# decrease

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```The Center of Mass
1
xcm
m1 x1  m2 x2

m1  m2
x1  x2
xcm 
2
m1  m2
2
xcm
m1 x1  m2 x2

m1  m2
3
4
5
6
(Omit)
Finding the Center of Mass
by Integration
7
Motion of the Center of Mass
8
Center of Mass Motion
d
dt
d
dt
Definitions of center of mass motion.
9
10
Conservation of Linear
Momentum
11
12
13
Kinetic Energy of a System
14
System Kinetic Energy:
K
sys
 m v  m v  m v 
1
2
2
1 1
1
2
2
2 2
1
2
2
3 3
(Eq. 8-18 frequently used in rotational dynamics.)
15
Practice: Momentum and Kinetic Energy
16
08-2. Two masses move on a frictionless horizontal surface.
M1 = 1kg, v1i = 4m/s. M2 = 2kg, v2i = 1m/s.
a) Find the center of mass speed.
vcm 
(1kg)( 4m / s)  (2kg)(1m / s)
 2m / s
3kg
b) The masses collide along a straight line. Find v1f if v2f =
2.3 m/s and no other external forces act.
M 1v1i  M 2 v2i  M 1v1 f  M 2 v2 f
(1kg)( 4m / s)  (2kg)(1m / s )  (1kg)(v1 f )  (2kg)( 2.3m / s)
6kg  m / s  (1kg)(v1 f )  4.6kg  m / s
1.4kg  m / s  (1kg)(v1 f )
v1 f  1.4m / s
17
c) Calculate the initial and final kinetic energies. Is the collision
energetically possible?
K isys  12 m1v12i  12 m2v22i  12 (1)( 4) 2  12 (2)(1) 2  9 J
2
2
2
2
1
1
1
1
K sys

m
v

m
v

(
1
)(
1
.
4
)

(
2
)(
2
.
3
)
 6.27 J
f
2 2f
2 1 1f
2
2
2
It is possible for kinetic energy to decrease due to the
production of thermal energy in a collision.
18
Collisions and Impulse
19
Impulse is Area under F(t)
20
21
Types of Collisions:
●Complete Inelastic:
K  Thermal
(v1f = v2f)
● Inelastic:
K  Thermal
(v1f ≠ v2f)
● Elastic:
Ki = Kf
(v1f ≠ v2f)
22
23
The Center-of-Mass
Reference Frame
24
25
26
27
28
29
Problems
30
Practice: Collisions and Impulse
31
08-4. A bullet of 230 grains moves horizontally at 830 feet per
second and strikes a 10lb wood block lying at rest on a horizontal
surface. The bullet takes 1.0 millisecond to stop inside the block.
a) Convert the data to SI units.
6.48 105 kg
230 grain
 0.0149kg
grain
830 ft 1m
 253m / s
s 3.28 ft
1kg
10lb
 4.55kg
2.2lb
32
08-4 b) Calculate the speed the block moves at just after the bullet
stops in the block.
Pi sys  (0.0149kg)(253m / s)  (4.55  0.0149)v f  Pfsys
System momentum conserved when
external impulse is negligible.
v f  0.825m / s
33
08-4 c) Calculate the kinetic energy of the bullet before the collision
and of the moving block + bullet after the collision. What percent of
the original kinetic energy is converted to other energies? What
percent is retained as kinetic?
K isys  12 (0.0149kg)( 253m / s) 2  477 J
K
sys
f
 (5.5  0.0149kg)(0.825m / s)  1.55J
1
2
2
sys
U other  K  K sys

K
 1.55  477  475.45 J
f
i
475.47
%converted 
100%  99.7%
477
1.55
%retained 
100%  0.3%
477
34
08-4 d) Calculate the impulse received by the block.
I F
Block
avg
t  p
Block
p
Block
f
p
Block
i
 (4.55kg)(0.825m / s)  (4.55kg)(0m / s)  3.75 Ns
e) If the collision lasts 1.0 millisecond, calculate the average force
exerted on each object.
Block

p
3.75 Ns
Block
Favg 

 3750 N
t
0.001s
bullet

p
 3.75 Ns
bullet
Favg 

 3750 N
t
0.001s
bullet
p bullet  p bullet

p
 mv f  mvi
f
i
 (0.0149)(0.825  253)  3.75
35
Practice
36
08-5. Two masses move on a frictionless horizontal surface.
M1 = 1kg, v1i = 4m/s. M2 = 4kg, v2i = 1m/s in a laboratory.
The masses collide elastically along a straight line.
a) Show that in the center of mass frame that the initial velocities
are +2m/s and -1m/s. (vcm = +2 from previous example)
v
cm
1i
lab
1i
v
v
cm
2i
v
lab
2i
 vcm  4  2  2m / s
cm
1f
 2m / s
cm
2f
 1m / s
v
 vcm  1  2  1m / s
v
b) What are the final velocities in the lab frame?
cm
v1lab

v
f
1 f  vcm  2  2  0m / s
lab
2f
v
 v  vcm  1  2  3m / s
cm
2f
37
08-5 c) Calculate the system-momentum before and after the
collision in the lab-frame.
Pi sys  (1kg)( 4m / s)  (2kg)(1m / s)  6kg  m / s
Pfsys  (1kg)(0m / s)  (2kg)(3m / s)  6kg  m / s
d) Calculate the initial and final kinetic energies of the system in
the lab-frame. Are these energies consistent with the definition
of an ‘elastic collision’?
K
sys
i
 (1kg)( 4m / s)  (2kg)(1m / s)  9 J
K
sys
i
 (1kg)(0m / s)  (2kg)(3m / s )  9 J
1
2
1
2
2
2
1
2
1
2
2
2
This is consistent with an elastic collision
38
39
08-3. Two objects collide in two dimensions. No external
forces act at any time.
In SI units: p1i = (3, 4)
p2i = (2, 1)
a) If p1f = (3, 2), then calculate p2f.
b) Make a sketch of the momentum vectors before and after the
collision.
a) Pi = (3, 4) + (2, 1) = (3, 2) + (px, py) = Pf
(5, 5) = (3+px, 2+py)
px = 2
py =3
p2f = (2, 3)
b)
Pi
Pf
40
c) Calculate the angle each momentum vector makes with the
x-axis.
b)
1i  tan 1 ( 43 )  53.1
 2i  tan 1 ( 12 )  26.6
1 f  tan 1 ( 23 )  33.7
 2 f  tan 1 ( 32 )  56.3
d) Angle of final total momentum vector Pf = Pi = (5, 5):
1,2 f  tan 1 ( 55 )  45
Pf
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42
43
44
45
46
47
48
49
50
51
52
53
```