Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 19
Today’s Agenda

Topics
Example problem (conservation of momentum)
Impulse
Ch. 9.2
Center of Mass
Ch. 9.6
Physics 151: Lecture 19, Pg 1
See text: 9-2
Conservation of Momentum
rocket science
final
v+Dv
video
See Figure 12-
Physics 151: Lecture 19, Pg 2
See text: 9-2

Example Problem :
A object of mass
m=0.200 kg is dropped
from l =30.0 cm height
above a basket base
(M=0.200 kg) which is
attached to the ceiling
with a spring. If the
spring is stretched by
x = 0.05 m before, find
the maximum distance
the spring will stretch ?
m
M
l
x
d
m
d = 0.182 m
See Figure 12-
Physics 151: Lecture 19, Pg 3
See text: 9-2
Force and Impulse

The diagram shows the force vs time for a typical collision.
The impulse, I, of the force is a vector defined as the
integral of the force during the collision.
F
t
I   F dt
Impulse I = area under this curve !
t
Dt
Impulse has units of Ns.
See Figure 12-
ti
tf
Physics 151: Lecture 19, Pg 4
See text: 9-2
Force and Impulse

F
Using
dP
dt
the impulse becomes:
t
I   F dt  
t
dP
dt
dt
F
t
  dP  Pf  Pi  DP
I  DP
impulse = change in momentum !
See Figure 12-
Dt
t
Physics 151: Lecture 19, Pg 5
See text: 9-2
Force and Impulse

Two different collisions can have
the same impulse since I depends
only on the change in momentum,
not the nature of the collision.
F
same area
F
Dt
Dt big, F small
t
Dt
t
Dt small, F big
Physics 151: Lecture 19, Pg 6
See text: 9-2
Force and Impulse
soft spring
F
F
stiff spring
Dt
t
Dt big, F small
Animation
Dt
t
Dt small, F big
Physics 151: Lecture 19, Pg 7
Lecture 19, ACT 1
Force & Impulse

Two boxes, one heavier than the other, are initially at rest on a
horizontal frictionless surface. The same constant force F acts
on each one for exactly 1 second.
Which box has the most momentum after the force acts ?
(a) heavier
F
light
(b)
lighter
F
(c) same
heavy
Physics 151: Lecture 19, Pg 8
Lecture 19, ACT 2
Force & Impulse

What is the average force that wall exerts on ball (0.40kg)
if duration of wall-ball contact is 0.01 s ?
a) 20 N
b) 200 N
c) 2,000 N
d) 20,000 N
vi = 30m/s
before collision
vf = 20m/s
after collision
Physics 151: Lecture 19, Pg 9
Lecture 19, ACT 3






A 0.20 kg stone you throw rises 20.3 m in the air. The
magnitude of the impulse the stone received from your
hand while being thrown is
a. 0.27 Ns.
b . 2.7 Ns.
c . 4.0 Ns.
d . 9.6 Ns.
e . 34.3 Ns.
Physics 151: Lecture 19, Pg 10
System of Particles:




Until now, we have considered the behavior of very simple
systems (one or two masses).
But real life is usually much more interesting !
For example, consider a simple rotating disk.
An extended solid object (like a disk) can be thought of as
a collection of parts. The motion of each little part depends
on where it is in the object!
Physics 151: Lecture 19, Pg 11
System of Particles: Center of Mass


How do we describe the “position” of a system made up of
many parts ?
Define the Center of Mass (average position):
For a collection of N individual pointlike particles whose
masses and positions we know:
N
 mi ri
RCM  i 1
M
m2
m1
r1
RCM
r2
y
x
(In this case, N = 2)
Physics 151: Lecture 19, Pg 12
System of Particles: Center of Mass

If the system is made up of only two particles:
N
 mi ri
RCM  i 1
M

So:

m1r1  m2 r2
m1  m2
m1  m2 r1  m2 r2  r1 
m1  m2 
RCM  r1 
m2
r2  r1 
M
where M = m1 + m2
(r1 - r2)
m1
r1
RCM
m2
r2
y
x
Physics 151: Lecture 19, Pg 13
System of Particles: Center of Mass

If the system is made up of only two particles:
RCM  r1 
m2
r2  r1 
M
where M = m1 + m2
If m1 = m2
RCM
1
 r1  r2  r1 
2
the CM is halfway between
the masses.
r2 - r1
+
m1
r1
RCM
m2
r2
y
x
Physics 151: Lecture 19, Pg 14
System of Particles: Center of Mass

The center of mass is where the system is balanced !
m1
+
m2
m1
+
m2
Physics 151: Lecture 19, Pg 15
See text: 9-6
Example Calculation:

Consider the following mass distribution:
i mi xi (m)0  (2 m)12  (m)24

 12
M
4m
i mi y i (m)0  (2 m)12  (m)0


6
M
4m
X CM 
YCM
RCM = (12,6)
(12,12)
2m
m
(0,0)
m
(24,0)
Physics 151: Lecture 19, Pg 16
See text: 9-6
System of Particles: Center of Mass

For a continuous solid, we have to do an integral.
dm
y
RCM 
r
x
 rdm   rdm
M
 dm
where dm is an infinitesimal
mass element.
video
See example 8-4, A Triangular Plate
Physics 151: Lecture 19, Pg 17
Example: Astronauts & Rope
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:-)

:-(
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A male astronaut and a female astronaut are at rest in outer
space and 20 meters apart. The male has 1.5 times the mass of
the female. The female is right by the ship and the male is out in
space a bit. The male wants to get back to the ship but his jet
pack is broken. Conveniently, there is a rope connected
between the two. So the guy starts pulling in the rope.
Does he get back to the ship?
Does he at least get to meet the woman?
m
M = 1.5m
Physics 151: Lecture 19, Pg 18
Lecture 19, ACT 4
Center of Mass Motion
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A woman weighs exactly as much as her 20 foot long boat.
Initially she stands in the center of the motionless boat, a distance of 20
feet from shore. Next she walks toward the shore until she gets to the
end of the boat.
What is her new distance from the shore.
(There is no horizontal force on the boat by the water).
20 ft
(a) 10 ft
before
20 ft
(b) 15 ft
(c) 16.7 ft
? ft
after
Physics 151: Lecture 19, Pg 19
See text: 9.6
Center of Mass Motion: Review
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We have the following law for CM motion:
FEXT  MACM
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This has several interesting implications:
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It tell us that the CM of an extended object behaves like a
simple point mass under the influence of external forces:
We can use it to relate F and A like we are used to doing.
It tells us that if FEXT = 0, the total momentum of the system
does not change.
As the woman moved forward in the boat, the boat went
backward to keep the center of mass at the same place.
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Physics 151: Lecture 19, Pg 20
Recap of today’s lecture

Chapter 9,
Center of Mass
Elastic Collisions
Impulse
Physics 151: Lecture 19, Pg 21