Download Physics 207: Lecture 2 Notes

Lecture 15

Goals
 Employ conservation of momentum in 1 D & 2D
 Introduce Momentum and Impulse
 Compare Force vs time to Force vs distance
 Introduce Center-of-Mass
Note: 2nd Exam, Monday, March 19th, 7:15 to 8:45 PM
Physics 201: Lecture 15, Pg 1
Comments on Momentum Conservation

More general than conservation of mechanical energy

Momentum Conservation occurs in systems with no net
external forces (as a vector quantity)
Physics 201: Lecture 15, Pg 2
Explosions: A collision in reverse

A two piece assembly is hanging vertically
at rest at the end of a 20 m long massless
string. The mass of the two pieces are 60
and 20 kg respectively. Suddenly you
observe that the 20 kg is ejected
horizontally at 30 m/s. The time of the
“explosion” is short compared to the
swing of the string.

Does the tension in the string increase or
decrease after the explosion?

After
If the time of the explosion is short then Before
momentum is conserved in the x-direction
because there is no net x force. This is
not true of the y-direction but this is what
we are interested in.
Physics 201: Lecture 15, Pg 3
Explosions: A collision in reverse
A two piece assembly is hanging
vertically at rest at the end of a 20 m long
massless string. The mass of the two
pieces are 60 and 20 kg respectively.
Suddenly you observe that the 20 kg
mass is ejected horizontally at 30 m/s.
 Decipher the physics:
1. The green ball recoils in the –x
direction (3rd Law) and, because there is
no net external force in the x-direction
the x-momentum is conserved.
2. The motion of the green ball is
Before
constrained to a circular path…there
must be centripetal (i.e., radial
acceleration)

After
Physics 201: Lecture 15, Pg 4
Explosions: A collision in reverse
A two piece assembly is hanging vertically at
rest at the end of a 20 m long massless string.
The mass of the two pieces are 60 & 20 kg
respectively. Suddenly you observe that the 20
kg mass is suddenly ejected horizontally at 30
m/s.
 Cons. of x-momentum
px before= px after = 0 = - M V + m v
V = m v / M = 20*30/ 60 = 10 m/s
Tbefore = Weight = (60+20) x 10 N = 800 N
Before

After
SFy = m acy = M V2/r = T – Mg
T = Mg + MV2 /r = 600 N + 60x(10)2/20 N = 900 N
Physics 201: Lecture 15, Pg 5
Exercise
Momentum is a Vector (!) quantity


A.
B.
C.
D.
A block slides down a frictionless ramp and then falls and
lands in a cart which then rolls horizontally without friction
In regards to the block landing in the cart is momentum
conserved?
Yes
No
Yes & No
Too little information given
Physics 201: Lecture 15, Pg 6
Exercise
Momentum is a Vector (!) quantity
x-direction: No net force so Px is conserved.
 y-direction: Net force, interaction with the ground so
depending on the system (i.e., do you include the Earth?)
py is not conserved (system is block and cart only)

2 kg
5.0 m
30°
Let a 2 kg block start at rest on a
30° incline and slide vertically a
distance 5.0 m and fall a distance 7.5 m
7.5 m into the 10 kg cart
10 kg
What is the final velocity of the cart?
Physics 201: Lecture 15, Pg 7
Exercise
Momentum is a Vector (!) quantity


1) ai = g sin 30°
= 5 m/s2
x-direction: No net force so Px is conserved
y-direction: vy of the cart + block will be zero
and we can ignore vy of the block when it
2) d = 5 m / sin 30°
lands in the cart.
j
= ½ ai Dt2
N
i 5.0 m
30° mg
30°
Initial
Final
Px: MVx + mvx = (M+m) V’x
M 0 + mvx = (M+m) V’x
V’x = m vx / (M + m)
= 2 (8.7)/ 12 m/s
V’x = 1.4 m/s
10 m = 2.5 m/s2 Dt2
2s = Dt
v = ai Dt = 10 m/s
vx= v cos 30°
= 8.7 m/s
7.5 m
y
x
Physics 201: Lecture 15, Pg 8
Impulse
(A variable external force applied for a given time)
Collisions often involve a varying force
F(t): 0  maximum  0
 We can plot 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 time of the collision.
 The impulse measures momentum transfer

Physics 201: Lecture 15, Pg 9
Force and Impulse
(A variable force applied for a given time)

J a vector that reflects momentum transfer
 t

t
p 
I   F dt   ( dp / dt ) dt   dp
F
Impulse I = area under this curve !
(Transfer of momentum !)
t
Dt
Impulse has units of Newton-seconds
ti
tf
Physics 201: Lecture 15, Pg 10
Force and Impulse

Two different collisions can have the same impulse since
I depends only on the momentum transfer, NOT the
nature of the collision.
F
same area
F
Dt
Dt big, F small
t
Dt
t
Dt small, F big
Physics 201: Lecture 15, Pg 11
Average Force and Impulse
F
Fav
F
Fav
Dt
Dt big, Fav small
t
Dt
t
Dt small, Fav big
Physics 201: Lecture 15, Pg 12
Exercise
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 ?
F
A.
B.
C.
D.
light
F
heavy
heavier
lighter
same
can’t tell
Physics 201: Lecture 15, Pg 13
A perfectly inelastic collision in 2-D

Consider a collision in 2-D (cars crashing at a slippery
intersection...no friction).
V
v1
q
m1 + m2
m1
m2
v2
before


after
If no external force momentum is conserved.
Momentum is a vector so px, py and pz
Physics 201: Lecture 15, Pg 15
A perfectly inelastic collision in 2-D


If no external force momentum is conserved.
Momentum is a vector so px, py and pz are conseved
V
v1
q
m1 + m2
m1
m2
v2
before
after
x-dir px : m1 v1 = (m1 + m2 ) V cos q
 y-dir py : m2 v2 = (m1 + m2 ) V sin q

Physics 201: Lecture 15, Pg 16
2D Elastic Collisions

Perfectly elastic means that the objects do not stick and,
by stipulation, mechanical energy is conservsed.

There are many more possible outcomes but, if no
external force, then momentum will always be conserved
Before
After
Physics 201: Lecture 15, Pg 17
Billiards

Consider the case where one ball is initially at rest.
after
before
pa q
pb
vcm
Pa f
F
The final direction of the red ball will
depend on where the balls hit.
Physics 201: Lecture 15, Pg 18
Billiards: Without external forces, conservation of
both momentum & mech. energy



Conservation of Momentum
x-dir Px : m vbefore = m vafter cos q + m Vafter cos f
y-dir Py :
0
= m vafter sin q + m Vafter sin f
after
before
pafter q
pb
F
Pafter f
If the masses of the two balls are equal then there will always
be a 90° angle between the paths of the outgoing balls
Physics 201: Lecture 15, Pg 19
Center of Mass
Most objects are not point-like but have a mass
density and are often deformable.
 So how does one account for this complexity in a
straightforward way?

Example
In football coaches often tell players attempting to
tackle the ball carrier to look at their navel.
 So why is this so?

Physics 201: Lecture 15, Pg 20
System of Particles: Center of Mass (CM)
If an object is not held then it will rotate about the
center of mass.
 Center of mass: Where the system is balanced !
 Building a mobile is an exercise in finding
centers of mass.

m1
+
m2
m1
+
m2
mobile
Physics 201: Lecture 15, Pg 21
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 point-like particles whose
masses and positions we know:
RCM
m2
m1
(In this case, N = 2)

 mi ri
N

rCM
r1
r2
y
x



m1r1  m2 r2  m3r3  
i 1
 N

M
 mi
i 1
Physics 201: Lecture 15, Pg 22
Momentum of the center-of-mass is just the total momentum

Notice

rCM 
1
M

 mi ri
N
i 1
N


d
d
1
r

(
dt CM
dt M  mi ri )
i 1
N

d 
1
vCM  ( M  mi dt ri )
i 1
N






MvCM  pCM   mi vi  p1  p2  p3  ...
i 1

Impulse and momentum conservation applies to
the center-of-mass
Physics 201: Lecture 15, Pg 23
Sample calculation:

Consider the following mass distribution:

 mi ri
N

rCM  i 1
M
 XCM î  YCM ĵ  ZCM k̂
XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters
RCM = (12,6)
YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters
(12,12)
2m
XCM = 12 meters
YCM = 6 meters
m
(0,0)
m
(24,0)
Physics 201: Lecture 15, Pg 24
A classic example
There is a disc of uniform mass and radius r.
However there is a hole of radius a a distance b
(along the x-axis) away from the center.
 Where is the center of mass for this object?


rgreen disk CM  (0,0)

rhole CM  (b,0)
m
  2
r
yCM 
xCM 
0 m b (  ) a 2
m  (  ) a 2
0 m  0 (  ) a 2
m  (  ) a 2

bma 2 / r 2
m  ma 2 / r 2
0

ba 2
b 2 a 2
Physics 201: Lecture 15, Pg 25
System of Particles: Center of Mass

For a continuous solid, convert sums to an integral.

rCM
dm
y
r


 r dm  r dm


M
 dm
where dm is an infinitesimal
mass element (see text for
an example).
x
Physics 201: Lecture 15, Pg 26
Recap


Thursday, Review for exam
For Tuesday, Read Chapter 10.1-10.5
Physics 201: Lecture 15, Pg 27