someone in the rocket,
but the rocket is an accelerating reference frame.
Physics 151
26
Consider the push from a little bit of gas released
during time dt.
Rocket algebra I
dM, u
M – dM, v+dv
dM u + (M – dM)(v+dv) = M v (conservation of momentum)
u = vexhaust+ (v+dv)
(relative motion)
dM(vexhaust+v+dv) + (M – dM)(v+dv) = M v
M dv = – dM vexhaust
M dv/dt = – vexhaust dM/dt
Physics 151
27
dv
dM
Rocket
M = -vexhaust
dt
dt
algebra II
Acceleration…
dv
vexhaust dM
a= =dt
M dt
So is a constant?
M changes
Force…
v changes
dv
dM
dM
dM
F = M +v
= -vexhaust
+v
dt
dt
dt
dt
Physics 151
28
Howequation
much change in velocity
Tsiolkovsky
will a certain mass of fuel
produce?
Mdv = -vexhaust dM
dv = -vexhaust
dM
M
dM
ò dv = -vexhaust ò
M
æM ö
v f = vi + vexhaust lnç i ÷
èMf ø
Tsiolkovsky equation
Physics 151
29
Two eggs areegg
thrown
with
equal velocity.
and
sheet
One hits a hanging sheet …
and one hits a blackboard.
Which feels the larger
impulse?
Which feels the larger
average force?
Physics 151
30
In a crash test, a 1500 kg auto collides with and recoils
Example
car
and
brick
wall
from a brick wall. If the collision lasts 0.15 s, what
average force was exerted on the auto by the wall?
Impulse = J = p = pf – pi = Fdt = Favgt
Physics 151
31
INELASTIC COLLISIONS in 2-D
A 1500 kg car traveling east at 25 m/s collides with a 2500 kg van
moving north at 20 m/s. If the car and van stick together following the
collision, what is their velocity (magnitude and direction)
immediately after the impact?
*Example car & van collision
å p y i = å p yf
(1) m v v v = ( mc +m v ) v f sin q
å p x i = å p xf
( 2 ) mc vc = ( mc +m v ) v f cos q
Divide Eq. (1) by Eq. (2)
Substitute in Eq. (1) :
vf =
mv v v
( mc +m v ) sin q
= 15.6 m/s
(1) m v v v sin q
4
=
= tan q =
Þ q = 53°
( 2) mc vc cos q
3
Physics 151
32
*Car & van
collision II
FAQ
Just one moment! We solved this problem using conservation of
momentum, which is valid so long as no net external forces act on
the system. What about friction– external to the car-van system–
between the vehicles and road?
Good point! We were asked about the combined motion immediately
after impact, so the effects of external friction at some later time are
of no concern. External friction is also present during the collision,
but can be neglected provided (i) the impact occurs during a very
short time, and/or (ii) the forces that the two colliding objects exert
on each other are much greater than any external forces such as
friction.
Physics 151
33
INELASTIC COLLISIONS in 2-D
*Car & van collision III
A more elegant solution expresses the components of the momentum
vector p = ( px , py ) in one equation:
å pi = å p f
(
)
(
å p x i , p y i = å p xf , p yf
)
( mc vc ,mv vv ) = ( ( mc + mv )v xf , ( mc + mv ) vyf )
Þ
ì
mc vc
ï vxf = ( m + m ) = 9.3 m/s
ï
c
v
í
ï v = mv vv = 12.5 m/s
ïî yf ( mc + mv )
Physics 151
34
end
Physics 151
35
Physics 151
36
*angular momentum
The conservation of linear momentum follows from
Newton’s laws in the case of no external force:
dptot
0 = Fnet =
Þ p = constant
dt
Things will also continue to rotate…
if you leave them alone.
Look for some quantity that describes
rotational motion and is conserved if
there is no tangential force.
Guess: mvt is conserved
No, there’s a problem with ice skaters.
Physics 151
37
*Ice skaters
Ice skaters teach us that
as the radius gets smaller,
vt gets bigger.
OK
Revise:
mrvt is conserved.
L = mrvt is known as “angular momentum”
Physics 151
38
Bouncing ball
A hard rubber ball
approaches a smooth
concrete floor at an angle
of 50o from horizontal and
a momentum of 12 kg-m/s.
It leaves at the same angle
with the same speed. If the
duration of the collision is
0.5 s, what is the average
force on the ball during this
time?
Favgt = p (impulse)
p = pyf – pyi
(px doesn’t change)
Favg = p/t = 212 sin(50)/0.5 = 37 N
Physics 151
39
Angular momentum
Why are pulsars so fast?
During the final stages of the collapse of a star’s core to
form a neutron star, angular momentum is conserved,
and the neutron star ends up rotating very rapidly.
The same principle
applies to the ice
skater who pulls in
her arms to spin
faster.
Physics 151
40
CENTER OF MASS
Rocket propulsion
Rocket propulsion can also be understood in
terms of momentum considerations
Physics 151
41
Perpetual motion machine
Physics 151
42
Rocket problem old
A rocket engine provides a constant
thrust*, Fth, for a duration t until
the fuel is spent. The fuel has a
mass mfuel and the rocket has a mass
mwagon. What is the velocity of the
rocket at the end of the burn?
Is this constant acceleration?
No. Mass is not constant.
Impulse problem:
Fth t = p = mwagonvf – (mwagon + mfuel ) vi
*nb: constant thrust is not realistic
Physics 151
0
Fth t / mwagon = vf
43